#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200000 + 50;
int as[MAX_N], bs[MAX_N], bnums[MAX_N];
void printv(int n, int v[]) {
  for (int i = 0; i <= n; i++) printf("%d ", v[i]);
  putchar('\n');
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i <= n; i++) scanf("%d", as + i);
  int m = 0;
  long long co = 0;
  for (; m <= n || co != 0; m++) {
    long long a = co + (m <= n ? as[m] : 0);
    bs[m] = a - (a / 2) * 2;
    co = a / 2;
    bnums[m + 1] = bnums[m] + (bs[m] != 0 ? 1 : 0);
  }
  long long r = 0;
  int c = 0;
  for (int i = m - 1; i >= 0; i--) {
    r = r * 2 + bs[i];
    if (abs(r) > k * 2) break;
    if (i <= n && bnums[i] == 0) {
      long long d = r - as[i];
      if (abs(d) <= k && (i < n || d != 0)) c++;
    }
  }
  printf("%d\n", c);
  return 0;
}

### Prompt

Please create a solution in cpp 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 int MAX_N = 200000 + 50;
int as[MAX_N], bs[MAX_N], bnums[MAX_N];
void printv(int n, int v[]) {
  for (int i = 0; i <= n; i++) printf("%d ", v[i]);
  putchar('\n');
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i <= n; i++) scanf("%d", as + i);
  int m = 0;
  long long co = 0;
  for (; m <= n || co != 0; m++) {
    long long a = co + (m <= n ? as[m] : 0);
    bs[m] = a - (a / 2) * 2;
    co = a / 2;
    bnums[m + 1] = bnums[m] + (bs[m] != 0 ? 1 : 0);
  }
  long long r = 0;
  int c = 0;
  for (int i = m - 1; i >= 0; i--) {
    r = r * 2 + bs[i];
    if (abs(r) > k * 2) break;
    if (i <= n && bnums[i] == 0) {
      long long d = r - as[i];
      if (abs(d) <= k && (i < n || d != 0)) c++;
    }
  }
  printf("%d\n", c);
  return 0;
}

```