Consider the 3x3 Hill cipher working modulo 2, with the key M= 1 0 1 0 1 1 0 0 1 (a) (10 points) Encrypt the following plaintext (spaces added for clarity) in CBC mode using this key with initial value Co = 111: 100 100 100 (b) (10 points) The following ciphertext was encrypted in CBC mode using this key with initial value Co = 010 (not displayed below). 110 000 111 Given that M-1 = M (mod 2), decrypt this message and obtain the original plaintext