(1 point) Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl. We conduct a known-plaintext attack against an LFSR. Through trial and error we have determined that the number of states is m= 4. The plaintext given by 10101100 = 1021121314151617 when encrypted by the LFSR produced the ciphertext 01110000 = yoy1929393959697 What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent p3 = 0,P2 = 1, P1 = 0, po = 1)