#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using vi = vector<int>;
using vii = vector<pii>;
using ll = long long;
using ld = long double;
const int maxn = 4e5;
const int lo = 40;
int a[maxn];
int b[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  ll k;
  cin >> n >> k;
  ++n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  ll c = 0;
  int p;
  for (p = 0; p < n; ++p) {
    c += a[p];
    if (c % 2 != 0) {
      b[p] = 1;
      --c;
    }
    c /= 2;
  }
  if (c < 0) {
    for (int i = 0; i < n; ++i) {
      a[i] = -a[i];
    }
    c = 0;
    for (p = 0; p < n; ++p) {
      c += a[p];
      if (c % 2 != 0) {
        b[p] = 1;
        --c;
      } else {
        b[p] = 0;
      }
      c /= 2;
    }
  }
  while (c > 0) {
    if (c % 2 != 0) {
      b[p] = 1;
      --c;
    }
    c /= 2;
    ++p;
  }
  int p1 = -1, p2 = -1;
  for (int i = 0; i < p; ++i) {
    if (b[i] == 1) {
      if (p1 == -1) {
        p1 = i;
      }
      p2 = i;
    }
  }
  if (p2 - p1 > lo) {
    cout << "0\n";
    return 0;
  }
  int res = 0;
  for (int i = max(p2 - lo, 0); i <= min(p1, n - 1); ++i) {
    ll t = 0, pc = 1;
    for (int j = i; j <= p2; ++j) {
      if (b[j] == 1) {
        t += pc;
      }
      pc *= 2;
    }
    if (abs(ll(a[i]) - t) <= k && (i != n - 1 || ll(a[i]) - t != 0)) {
      ++res;
    }
  }
  cout << res << '\n';
  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;
using pii = pair<int, int>;
using vi = vector<int>;
using vii = vector<pii>;
using ll = long long;
using ld = long double;
const int maxn = 4e5;
const int lo = 40;
int a[maxn];
int b[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  ll k;
  cin >> n >> k;
  ++n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  ll c = 0;
  int p;
  for (p = 0; p < n; ++p) {
    c += a[p];
    if (c % 2 != 0) {
      b[p] = 1;
      --c;
    }
    c /= 2;
  }
  if (c < 0) {
    for (int i = 0; i < n; ++i) {
      a[i] = -a[i];
    }
    c = 0;
    for (p = 0; p < n; ++p) {
      c += a[p];
      if (c % 2 != 0) {
        b[p] = 1;
        --c;
      } else {
        b[p] = 0;
      }
      c /= 2;
    }
  }
  while (c > 0) {
    if (c % 2 != 0) {
      b[p] = 1;
      --c;
    }
    c /= 2;
    ++p;
  }
  int p1 = -1, p2 = -1;
  for (int i = 0; i < p; ++i) {
    if (b[i] == 1) {
      if (p1 == -1) {
        p1 = i;
      }
      p2 = i;
    }
  }
  if (p2 - p1 > lo) {
    cout << "0\n";
    return 0;
  }
  int res = 0;
  for (int i = max(p2 - lo, 0); i <= min(p1, n - 1); ++i) {
    ll t = 0, pc = 1;
    for (int j = i; j <= p2; ++j) {
      if (b[j] == 1) {
        t += pc;
      }
      pc *= 2;
    }
    if (abs(ll(a[i]) - t) <= k && (i != n - 1 || ll(a[i]) - t != 0)) {
      ++res;
    }
  }
  cout << res << '\n';
  return 0;
}

```