#include <bits/stdc++.h>
using namespace std;
const int N = 400500;
long long A[N];
long long B[N];
inline long long mymod2(long long x) {
  if (x % 2 == 0)
    return 0;
  else
    return 1;
}
inline long long mydiv2(long long x) {
  if (x >= 0)
    return x / 2;
  else {
    if (x % 2 == 0)
      return x / 2;
    else
      return x / 2 - 1;
  }
}
void buildB() {
  for (int i = 0; i < N; i++) B[i] = A[i];
  for (int i = 0; i < N - 1; i++) {
    B[i + 1] += mydiv2(B[i]);
    B[i] = mymod2(B[i]);
  }
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i <= n; i++) {
    int x;
    scanf("%d", &x);
    A[i] = x;
  }
  buildB();
  if (B[N - 1]) {
    for (int i = 0; i <= n; i++) A[i] = -A[i];
    buildB();
  }
  int fst = 0;
  while (fst != N && B[fst] == 0) {
    fst++;
  }
  assert(fst != N);
  long long cur = 0;
  bool bad = false;
  for (int i = fst; i < N; i++) {
    if (B[i]) {
      int p = i - fst;
      if (p > 40) {
        bad = true;
        break;
      }
      cur += 1ll << p;
    }
  }
  int cnt = 0;
  if (!bad) {
    for (int i = fst; i >= 0; i--) {
      if (cur > (1ll << 32)) break;
      long long nA = A[i] - cur;
      if (abs(nA) <= k && (i != n || nA != 0) && (i <= n)) cnt++;
      cur <<= 1;
    }
  }
  printf("%d\n", cnt);
}

### Prompt

Develop a solution in CPP 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>
using namespace std;
const int N = 400500;
long long A[N];
long long B[N];
inline long long mymod2(long long x) {
  if (x % 2 == 0)
    return 0;
  else
    return 1;
}
inline long long mydiv2(long long x) {
  if (x >= 0)
    return x / 2;
  else {
    if (x % 2 == 0)
      return x / 2;
    else
      return x / 2 - 1;
  }
}
void buildB() {
  for (int i = 0; i < N; i++) B[i] = A[i];
  for (int i = 0; i < N - 1; i++) {
    B[i + 1] += mydiv2(B[i]);
    B[i] = mymod2(B[i]);
  }
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i <= n; i++) {
    int x;
    scanf("%d", &x);
    A[i] = x;
  }
  buildB();
  if (B[N - 1]) {
    for (int i = 0; i <= n; i++) A[i] = -A[i];
    buildB();
  }
  int fst = 0;
  while (fst != N && B[fst] == 0) {
    fst++;
  }
  assert(fst != N);
  long long cur = 0;
  bool bad = false;
  for (int i = fst; i < N; i++) {
    if (B[i]) {
      int p = i - fst;
      if (p > 40) {
        bad = true;
        break;
      }
      cur += 1ll << p;
    }
  }
  int cnt = 0;
  if (!bad) {
    for (int i = fst; i >= 0; i--) {
      if (cur > (1ll << 32)) break;
      long long nA = A[i] - cur;
      if (abs(nA) <= k && (i != n || nA != 0) && (i <= n)) cnt++;
      cur <<= 1;
    }
  }
  printf("%d\n", cnt);
}

```