codeforces 658D D. Bear and Polynomials(数学)

题目链接：

D. Bear and Polynomials

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

 

题意：

问能不能改变一个系数使Q(2)=0;

 

思路：

 像二进制那样都变成0，-1，+1,都转移到n位上;再从高位到低位计算; 

 

AC代码：

 

/*
2014300227    658D - 26    GNU C++11    Accepted    576 ms    5308 KB
*/
#include <bits/stdc++.h>
using namespace std;
const int N=2e5+4;
long long a[N],b[N+100];
int n,l=0,k;
int main()
{

    scanf("%d%d",&n,&k);
    for(int i=0;i<=n;i++)
    {
        cin>>a[i];
        b[i]=a[i];
    }
    for(int i=0;i<n;i++)
    {
        a[i+1]+=a[i]/2;
        a[i]=a[i]%2;
    }
    for(int i=0;i<=n;i++)
    {
        if(a[i])
        {
            l=i;
            break;
        }
    }
    int answ=0;
    long long sum=0;
    for(int i=n;i>=0;i--)
    {
        sum=sum*2+a[i];
        if(abs(sum)>1e10)break;
        if(i<=l)
        {
            long long x=abs(sum-b[i]);
            if(x==0&&i==n)continue;
            if(x<=k)answ++;
        }
    }
    cout<<answ<<"\n";
    return 0;
}