#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
const long long inf = 1e10;
long long a[N], b[N], n, k, v, res, cur;
void solve() {
  for (int i = 0; i <= n + v; ++i) {
    b[i] += a[i], b[i + 1] += b[i] / 2;
    if (!(b[i] & 1))
      b[i] = 0;
    else {
      if (b[i] < 0)
        b[i] = -1;
      else if (b[i] > 0)
        b[i] = 1;
    }
  }
  bool check = 0;
  for (int i = n + v; i >= 0; --i)
    if (b[i] == -1) {
      check = 1;
      break;
    } else if (b[i] == 1)
      break;
  if (check)
    for (int i = 0; i <= n + v; ++i) b[i] = -b[i], a[i] = -a[i];
  for (int i = 0; i <= n + v; ++i)
    if (b[i] < 0) b[i] = b[i] / 2 + 1, b[i + 1]--;
  int start = -1;
  for (int i = 0; i <= n + v; ++i) {
    if (!b[i]) continue;
    if (start == -1) {
      start = i, cur = 1;
      continue;
    }
    if (i - start >= 32) return;
    cur += (1LL << (i - start));
  }
  for (int i = start; i >= 0; --i) {
    if (cur > inf) return;
    if (i == n && a[i] == cur) {
      cur <<= 1;
      continue;
    }
    if (fabs(a[i] - cur) <= k && i <= n) res++;
    cur <<= 1;
  }
}
void print() {
  for (int i = 0; i <= n + v; ++i) cout << b[i] << ' ';
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  v = log2(k) + 1;
  for (int i = 0; i <= n; ++i) cin >> a[i];
  solve();
  cout << res;
}

### Prompt

Generate a CPP solution 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 N = 2e5 + 100;
const long long inf = 1e10;
long long a[N], b[N], n, k, v, res, cur;
void solve() {
  for (int i = 0; i <= n + v; ++i) {
    b[i] += a[i], b[i + 1] += b[i] / 2;
    if (!(b[i] & 1))
      b[i] = 0;
    else {
      if (b[i] < 0)
        b[i] = -1;
      else if (b[i] > 0)
        b[i] = 1;
    }
  }
  bool check = 0;
  for (int i = n + v; i >= 0; --i)
    if (b[i] == -1) {
      check = 1;
      break;
    } else if (b[i] == 1)
      break;
  if (check)
    for (int i = 0; i <= n + v; ++i) b[i] = -b[i], a[i] = -a[i];
  for (int i = 0; i <= n + v; ++i)
    if (b[i] < 0) b[i] = b[i] / 2 + 1, b[i + 1]--;
  int start = -1;
  for (int i = 0; i <= n + v; ++i) {
    if (!b[i]) continue;
    if (start == -1) {
      start = i, cur = 1;
      continue;
    }
    if (i - start >= 32) return;
    cur += (1LL << (i - start));
  }
  for (int i = start; i >= 0; --i) {
    if (cur > inf) return;
    if (i == n && a[i] == cur) {
      cur <<= 1;
      continue;
    }
    if (fabs(a[i] - cur) <= k && i <= n) res++;
    cur <<= 1;
  }
}
void print() {
  for (int i = 0; i <= n + v; ++i) cout << b[i] << ' ';
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  v = log2(k) + 1;
  for (int i = 0; i <= n; ++i) cin >> a[i];
  solve();
  cout << res;
}

```