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 a₀, a₁, ..., a_n denote the coefficients, so $\text{[math]}$ . 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 a₀, a₁, a₂, ..., 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 ≤ 10⁹) — the degree of the polynomial and the limit for absolute values of coefficients.

The second line contains n + 1 integers a₀, a₁, ..., a_n (|a_i| ≤ k, a_n ≠ 0) — describing a valid polynomial $\text{[math]}$ . 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.

Note

In the first sample, we are given a polynomial P(x) = 10 - 9x - 3x² + 5x³.

Limak can change one coefficient in three ways:

He can set a₀ = - 10. Then he would get Q(x) = - 10 - 9x - 3x² + 5x³ and indeed Q(2) = - 10 - 18 - 12 + 40 = 0.
Or he can set a₂ = - 8. Then Q(x) = 10 - 9x - 8x² + 5x³ and indeed Q(2) = 10 - 18 - 32 + 40 = 0.
Or he can set a₁ = - 19. Then Q(x) = 10 - 19x - 3x² + 5x³ 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 10⁹. Two first of ways listed above are still valid but in the third way we would get |a₁| > k what is not allowed. Thus, the answer is 2 this time.

Bear and Polynomials

Examples