#include <bits/stdc++.h>
long long n, k, a[200100], b[200100], sgn[2] = {1, -1}, ans;
long long sum;
template <class T>
T abs(T x) {
  return (x > 0) ? x : (-x);
}
void add(int i, long long x) {
  long long s = sgn[x < 0];
  x = abs<long long>(x);
  while (x) {
    b[i] += s * (x & 1);
    x >>= 1;
    ++i;
  }
}
void push(int i) {
  long long t = b[i];
  b[i] = 0;
  add(i, t);
}
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (int i = 0; i <= n; ++i) {
    scanf("%I64d", a + i);
    add(i, a[i]);
    push(i);
  }
  for (int i = n + 1; i < n + 32; ++i) push(i);
  int p = 0;
  while (!b[p]) ++p;
  for (int i = n + 31; i > 0; --i) {
    if (!b[i]) continue;
    if (b[i] * b[i - 1] < 0) {
      b[i - 1] = b[i];
      b[i] = 0;
    } else
      break;
  }
  for (int i = p; abs<long long>(sum) <= 2ll * 0x3f3f3f3f && i < n + 32; ++i) {
    if (b[i] && i - p > 31) {
      sum = 0x3f3f3f3f;
      break;
    }
    sum += b[i] * (1ll << (i - p));
  }
  while (abs<long long>(sum) <= 2 * k && p >= 0) {
    if (p <= n) {
      if (abs<long long>(a[p] - sum) <= k) ++ans;
      if (p == n && a[p] == sum) --ans;
    }
    sum *= 2;
    --p;
  }
  printf("%I64d\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>
long long n, k, a[200100], b[200100], sgn[2] = {1, -1}, ans;
long long sum;
template <class T>
T abs(T x) {
  return (x > 0) ? x : (-x);
}
void add(int i, long long x) {
  long long s = sgn[x < 0];
  x = abs<long long>(x);
  while (x) {
    b[i] += s * (x & 1);
    x >>= 1;
    ++i;
  }
}
void push(int i) {
  long long t = b[i];
  b[i] = 0;
  add(i, t);
}
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (int i = 0; i <= n; ++i) {
    scanf("%I64d", a + i);
    add(i, a[i]);
    push(i);
  }
  for (int i = n + 1; i < n + 32; ++i) push(i);
  int p = 0;
  while (!b[p]) ++p;
  for (int i = n + 31; i > 0; --i) {
    if (!b[i]) continue;
    if (b[i] * b[i - 1] < 0) {
      b[i - 1] = b[i];
      b[i] = 0;
    } else
      break;
  }
  for (int i = p; abs<long long>(sum) <= 2ll * 0x3f3f3f3f && i < n + 32; ++i) {
    if (b[i] && i - p > 31) {
      sum = 0x3f3f3f3f;
      break;
    }
    sum += b[i] * (1ll << (i - p));
  }
  while (abs<long long>(sum) <= 2 * k && p >= 0) {
    if (p <= n) {
      if (abs<long long>(a[p] - sum) <= k) ++ans;
      if (p == n && a[p] == sum) --ans;
    }
    sum *= 2;
    --p;
  }
  printf("%I64d\n", ans);
  return 0;
}

```