#include <bits/stdc++.h>
using namespace std;
int z[200001];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i <= n; i++) scanf("%d", z + i);
  long long ss = 0;
  vector<int> pos;
  bool neg = false;
  {
    long long s = 0;
    for (int i = 0; i <= n; i++) {
      s += z[i];
      if (s % 2) s--, pos.push_back(i);
      s /= 2;
    }
    if (s >= 0) {
      ss = s;
    } else {
      pos.clear();
      neg = true;
    }
  }
  if (neg) {
    long long s = 0;
    for (int i = 0; i <= n; i++) {
      s += z[i];
      if (s % 2) s++, pos.push_back(i);
      s /= 2;
    }
    if (s <= 0) {
      ss = -s;
    }
  }
  for (int i = n + 1; ss; i++) {
    if (ss % 2) pos.push_back(i);
    ss /= 2;
  }
  int r = pos[0];
  if (pos[pos.size() - 1] - r > 31) {
    cout << 0;
    return 0;
  }
  if (r > n) r = n;
  long long q = 0;
  for (int p : pos) q |= 1ll << p - r;
  if (neg) q = -q;
  int s = 0;
  while (abs(q) <= 2 * k && r >= 0) {
    if (abs(z[r] - q) <= k && !(r == n && z[r] == q)) s++;
    r--;
    q *= 2;
  }
  cout << s;
}

### Prompt

Please formulate 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;
int z[200001];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i <= n; i++) scanf("%d", z + i);
  long long ss = 0;
  vector<int> pos;
  bool neg = false;
  {
    long long s = 0;
    for (int i = 0; i <= n; i++) {
      s += z[i];
      if (s % 2) s--, pos.push_back(i);
      s /= 2;
    }
    if (s >= 0) {
      ss = s;
    } else {
      pos.clear();
      neg = true;
    }
  }
  if (neg) {
    long long s = 0;
    for (int i = 0; i <= n; i++) {
      s += z[i];
      if (s % 2) s++, pos.push_back(i);
      s /= 2;
    }
    if (s <= 0) {
      ss = -s;
    }
  }
  for (int i = n + 1; ss; i++) {
    if (ss % 2) pos.push_back(i);
    ss /= 2;
  }
  int r = pos[0];
  if (pos[pos.size() - 1] - r > 31) {
    cout << 0;
    return 0;
  }
  if (r > n) r = n;
  long long q = 0;
  for (int p : pos) q |= 1ll << p - r;
  if (neg) q = -q;
  int s = 0;
  while (abs(q) <= 2 * k && r >= 0) {
    if (abs(z[r] - q) <= k && !(r == n && z[r] == q)) s++;
    r--;
    q *= 2;
  }
  cout << s;
}

```