During summer vacation,Alice stay at home for a long time, with nothing to do. She went out and bought m pokers, tending to play poker. But she hated the traditional gameplay. She wants to change. She puts these pokers face down, she decided to flip poker n times, and each time she can flip Xi pokers. She wanted to know how many the results does she get. Can you help her solve this problem?

Input The input consists of multiple test cases. 

 Each test case begins with a line containing two non-negative integers n and m(0<n,m<=100000). 


 The next line contains n integers Xi(0<=Xi<=m).

Output Output the required answer modulo 1000000009 for each test case, one per line. Sample Input

3 4
3 2 3
3 3
3 2 3

Sample Output

8
3

Hint

For the second example:

0 express face down,1 express face up Initial state 000 The first result:000->111->001->110 The second result:000->111->100->011 The third result:000->111->010->101 So, there are three kinds of results(110,011,101)


给出m张牌，每次可以翻动一些牌，问最后牌的可能情况。因为每次都是随便翻的，只要能确定最后有几张牌是正面朝上，那么就可以


用组合数直接计算，然后我们可以考虑，正面朝上的可能情况，分为上下界[L,R]显然，LR是同奇偶的，并且中间的间隔都是2， 


#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<algorithm>
#include<iostream>
using namespace std;
const int mod=1e9+9;
const int N=1e5+10;
int n, m, x, ans;
int inv[N],g[N],f[N];

int c(int x,int y)
{
    return 1LL*f[x]*g[y]%mod*g[x-y]%mod;
}

int main()
{
    f[0]=g[0]=1;
    for (int i=1;i<N;i++)
    {
        f[i]=1LL*f[i-1]*i%mod;
        inv[i]=i==1?1:1LL*inv[mod%i]*(mod-mod/i)%mod;
        g[i]=1LL*g[i-1]*inv[i]%mod;
    }
    while (~scanf("%d%d",&n,&m))
    {
        int l=m,r=m;
        for (int i=1,j,k;i<=n;i++)
        {
            scanf("%d",&x);
            if (r-x<=0) j=x-r;
            else if (l-x>=0) j=l-x;
            else j=(r-x)%2;
            if (r+x<=m) k=r+x;
            else if (l+x>=m) k=2*m-l-x;
            else k=(m+j)%2?m-1:m;
            l=j; r=k;
        }
        ans=0;
        for (int i=l;i<=r;i+=2) ans=(ans+c(m,i))%mod;
        printf("%d\n",ans);
    }
    return 0;
}