#include <bits/stdc++.h>
const long long Inf = 1ll << 50;
const long long N = 200010;
using namespace std;
long long cl[2 * N], b, c, d, i, j, k, l, n, ans, a[N], y, kk;
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (i = n + 1; i > 0; i--) scanf("%I64d", &a[i]);
  memset(cl, 0, sizeof(cl));
  cl[0] = a[0] = 0;
  for (i = 1; i <= n + 1; i++) {
    cl[i] = (cl[i - 1] + a[i - 1]) << 1ll;
    if ((cl[i] > Inf) || (cl[i] < -Inf)) break;
  }
  y = 0;
  ans = 0;
  i--;
  for (j = n + 1; j > i; j--) {
    if ((y & 1) == 0)
      y >>= 1ll;
    else
      break;
    y += a[j];
  }
  if (i == j) {
    for (kk = i; kk > 0; kk--) {
      if ((y % 2) == 0)
        y /= 2ll;
      else
        break;
      if (cl[kk] + y - y == cl[kk] && abs(cl[kk] + y) <= k &&
          (!(kk == 1 && (cl[kk] + y == 0))))
        ans++;
      y += a[kk];
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
const long long Inf = 1ll << 50;
const long long N = 200010;
using namespace std;
long long cl[2 * N], b, c, d, i, j, k, l, n, ans, a[N], y, kk;
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (i = n + 1; i > 0; i--) scanf("%I64d", &a[i]);
  memset(cl, 0, sizeof(cl));
  cl[0] = a[0] = 0;
  for (i = 1; i <= n + 1; i++) {
    cl[i] = (cl[i - 1] + a[i - 1]) << 1ll;
    if ((cl[i] > Inf) || (cl[i] < -Inf)) break;
  }
  y = 0;
  ans = 0;
  i--;
  for (j = n + 1; j > i; j--) {
    if ((y & 1) == 0)
      y >>= 1ll;
    else
      break;
    y += a[j];
  }
  if (i == j) {
    for (kk = i; kk > 0; kk--) {
      if ((y % 2) == 0)
        y /= 2ll;
      else
        break;
      if (cl[kk] + y - y == cl[kk] && abs(cl[kk] + y) <= k &&
          (!(kk == 1 && (cl[kk] + y == 0))))
        ans++;
      y += a[kk];
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```