Turn the pokers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description

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 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 43 2 33 33 2 3

Sample Output

83

Hint

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

区间维护很绕人！！！！！前面把它想简单了，无限跪！！

还有组合数快速幂求模，以前都没记这东西！！！

按题解在赛后总算把它A了！

AC代码如下：

#include
  
   
#include
   
     #include
    
      #define mod 1000000009 #define ll long long #define M 100005 using namespace std; ll n,m; ll a[M],c[M]; ll pow_mod(ll a,ll b) { ll s=1; while(b) { if(b&1)s=s*a%mod; a=a*a%mod; b=b>>1; } return s; } int main() { ll i,j; ll ans,minn,maxx; while(~scanf("%lld%lld",&n,&m)) { ans=0; ll l=0,r=0; for(i=0;i
     
      =0)//左区间的改变 minn=l-a[i]; else if(r-a[i]>=0) minn=((l+a[i])&1)?1:0; else minn=a[i]-r; l=minn;r=maxx; } //cout<