Discrete Superlog

We define a^^b to be such that a^^0 = 1 and a^^b = a^(a^^(b-1)), where x^y represents x to the power of y.
Given this, find a positive integer x such that a^^x = b mod p.
Generating challenge 1 of 10...
p = 1169174076520537727987
a = 349507167920263434300
b = 349507167920263434300
Enter x: 

import pwn, random

got_flag = False
while not got_flag:
    conn = pwn.remote('crypto.2020.chall.actf.co', 20603)
    conn.recvuntil(b'a^^x = b mod p.\n')
    for i in range(1, 11):
        res = conn.recv()
        if b'Wrong!' in res:
            print(f'Challenge {i-1} failed..')
            break
        sp = res.decode().split('\n')
        if len(sp) <= 4:
            res = conn.recv()
            sp = res.decode().split('\n')
        p = int(sp[-4].split(' ')[2])
        a = int(sp[-3].split(' ')[2])
        b = int(sp[-2].split(' ')[2])
        if (a == b):
            choice = 1
        else:
            choice = pow(b, a, p)
        print(f'Challenge {i}: Sending {choice}')
        conn.send((str(choice) + '\n').encode())
    if i == 10:
        res = conn.recv()
        print(res)
        if not b'Wrong!' in res:
            got_flag = True
    conn.close()