#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  long long C = -1e13;
  C--;
  vector<int> A(n + 1);
  for (int i = 0; i <= n; i++) cin >> A[i];
  vector<long long> L(n, C), R(n, C);
  if (A[0] % 2 == 0) L[0] = -A[0] / 2;
  R[n - 1] = A[n];
  for (int i = 1; i < n; i++) {
    R[n - 1 - i] = A[n - i] + 2 * R[n - i];
    if (R[n - 1 - i] > 1e13 || R[n - 1 - i] < -1e13) R[n - 1 - i] = C;
    if (L[i - 1] == C)
      L[i] = C;
    else {
      L[i] = -A[i] + L[i - 1];
      if (L[i] % 2 == 0)
        L[i] = L[i] / 2;
      else
        L[i] = C;
    }
  }
  int ans = 0;
  long long newc = -2 * R[0];
  if (llabs(newc) <= k) ans++;
  for (int i = 1; i < n; i++) {
    long long newc = -2 * R[i] + L[i - 1];
    if (llabs(newc) <= k) ans++;
  }
  newc = L[n - 1];
  if (llabs(newc) <= k && newc != 0) ans++;
  printf("%d", ans);
  return 0;
}

### Prompt

Please formulate a CPP solution 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() {
  int n, k;
  cin >> n >> k;
  long long C = -1e13;
  C--;
  vector<int> A(n + 1);
  for (int i = 0; i <= n; i++) cin >> A[i];
  vector<long long> L(n, C), R(n, C);
  if (A[0] % 2 == 0) L[0] = -A[0] / 2;
  R[n - 1] = A[n];
  for (int i = 1; i < n; i++) {
    R[n - 1 - i] = A[n - i] + 2 * R[n - i];
    if (R[n - 1 - i] > 1e13 || R[n - 1 - i] < -1e13) R[n - 1 - i] = C;
    if (L[i - 1] == C)
      L[i] = C;
    else {
      L[i] = -A[i] + L[i - 1];
      if (L[i] % 2 == 0)
        L[i] = L[i] / 2;
      else
        L[i] = C;
    }
  }
  int ans = 0;
  long long newc = -2 * R[0];
  if (llabs(newc) <= k) ans++;
  for (int i = 1; i < n; i++) {
    long long newc = -2 * R[i] + L[i - 1];
    if (llabs(newc) <= k) ans++;
  }
  newc = L[n - 1];
  if (llabs(newc) <= k && newc != 0) ans++;
  printf("%d", ans);
  return 0;
}

```