#include <bits/stdc++.h>
using namespace std;
template <class T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
int sgn(T x) {
  return x > 0 ? 1 : (x < 0 ? -1 : 0);
}
const long double pi = acos(-1.0);
const long long MAX = 1e18;
int n;
int m;
int a[200001];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i <= n; i++) scanf("%d", &a[i]);
  long long cur = 0;
  int pw = 0;
  for (int i = 0; i < n; i++) {
    if ((cur + a[i]) % 2 != 0) break;
    cur = (cur + a[i]) / 2;
    pw++;
  }
  long long delta = 0;
  for (int i = n; i >= pw; i--) {
    delta = delta * 2 + a[i];
    if (abs(delta) > MAX) {
      printf("0\n");
      return 0;
    }
  }
  cur += delta;
  int ans = 0;
  while (pw >= 0 && abs(cur) < MAX) {
    long long na = a[pw] - cur;
    if (abs(na) <= m && (na != 0 || pw != n)) ans++;
    cur *= 2;
    pw--;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
template <class T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
int sgn(T x) {
  return x > 0 ? 1 : (x < 0 ? -1 : 0);
}
const long double pi = acos(-1.0);
const long long MAX = 1e18;
int n;
int m;
int a[200001];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i <= n; i++) scanf("%d", &a[i]);
  long long cur = 0;
  int pw = 0;
  for (int i = 0; i < n; i++) {
    if ((cur + a[i]) % 2 != 0) break;
    cur = (cur + a[i]) / 2;
    pw++;
  }
  long long delta = 0;
  for (int i = n; i >= pw; i--) {
    delta = delta * 2 + a[i];
    if (abs(delta) > MAX) {
      printf("0\n");
      return 0;
    }
  }
  cur += delta;
  int ans = 0;
  while (pw >= 0 && abs(cur) < MAX) {
    long long na = a[pw] - cur;
    if (abs(na) <= m && (na != 0 || pw != n)) ans++;
    cur *= 2;
    pw--;
  }
  printf("%d\n", ans);
  return 0;
}

```