#include <bits/stdc++.h>
using namespace std;
void read(long long &digit) {
  digit = 0;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  bool type = false;
  if (c == '-') type = true, c = getchar();
  for (; c >= '0' && c <= '9'; digit = digit * 10 + c - '0', c = getchar())
    ;
  if (type == true) digit = -digit;
}
long long a[200010], b[200010 * 10];
long long n, k;
int main() {
  read(n), read(k);
  for (int i = 0; i <= n; i++) read(a[i]), b[i] = a[i];
  long long m = 0;
  for (int i = 0; i <= n; i++) {
    b[i] += m;
    if (b[i] % 2 == 0)
      m = b[i] / 2, b[i] = 0;
    else if (i == n && b[i] == -1)
      b[i] = -1, m = 0;
    else {
      m = (b[i] - 1) / 2;
      b[i] = 1;
    }
  }
  int len = n;
  while (m != 0) {
    len++;
    if (m % 2 == 0) {
      b[len] = 0;
      m = m / 2;
    } else {
      if (m == -1)
        b[len] = -1, m = 0;
      else
        m = (m - 1) / 2, b[len] = 1;
    }
  }
  int pos = 0;
  while (b[pos] == 0) pos++;
  long long now = 0;
  int mjy = 0;
  for (int i = len; i >= 0; i--) {
    now = now * 2 + b[i];
    if (abs(now) > 2 * k) break;
    if (i <= n && i <= pos)
      if (abs(-now + a[i]) <= k)
        if (i != n || a[i] - now != 0) mjy++;
  }
  printf("%d\n", mjy);
  return 0;
}

### Prompt

Generate a cpp solution to 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;
void read(long long &digit) {
  digit = 0;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  bool type = false;
  if (c == '-') type = true, c = getchar();
  for (; c >= '0' && c <= '9'; digit = digit * 10 + c - '0', c = getchar())
    ;
  if (type == true) digit = -digit;
}
long long a[200010], b[200010 * 10];
long long n, k;
int main() {
  read(n), read(k);
  for (int i = 0; i <= n; i++) read(a[i]), b[i] = a[i];
  long long m = 0;
  for (int i = 0; i <= n; i++) {
    b[i] += m;
    if (b[i] % 2 == 0)
      m = b[i] / 2, b[i] = 0;
    else if (i == n && b[i] == -1)
      b[i] = -1, m = 0;
    else {
      m = (b[i] - 1) / 2;
      b[i] = 1;
    }
  }
  int len = n;
  while (m != 0) {
    len++;
    if (m % 2 == 0) {
      b[len] = 0;
      m = m / 2;
    } else {
      if (m == -1)
        b[len] = -1, m = 0;
      else
        m = (m - 1) / 2, b[len] = 1;
    }
  }
  int pos = 0;
  while (b[pos] == 0) pos++;
  long long now = 0;
  int mjy = 0;
  for (int i = len; i >= 0; i--) {
    now = now * 2 + b[i];
    if (abs(now) > 2 * k) break;
    if (i <= n && i <= pos)
      if (abs(-now + a[i]) <= k)
        if (i != n || a[i] - now != 0) mjy++;
  }
  printf("%d\n", mjy);
  return 0;
}

```