#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, k;
  cin >> n >> k;
  n++;
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<long long> res = arr;
  long long h = n - 1;
  for (long long i = 0; i < n - 1; i++) {
    if (res[i] % 2 != 0) {
      h = i;
      break;
    }
    res[i + 1] = res[i + 1] + (res[i] / 2);
    res[i] %= 2;
  }
  long long sum = 0;
  bool is = 1;
  for (long long i = n - 1; i > h; i--) {
    sum += arr[i];
    if (abs(sum) > 20 * k) {
      is = 0;
      break;
    }
    sum *= 2;
    if (abs(sum) > 20 * k) {
      is = 0;
      break;
    }
  }
  long long ans = 0;
  if (is) {
    for (long long i = h; i >= 0; i--) {
      if (i == n - 1) {
        if (arr[h] == res[h] && (abs(sum - arr[i]) <= k)) {
          ans--;
        }
      }
      sum += res[i];
      if (abs(sum - arr[i]) <= k) {
        ans++;
      }
      sum *= 2;
      if (abs(sum) > 20 * k) {
        break;
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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 main() {
  long long n, k;
  cin >> n >> k;
  n++;
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<long long> res = arr;
  long long h = n - 1;
  for (long long i = 0; i < n - 1; i++) {
    if (res[i] % 2 != 0) {
      h = i;
      break;
    }
    res[i + 1] = res[i + 1] + (res[i] / 2);
    res[i] %= 2;
  }
  long long sum = 0;
  bool is = 1;
  for (long long i = n - 1; i > h; i--) {
    sum += arr[i];
    if (abs(sum) > 20 * k) {
      is = 0;
      break;
    }
    sum *= 2;
    if (abs(sum) > 20 * k) {
      is = 0;
      break;
    }
  }
  long long ans = 0;
  if (is) {
    for (long long i = h; i >= 0; i--) {
      if (i == n - 1) {
        if (arr[h] == res[h] && (abs(sum - arr[i]) <= k)) {
          ans--;
        }
      }
      sum += res[i];
      if (abs(sum - arr[i]) <= k) {
        ans++;
      }
      sum *= 2;
      if (abs(sum) > 20 * k) {
        break;
      }
    }
  }
  cout << ans;
  return 0;
}

```