#include <bits/stdc++.h>
using namespace std;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
int N, K, A[(1 << 18)], I, C, S, W;
long long dp[(1 << 18)];
int main(void) {
  scanf("%d%d", &N, &K), ga(I = ++N, A);
  while (~--I) dp[I] = llabs(dp[I + 1]) > K ? dp[I + 1] : A[I] + 2 * dp[I + 1];
  for (int i(0); i < N && (C + 1) & 1; i++) {
    C >>= 1;
    if (llabs(dp[i + 1]) <= K)
      W = C + dp[i] - A[i], S += llabs(W) <= K && (i ^ (N - 1) || W);
    C += A[i];
  }
  printf("%d\n", S);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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, ..., an denote the coefficients, so <image>. Then, a polynomial P(x) is valid if all the following conditions are satisfied:

  * ai is integer for every i; 
  * |ai| ≤ k for every i; 
  * an ≠ 0. 



Limak has recently got a valid polynomial P with coefficients a0, a1, a2, ..., an. 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, ..., an (|ai| ≤ k, an ≠ 0) — describing a valid polynomial <image>. 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 - 3x2 + 5x3.

Limak can change one coefficient in three ways:

  1. He can set a0 = - 10. Then he would get Q(x) = - 10 - 9x - 3x2 + 5x3 and indeed Q(2) = - 10 - 18 - 12 + 40 = 0. 
  2. Or he can set a2 = - 8. Then Q(x) = 10 - 9x - 8x2 + 5x3 and indeed Q(2) = 10 - 18 - 32 + 40 = 0. 
  3. 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
int N, K, A[(1 << 18)], I, C, S, W;
long long dp[(1 << 18)];
int main(void) {
  scanf("%d%d", &N, &K), ga(I = ++N, A);
  while (~--I) dp[I] = llabs(dp[I + 1]) > K ? dp[I + 1] : A[I] + 2 * dp[I + 1];
  for (int i(0); i < N && (C + 1) & 1; i++) {
    C >>= 1;
    if (llabs(dp[i + 1]) <= K)
      W = C + dp[i] - A[i], S += llabs(W) <= K && (i ^ (N - 1) || W);
    C += A[i];
  }
  printf("%d\n", S);
  return 0;
}

```