Here `p` is small, and so is `k`. We [bruteforce](https://github.com/CTF-STeam/ctf-writeups/blob/master/2021/UIUCTF/dhkectfintro_solve.py) all values of `k` from 0 to 30 and see which one gives us the flag.
```pythonciphertext = unhexlify('b31699d587f7daf8f6b23b30cfee0edca5d6a3594cd53e1646b9e72de6fc44fe7ad40f0ea6')
iv = bytes("kono DIO daaaaaa", encoding = 'ascii')for k in range(31):    key = pad_key(str(k))    cipher = AES.new(key, AES.MODE_CFB, iv)    flag = cipher.decrypt(ciphertext)    if b'uiuctf' in flag:        print(flag)```
Flag: `uiuctf{omae_ha_mou_shindeiru_b9e5f9}`