Limak is a little polar bear. He doesn't have many toys and thus he often plays with polynomials. He considers a polynomial valid if its degree is n and its coefficients are integers not exceeding k by the absolute value. More formally: Let a 0 , a 1 , ..., a n denote the coefficients, so . Then, a polynomial P ( x ) is valid if all the following conditions are satisfied: a i is integer for every i ; | a i | ≤ k for every i ; a n ≠ 0 . Limak has recently got a valid polynomial P with coefficients a 0 , a 1 , a 2 , ..., a n . He noticed that P (2) ≠ 0 and he wants to change it. He is going to change one coefficient to get a valid polynomial Q of degree n that Q (2) = 0 . Count the number of ways to do so. You should count two ways as a distinct if coefficients of target polynoms differ. The first line contains two integers n and k ( 1 ≤ n ≤ 200 000, 1 ≤ k ≤ 10 9 ) — the degree of the polynomial and the limit for absolute values of coefficients. The second line contains n + 1 integers a 0 , a 1 , ..., a n ( | a i | ≤ k , a n ≠ 0 ) — describing a valid polynomial . It's guaranteed that P (2) ≠ 0 . Print the number of ways to change one coefficient to get a valid polynomial Q that Q (2) = 0 . In the first sample, we are given a polynomial P ( x ) = 10 - 9 x - 3 x 2 + 5 x 3 . Limak can change one coefficient in three ways: He can set a 0 = - 10 . Then he would get Q ( x ) = - 10 - 9 x - 3 x 2 + 5 x 3 and indeed Q (2) = - 10 - 18 - 12 + 40 = 0 . Or he can set a 2 = - 8 . Then Q ( x ) = 10 - 9 x - 8 x 2 + 5 x 3 and indeed Q (2) = 10 - 18 - 32 + 40 = 0 . Or he can set a 1 = - 19 . Then Q ( x ) = 10 - 19 x - 3 x 2 + 5 x 3 and indeed Q (2) = 10 - 38 - 12 + 40 = 0 . In the second sample, we are given the same polynomial. This time though, k is equal to 12 instead of 10 9 . Two first of ways listed above are still valid but in the third way we would get | a 1 | > k what is not allowed. Thus, the answer is 2 this time.