#include <bits/stdc++.h>
using namespace std;
int arr[200010], n, k, init[200010];
void work() {
  for (int i = 0; i < n; ++i) {
    arr[i + 1] += arr[i] / 2;
    arr[i] %= 2;
    if (arr[i] == -1) arr[i] = 1, arr[i + 1] -= 1;
  }
  if (arr[n] < 0) {
    for (int i = 0; i < n + 1; ++i) arr[i] *= -1, init[i] *= -1;
    work();
  }
}
int main() {
  scanf("%d", &n), scanf("%d", &k);
  for (int i = 0; i < n + 1; ++i) scanf("%d", &arr[i]), init[i] = arr[i];
  work();
  int fnz = 0;
  while (!arr[fnz]) fnz++;
  int out = 0;
  for (long long i = n, val = 0; i >= 0; i--) {
    val = 2 * val + arr[i];
    if (val > 1e10) break;
    if (abs(init[i] - val) > k) continue;
    if (fnz >= i && (i != n || val != init[n])) out++;
  }
  cout << out;
}

### Prompt

Generate 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 arr[200010], n, k, init[200010];
void work() {
  for (int i = 0; i < n; ++i) {
    arr[i + 1] += arr[i] / 2;
    arr[i] %= 2;
    if (arr[i] == -1) arr[i] = 1, arr[i + 1] -= 1;
  }
  if (arr[n] < 0) {
    for (int i = 0; i < n + 1; ++i) arr[i] *= -1, init[i] *= -1;
    work();
  }
}
int main() {
  scanf("%d", &n), scanf("%d", &k);
  for (int i = 0; i < n + 1; ++i) scanf("%d", &arr[i]), init[i] = arr[i];
  work();
  int fnz = 0;
  while (!arr[fnz]) fnz++;
  int out = 0;
  for (long long i = n, val = 0; i >= 0; i--) {
    val = 2 * val + arr[i];
    if (val > 1e10) break;
    if (abs(init[i] - val) > k) continue;
    if (fnz >= i && (i != n || val != init[n])) out++;
  }
  cout << out;
}

```