#include <bits/stdc++.h>
using namespace std;
void fre() {
  freopen("c://test//input.in", "r", stdin);
  freopen("c://test//output.out", "w", stdout);
}
template <class T1, class T2>
inline void gmax(T1 &a, T2 b) {
  if (b > a) a = b;
}
template <class T1, class T2>
inline void gmin(T1 &a, T2 b) {
  if (b < a) a = b;
}
const int N = 2e5 + 1000, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
long long k;
long long o[N];
long long a[N];
bool v[N];
const long long big = 4e9;
int solve() {
  int ret = 0;
  for (int i = n + 100; i >= 1; --i) {
    if (abs(a[i] - o[i]) > big) return ret;
    if (abs(a[i] - o[i]) <= k) {
      if (i <= n + 1)
        if (v[i - 1] == 0)
          if (i != n + 1 || abs(a[i] - o[i]) != 0) ++ret;
    }
    a[i - 1] += a[i] * 2;
  }
  return ret;
}
int main() {
  while (~scanf("%d%lld", &n, &k)) {
    memset(o, 0, sizeof(o));
    for (int i = 1; i <= n + 1; ++i) scanf("%lld", &o[i]);
    memcpy(a, o, sizeof(a));
    for (int i = 1; i <= n + 100; ++i) {
      a[i + 1] += a[i] / 2;
      a[i] %= 2;
      v[i] = v[i - 1] | a[i];
    }
    printf("%d\n", solve());
  }
  return 0;
}

### 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;
void fre() {
  freopen("c://test//input.in", "r", stdin);
  freopen("c://test//output.out", "w", stdout);
}
template <class T1, class T2>
inline void gmax(T1 &a, T2 b) {
  if (b > a) a = b;
}
template <class T1, class T2>
inline void gmin(T1 &a, T2 b) {
  if (b < a) a = b;
}
const int N = 2e5 + 1000, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
long long k;
long long o[N];
long long a[N];
bool v[N];
const long long big = 4e9;
int solve() {
  int ret = 0;
  for (int i = n + 100; i >= 1; --i) {
    if (abs(a[i] - o[i]) > big) return ret;
    if (abs(a[i] - o[i]) <= k) {
      if (i <= n + 1)
        if (v[i - 1] == 0)
          if (i != n + 1 || abs(a[i] - o[i]) != 0) ++ret;
    }
    a[i - 1] += a[i] * 2;
  }
  return ret;
}
int main() {
  while (~scanf("%d%lld", &n, &k)) {
    memset(o, 0, sizeof(o));
    for (int i = 1; i <= n + 1; ++i) scanf("%lld", &o[i]);
    memcpy(a, o, sizeof(a));
    for (int i = 1; i <= n + 100; ++i) {
      a[i + 1] += a[i] / 2;
      a[i] %= 2;
      v[i] = v[i - 1] | a[i];
    }
    printf("%d\n", solve());
  }
  return 0;
}

```