#include <bits/stdc++.h>
using namespace std;
const long long inf = 1LL << 60;
long long a[200005], b[200005], c[200005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i <= n; ++i) scanf("%lld", a + i);
  memcpy(b, a, sizeof a);
  int r = n;
  for (int i = 0; i < n; ++i) {
    if (b[i] % 2 == 0) {
      b[i + 1] += b[i] / 2;
    } else {
      r = i;
      break;
    }
  }
  memcpy(c, a, sizeof a);
  int l = 0;
  for (int i = n; i > 0; --i) {
    if (abs(c[i]) < inf) {
      c[i - 1] += c[i] * 2;
    } else {
      l = i;
      break;
    }
  }
  int cnt = 0;
  for (int i = l; i <= r; ++i) {
    if (abs(c[i] + b[i] - 2 * a[i]) <= k) ++cnt;
    if (i == n && b[i] + c[i] - 2 * a[i] == 0) --cnt;
  }
  printf("%d\n", cnt);
  return 0;
}

### Prompt

Your task is to create 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;
const long long inf = 1LL << 60;
long long a[200005], b[200005], c[200005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i <= n; ++i) scanf("%lld", a + i);
  memcpy(b, a, sizeof a);
  int r = n;
  for (int i = 0; i < n; ++i) {
    if (b[i] % 2 == 0) {
      b[i + 1] += b[i] / 2;
    } else {
      r = i;
      break;
    }
  }
  memcpy(c, a, sizeof a);
  int l = 0;
  for (int i = n; i > 0; --i) {
    if (abs(c[i]) < inf) {
      c[i - 1] += c[i] * 2;
    } else {
      l = i;
      break;
    }
  }
  int cnt = 0;
  for (int i = l; i <= r; ++i) {
    if (abs(c[i] + b[i] - 2 * a[i]) <= k) ++cnt;
    if (i == n && b[i] + c[i] - 2 * a[i] == 0) --cnt;
  }
  printf("%d\n", cnt);
  return 0;
}

```