#include <bits/stdc++.h>
using namespace std;
const int nmax = 3 + 2 * 1e5;
const long long lim = 1e16 + 1;
long long v[nmax + 1];
long long st[nmax + 1], dr[nmax + 1];
inline long long valabs(long long x) { return (x < 0 ? -x : x); }
int main() {
  int n, k;
  fscanf(stdin, "%d%d", &n, &k);
  ++n;
  for (int i = 1; i <= n; ++i) {
    fscanf(stdin, "%I64d", &v[i]);
  }
  bool ok = 1;
  for (int i = 1; i <= n; ++i) {
    if (ok) {
      st[i] = (st[i - 1] >> 1) + v[i];
      if (st[i] & 1) {
        ok = 0;
      }
    } else {
      st[i] = lim;
    }
  }
  ok = 1;
  for (int i = n; i > 0; --i) {
    if (ok) {
      dr[i] = (dr[i + 1] << 1) + v[i];
      if (valabs(dr[i] << 1) >= lim) {
        ok = 0;
      }
    } else {
      dr[i] = lim;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (st[i] != lim && dr[i] != lim) {
      long long x = -(st[i] + dr[i] - v[i]);
      if (valabs(v[i] + x) <= k && (i != n || v[i] + x != 0)) {
        ++ans;
      }
    }
  }
  fprintf(stdout, "%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 int nmax = 3 + 2 * 1e5;
const long long lim = 1e16 + 1;
long long v[nmax + 1];
long long st[nmax + 1], dr[nmax + 1];
inline long long valabs(long long x) { return (x < 0 ? -x : x); }
int main() {
  int n, k;
  fscanf(stdin, "%d%d", &n, &k);
  ++n;
  for (int i = 1; i <= n; ++i) {
    fscanf(stdin, "%I64d", &v[i]);
  }
  bool ok = 1;
  for (int i = 1; i <= n; ++i) {
    if (ok) {
      st[i] = (st[i - 1] >> 1) + v[i];
      if (st[i] & 1) {
        ok = 0;
      }
    } else {
      st[i] = lim;
    }
  }
  ok = 1;
  for (int i = n; i > 0; --i) {
    if (ok) {
      dr[i] = (dr[i + 1] << 1) + v[i];
      if (valabs(dr[i] << 1) >= lim) {
        ok = 0;
      }
    } else {
      dr[i] = lim;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (st[i] != lim && dr[i] != lim) {
      long long x = -(st[i] + dr[i] - v[i]);
      if (valabs(v[i] + x) <= k && (i != n || v[i] + x != 0)) {
        ++ans;
      }
    }
  }
  fprintf(stdout, "%d\n", ans);
  return 0;
}

```