Codeforces 658D Bear and Polynomials

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 . 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 . 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.

题目大意：给你一个n次多项式，问你能不能通过改变其中一项的系数来使整个多项式值为0，其中系数的绝对值不能超过k。

分析：我们通过把整个多项式转换成二进制形式，然后解只能通过第一个先导非零项和它前边的几项产生，因为一个2的整数次幂无法表示比它还小的项，倒着枚举并统计前边几项的和并统计答案。

#include 
#include 
#define MAXN 2147483647
using namespace std;
long long n,k,tot,num,now,a[200001],f[200001];
int main()
{
	cin.sync_with_stdio(false);
	cin>>n>>k;
	for(int i = 0;i <= n;i++) 
	 cin>>a[i];	
	now = -1;
	for(int i = 0;i <= n;i++)
	{
		f[i] += a[i];
		if(i != n)
		{
			f[i+1] += f[i]/2;
			f[i] %= 2;
		}
		if(now == -1 && f[i] != 0) now = i;
	}
	for(int i = n;i >= 0;i--)
	{
		tot *= 2;
		tot += f[i];
		if(tot > MAXN || tot < -MAXN) break;
		if(i <= now)
		{
			long long ans = tot - a[i];
			if(ans <= k && ans >= -k && !(i == n && ans == 0)) 
			 num++;
		}
	}
	cout<