from pwn import *
frontPadding = '.' * 43
backPadding = '.' * 3

# implementation of the poodle attack
def send(frontPadding, backPadding):
# returns false and an empty array if unsuccessful, if unsuccessful, returns true and an array holding the char of interest
	p = remote('2018shell1.picoctf.com', 14263)
	p.recvuntil('(S)')
	p.sendline('e')
	p.recvuntil('report: ')
	p.sendline(frontPadding)
	p.recvuntil('else? ')
	p.sendline(backPadding)
	outputStr = p.recvuntil('(S)').split('\n')[0][11:] # getting the hex output
	output = []
	for i in range(0, len(outputStr) - 32, 32):
		output.append(outputStr[i:i+32])
	# we are interested in block 8 and its previous block, 7, remember first block is the randomized iv
	interestBlock = output[8]
	interestBlockPrev = output[7]
	# also need second to last block for decryption if successful
	interestBlockLast = output[13]
	# replace padding block (block 14) with interestBlock
	exploit = outputStr[0:448] + interestBlock
	# sending in the probable exploit
	p.sendline('s')
	p.recvuntil('message: ')
	p.sendline(exploit)
	result = p.recvall()
	p.close()
	if 'Successful decryption' not in result:
		return False, []
	else:
		return True, [(xor(xor(0x10, interestBlockLast[-1].decode('hex')), interestBlockPrev[-1].decode('hex'))]

#exploiting
flag = ''
result = False
output = []
# Exception catching
repeat = True
for i in range(29):
	while repeat:
		try:
			result, output = send(frontPadding[i:], backPadding + '.' * i)
			repeat = False
			if result:
				print '\033[92mSUCCESS ON ONE CHARACTER\033[0m'
				print '\033[92m' + chr(output[0]) + '\033[0m'
				break
			else:
				repeat = True
		except Exception as e:
			print 'Error... retrying'
	output += chr(output[0])
	print output
print output