#include <bits/stdc++.h>
using namespace std;
long long a[444444];
long long oa[444444];
void gen() {
  ofstream out("input.txt");
  int n = 20;
  int k = 10000;
  out << n << " " << k << endl;
  for (int i = 0; i < n - 10; i++) {
    out << 0 << " ";
  }
  for (int i = max(0, n - 10); i <= n; i++) {
    out << rand() % 11 - 5 << " ";
  }
  out.close();
}
int stupid() {
  ifstream in("input.txt");
  int n, k;
  in >> n >> k;
  n++;
  vector<int> a(n);
  long long s = 0;
  for (int i = 0; i < n; i++) {
    in >> a[i];
    s += a[i] * (1LL << i);
  }
  if (s == 0) {
    return -1;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = -k; j <= k; j++) {
      long long add = 1LL * (j - a[i]) * (1LL << i);
      if (s + add == 0) {
        if (i + 1 != n || j != 0) ans++;
      }
    }
  }
  in.close();
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  n++;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    oa[i] = a[i];
  }
lstart:
  bool cool0 = true;
  for (int i = 0; i < n; i++) {
    if (i + 2 == n) {
      for (int j = 0; j <= i; j++) {
        if (a[j] != 0) {
          cool0 = false;
        }
      }
    }
    if (a[i] > 0) {
      a[i + 1] += a[i] / 2;
      a[i] %= 2;
    } else {
      long long t = (-a[i] + 1) / 2;
      a[i] += t * 2;
      a[i + 1] -= t;
    }
  }
  int x = n + 5;
  while (a[x] == 0) {
    x--;
  }
  if (a[x] < 0) {
    memset(a, 0, sizeof(a));
    for (int i = 0; i < n; i++) {
      oa[i] = -oa[i];
      a[i] = oa[i];
    }
    cerr << "goto!!" << endl;
    goto lstart;
  }
  while (a[x] > 1) {
    a[x + 1] += a[x] / 2;
    a[x] %= 2;
    x++;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    long long c = oa[i] + k;
    long long pp = 1;
    int pc = 0;
    int j;
    for (j = i; pc <= 40 && pp <= c && c > 0 && j <= x; j++, pc++) {
      if (a[j] == 1) {
        c -= pp;
      }
      pp *= 2;
    }
    if (j > x && c >= 0) {
      ans++;
    }
    if (a[i] == 1) {
      break;
    }
  }
  if (cool0) {
    ans--;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long a[444444];
long long oa[444444];
void gen() {
  ofstream out("input.txt");
  int n = 20;
  int k = 10000;
  out << n << " " << k << endl;
  for (int i = 0; i < n - 10; i++) {
    out << 0 << " ";
  }
  for (int i = max(0, n - 10); i <= n; i++) {
    out << rand() % 11 - 5 << " ";
  }
  out.close();
}
int stupid() {
  ifstream in("input.txt");
  int n, k;
  in >> n >> k;
  n++;
  vector<int> a(n);
  long long s = 0;
  for (int i = 0; i < n; i++) {
    in >> a[i];
    s += a[i] * (1LL << i);
  }
  if (s == 0) {
    return -1;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = -k; j <= k; j++) {
      long long add = 1LL * (j - a[i]) * (1LL << i);
      if (s + add == 0) {
        if (i + 1 != n || j != 0) ans++;
      }
    }
  }
  in.close();
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  n++;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    oa[i] = a[i];
  }
lstart:
  bool cool0 = true;
  for (int i = 0; i < n; i++) {
    if (i + 2 == n) {
      for (int j = 0; j <= i; j++) {
        if (a[j] != 0) {
          cool0 = false;
        }
      }
    }
    if (a[i] > 0) {
      a[i + 1] += a[i] / 2;
      a[i] %= 2;
    } else {
      long long t = (-a[i] + 1) / 2;
      a[i] += t * 2;
      a[i + 1] -= t;
    }
  }
  int x = n + 5;
  while (a[x] == 0) {
    x--;
  }
  if (a[x] < 0) {
    memset(a, 0, sizeof(a));
    for (int i = 0; i < n; i++) {
      oa[i] = -oa[i];
      a[i] = oa[i];
    }
    cerr << "goto!!" << endl;
    goto lstart;
  }
  while (a[x] > 1) {
    a[x + 1] += a[x] / 2;
    a[x] %= 2;
    x++;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    long long c = oa[i] + k;
    long long pp = 1;
    int pc = 0;
    int j;
    for (j = i; pc <= 40 && pp <= c && c > 0 && j <= x; j++, pc++) {
      if (a[j] == 1) {
        c -= pp;
      }
      pp *= 2;
    }
    if (j > x && c >= 0) {
      ans++;
    }
    if (a[i] == 1) {
      break;
    }
  }
  if (cool0) {
    ans--;
  }
  cout << ans << "\n";
  return 0;
}

```