#include <bits/stdc++.h>
int n, k, ans;
long long s[200005], a[200005], b[200005];
void neg() {
  for (int i = 0; i <= n; i++) {
    a[i] = -a[i];
    b[i] = -b[i];
  }
}
void trans() {
  b[n]--;
  b[0] += 2;
  for (int i = 1; i < n; i++) b[i]++;
  for (int i = 0; i < n; i++) {
    b[i + 1] += b[i] / 2;
    b[i] %= 2;
  }
}
void print(long long x[]) {
  for (int i = 0; i <= n; i++) printf("%I64d ", x[i]);
  printf("\n");
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i <= n; i++) scanf("%I64d", &a[i]);
  for (int i = 0; i < n; i++) {
    b[i] += a[i];
    b[i + 1] += b[i] / 2;
    b[i] %= 2;
    if (b[i] < 0) {
      b[i + 1]--;
      b[i] += 2;
    }
  }
  b[n] += a[n];
  if (b[n] < 0) neg();
  trans();
  for (int i = n; i >= 0; i--) {
    if (s[i + 1] == 10ll * k + 1)
      s[i] = 10ll * k + 1;
    else
      s[i] = s[i + 1] * 2 + b[i];
    if (s[i] > 10ll * k) s[i] = 10ll * k + 1;
  }
  for (int i = 0; i <= n; i++) {
    if (i == n && llabs(s[i] - a[i]) <= k && s[i] != a[i]) ans++;
    if (i != n && llabs(s[i] - a[i]) <= k) ans++;
    if (b[i]) break;
  }
  printf("%d\n", ans);
  scanf("\n");
  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>
int n, k, ans;
long long s[200005], a[200005], b[200005];
void neg() {
  for (int i = 0; i <= n; i++) {
    a[i] = -a[i];
    b[i] = -b[i];
  }
}
void trans() {
  b[n]--;
  b[0] += 2;
  for (int i = 1; i < n; i++) b[i]++;
  for (int i = 0; i < n; i++) {
    b[i + 1] += b[i] / 2;
    b[i] %= 2;
  }
}
void print(long long x[]) {
  for (int i = 0; i <= n; i++) printf("%I64d ", x[i]);
  printf("\n");
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i <= n; i++) scanf("%I64d", &a[i]);
  for (int i = 0; i < n; i++) {
    b[i] += a[i];
    b[i + 1] += b[i] / 2;
    b[i] %= 2;
    if (b[i] < 0) {
      b[i + 1]--;
      b[i] += 2;
    }
  }
  b[n] += a[n];
  if (b[n] < 0) neg();
  trans();
  for (int i = n; i >= 0; i--) {
    if (s[i + 1] == 10ll * k + 1)
      s[i] = 10ll * k + 1;
    else
      s[i] = s[i + 1] * 2 + b[i];
    if (s[i] > 10ll * k) s[i] = 10ll * k + 1;
  }
  for (int i = 0; i <= n; i++) {
    if (i == n && llabs(s[i] - a[i]) <= k && s[i] != a[i]) ans++;
    if (i != n && llabs(s[i] - a[i]) <= k) ans++;
    if (b[i]) break;
  }
  printf("%d\n", ans);
  scanf("\n");
  return 0;
}

```