#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9;
const long long INF64 = (long long)1e18;
#pragma comment(linker, "/STACK:16777216")
#pragma warning(disable : 4786)
int Set(int N, int pos) { return N = N | (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
const int mx = 1 << 29;
long long ans, sum, t, n, k, m, ar[200005], lft[200005], rght[200005];
int dx[] = {0, 1, -1, 0, 0};
int dy[] = {0, 0, 0, 1, -1};
int main() {
  cin >> n >> k;
  n++;
  for (int i = 1; i <= n; i++) scanf("%lld", &ar[i]);
  for (int i = 0; i <= n; i++) {
    lft[i] = 9223372036854775807LL;
    rght[i] = 9223372036854775807LL;
  }
  lft[1] = 0;
  sum = ar[1];
  for (int i = 2; i <= n; i++) {
    if (abs(sum + lft[i - 1]) % 2 == 1) break;
    lft[i] = (sum + lft[i - 1]) / 2;
    sum = ar[i];
  }
  rght[n] = 0;
  sum = ar[n];
  for (int i = n - 1; i >= 1; i--) {
    if (abs(sum + rght[i + 1]) >= 2LL * k) break;
    rght[i] = (sum + rght[i + 1]) * 2;
    sum = ar[i];
  }
  for (int i = 1; i <= n - 1; i++) {
    if (lft[i] == 9223372036854775807LL || rght[i] == 9223372036854775807LL)
      continue;
    if (abs(lft[i] + rght[i]) <= k) ans++;
  }
  if (lft[n] != 0) {
    if (abs(lft[n]) <= k) ans++;
  }
  cout << ans << "\n";
  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>
using namespace std;
const int INF = (int)1e9;
const long long INF64 = (long long)1e18;
#pragma comment(linker, "/STACK:16777216")
#pragma warning(disable : 4786)
int Set(int N, int pos) { return N = N | (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
const int mx = 1 << 29;
long long ans, sum, t, n, k, m, ar[200005], lft[200005], rght[200005];
int dx[] = {0, 1, -1, 0, 0};
int dy[] = {0, 0, 0, 1, -1};
int main() {
  cin >> n >> k;
  n++;
  for (int i = 1; i <= n; i++) scanf("%lld", &ar[i]);
  for (int i = 0; i <= n; i++) {
    lft[i] = 9223372036854775807LL;
    rght[i] = 9223372036854775807LL;
  }
  lft[1] = 0;
  sum = ar[1];
  for (int i = 2; i <= n; i++) {
    if (abs(sum + lft[i - 1]) % 2 == 1) break;
    lft[i] = (sum + lft[i - 1]) / 2;
    sum = ar[i];
  }
  rght[n] = 0;
  sum = ar[n];
  for (int i = n - 1; i >= 1; i--) {
    if (abs(sum + rght[i + 1]) >= 2LL * k) break;
    rght[i] = (sum + rght[i + 1]) * 2;
    sum = ar[i];
  }
  for (int i = 1; i <= n - 1; i++) {
    if (lft[i] == 9223372036854775807LL || rght[i] == 9223372036854775807LL)
      continue;
    if (abs(lft[i] + rght[i]) <= k) ans++;
  }
  if (lft[n] != 0) {
    if (abs(lft[n]) <= k) ans++;
  }
  cout << ans << "\n";
  return 0;
}

```