CTF: FluxFingers4Future - Evil Corp Solution

For this years hack.lu CTF I felt like creating a challenge. Since I work a lot with TLS it was only natural for me to create a TLS challenge. I was informed that TLS challenges are quite uncommon but nevertheless I thought it would be nice to spice the competition up with something "unusual". The challenge mostly requires you to know a lot of details on how the TLS record layer and the key derivation works. The challenge was only solved by one team (0ops from China) during the CTF. Good job!



So let me introduce the challenge first.

The Challenge

You were called by the incident response team of Evil-Corp, the urgently need your help. Somebody broke into the main server of the company, bricked the device and stole all the files! Nothing is left! This should have been impossible. The hacker used some secret backdoor to bypass authentication. Without the knowledge of the secret backdoor other servers are at risk as well! The incident response team has a full packet capture of the incident and performed an emergency cold boot attack on the server to retrieve the contents of the memory (its a really important server, Evil Corp is always ready for such kinds of incidents). However they were unable to retrieve much information from the RAM, what's left is only some parts of the "key_block" of the TLS server. Can you help Evil-Corp to analyze the exploit the attacker used?

(Flag is inside of the attackers' secret message).


TT = Could not recover

key_block:
6B 4F 93 6A TT TT TT TT  TT TT 00 D9 F2 9B 4C B0
2D 88 36 CF B0 CB F1 A6  7B 53 B2 00 B6 D9 DC EF
66 E6 2C 33 5D 89 6A 92  ED D9 7C 07 49 57 AD E1
TT TT TT TT TT TT TT TT  56 C6 D8 3A TT TT TT TT
TT TT TT TT TT TT TT TT  94 TT 0C EB 50 8D 81 C4
E4 40 B6 26 DF E3 40 9A  6C F3 95 84 E6 C5 86 40
49 FD 4E F2 A0 A3 01 06

Download PCAPNG

If you are not interested in the solution and want to try the challenge on your own first, do not read past this point. Spoilers ahead.

The Solution

So lets analyze first what we got. We have something called a "key_block" but we do not have all parts of it. Some of the bytes have been destroyed and are unknown to us. Additionally, we have a PCAP file with some weird messages in them. Lets look at the general structure of the message exchange first.

So looking at the IP address and TCP ports we see that the attacker/client was 127.0.0.1:36674 and was talking with the Server 127.0.0.1:4433. When looking at the individual messages we can see that the message exchange looked something like this:

ENC HS MESSAGE .... ENC HS MESSAGE ->
<- SERVER HELLO, CERTIFICATE, SERVER HELLO DONE
ENC HS MESSAGE .... ENC HS MESSAGE CCS ENC HS MESSAGE, ENC HS MESSAGE ->
<-CCS, ENC HS MESSAGE
ENC HEARTBEAT ->
<- ENC HEARTBEAT
-> ENC APPLICATION DATA
<- INTERNAL ERROR ... INTERNAL ERROR

So this message exchange appears weird. Usually the client is supposed to send a ClientHello in the beginning of the connection, and not encrypted handshake messages. The same is true for the second flight of the client. Usually it transmits its ClientKeyExchange message here, then a ChangeCipherSpec message and finally its Finished message. If we click at the first flight of the client, we can also see some ASCII text fragments in its messages.

Furthermore we can assume that the message sent after the ChangeCipherSpec from the server is actually a TLS Finished message.

Since we cannot read a lot from the messages the client is sending (in Wireshark at least), we can look at the messages the server is sending to get a better hold of what is going on. In the ServerHello message the server selects the parameters for the connection. This reveals that this is indeed a TLS 1.1 connection with TLS_RSA_WITH_AES_256_CBC_SHA , no compression and the Heartbeat Extension negotiated. We can also see that the ServerRandom is: 1023047c60b420bb3321d9d47acb933dbe70399bf6c92da33af01d4fb770e98c (note that it is always 32 bytes long, the UNIX time is part of the ServerRandom).

Looking at the certificate the server sent we can see that the server used a self-signed certificate for Evil.corp.com with an 800-bit RSA modulus:

00ad87f086a4e1acd255d1d77324a05ea7d250f285f3a6de35b9f07c5d083add5166677425b8335328255e7b562f944d55c56ff084f4316fdc9e3f5b009fefd65015a5ca228c94e3fd35c6aba83ea4e20800a34548aa36a5d40e3c7496c65bdbc864e8f161

and the public exponent 65537.


If you pay very close attention to the handshake you can see another weird thing. The size of the exchanged HeartbeatMessages is highly uneven. The client/attacker sent 3500 bytes, the server is supposed to decrypt these messages, and reflect the contents of them. However, the Server sent ~64000 bytes instead. The heartbeat extension became surprisingly well known in 2014, due to the Heartbleed bug in OpenSSL. The bug causes a buffer over-read on the server, causing it to reflect parts of its memory content in return to malicious heartbeat requests. This is a good indicator that this bug might play a role in this challenge.

But what is this key_block thing we got from the incident response team? TLS 1.1 CBC uses 4 symmetric keys in total. Both parties derive these keys from the "master secret" as the key_block. This key_block is then chunked into the individual keys. You can imagine the key_block as some PRF output and both parties knowing which parts of the output to use for which individual key. In TLS 1.1 CBC the key_block is chunked as follows: The first N bytes are the client_write_MAC key, the next N bytes are the server_write_MAC key, the next P bytes are the client_write key and the last P bytes are the server_write key. N is the length of the HMAC key (which is at the time of writing for all cipher suites the length of the HMAC) and P is the length of the key for the block cipher.

In the present handshake AES-256 was negotiated as the block cipher and SHA (SHA-1) was negotiated for the HMAC. This means that N is 20 (SHA-1 is 20 bytes) and P is 32 (AES-256 requires 32 bytes of key material).

Looking at the given key_block we can chunk it into the individual keys:
client_write_MAC = 6B4F936ATTTTTTTTTTTT00D9F29B4CB02D8836CF
server_write_MAC = B0CBF1A67B53B200B6D9DCEF66E62C335D896A92
client_write = EDD97C074957ADE1TTTTTTTTTTTTTTTT56C6D83ATTTTTTTTTTTTTTTTTTTTTTTT
server_write = 94TT0CEB508D81C4E440B626DFE3409A6CF39584E6C5864049FD4EF2A0A30106

Since not all parts of the key_block are present, we can see that we actually have 14/20 bytes of the client_write_MAC key, the whole server_write_MAC key, 12/32 bytes of the client_write key and 31/32 bytes of the server_write key.

The client_write_MAC key is used in the HMAC computations from the client to the server (the server uses the same key to verify the HMAC),
The server_write_MAC key is used in the HMAC computations from the server to the client (the client uses the same key to verify the HMAC),
The client_write key is used to encrypt messages from the client to the server, while the server_write key is used to encrypt messages from the server to the client.

So looking at the keys we could compute HMAC's from the client if we could guess the remaining 6 bytes. We could compute HMAC's from the server directly, we have not enough key material to decrypt the client messages, but we could decrypt server messages if we brute-forced one byte of the server_write key. But how would you brute force this byte? When do we know when we got the correct key? Lets look at how the TLS record layer works to find out :)

The Record Layer

TLS consists out of multiple protocols (Handshake, Alert, CCS, Application (and Heartbeat)). If one of those protocols wants to send any data, it has to pass this data to the record layer. The record layer will chunk this data, compress it if necessary, encrypt it and attach a "record header" to it.

This means, that if we want to decrypt a message we know that if we used the correct key the message should always have a correct padding. If we are unsure we could even check the HMAC with the server_write_MAC key.

In TLS 1.0 - TLS 1.2 the padding looks like this:

1 byte padding  : 00
2 bytes padding: 01 01
3 bytes padding: 02 02 02
4 bytes padding: 03 03 03 03
...

So if we guessed the correct key we know that the plaintext has to have valid padding.
An ideal candidate for our brute force attack is the server Finished message. So lets use that to check our key guesses.
The ciphertext looks like this:
0325f41d3ebaf8986da712c82bcd4d55c3bb45c1bc2eacd79e2ea13041fc66990e217bdfe4f6c25023381bab9ddc8749535973bd4cacc7a4140a14d78cc9bddd

The first 16 bytes of the ciphertext are the IV:
IV: 0325f41d3ebaf8986da712c82bcd4d55
Therefore the actual ciphertext is:
Ciphertext: c3bb45c1bc2eacd79e2ea13041fc66990e217bdfe4f6c25023381bab9ddc8749535973bd4cacc7a4140a14d78cc9bddd

The 256 key candidates are quick to check, and it is revealed that 0xDC was the missing byte.
(The plaintext of the Finished is 1400000C455379AAA141E1B9410B413320C435DEC948BFA451C64E4F30FE5F6928B816CA0B0B0B0B0B0B0B0B0B0B0B0B)

Now that we have the full server_write key we can use it to decrypt the heartbeat records.

This is done in the same way as with the Finished. Looking at the decrypted heartbeat messages we can see a lot of structured data, which is an indicator that we are actually dealing
with the Heartbleed bug. If we convert the content of the heartbeat messages to ASCII we can actually see that the private key of the server is PEM encoded in the first heartbeat message.

Note: This is different to a real heartbeat exploit. Here you don't usually get the private key nicely encoded but have to extract it using the coppersmith's attack or similar things. I did not want to make this challenge even harder so I was so nice to write it to the memory for you :)


The private key within the Heartbeat messages looks like this:
-----BEGIN RSA PRIVATE KEY-----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-----END RSA PRIVATE KEY-----

We should store it in a file and decode it with OpenSSL to get the actual key material.

>> openssl rsa -in key.pem -text -noout
RSA Private-Key: (800 bit, 2 primes)
modulus:
    00:ad:87:f0:86:a4:e1:ac:d2:55:d1:d7:73:24:a0:
    5e:a7:d2:50:f2:85:f3:a6:de:35:b9:f0:7c:5d:08:
    3a:dd:51:66:67:74:25:b8:33:53:28:25:5e:7b:56:
    2f:94:4d:55:c5:6f:f0:84:f4:31:6f:dc:9e:3f:5b:
    00:9f:ef:d6:50:15:a5:ca:22:8c:94:e3:fd:35:c6:
    ab:a8:3e:a4:e2:08:00:a3:45:48:aa:36:a5:d4:0e:
    3c:74:96:c6:5b:db:c8:64:e8:f1:61
publicExponent: 65537 (0x10001)
privateExponent:
    26:3f:79:3f:64:26:2d:be:6a:96:06:e3:e5:25:c7:
    d7:3b:9f:05:e5:8a:6f:b4:38:a9:54:1d:45:30:24:
    31:55:d8:b9:62:bb:51:9f:56:6b:d9:d8:ba:5c:a3:
    be:0f:51:a1:63:8e:f9:26:83:36:a8:62:4d:62:a8:
    b1:0f:99:ae:f1:b7:03:54:c3:31:c5:59:4a:43:f1:
    6d:70:ae:86:14:69:a7:88:f4:4f:d9:5e:bd:08:b4:
    b3:cb:51:d5:8b:85:cd:51:62:f1
prime1:
    00:d3:4b:ae:02:b6:60:e2:62:0e:5c:ae:ef:cb:7f:
    79:56:c1:7b:2e:26:ba:1d:b1:6b:1b:c4:77:cc:46:
    5f:86:c8:06:54:ee:5a:52:88:ab:03:cd:c4:4f:1f:
    ca:09:af:e2:e1:c5
prime2:
    00:d2:3e:d4:93:54:e1:63:e5:02:61:d7:82:9b:8f:
    cf:79:5a:a4:3d:8b:bb:8c:d4:c7:1b:64:c6:af:f0:
    7a:86:cb:71:c8:5a:57:06:5a:ba:30:94:3f:a8:f7:
    51:5c:b9:0b:16:ed
exponent1:
    00:9f:fc:e2:c2:4d:03:e9:06:24:27:cb:91:e8:3d:
    1a:4c:35:6e:26:d0:ce:05:e3:ab:dd:37:93:1a:0a:
    83:14:53:ea:6f:6e:96:d7:7d:82:37:fc:1a:d3:6a:
    97:99:64:23:5f:9d
exponent2:
    00:9e:d7:cd:6f:4a:8f:c7:13:3c:8b:83:71:2f:ea:
    a5:0b:c0:89:99:de:3a:62:9a:57:9b:c0:b5:c4:33:
    61:be:f9:72:0b:b7:05:4c:cd:bb:21:fc:bf:63:ff:
    06:bf:91:26:69:b9
coefficient:
    00:8e:9f:4d:fb:17:44:9b:87:56:e7:59:72:52:9e:
    9b:a1:55:3a:7f:16:62:99:f6:11:7d:80:0d:53:66:
    d5:16:66:ec:48:05:3d:8f:38:d6:6f:40:6e:5d:ec:
    4a:29:c5:fb:5c:6a

So now we got the private key. But what do we do with it? Since this is an RSA handshake we should be able to decrypt the whole session (RSA is not perfect forward secure). Loading it into Wireshark does not work, as Wireshark is unable to read the messages sent by the client. What is going on there?

De-fragmentation

So if you do not yet have a good idea of what the record layer is for, you can imagine it like envelops. If someone wants to send some bytes, you have to put them in an envelop and transmit them. Usually implementations use one big envelop for every message, however you can also send a single message in multiple envelops.

The attacker did exactly that. He fragmented its messages into multiple records. This is not very common for handshake messages but fine according to the specification and accepted by almost all implementations. However, Wireshark is unable to decode these kinds of messages and therefore unable to use our private key to decrypt the connection. So we have to do this step manually.

So each record has the following fields:
Type | Version | Length | Data
If we want to reconstruct the ClientHello message we have to get all the data fields of the records of the first flight and decode them.
This is simply done by clicking on each record in Wireshark and concatenating the data fields. This step is at least on my Wireshark version (3.0.5) not very easy as the copying is actually buggy, and Wireshark is not copying the correct bytes.

 As you can see in the image, the record is supposed to have a length of 8 bytes, but Wireshark is only copying 4 bytes. I am not sure if this bug is actually only in my version or affects all Wireshark versions. Instead of copying the records individually I therefore copied the whole TCP payload and chunked it manually into the individual records.

16030200080100009e03020000
160302000800000000004e6f62
16030200086f64796b6e6f7773
1603020008696d616361740000
16030200080000000000002053
1603020008746f70206c6f6f6b
1603020008696e67206e6f7468
1603020008696e6720746f2066
1603020008696e646865726500
16030200080200350100005300
16030200080f00010113370015
16030200084576696c436f7270
1603020008206b696c6c732070
1603020008656f706c65000d00
16030200082c002a0102020203
16030200080204020502060201
16030200080102010301040105
16030200080106010103020303
160302000803040305030603ed
1603020008edeeeeefefff0100
16030200020100

If we structure this data it looks like this:
Type  Version Length  Payload
16    0302    0008    0100009e03020000
16    0302    0008    00000000004e6f62
16    0302    0008    6f64796b6e6f7773
16    0302    0008    696d616361740000
16    0302    0008    0000000000002053
16    0302    0008    746f70206c6f6f6b
16    0302    0008    696e67206e6f7468
16    0302    0008    696e6720746f2066
16    0302    0008    696e646865726500
16    0302    0008    0200350100005300
16    0302    0008    0f00010113370015
16    0302    0008    4576696c436f7270
16    0302    0008    206b696c6c732070
16    0302    0008    656f706c65000d00
16    0302    0008    2c002a0102020203
16    0302    0008    0204020502060201
16    0302    0008    0102010301040105
16    0302    0008    0106010103020303
16    0302    0008    03040305030603ed
16    0302    0008    edeeeeefefff0100
16    0302    0002    0100

The actual message is the concatenation of the record payloads:

0100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

So what is left is to parse this message. There is an easy way on how to do this an a labor intensive manual way. Lets do the manual process first :) .
We know from the record header that his message is in fact a handshake message (0x16).
According to the specification handshake messages look like this:
    
      struct {
          HandshakeType msg_type;    /* handshake type */
          uint24 length;             /* bytes in message */
          select (HandshakeType) {
              case hello_request:       HelloRequest;
              case client_hello:        ClientHello;
              case server_hello:        ServerHello;
              case certificate:         Certificate;
              case server_key_exchange: ServerKeyExchange;
              case certificate_request: CertificateRequest;
              case server_hello_done:   ServerHelloDone;
              case certificate_verify:  CertificateVerify;
              case client_key_exchange: ClientKeyExchange;
              case finished:            Finished;
          } body;
      } Handshake;
    
This is RFC speak for: Each handshake message starts with a type field which says which handshake message this is, followed by a 3 byte length field which determines the length of rest of the handshake message.
So in our case the msg_type is 0x01 , followed by a 3 Byte length field (0x00009e, 158[base10]). 0x01 means ClientHello (https://www.iana.org/assignments/tls-parameters/tls-parameters.xhtml#tls-parameters-7). This means we have to parse the bytes after the length field as a ClientHello.
    
      {
          ProtocolVersion client_version;
          Random random;
          SessionID session_id;
          CipherSuite cipher_suites<2..2^16-2>;
          CompressionMethod compression_methods<1..2^8-1>;
          select (extensions_present) {
              case false:
                  struct {};
              case true:
                  Extension extensions<0..2^16-1>;
          };
      } ClientHello;

This means: The next 2 bytes are the ProtocolVersion, the next 32 bytes are the ClientRandom, the next byte is the SessionID Length, the next SessionID Length many bytes are the SessionID, the next 2 bytes are the CipherSuite Length bytes, followed by CipherSuite Length many CipherSuites, followed by a 1 byte Compression Length field, followed by Compression Length many CompressionBytes followed by a 2 byte Extension Length field followed by extension length many ExtensionBytes. So lets try to parse this:

Handshakye Type   : 01
Handshake Length  : 00009e
ProtocolVersion   : 0302
ClientRandom      : 000000000000004e6f626f64796b6e6f7773696d616361740000000000000000
SessionID Length  : 20
SessionID         : 53746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e6468657265
CipherSuite Length: 0002
CipherSuites      : 0035
Compression Length: 01
CompressionBytes  : 00
Extension Length  : 0053
ExtensionBytes:   : 000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

This is manual parsing is the slow method of dealing with this. Instead of looking at the specification to parse this message we could also compare the message structure to another ClientHello. This eases this process a lot. What we could also do is record the transmission of this message as a de-fragmented message to something and let Wireshark decode it for us. To send the de-fragmented message we need to create a new record header ourselves. The record should look like this:

Type   : 16
Version: 0302
Length : 00A2
Payload: 0100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

To send this record we can simply use netcat:


echo '16030200A20100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100' | xxd -r -p | nc localhost 4433


Now we can use Wireshark to parse this message. As we can see now, the weired ASCII fragments we could see in the previous version are actually the ClientRandom, the SessionID, and a custom extension from the attacker. Now that we have de-fragmented the message, we know the ClientRandom: 000000000000004e6f626f64796b6e6f7773696d616361740000000000000000

De-fragmenting the ClientKeyExchange message

Now that we have de-fragmented the first flight from the attacker, we can de-fragment the second flight from the client. We can do this in the same fashion as we de-fragmented the ClientHello.

16    0302    0008    1000006600645de1
16    0302    0008    66a6d3669bf21936
16    0302    0008    5ef3d35410c50283
16    0302    0008    c4dd038a1b6fedf5
16    0302    0008    26d5b193453d796f
16    0302    0008    6e63c144bbda6276
16    0302    0008    3740468e21891641
16    0302    0008    0671318e83da3c2a
16    0302    0008    de5f6da6482b09fc
16    0302    0008    a5c823eb4d9933fe
16    0302    0008    ae17d165a6db0e94
16    0302    0008    bb09574fc1f7b8ed
16    0302    0008    cfbcf9e9696b6173
16    0302    0002    f4b6

14    0302    0001    01

16    0302    0030    cbe6bf1ae7f2bc40a49709a06c0e3149a65b8cd93c2525b5bfa8f696e29880d3447aef3dc9a996ca2aff8be99b1a4157
16    0302    0030    9bf02969ca42d203e566bcc696de08fa80e0bfdf44b1b315aed17fe867aed6d0d600c73de59c14beb74b0328eacadcf9

Note that his time we have 3 record groups. First there is chain of handshake records, followed by a ChangeCipherSpec record, followed by 2 more handshake records. The TLS specification forbids that records of different types are interleaved. This means that the first few records a probably forming a group of messages. The ChangeCipherSpec record is telling the server that subsequent messages are encrypted. This seems to be true, since the following records do not appear to be plaintext handshake messages.

So lets de-fragment the first group of records by concatenating their payloads:

1000006600645de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b6

Since this is a handshake message, we know that the first byte should tell us which handshake message this is. 0x10 means this is a ClientKeyExchange message. Since we already know that TLS_RSA_WITH_AES_256_CBC_SHA was negotiated for this connection, we know that this is an RSA ClientKeyExchange message.

These messages are supposed to look like this (I will spare you the lengthy RFC definition):

Type (0x10)
Length (Length of the content) (3 bytes)
EncryptedPMS Length(Length of the encrypted PMS) (2 bytes)
EncrpytedPMS  (EncryptedPMS Length many bytes)
    
For our message this should look like this:

Type: 10
Length: 000066
Encrypted PMS Length: 0064
Encrypted PMS: 5de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b6

Now that we got the Encrypted PMS we can decrypt it with the private key. Since the connection negotiated RSA as the key exchange algorithm this is done with:

encPMS^privKey mod modulus = plainPMS

We can solve this equation with the private key from the leaked PEM file.

2445298227328938658090475430796587247849533931395726514458166123599560640691186073871766111778498132903314547451268864032761115999716779282639547079095457185023600638251088359459150271827705392301109265654638212139757207501494756926838535350 ^ 996241568615939319506903357646514527421543094912647981212056826138382708603915022492738949955085789668243947380114192578398909946764789724993340852568712934975428447805093957315585432465710754275221903967417599121549904545874609387437384433 mod 4519950410687629988405948449295924027942240900746985859791939647949545695657651701565014369207785212702506305257912346076285925743737740481250261638791708483082426162177210620191963762755634733255455674225810981506935192783436252402225312097

Solving this equation gives us:

204742908894949049937193473353729060739551644014729690547520028508481967333831905155391804994508783506773012994170720979080285416764402813364718099379387561201299457228993584122400808905739026823578773289385773545222916543755807247900961

in hexadecimal this is:

00020325f41d3ebaf8986da712c82bcd4d554bf0b54023c29b624de9ef9c2f931efc580f9afb081b12e107b1e805f2b4f5f0f1000302476574204861636b6564204e6f6f622c20796f752077696c6c206e65766572206361746368206d65212121212121

The PMS is PKCS#1.5 encoded. This means that it is supposed to start with 0x0002 followed by a padding which contains no 0x00 bytes, followed by a separator 0x00 byte followed by a payload. In TLS, the payload has to be exactly 48 bytes long and has to start with the highest proposed protocol version of the client. We can see that this is indeed the case for our decrypted payload. The whole decrypted payload is the PMS for the connection.

This results in the PMS: 0302476574204861636b6564204e6f6f622c20796f752077696c6c206e65766572206361746368206d65212121212121 (which besides the protocol version is also ASCII :) )

Now that we have the PMS its time to revisit the key scheduling in TLS. We already briefly touched it but here is a overview:

As you can see, we first have to compute the master secret. With the master secret we can reconstruct the key_block. If we have computed the key_block, we can finally get the client_write key and decrypt the message from the attacker.


master secret = PRF ( PMS, "master secret", ClientRandom | ServerRandom)

key_block = PRF (master_secret, "key expansion", ServerRandom |  ClientRandom )

Where "master secret" and "key expansion" are literally ASCII Strings.

Note that in the key_block computation ClientRandom and ServerRandom are exchanged.



To do this computation we can either implement the PRF ourselfs, or easier, steal it from somewhere. The PRF in TLS 1.1 is the same as in TLS 1.0. Good places to steal from are for example openssl (C/C++), the scapy project (python), the TLS-Attacker project (java) or your favourite TLS library. The master secret is exactly 48 bytes long. The length of the key_block varies depending on the selected cipher suite and protocol version. In our case we need 2 * 20 bytes (for the 2 HMAC keys) + 2 * 32 bytes (for the 2 AES keys) = 104 bytes.

I will use the TLS-Attacker framework for this computation. The code will look like this:


This results in the following master secret: 292EABADCF7EFFC495825AED17EE7EA575E02DF0BAB7213EC1B246BE23B2E0912DA2B99C752A1F8BD3D833E8331D649F  And the following key_block:
6B4F936ADE9B4010393B00D9F29B4CB02D8836CFB0CBF1A67B53B200B6D9DCEF66E62C335D896A92EDD97C074957ADE136D6BAE74AE8193D56C6D83ACDE6A3B365679C5604312A1994DC0CEB508D81C4E440B626DFE3409A6CF39584E6C5864049FD4EF2A0A30106

Now we can chunk our resulting key_block into its individual parts. This is done analogously to the beginning of the challenge.

client_write_mac key = 6B4F936ADE9B4010393B00D9F29B4CB02D8836CF
server_write_mac key = B0CBF1A67B53B200B6D9DCEF66E62C335D896A92
client_write key = EDD97C074957ADE136D6BAE74AE8193D56C6D83ACDE6A3B365679C5604312A19
server_write key = 94DC0CEB508D81C4E440B626DFE3409A6CF39584E6C5864049FD4EF2A0A30106

Now that we have the full client_write key we can use that key to decrypt the application data messages. But these messages are also fragmented. But since the messages are now encrypted, we cannot simply concatenate the payloads of the records, but we have to decrypt them individually and only concatenate the resulting plaintext.

Analogue to the decryption of the heartbeat message, the first 16 bytes of each encrypted record payload are used as an IV

IV, Ciphertext Plaintext
6297cb6d9afba63ec4c0dd7ac0184570    a9c605307eb5f8ccbe8bbc210ff1ff14943873906fad3eca017f49af8feaec87      557365723A20726FB181CF546350A88ACBE8D0248D6FF07675D1514E03030303
063c60d43e08c4315f261f8a4f06169a    cdb5818d80075143afe83c79b570ab0b349b2e8748f8b767c54c0133331fb886    6F743B0A50617373D6F734D45FB99850CCAF32DF113914FC412C523603030303
cd839b95954fcadf1e60ee983cbe5c21    ac6f6e1fe34ae4b1214cded895db4746b8e38d7960d7d45cb001aab8e18c7fc7    3A20726F6F743B0A937048A265327642BD5626E00E4BC79618F9A95C03030303
8092d75f72b16cb23a856b00c4c39898    8df099441e10dca5e850398e616e4597170796b7202e2a8726862cd760ebacdf    6563686F20224F7769EACFBEEB5EE5D1F0B72306F8C78AD86CB4835003030303
8e9f83b015fce7f9c925b8b64abee426    224a5fbd2d9b8fc6ded34222a943ec0e8e973bcf125b81f918e391a22b4b0e65    6E6564206279204061736E93BFDC5103C8C2FE8C543A72B924212E8403030303
0e24ba11e41bfcf66452dc80221288ce    a66fb3aed9bdc7e08a31a0e7f14e11ce0983ec3d20dd47c179425243b14b08c9    6963306E7A31223B84A3CAFA7980B461DE0A6410D6251551AE401DD903030303
0465fdb05b121cdc08fa01cdacb2c8f4    eff59402f4dbf35a85cc91a6d1264a895cd1b3d2014c91fbba03ec4c85d058c9     0A7375646F20726DB97422D8B30C54CC672FFEC3E9D771D4743D96B903030303
e2ddbbb83fe8318c41c26d57a5813fab    89549a874ff74d83e182de34ecf55fff1a57008afd3a29ef0d839b991143cd2a      202F202D72663B0A996F3F1789CB9B671223E73C66A0BA578D0C0F3203030303
524f5210190f73c984bd6a59b9cf424c    b7f30fafe5ea3ac51b6757c51911e86b0aa1a6bbf4861c961f8463154acea315    0A666C61677B436868BF764B01D2CDCB2C06EA0DFC5443DABB6EC9AE03030303
32765985e2e594cddca3d0f45bd21f49    a5edfe89fdb3782e2af978585c0e27ba3ef90eb658304716237297f97e4e72bc    696D696368616E67FBF32127FA3AF2F97770DE5B9C6D376A254EF51E03030303
e0ae69b1fa54785dc971221fd92215fb    14e918a9e6e37139153be8cb9c16d2a787385746f9a80d0596580ba22eaf254e    61467233346B7D8BE8B903A167C44945E7676BF99D888A4B86FA8E0404040404

The plaintext then has to be de-padded and de-MACed.

Data HMAC Pad:
557365723A20726F    B181CF546350A88ACBE8D0248D6FF07675D1514E    03030303
6F743B0A50617373    D6F734D45FB99850CCAF32DF113914FC412C5236    03030303
3A20726F6F743B0A    937048A265327642BD5626E00E4BC79618F9A95C    03030303
6563686F20224F77    69EACFBEEB5EE5D1F0B72306F8C78AD86CB48350    03030303
6E65642062792040    61736E93BFDC5103C8C2FE8C543A72B924212E84    03030303
6963306E7A31223B    84A3CAFA7980B461DE0A6410D6251551AE401DD9    03030303
0A7375646F20726D    B97422D8B30C54CC672FFEC3E9D771D4743D96B9    03030303
202F202D72663B0A    996F3F1789CB9B671223E73C66A0BA578D0C0F32    03030303
0A666C61677B4368    68BF764B01D2CDCB2C06EA0DFC5443DABB6EC9AE    03030303
696D696368616E67    FBF32127FA3AF2F97770DE5B9C6D376A254EF51E    03030303
61467233346B7D      8BE8B903A167C44945E7676BF99D888A4B86FA8E    0404040404

This then results in the following data:

Data:
557365723A20726F6F743B0A506173733A20726F6F743B0A6563686F20224F776E656420627920406963306E7A31223B0A7375646F20726D202F202D72663B0A0A666C61677B4368696D696368616E6761467233346B7D8B

Which is ASCII for:

User: root;
Pass: root;
echo "Owned by @ic0nz1";
sudo rm / -rf;

flag{ChimichangaFr34k}


Honestly this was quite a journey. But this presented solution is the tedious manual way. There is also a shortcut with which you can skip most of the manual cryptographic operations.

The Shortcut

After you de-fragmented the messages you can patch the PCAP file and then use Wireshark to decrypt the whole session. This way you can get the flag without performing any cryptographic operation after you got the private key. Alternatively you can replay the communication and record it with Wireshark. I will show you the replay of the messages. To recap the de-fragmented messages looks like this:

ClientHello
0100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

ClientKeyExchange:
1000006600645de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b6

We should now add new (not fragmented) record header to the previously fragmented message. The messages sent from the server can stay as they are. The ApplicationData from the client can also stay the same. The messages should now look like this

ClientHello
16030200A20100009e0302000000000000004e6f626f64796b6e6f7773696d6163617400000000000000002053746f70206c6f6f6b696e67206e6f7468696e6720746f2066696e64686572650002003501000053000f000101133700154576696c436f7270206b696c6c732070656f706c65000d002c002a010202020302040205020602010102010301040105010601010302030303040305030603ededeeeeefefff01000100

ServerHello / Certificate / ServerHelloDone
160302006A1000006600645de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b61403020001011603020030cbe6bf1ae7f2bc40a49709a06c0e3149a65b8cd93c2525b5bfa8f696e29880d3447aef3dc9a996ca2aff8be99b1a415716030200309bf02969ca42d203e566bcc696de08fa80e0bfdf44b1b315aed17fe867aed6d0d600c73de59c14beb74b0328eacadcf9

ClientKeyExchange / ChangeCipherSpec / Finished
160302006A1000006600645de166a6d3669bf219365ef3d35410c50283c4dd038a1b6fedf526d5b193453d796f6e63c144bbda62763740468e218916410671318e83da3c2ade5f6da6482b09fca5c823eb4d9933feae17d165a6db0e94bb09574fc1f7b8edcfbcf9e9696b6173f4b61403020001011603020030cbe6bf1ae7f2bc40a49709a06c0e3149a65b8cd93c2525b5bfa8f696e29880d3447aef3dc9a996ca2aff8be99b1a415716030200309bf02969ca42d203e566bcc696de08fa80e0bfdf44b1b315aed17fe867aed6d0d600c73de59c14beb74b0328eacadcf9')

ApplicationData
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

What we want to do now is create the following conversation:
CH->
<-SH, CERT, SHD
-> CKE, CCS, FIN
-> APP, APP ,APP

This will be enough for Wireshark to decrypt the traffic. However, since we removed some messages (the whole HeartbeatMessages) our HMAC's will be invalid.

We need to record an interleaved transmission of these message with Wireshark. I will use these simple python programs to create the traffic:




If we record these transmissions and tick the flag in Wireshark to ignore invalid HMAC's we can see the plaintext (if we added the private key in Wireshark).

Challenge Creation

I used our TLS-Attacker project to create this challenge. With TLS-Attacker you can send arbitrary TLS messages with arbitrary content in an arbitrary order, save them in XML and replay them. The communication between the peers are therefore only two XML files which are loaded into TLS-Attacker talking to each other. I then copied parts of the key_block from the debug output and the challenge was completed :)

If you have question in regards to the challenge you can DM me at @ic0nz1
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CVE-2020-2655 JSSE Client Authentication Bypass

During our joint research on DTLS state machines, we discovered a really interesting vulnerability (CVE-2020-2655) in the recent versions of Sun JSSE (Java 11, 13). Interestingly, the vulnerability does not only affect DTLS implementations but does also affects the TLS implementation of JSSE in a similar way. The vulnerability allows an attacker to completely bypass client authentication and to authenticate as any user for which it knows the certificate WITHOUT needing to know the private key. If you just want the PoC's, feel free to skip the intro.

DTLS

I guess most readers are very familiar with the traditional TLS handshake which is used in HTTPS on the web.

DTLS is the crayon eating brother of TLS. It was designed to be very similar to TLS, but to provide the necessary changes to run TLS over UDP. DTLS currently exists in 2 versions (DTLS 1.0 and DTLS 1.2), where DTLS 1.0 roughly equals TLS 1.1 and DTLS 1.2 roughly equals TLS 1.2. DTLS 1.3 is currently in the process of being standardized. But what exactly are the differences? If a protocol uses UDP instead of TCP, it can never be sure that all messages it sent were actually received by the other party or that they arrived in the correct order. If we would just run vanilla TLS over UDP, an out of order or dropped message would break the connection (not only during the handshake). DTLS, therefore, includes additional sequence numbers that allow for the detection of out of order handshake messages or dropped packets. The sequence number is transmitted within the record header and is increased by one for each record transmitted. This is different from TLS, where the record sequence number was implicit and not transmitted with each record. The record sequence numbers are especially relevant once records are transmitted encrypted, as they are included in the additional authenticated data or HMAC computation. This allows a receiving party to verify AEAD tags and HMACs even if a packet was dropped on the transport and the counters are "out of sync".
Besides the record sequence numbers, DTLS has additional header fields in each handshake message to ensure that all the handshake messages have been received. The first handshake message a party sends has the message_seq=0 while the next handshake message a party transmits gets the message_seq=1 and so on. This allows a party to check if it has received all previous handshake messages. If, for example, a server received message_seq=2 and message_seq=4 but did not receive message_seq=3, it knows that it does not have all the required messages and is not allowed to proceed with the handshake. After a reasonable amount of time, it should instead periodically retransmit its previous flight of handshake message, to indicate to the opposing party they are still waiting for further handshake messages. This process gets even more complicated by additional fragmentation fields DTLS includes. The MTU (Maximum Transmission Unit) plays a crucial role in UDP as when you send a UDP packet which is bigger than the MTU the IP layer might have to fragment the packet into multiple packets, which will result in failed transmissions if parts of the fragment get lost in the transport. It is therefore desired to have smaller packets in a UDP based protocol. Since TLS records can get quite big (especially the certificate message as it may contain a whole certificate chain), the messages have to support fragmentation. One would assume that the record layer would be ideal for this scenario, as one could detect missing fragments by their record sequence number. The problem is that the protocol wants to support completely optional records, which do not need to be retransmitted if they are lost. This may, for example, be warning alerts or application data records. Also if one party decides to retransmit a message, it is always retransmitted with an increased record sequence number. For example, the first ClientKeyExchange message might have record sequence 2, the message gets dropped, the client decides that it is time to try again and might send it with record sequence 5. This was done as retransmissions are only part of DTLS within the handshake. After the handshake, it is up to the application to deal with dropped or reordered packets. It is therefore not possible to see just from the record sequence number if handshake fragments have been lost. DTLS, therefore, adds additional handshake message fragment information in each handshake message record which contains information about where the following bytes are supposed to be within a handshake message.


If a party has to replay messages, it might also refragment the messages into bits of different (usually smaller) sizes, as dropped packets might indicate that the packets were too big for the MTU). It might, therefore, happen that you already have received parts of the message, get a retransmission which contains some of the parts you already have, while others are completely new to you and you still do not have the complete message. The only option you then have is to retransmit your whole previous flight to indicate that you still have missing fragments. One notable special case in this retransmission fragmentation madness is the ChangecipherSpec message. In TLS, the ChangecipherSpec message is not a handshake message, but a message of the ChangeCipherSpec protocol. It, therefore, does not have a message_sequence. Only the record it is transmitted in has a record sequence number. This is important for applications that have to determine where to insert a ChangeCipherSpec message in the transcript.

As you might see, this whole record sequence, message sequence, 2nd layer of fragmentation, retransmission stuff (I didn't even mention epoch numbers) which is within DTLS, complicates the whole protocol a lot. Imagine being a developer having to implement this correctly and secure...  This also might be a reason why the scientific research community often does not treat DTLS with the same scrutiny as it does with TLS. It gets really annoying really fast...

Client Authentication

In most deployments of TLS only the server authenticates itself. It usually does this by sending an X.509 certificate to the client and then proving that it is in fact in possession of the private key for the certificate. In the case of RSA, this is done implicitly the ability to compute the shared secret (Premaster secret), in case of (EC)DHE this is done by signing the ephemeral public key of the server. The X.509 certificate is transmitted in plaintext and is not confidential. The client usually does not authenticate itself within the TLS handshake, but rather authenticates in the application layer (for example by transmitting a username and password in HTTP). However, TLS also offers the possibility for client authentication during the TLS handshake. In this case, the server sends a CertificateRequest message during its first flight. The client is then supposed to present its X.509 Certificate, followed by its ClientKeyExchange message (containing either the encrypted premaster secret or its ephemeral public key). After that, the client also has to prove to the server that it is in possession of the private key of the transmitted certificate, as the certificate is not confidential and could be copied by a malicious actor. The client does this by sending a CertificateVerify message, which contains a signature over the handshake transcript up to this point, signed with the private key which belongs to the certificate of the client. The handshake then proceeds as usual with a ChangeCipherSpec message (which tells the other party that upcoming messages will be encrypted under the negotiated keys), followed by a Finished message, which assures that the handshake has not been tampered with. The server also sends a CCS and Finished message, and after that handshake is completed and both parties can exchange application data. The same mechanism is also present in DTLS.

But what should a Client do if it does not possess a certificate? According to the RFC, the client is then supposed to send an empty certificate and skip the CertificateVerify message (as it has no key to sign anything with). It is then up to the TLS server to decide what to do with the client. Some TLS servers provide different options in regards to client authentication and differentiate between REQUIRED and WANTED (and NONE). If the server is set to REQUIRED, it will not finish the TLS handshake without client authentication. In the case of WANTED, the handshake is completed and the authentication status is then passed to the application. The application then has to decide how to proceed with this. This can be useful to present an error to a client asking him to present a certificate or insert a smart card into a reader (or the like). In the presented bugs we set the mode to REQUIRED.

State machines

As you might have noticed it is not trivial to decide when a client or server is allowed to receive or send each message. Some messages are optional, some are required, some messages are retransmitted, others are not. How an implementation reacts to which message when is encompassed by its state machine. Some implementations explicitly implement this state machine, while others only do this implicitly by raising errors internally if things happen which should not happen (like setting a master_secret when a master_secret was already set for the epoch). In our research, we looked exactly at the state machines of DTLS implementations using a grey box approach. The details to our approach will be in our upcoming paper (which will probably have another blog post), but what we basically did is carefully craft message flows and observed the behavior of the implementation to construct a mealy machine which models the behavior of the implementation to in- and out of order messages. We then analyzed these mealy machines for unexpected/unwanted/missing edges. The whole process is very similar to the work of Joeri de Ruiter and Erik Poll.

JSSE Bugs

The bugs we are presenting today were present in Java 11 and Java 13 (Oracle and OpenJDK). Older versions were as far as we know not affected. Cryptography in Java is implemented with so-called SecurityProvider. Per default SUN JCE is used to implement cryptography, however, every developer is free to write or add their own security provider and to use them for their cryptographic operations. One common alternative to SUN JCE is BouncyCastle. The whole concept is very similar to OpenSSL's engine concept (if you are familiar with that). Within the JCE exists JSSE - the Java Secure Socket Extension, which is the SSL/TLS part of JCE. The presented attacks were evaluated using SUN JSSE, so the default TLS implementation in Java. JSSE implements TLS and DTLS (added in Java 9). However, DTLS is not trivial to use, as the interface is quite complex and there are not a lot of good examples on how to use it. In the case of DTLS, only the heart of the protocol is implemented, how the data is moved from A to B is left to the developer. We developed a test harness around the SSLEngine.java to be able to speak DTLS with Java. The way JSSE implemented a state machine is quite interesting, as it was completely different from all other analyzed implementations. JSSE uses a producer/consumer architecture to decided on which messages to process. The code is quite complex but worth a look if you are interested in state machines.

So what is the bug we found? The first bug we discovered is that a JSSE DTLS/TLS Server accepts the following message sequence, with client authentication set to required:


JSSE is totally fine with the messages and finishes the handshake although the client does NOT provide a certificate at all (nor a CertificateVerify message). It is even willing to exchange application data with the client. But are we really authenticated with this message flow? Who are we? We did not provide a certificate! The answer is: it depends. Some applications trust that needClientAuth option of the TLS socket works and that the user is *some* authenticated user, which user exactly does not matter or is decided upon other authentication methods. If an application does this - then yes, you are authenticated. We tested this bug with Apache Tomcat and were able to bypass ClientAuthentication if it was activated and configured to use JSSE. However, if the application decides to check the identity of the user after the TLS socket was opened, an exception is thrown:

The reason for this is the following code snippet from within JSSE:


As we did not send a client certificate the value of peerCerts is null, therefore an exception is thrown. Although this bug is already pretty bad, we found an even worse (and weirder) message sequence which completely authenticates a user to a DTLS server (not TLS server though). Consider the following message sequence:

If we send this message sequence the server magically finishes the handshake with us and we are authenticated.

First off: WTF
Second off: WTF!!!111

This message sequence does not make any sense from a TLS/DTLS perspective. It starts off as a "no-authentication" handshake but then weird things happen. Instead of the Finished message, we send a Certificate message, followed by a Finished message, followed by a second(!) CCS message, followed by another Finished message. Somehow this sequence confuses JSSE such that we are authenticated although we didn't even provide proof that we own the private key for the Certificate we transmitted (as we did not send a CertificateVerify message).
So what is happening here? This bug is basically a combination of multiple bugs within JSSE. By starting the flight with a ClientKeyExchange message instead of a Certificate message, we make JSSE believe that the next messages we are supposed to send are ChangeCipherSpec and Finished (basically the first exploit). Since we did not send a Certificate message we are not required to send a CertificateVerify message. After the ClientKeyExchange message, JSSE is looking for a ChangeCipherSpec message followed by an "encrypted handshake message". JSSE assumes that the first encrypted message it receives will be the Finished message. It, therefore, waits for this condition. By sending ChangeCipherSpec and Certificate we are fulfilling this condition. The Certificate message really is an "encrypted handshake message" :). This triggers JSSE to proceed with the processing of received messages, ChangeCipherSpec message is consumed, and then the Certifi... Nope, JSSE notices that this is not a Finished message, so what JSSE does is buffer this message and revert to the previous state as this step has apparently not worked correctly. It then sees the Finished message - this is ok to receive now as we were *somehow* expecting a Finished message, but JSSE thinks that this Finished is out of place, as it reverted the state already to the previous one. So this message gets also buffered. JSSE is still waiting for a ChangeCipherSpec, "encrypted handshake message" - this is what the second ChangeCipherSpec & Finished is for. These messages trigger JSSE to proceed in the processing. It is actually not important that the last message is a Finished message, any handshake message will do the job. Since JSSE thinks that it got all required messages again it continues to process the received messages, but the Certificate and Finished message we sent previously are still in the buffer. The Certificate message is processed (e.g., the client certificate is written to the SSLContext.java). Then the next message in the buffer is processed, which is a Finished message. JSSE processes the Finished message (as it already had checked that it is fine to receive), it checks that the verify data is correct, and then... it stops processing any further messages. The Finished message basically contains a shortcut. Once it is processed we can stop interpreting other messages in the buffer (like the remaining ChangeCipherSpec & "encrypted handshake message"). JSSE thinks that the handshake has finished and sends ChangeCipherSpec Finished itself and with that the handshake is completed and the connection can be used as normal. If the application using JSSE now decides to check the Certificate in the SSLContext, it will see the certificate we presented (with no possibility to check that we did not present a CertificateVerify). The session is completely valid from JSSE's perspective.

Wow.

The bug was quite complex to analyze and is totally unintuitive. If you are still confused - don't worry. You are in good company, I spent almost a whole day analyzing the details... and I am still confused. The main problem why this bug is present is that JSSE did not validate the received message_sequence numbers of incoming handshake message. It basically called receive, sorted the received messages by their message_sequence, and processed the message in the "intended" order, without checking that this is the order they are supposed to be sent in.
For example, for JSSE the following message sequence (Certificate and CertificateVerify are exchanged) is totally fine:

Not sending a Certificate message was fine for JSSE as the REQUIRED setting was not correctly evaluated during the handshake. The consumer/producer architecture of JSSE then allowed us to cleverly bypass all the sanity checks.
But fortunately (for the community) this bypass does not work for TLS. Only the less-used DTLS is vulnerable. And this also makes kind of sense. DTLS has to be much more relaxed in dealing with out of order messages then TLS as UDP packets can get swapped or lost on transport and we still want to buffer messages even if they are out of order. But unfortunately for the community, there is also a bypass for JSSE TLS - and it is really really trivial:

Yep. You can just not send a CertificateVerify (and therefore no signature at all). If there is no signature there is nothing to be validated. From JSSE's perspective, you are completely authenticated. Nothing fancy, no complex message exchanges. Ouch.

PoC

A vulnerable java server can be found _*here*_. The repository includes a pre-built JSSE server and a Dockerfile to run the server in a vulnerable Java version. (If you want, you can also build the server yourself).
You can build the docker images with the following commands:

docker build . -t poc

You can start the server with docker:

docker run -p 4433:4433 poc tls

The server is configured to enforce client authentication and to only accept the client certificate with the SHA-256 Fingerprint: B3EAFA469E167DDC7358CA9B54006932E4A5A654699707F68040F529637ADBC2.

You can change the fingerprint the server accepts to your own certificates like this:

docker run -p 4433:4433 poc tls f7581c9694dea5cd43d010e1925740c72a422ff0ce92d2433a6b4f667945a746

To exploit the described vulnerabilities, you have to send (D)TLS messages in an unconventional order or have to not send specific messages but still compute correct cryptographic operations. To do this, you could either modify a TLS library of your choice to do the job - or instead use our TLS library TLS-Attacker. TLS-Attacker was built to send arbitrary TLS messages with arbitrary content in an arbitrary order - exactly what we need for this kind of attack. We have already written a few times about TLS-Attacker. You can find a general tutorial __here__, but here is the TLDR (for Ubuntu) to get you going.

Now TLS-Attacker should be built successfully and you should have some built .jar files within the apps/ folder.
We can now create a custom workflow as an XML file where we specify the messages we want to transmit:

This workflow trace basically tells TLS-Attacker to send a default ClientHello, wait for a ServerHelloDone message, then send a ClientKeyExchange message for whichever cipher suite the server chose and then follow it up with a ChangeCipherSpec & Finished message. After that TLS-Attacker will just wait for whatever the server sent. The last action prints the (eventually) transmitted application data into the console. You can execute this WorkflowTrace with the TLS-Client.jar:

java -jar TLS-Client.jar -connect localhost:4433 -workflow_input exploit1.xml

With a vulnerable server the result should look something like this:

and from TLS-Attackers perspective:

As mentioned earlier, if the server is trying to access the certificate, it throws an SSLPeerUnverifiedException. However, if the server does not - it is completely fine exchanging application data.
We can now also run the second exploit against the TLS server (not the one against DTLS). For this case I just simply also send the certificate of a valid client to the server (without knowing the private key). The modified WorkflowTrace looks like this:

Your output should now look like this:

As you can see, when accessing the certificate, no exception is thrown and everything works as if we would have the private key. Yep, it is that simple.
To test the DTLS specific vulnerability we need a vulnerable DTLS-Server:

docker run -p 4434:4433/udp poc:latest dtls

A WorkflowTrace which exploits the DTLS specific vulnerability would look like this:

To execute the handshake we now need to tell TLS-Attacker additionally to use UDP instead of TCP and DTLS instead of TLS:

java -jar TLS-Client.jar -connect localhost:4434 -workflow_input exploit2.xml -transport_handler_type UDP -version DTLS12

Resulting in the following handshake:

As you can see, we can exchange ApplicationData as an authenticated user. The server actually sends the ChangeCipherSpec,Finished messages twice - to avoid retransmissions from the client in case his ChangeCipherSpec,Finished is lost in transit (this is done on purpose).

Conclusion

These bugs are quite fatal for client authentication. The vulnerability got CVSS:4.8 as it is "hard to exploit" apparently. It's hard to estimate the impact of the vulnerability as client authentication is often done in internal networks, on unusual ports or in smart-card setups. If you want to know more about how we found these vulnerabilities you sadly have to wait for our research paper. Until then ~:)

Credits

Paul Fiterau Brostean (@PaulTheGreatest) (Uppsala University)
Robert Merget (@ic0nz1) (Ruhr University Bochum)
Juraj Somorovsky (@jurajsomorovsky) (Ruhr University Bochum)
Kostis Sagonas (Uppsala University)
Bengt Jonsson (Uppsala University)
Joeri de Ruiter (@cypherpunknl)  (SIDN Labs)

 

 Responsible Disclosure

We reported our vulnerabilities to Oracle in September 2019. The patch for these issues was released on 14.01.2020.

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