D

题目链接：https://codeforces.com/problemset/problem/658/D

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, ..., a_n denote the coefficients, so . Then, a polynomial P(x) is valid if all the following conditions are satisfied:

• a_i is integer for every i;
• |a_i| ≤ k for every i;
• a_n ≠ 0.

Limak has recently got a valid polynomial P with coefficients a₀, a₁, a₂, ..., a_n. 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, ..., a_n (|a_i| ≤ k, a_n ≠ 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：问能不能改变一个系数使Q(2)=0;

思路：很容易想到暴力，直接求个总的值，再减去每个位置上的值不就好了吗，但是看清楚数据范围，这样写稳稳的错，2^n，用大数也不行，稳稳超时

所以我们要解决的就是怎么处理这么大的数的问题。可以把每个系数看做类似在二进制下的值，先把能进位的数进位，这样到最后所有能进位的全都进到第N位了

其它位要么为0 要么为1 要么为-1 。 现在我们想想哪些位置的系数改变会得到解，我们把能进位的全部进位了，然后从0遍历到N，找到第一个不为0的位置

只有在这个位置之前的位置改变系数才有效。 为啥呢？ 因为当前为不为0，只能通过前面的改变系数来抵消这一位的值，后面的都比自己大，怎么改变系数也没用吧，不可能

把这一位消掉的。 看代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<list>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int maxn=2e5+5;
const int maxm=100000+10;
const ull base=2333;
const LL INF=1e10;
LL a[maxn],b[maxn];
int main()
{
    LL N,K,sum=0,l;scanf("%lld%lld",&N,&K);
    for(int i=0;i<=N;i++)
    {
        scanf("%lld",&a[i]);b[i]=a[i];
    }
    for(int i=0;i<N;i++)
    {
        a[i+1]+=a[i]/2;a[i]%=2;
    }
    for(int i=0;i<=N;i++)
    {
        if(a[i])
        {
            l=i;break;
        }
        //找到第一个不为0的地方 只有从0-这个地方改变值才有效
        //因为只能改一个位置 如果改了后面的位置 这个位置的值就无法清空了 因为后面的都比它大啊
    }
    LL ans=0;
    for(int i=N;i>=0;i--)
    {
        sum=sum*2+a[i];
        if(abs(sum)>INF) break;
        if(i<=l)//修改的位置有效 而此时的sum也就是总的sum
        {
            LL x=abs(sum-b[i]);
            if(x==0&&i==N) continue;//题目中的限制条件 N为不能为0
            if(x<=K) ans++;
        }
    }
    printf("%lld
",ans);
    return 0;
}
/**

*/