#include <bits/stdc++.h>
using namespace std;
int mas[(int)3e5];
long long a[(int)3e5];
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cin.tie(0);
  int n;
  long long mx;
  cin >> n >> mx;
  n++;
  for (int i = 0; i < (int)3e5; i++) mas[i] = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    int t = 1;
    if (a[i] < 0) {
      t = -1;
      a[i] *= -1;
    }
    for (int j = 0; j < 30; j++) {
      mas[i + j] += t * (int)((a[i] & (1 << j)) > 0);
    }
    a[i] *= t;
  }
  bool d = 0;
  for (int i = 0; i < 3e5; i++) {
    if (mas[i] > 1) {
      mas[i + 1] += (mas[i] >> 1);
      mas[i] = (mas[i] & 1);
    } else if (mas[i] < 0) {
      if (i > (int)2e5 + 40) {
        for (int j = 0; j < n; j++) a[j] *= -1;
        for (int j = 0; j <= i; j++) mas[j] *= -1;
        d = 1;
        break;
      }
      if ((mas[i] % 2) == 0) {
        mas[i + 1] += mas[i] / 2;
        mas[i] = 0;
      } else {
        mas[i + 1] -= (1 - mas[i]) / 2;
        mas[i] = 1;
      }
    }
  }
  if (d) {
    for (int i = 0; i < 3e5; i++) {
      if (mas[i] > 1) {
        mas[i + 1] += (mas[i] >> 1);
        mas[i] = (mas[i] & 1);
      } else if (mas[i] < 0) {
        if (i > (int)2e5 + 40) {
          for (int j = 0; j < n; j++) a[j] *= -1;
          for (int j = 0; j <= i; j++) mas[j] *= -1;
          d = 1;
          break;
        }
        if (mas[i] % 2 == 0) {
          mas[i + 1] += mas[i] / 2;
          mas[i] = 0;
        } else {
          mas[i + 1] -= (1 - mas[i]) / 2;
          mas[i] = 1;
        }
      }
    }
  }
  int q = -1;
  for (int i = 0; i < (int)3e5; i++)
    if (mas[i] == 1) q = i;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (q - i + 1 < 33) {
      long long b = 0;
      for (int j = 0; j < 33; j++) {
        b += ((long long)mas[i + j] << j);
      }
      if (abs(a[i] - b) <= mx && (i != n - 1 || a[i] - b != 0)) {
        ans++;
      }
    }
    if (mas[i] != 0) break;
  }
  cout << ans << endl;
  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;
int mas[(int)3e5];
long long a[(int)3e5];
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cin.tie(0);
  int n;
  long long mx;
  cin >> n >> mx;
  n++;
  for (int i = 0; i < (int)3e5; i++) mas[i] = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    int t = 1;
    if (a[i] < 0) {
      t = -1;
      a[i] *= -1;
    }
    for (int j = 0; j < 30; j++) {
      mas[i + j] += t * (int)((a[i] & (1 << j)) > 0);
    }
    a[i] *= t;
  }
  bool d = 0;
  for (int i = 0; i < 3e5; i++) {
    if (mas[i] > 1) {
      mas[i + 1] += (mas[i] >> 1);
      mas[i] = (mas[i] & 1);
    } else if (mas[i] < 0) {
      if (i > (int)2e5 + 40) {
        for (int j = 0; j < n; j++) a[j] *= -1;
        for (int j = 0; j <= i; j++) mas[j] *= -1;
        d = 1;
        break;
      }
      if ((mas[i] % 2) == 0) {
        mas[i + 1] += mas[i] / 2;
        mas[i] = 0;
      } else {
        mas[i + 1] -= (1 - mas[i]) / 2;
        mas[i] = 1;
      }
    }
  }
  if (d) {
    for (int i = 0; i < 3e5; i++) {
      if (mas[i] > 1) {
        mas[i + 1] += (mas[i] >> 1);
        mas[i] = (mas[i] & 1);
      } else if (mas[i] < 0) {
        if (i > (int)2e5 + 40) {
          for (int j = 0; j < n; j++) a[j] *= -1;
          for (int j = 0; j <= i; j++) mas[j] *= -1;
          d = 1;
          break;
        }
        if (mas[i] % 2 == 0) {
          mas[i + 1] += mas[i] / 2;
          mas[i] = 0;
        } else {
          mas[i + 1] -= (1 - mas[i]) / 2;
          mas[i] = 1;
        }
      }
    }
  }
  int q = -1;
  for (int i = 0; i < (int)3e5; i++)
    if (mas[i] == 1) q = i;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (q - i + 1 < 33) {
      long long b = 0;
      for (int j = 0; j < 33; j++) {
        b += ((long long)mas[i + j] << j);
      }
      if (abs(a[i] - b) <= mx && (i != n - 1 || a[i] - b != 0)) {
        ans++;
      }
    }
    if (mas[i] != 0) break;
  }
  cout << ans << endl;
  return 0;
}

```