#include <bits/stdc++.h>
using namespace std;
template <typename T>
T sqr(T x) {
  return x * x;
}
const int MAXBIT = 32;
int main() {
  int n;
  long long k;
  cin >> n >> k;
  ++n;
  vector<long long> v(n);
  for (int i = 0; i < int(n); i++) {
    cin >> v[i];
  }
  vector<long long> p = v;
  p.resize(2 * n + MAXBIT + 1, 0);
  for (int i = 0; i < int(p.size() - 1); i++) {
    long long sign = p[i] < 0 ? -1 : 1;
    long long a = abs(p[i]);
    p[i + 1] += (a >> 1) * sign;
    p[i] = (a & 1) * sign;
    if (p[i] < 0) {
      p[i] = 1;
      p[i + 1] -= 1;
    }
  }
  long long sign = p.back() < 0 ? -1 : 1;
  if (p.back() < 0) {
    p.back() = 0;
    for (int i = 0; i < int(p.size() - 1); i++) {
      p[i] = !p[i];
    }
    p[0] += 1;
    for (int i = 0; i < int(p.size() - 1); i++) {
      p[i + 1] += p[i] >> 1;
      p[i] = p[i] & 1;
    }
  }
  int total = 0;
  for (int i = 0; i < int(p.size()); i++) {
    total += p[i];
  }
  int ilast = 0;
  int inwnd = 0;
  int res = 0;
  for (int i = 0; i < int(n); i++) {
    while (ilast - i < MAXBIT) {
      inwnd += p[ilast++];
    }
    if (inwnd != total) continue;
    long long val = 0;
    for (int j = 0; j < int(MAXBIT); j++) {
      if (p[i + j]) {
        val |= 1ll << j;
      }
    }
    val *= sign;
    val -= v[i];
    if (abs(val) <= k && !(i == n - 1 && val == 0)) ++res;
    inwnd -= p[i];
  }
  cout << res << endl;
  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;
template <typename T>
T sqr(T x) {
  return x * x;
}
const int MAXBIT = 32;
int main() {
  int n;
  long long k;
  cin >> n >> k;
  ++n;
  vector<long long> v(n);
  for (int i = 0; i < int(n); i++) {
    cin >> v[i];
  }
  vector<long long> p = v;
  p.resize(2 * n + MAXBIT + 1, 0);
  for (int i = 0; i < int(p.size() - 1); i++) {
    long long sign = p[i] < 0 ? -1 : 1;
    long long a = abs(p[i]);
    p[i + 1] += (a >> 1) * sign;
    p[i] = (a & 1) * sign;
    if (p[i] < 0) {
      p[i] = 1;
      p[i + 1] -= 1;
    }
  }
  long long sign = p.back() < 0 ? -1 : 1;
  if (p.back() < 0) {
    p.back() = 0;
    for (int i = 0; i < int(p.size() - 1); i++) {
      p[i] = !p[i];
    }
    p[0] += 1;
    for (int i = 0; i < int(p.size() - 1); i++) {
      p[i + 1] += p[i] >> 1;
      p[i] = p[i] & 1;
    }
  }
  int total = 0;
  for (int i = 0; i < int(p.size()); i++) {
    total += p[i];
  }
  int ilast = 0;
  int inwnd = 0;
  int res = 0;
  for (int i = 0; i < int(n); i++) {
    while (ilast - i < MAXBIT) {
      inwnd += p[ilast++];
    }
    if (inwnd != total) continue;
    long long val = 0;
    for (int j = 0; j < int(MAXBIT); j++) {
      if (p[i + j]) {
        val |= 1ll << j;
      }
    }
    val *= sign;
    val -= v[i];
    if (abs(val) <= k && !(i == n - 1 && val == 0)) ++res;
    inwnd -= p[i];
  }
  cout << res << endl;
  return 0;
}

```