#include <bits/stdc++.h>
const long double pi = 3.14159265359;
const long double Exp = 2.71828182846;
using namespace std;
void Prepere_streams() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long t[200009];
long long sum[200009];
long long cq[200009];
int main() {
  long long n, k;
  long long first_error = -1;
  cin >> n >> k;
  long long tk = k;
  ++n;
  int non_zero = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%I64d", &t[i]);
    sum[i] += t[i];
    if (t[i]) ++non_zero;
    if (i < n - 1) {
      sum[i + 1] = sum[i] / 2;
      sum[i] %= 2;
      if (sum[i] % 2 != 0 && first_error == -1) first_error = i;
    }
  }
  long long curent_sum = 0;
  long long res = 0;
  long long l, r;
  if (first_error == -1) first_error = n - 1;
  int temp;
  for (int i = n - 1;
       i >= 0 && ((curent_sum > 0) ? (curent_sum) : (-curent_sum)) <= 100 * k;
       --i) {
    curent_sum = 2 * curent_sum + sum[i];
    l = -k - t[i];
    r = k - t[i];
    if (i <= first_error && -curent_sum >= l && -curent_sum <= r &&
        (i != n - 1 || curent_sum != t[i]))
      ++res;
  }
  cout << res;
}

### Prompt

Please provide a CPP coded solution 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>
const long double pi = 3.14159265359;
const long double Exp = 2.71828182846;
using namespace std;
void Prepere_streams() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long t[200009];
long long sum[200009];
long long cq[200009];
int main() {
  long long n, k;
  long long first_error = -1;
  cin >> n >> k;
  long long tk = k;
  ++n;
  int non_zero = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%I64d", &t[i]);
    sum[i] += t[i];
    if (t[i]) ++non_zero;
    if (i < n - 1) {
      sum[i + 1] = sum[i] / 2;
      sum[i] %= 2;
      if (sum[i] % 2 != 0 && first_error == -1) first_error = i;
    }
  }
  long long curent_sum = 0;
  long long res = 0;
  long long l, r;
  if (first_error == -1) first_error = n - 1;
  int temp;
  for (int i = n - 1;
       i >= 0 && ((curent_sum > 0) ? (curent_sum) : (-curent_sum)) <= 100 * k;
       --i) {
    curent_sum = 2 * curent_sum + sum[i];
    l = -k - t[i];
    r = k - t[i];
    if (i <= first_error && -curent_sum >= l && -curent_sum <= r &&
        (i != n - 1 || curent_sum != t[i]))
      ++res;
  }
  cout << res;
}

```