#include <bits/stdc++.h>
using namespace std;
const int M1 = 1e9 + 7;
const int M2 = 1e9 + 33;
long long power(long long a, long long b, int MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b & 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = y * y;
    if (y > MOD) y %= MOD;
    b >>= 1;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(n + 1), t1(n + 1), t2(n + 1);
  for (int &x : a) cin >> x;
  t1[0] = t2[0] = 1;
  for (int i = int(1); i <= int(n); i++) {
    t1[i] = 2LL * t1[i - 1] % M1;
    t2[i] = 2LL * t2[i - 1] % M2;
  }
  long long all1 = 0, all2 = 0;
  for (int i = 0; i < int(n + 1); i++) {
    all1 += 1LL * t1[i] * a[i] % M1;
    all1 %= M1;
    all2 += 1LL * t2[i] * a[i] % M2;
    all2 %= M2;
  }
  int ans = 0;
  for (int i = 0; i < int(n + 1); i++) {
    long long tmp1 = all1 - 1LL * t1[i] * a[i] % M1;
    tmp1 += M1;
    tmp1 %= M1;
    long long tmp2 = all2 - 1LL * t2[i] * a[i] % M2;
    ;
    tmp2 += M2;
    tmp2 %= M2;
    bool flag = false;
    for (int r = -1; r <= 1 && !flag; r += 2) {
      for (int t = -1; t <= 1 && !flag; t += 2) {
        long long tm1 = (tmp1 + r * M1) % M1;
        long long tm2 = (tmp2 + r * M2) % M2;
        long long x1 = -tm1 * power(t1[i], M1 - 2, M1) % M1;
        long long x2 = -tm2 * power(t2[i], M2 - 2, M2) % M2;
        if (i < n)
          flag |= (x1 == x2) && (abs(x1) <= k);
        else
          flag |= (x1 == x2) && (abs(x1) <= k) && (x1 != 0);
      }
    }
    ans += flag;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 M1 = 1e9 + 7;
const int M2 = 1e9 + 33;
long long power(long long a, long long b, int MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b & 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = y * y;
    if (y > MOD) y %= MOD;
    b >>= 1;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(n + 1), t1(n + 1), t2(n + 1);
  for (int &x : a) cin >> x;
  t1[0] = t2[0] = 1;
  for (int i = int(1); i <= int(n); i++) {
    t1[i] = 2LL * t1[i - 1] % M1;
    t2[i] = 2LL * t2[i - 1] % M2;
  }
  long long all1 = 0, all2 = 0;
  for (int i = 0; i < int(n + 1); i++) {
    all1 += 1LL * t1[i] * a[i] % M1;
    all1 %= M1;
    all2 += 1LL * t2[i] * a[i] % M2;
    all2 %= M2;
  }
  int ans = 0;
  for (int i = 0; i < int(n + 1); i++) {
    long long tmp1 = all1 - 1LL * t1[i] * a[i] % M1;
    tmp1 += M1;
    tmp1 %= M1;
    long long tmp2 = all2 - 1LL * t2[i] * a[i] % M2;
    ;
    tmp2 += M2;
    tmp2 %= M2;
    bool flag = false;
    for (int r = -1; r <= 1 && !flag; r += 2) {
      for (int t = -1; t <= 1 && !flag; t += 2) {
        long long tm1 = (tmp1 + r * M1) % M1;
        long long tm2 = (tmp2 + r * M2) % M2;
        long long x1 = -tm1 * power(t1[i], M1 - 2, M1) % M1;
        long long x2 = -tm2 * power(t2[i], M2 - 2, M2) % M2;
        if (i < n)
          flag |= (x1 == x2) && (abs(x1) <= k);
        else
          flag |= (x1 == x2) && (abs(x1) <= k) && (x1 != 0);
      }
    }
    ans += flag;
  }
  cout << ans << '\n';
  return 0;
}

```