Consider a simple one-time pad for binary string of length 2 in which the message space, ciphertext space, and key space are defined as M=C=K={00,01,10,11}. It is known that the one-time pad is perfectly secret if the key k is chosen uniformly random, i.e., P[k=00]=P[k=01]=P[10]=P[11]=41. This can be proved by showing that, for any m1,m2,c, the method satisfies P[Enc(k1,m1)=c]=P[Enc(k2,m2)=c]. Let's look at the first probability. Given m1 and c, P[Enc(k,m1)=c]=P[km1=c]=P[k=cm1]=41. The above means that m1 will be encrypted into c if the specific key k=cm1 is chosen. Since in one-time pad all keys are equally likely, m1 has the probability of 41 to be encrypted into c. Observe that, with the same reasoning, m2 also has the same probability to be encrypted into c, i.e., P[Enc(k,m2)=c]=41. Suppose that you decided to remove the key k=00 from the key space as it sends the message in clear. When k=00, we have c=Enc(k,m)=00m=m. Prove or disprove whether the modified method is still perfectly secret. In other words, either show that (1) still holds true with k=00 removed or provide a counter example (m1,m2,c) for which P[Enc(k1,m1)=c]=P[Enc(k2,m2)=c]. Hint: given fixed m and c, there exists a unique k such that mk=c