#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 1;
const double pi = acos(-1);
int mas[200100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  int n;
  int k;
  cin >> n >> k;
  int t;
  vector<int> coef(n + 1);
  for (int i = 0; i <= n; ++i) {
    cin >> t;
    coef[i] = t;
    int modif = 1;
    if (t < 0) {
      modif = -1;
      t = -t;
    }
    int j = 0;
    while (t > 0) {
      mas[i + j] += (t % 2) * modif;
      j++;
      t /= 2;
    }
  }
  int end = 0;
  for (int i = 200099; i >= 0; --i) {
    if (mas[i] != 0) {
      end = i;
      break;
    }
  }
  int n_coeff = -1;
  int i = 0;
  while (i <= end) {
    if (mas[i] > 1) {
      int t = mas[i] / 2;
      mas[i + 1] += t;
      mas[i] = mas[i] % 2;
      if (i == end) end++;
    } else if (mas[i] < 0) {
      if (i == end && mas[i] == -1) {
        for (int j = 0; j <= end; ++j) {
          mas[j] = -mas[j];
          n_coeff = 1;
        }
        break;
      }
      int t = (-mas[i]) / 2 + (-mas[i]) % 2;
      mas[i + 1] -= t;
      mas[i] = (-mas[i]) % 2;
      if (i == end) end++;
    }
    i++;
  }
  if (n_coeff == 1) {
    i = 0;
    while (i <= end) {
      if (mas[i] > 1) {
        int t = mas[i] / 2;
        mas[i + 1] += t;
        mas[i] = mas[i] % 2;
        if (i == end) end++;
      } else if (mas[i] < 0) {
        int t = (-mas[i]) / 2 + (-mas[i]) % 2;
        mas[i + 1] -= t;
        mas[i] = (-mas[i]) % 2;
        if (i == end) end++;
      }
      i++;
    }
  }
  int first(200100), last;
  for (int i = 0; i < 200100; ++i) {
    if (mas[i] != 0) {
      first = min(first, i);
      last = i;
    }
  }
  if (last - first > 32)
    cout << 0 << endl;
  else {
    long long num = 0;
    for (int i = first, j = 0; i <= last; ++i, ++j)
      num += (mas[i] ? (1ll << j) : 0);
    while (n < first && num <= k) {
      first--;
      num <<= 1;
    }
    int ans = 0;
    if (num == abs(coef[n]) && first == n) ans--;
    while (num <= 2 * k && first >= 0) {
      if (abs(coef[first] + num * n_coeff) <= k) ans++;
      first--;
      num <<= 1;
    }
    cout << ans << 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;
const int INF = 1e9 + 1;
const double pi = acos(-1);
int mas[200100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  int n;
  int k;
  cin >> n >> k;
  int t;
  vector<int> coef(n + 1);
  for (int i = 0; i <= n; ++i) {
    cin >> t;
    coef[i] = t;
    int modif = 1;
    if (t < 0) {
      modif = -1;
      t = -t;
    }
    int j = 0;
    while (t > 0) {
      mas[i + j] += (t % 2) * modif;
      j++;
      t /= 2;
    }
  }
  int end = 0;
  for (int i = 200099; i >= 0; --i) {
    if (mas[i] != 0) {
      end = i;
      break;
    }
  }
  int n_coeff = -1;
  int i = 0;
  while (i <= end) {
    if (mas[i] > 1) {
      int t = mas[i] / 2;
      mas[i + 1] += t;
      mas[i] = mas[i] % 2;
      if (i == end) end++;
    } else if (mas[i] < 0) {
      if (i == end && mas[i] == -1) {
        for (int j = 0; j <= end; ++j) {
          mas[j] = -mas[j];
          n_coeff = 1;
        }
        break;
      }
      int t = (-mas[i]) / 2 + (-mas[i]) % 2;
      mas[i + 1] -= t;
      mas[i] = (-mas[i]) % 2;
      if (i == end) end++;
    }
    i++;
  }
  if (n_coeff == 1) {
    i = 0;
    while (i <= end) {
      if (mas[i] > 1) {
        int t = mas[i] / 2;
        mas[i + 1] += t;
        mas[i] = mas[i] % 2;
        if (i == end) end++;
      } else if (mas[i] < 0) {
        int t = (-mas[i]) / 2 + (-mas[i]) % 2;
        mas[i + 1] -= t;
        mas[i] = (-mas[i]) % 2;
        if (i == end) end++;
      }
      i++;
    }
  }
  int first(200100), last;
  for (int i = 0; i < 200100; ++i) {
    if (mas[i] != 0) {
      first = min(first, i);
      last = i;
    }
  }
  if (last - first > 32)
    cout << 0 << endl;
  else {
    long long num = 0;
    for (int i = first, j = 0; i <= last; ++i, ++j)
      num += (mas[i] ? (1ll << j) : 0);
    while (n < first && num <= k) {
      first--;
      num <<= 1;
    }
    int ans = 0;
    if (num == abs(coef[n]) && first == n) ans--;
    while (num <= 2 * k && first >= 0) {
      if (abs(coef[first] + num * n_coeff) <= k) ans++;
      first--;
      num <<= 1;
    }
    cout << ans << endl;
  }
  return 0;
}

```