#include <bits/stdc++.h>
using namespace std;
long long dx[] = {1, -1, 0, 0};
long long dy[] = {0, 0, 1, -1};
const long long sz = 1e5 + 5, mod = 1e9 + 7;
long long t, n, m, k, p;
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long Inv(long long a) { return power(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    cin >> n >> k;
    n++;
    vector<long long> a(n, 0), b(n, 0);
    for (int i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a[i] = x;
    }
    long long z = 0;
    int up = 0;
    for (int i = 0; i < n; i++) {
      z += a[i];
      b[i] = z;
      if (z % 2 != 0) {
        break;
      }
      up++;
      z /= 2;
    }
    long long value = 0;
    int ans = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (i <= up) {
        long long tmp = value * 2 + b[i];
        long long z = a[i] - tmp;
        if (-k <= z && z <= k && (i != n - 1 || z != 0)) {
          ans++;
        }
      }
      value = value * 2 + a[i];
      if (value < -1e18 || value > 1e18) {
        break;
      }
    }
    cout << ans;
  }
  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 dx[] = {1, -1, 0, 0};
long long dy[] = {0, 0, 1, -1};
const long long sz = 1e5 + 5, mod = 1e9 + 7;
long long t, n, m, k, p;
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long Inv(long long a) { return power(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    cin >> n >> k;
    n++;
    vector<long long> a(n, 0), b(n, 0);
    for (int i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a[i] = x;
    }
    long long z = 0;
    int up = 0;
    for (int i = 0; i < n; i++) {
      z += a[i];
      b[i] = z;
      if (z % 2 != 0) {
        break;
      }
      up++;
      z /= 2;
    }
    long long value = 0;
    int ans = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (i <= up) {
        long long tmp = value * 2 + b[i];
        long long z = a[i] - tmp;
        if (-k <= z && z <= k && (i != n - 1 || z != 0)) {
          ans++;
        }
      }
      value = value * 2 + a[i];
      if (value < -1e18 || value > 1e18) {
        break;
      }
    }
    cout << ans;
  }
  return 0;
}

```