#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
long long rlt, tmp, a[222222 + 1], b[222222 + 1], c[222222 + 1], n, K, i, j, u,
    d, sgn = 1, l;
int main() {
  scanf("%lld%lld", &n, &K);
  for (i = 0; i <= n; i++) scanf("%lld", &a[i]);
  for (i = 0; i <= n; i++) {
    if (a[i] >= 0)
      b[i] = a[i];
    else
      c[i] = -a[i];
  }
  for (i = 0; i < 222222; i++) {
    if (b[i] < 0 || c[i] < 0)
      while (1) {
      }
    b[i + 1] += b[i] / 2, b[i] %= 2;
    c[i + 1] += c[i] / 2, c[i] %= 2;
  }
  for (i = 222222 - 1; i >= 0; i--)
    if (b[i] != c[i]) break;
  if (b[i] < c[i]) {
    sgn = -1;
    for (j = 0; j < 222222; j++) swap(b[j], c[j]);
  }
  for (i = 0; i < 222222; i++) b[i] -= c[i];
  for (i = 0; i < 222222; i++) {
    b[i + 1] += ((b[i] + 20) / 2 - 10);
    b[i] = (b[i] + 20) % 2;
  }
  for (i = 222222 - 1; i >= 0; i--)
    if (b[i] != 0) break;
  u = i;
  if (b[u] < 0)
    while (1) {
    }
  for (i = 0; i < 222222; i++)
    if (b[i] != 0) break;
  d = min(n, i);
  for (i = 0; i <= d; i++) {
    l = u - i + 1;
    if (l > 40) continue;
    tmp = 0;
    for (j = u; j >= i; j--) tmp = tmp * 2 + b[j];
    tmp = tmp * sgn;
    tmp = abs(a[i] - tmp);
    if (i == n && tmp == 0) continue;
    if (tmp <= K) rlt++;
  }
  printf("%lld\n", rlt);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 <typename T, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
long long rlt, tmp, a[222222 + 1], b[222222 + 1], c[222222 + 1], n, K, i, j, u,
    d, sgn = 1, l;
int main() {
  scanf("%lld%lld", &n, &K);
  for (i = 0; i <= n; i++) scanf("%lld", &a[i]);
  for (i = 0; i <= n; i++) {
    if (a[i] >= 0)
      b[i] = a[i];
    else
      c[i] = -a[i];
  }
  for (i = 0; i < 222222; i++) {
    if (b[i] < 0 || c[i] < 0)
      while (1) {
      }
    b[i + 1] += b[i] / 2, b[i] %= 2;
    c[i + 1] += c[i] / 2, c[i] %= 2;
  }
  for (i = 222222 - 1; i >= 0; i--)
    if (b[i] != c[i]) break;
  if (b[i] < c[i]) {
    sgn = -1;
    for (j = 0; j < 222222; j++) swap(b[j], c[j]);
  }
  for (i = 0; i < 222222; i++) b[i] -= c[i];
  for (i = 0; i < 222222; i++) {
    b[i + 1] += ((b[i] + 20) / 2 - 10);
    b[i] = (b[i] + 20) % 2;
  }
  for (i = 222222 - 1; i >= 0; i--)
    if (b[i] != 0) break;
  u = i;
  if (b[u] < 0)
    while (1) {
    }
  for (i = 0; i < 222222; i++)
    if (b[i] != 0) break;
  d = min(n, i);
  for (i = 0; i <= d; i++) {
    l = u - i + 1;
    if (l > 40) continue;
    tmp = 0;
    for (j = u; j >= i; j--) tmp = tmp * 2 + b[j];
    tmp = tmp * sgn;
    tmp = abs(a[i] - tmp);
    if (i == n && tmp == 0) continue;
    if (tmp <= K) rlt++;
  }
  printf("%lld\n", rlt);
  return 0;
}

```