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 $P(x) = \sum_{i = 0}^{n} a_{i} \cdot x^{i}$. 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.

-----Input-----

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 $P(x) = \sum_{i = 0}^{n} a_{i} \cdot x^{i}$. It's guaranteed that P(2) ≠ 0.

-----Output-----

Print the number of ways to change one coefficient to get a valid polynomial Q that Q(2) = 0.

-----Examples-----

Input 3 1000000000 10 -9 -3 5

Output 3

Input 3 12 10 -9 -3 5

Output 2

Input 2 20 14 -7 19

Output 0

-----Note-----

In the first sample, we are given a polynomial P(x) = 10 - 9x - 3x^2 + 5x^3.

Limak can change one coefficient in three ways: He can set a_0 = - 10. Then he would get Q(x) = - 10 - 9x - 3x^2 + 5x^3 and indeed Q(2) = - 10 - 18 - 12 + 40 = 0. Or he can set a_2 = - 8. Then Q(x) = 10 - 9x - 8x^2 + 5x^3 and indeed Q(2) = 10 - 18 - 32 + 40 = 0. Or he can set a_1 = - 19. Then Q(x) = 10 - 19x - 3x^2 + 5x^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.