#include <bits/stdc++.h>
using namespace std;
long long pr[211111], la[211111], a[211111];
int n;
long long k;
int main() {
  while (scanf("%d%lld", &n, &k) == 2) {
    memset(pr, 0, sizeof pr);
    memset(la, 0, sizeof la);
    n++;
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    int l = 1, r = n;
    for (int i = 1; i <= n; i++) {
      if (i == 1) {
        pr[i] = a[i];
      } else if (pr[i - 1] % 2 == 0) {
        l = i;
        pr[i] = pr[i - 1] / 2 + a[i];
      } else
        break;
    }
    for (int i = n; i >= 1; i--) {
      if (i == n) {
        la[i] = a[i];
      } else if (abs(la[i + 1] * 2 + a[i]) >= k) {
        la[i] = la[i + 1] * 2 + a[i];
        break;
      } else {
        la[i] = la[i + 1] * 2 + a[i];
        r = i;
      }
    }
    pr[0] = la[n + 1] = 0;
    if (r > 1) r--;
    if (l < r) {
      printf("0\n");
      continue;
    }
    int cnt = 0;
    for (int i = min(l, r); i <= max(l, r); i++) {
      long long ans = -1 * pr[i - 1] / 2 - 2 * la[i + 1];
      if (pr[i - 1] % 2 != 0) continue;
      if (i == n && ans == 0) continue;
      if (abs(ans) <= k) {
        cnt++;
      }
    }
    printf("%d\n", cnt);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long pr[211111], la[211111], a[211111];
int n;
long long k;
int main() {
  while (scanf("%d%lld", &n, &k) == 2) {
    memset(pr, 0, sizeof pr);
    memset(la, 0, sizeof la);
    n++;
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    int l = 1, r = n;
    for (int i = 1; i <= n; i++) {
      if (i == 1) {
        pr[i] = a[i];
      } else if (pr[i - 1] % 2 == 0) {
        l = i;
        pr[i] = pr[i - 1] / 2 + a[i];
      } else
        break;
    }
    for (int i = n; i >= 1; i--) {
      if (i == n) {
        la[i] = a[i];
      } else if (abs(la[i + 1] * 2 + a[i]) >= k) {
        la[i] = la[i + 1] * 2 + a[i];
        break;
      } else {
        la[i] = la[i + 1] * 2 + a[i];
        r = i;
      }
    }
    pr[0] = la[n + 1] = 0;
    if (r > 1) r--;
    if (l < r) {
      printf("0\n");
      continue;
    }
    int cnt = 0;
    for (int i = min(l, r); i <= max(l, r); i++) {
      long long ans = -1 * pr[i - 1] / 2 - 2 * la[i + 1];
      if (pr[i - 1] % 2 != 0) continue;
      if (i == n && ans == 0) continue;
      if (abs(ans) <= k) {
        cnt++;
      }
    }
    printf("%d\n", cnt);
  }
  return 0;
}

```