#include <bits/stdc++.h>
using std::cin;
using std::cout;
using std::endl;
using std::vector;
void solve() {
  int n;
  long long k;
  cin >> n >> k;
  vector<int> a(n + 1);
  for (int i = 0; i <= n; ++i) {
    cin >> a[i];
  }
  vector<int> b(n + 42, 0);
  for (int i = 0; i <= n; ++i) {
    b[i] += a[i];
    b[i + 1] += b[i] / 2;
    b[i] = b[i] % 2;
  }
  int pos = n + 1;
  while (abs(b[pos]) > 1) {
    b[pos + 1] += b[pos] / 2;
    b[pos] = b[pos] % 2;
    ++pos;
  }
  int one = 0;
  while (one < b.size() && b[one] == 0) {
    ++one;
  }
  if (one == b.size()) {
    cout << 0;
    return;
  }
  long long su = 0, ans = 0;
  for (int i = b.size() - 1; i >= 0 && std::abs(su) <= 2 * k; --i) {
    if (i <= one && i <= n) {
      if (su + b[i] >= 0 && su + b[i] <= k + a[i] &&
          (su + b[i] != a[i] || i < n)) {
        ++ans;
      } else if (su + b[i] < 0 && std::abs(su + b[i]) <= k - a[i] &&
                 (su + b[i] != a[i] || i < n)) {
        ++ans;
      }
    }
    su = su + b[i];
    su = su * 2;
  }
  cout << ans;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.setf(std::ios_base::fixed);
  cout.precision(28);
  solve();
  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 std::cin;
using std::cout;
using std::endl;
using std::vector;
void solve() {
  int n;
  long long k;
  cin >> n >> k;
  vector<int> a(n + 1);
  for (int i = 0; i <= n; ++i) {
    cin >> a[i];
  }
  vector<int> b(n + 42, 0);
  for (int i = 0; i <= n; ++i) {
    b[i] += a[i];
    b[i + 1] += b[i] / 2;
    b[i] = b[i] % 2;
  }
  int pos = n + 1;
  while (abs(b[pos]) > 1) {
    b[pos + 1] += b[pos] / 2;
    b[pos] = b[pos] % 2;
    ++pos;
  }
  int one = 0;
  while (one < b.size() && b[one] == 0) {
    ++one;
  }
  if (one == b.size()) {
    cout << 0;
    return;
  }
  long long su = 0, ans = 0;
  for (int i = b.size() - 1; i >= 0 && std::abs(su) <= 2 * k; --i) {
    if (i <= one && i <= n) {
      if (su + b[i] >= 0 && su + b[i] <= k + a[i] &&
          (su + b[i] != a[i] || i < n)) {
        ++ans;
      } else if (su + b[i] < 0 && std::abs(su + b[i]) <= k - a[i] &&
                 (su + b[i] != a[i] || i < n)) {
        ++ans;
      }
    }
    su = su + b[i];
    su = su * 2;
  }
  cout << ans;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.setf(std::ios_base::fixed);
  cout.precision(28);
  solve();
  return 0;
}

```