#include <bits/stdc++.h>
using namespace std;
const int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const double pi = acos(-1);
const double eps = 1e-8;
const long long INF = 2e10;
template <class T>
void _print(T arg) {
  cout << arg << " ";
}
template <class... Args>
void log(Args... args) {
  int arr[] = {(_print(args), 0)...};
  cout << endl;
}
long long a[N], b[N];
int n, m, k;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 0; i <= n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  for (int i = 0; i < n; ++i) {
    a[i + 1] += a[i] / 2;
    a[i] %= 2;
  }
  int pos;
  for (int i = 0; i <= n; ++i) {
    if (a[i]) {
      pos = i;
      break;
    }
  }
  int res = 0;
  long long tmp = 0;
  for (int i = n; i >= 0; --i) {
    tmp = tmp * 2 + a[i];
    if (abs(tmp) > INF) break;
    if (i <= pos) {
      long long x = abs(tmp - b[i]);
      if (i == n & x == 0) continue;
      res += (x <= k);
    }
  }
  cout << res << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const double pi = acos(-1);
const double eps = 1e-8;
const long long INF = 2e10;
template <class T>
void _print(T arg) {
  cout << arg << " ";
}
template <class... Args>
void log(Args... args) {
  int arr[] = {(_print(args), 0)...};
  cout << endl;
}
long long a[N], b[N];
int n, m, k;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 0; i <= n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  for (int i = 0; i < n; ++i) {
    a[i + 1] += a[i] / 2;
    a[i] %= 2;
  }
  int pos;
  for (int i = 0; i <= n; ++i) {
    if (a[i]) {
      pos = i;
      break;
    }
  }
  int res = 0;
  long long tmp = 0;
  for (int i = n; i >= 0; --i) {
    tmp = tmp * 2 + a[i];
    if (abs(tmp) > INF) break;
    if (i <= pos) {
      long long x = abs(tmp - b[i]);
      if (i == n & x == 0) continue;
      res += (x <= k);
    }
  }
  cout << res << endl;
  return 0;
}

```