Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x. 
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); 
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

Input

The problem contains mutiple test cases.Please process to the end of file. 
In each case, there will be two lines. 
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

Output

For each case, output f(k) % m in one line.

Sample Input

10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0

Sample Output

45
104

代码：递推式套矩阵快速幂即可，注意多组输入

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>using namespace std;const int N=10;int n,mod;
int temp[N][N];
int res[N][N],a[N][N];
void mul(int a[][N],int b[][N]) {memset(temp,0,sizeof(temp));for(int i=0; i<N; i++)for(int j=0; j<N; j++)for(int k=0; k<N; k++)temp[i][j]=(temp[i][j]+a[i][k]*b[k][j]%mod)%mod;for(int i=0; i<N; i++)for(int j=0; j<N; j++)a[i][j]=temp[i][j];return ;
}
void QuickPow(int nn) {memset(res,0,sizeof(res));for(int i=0; i<N; i++)res[i][i]=1;while(nn) {if(nn&1)mul(res,a);mul(a,a);nn>>=1;}return ;
}
int main() {while(~scanf("%d %d",&n,&mod)) {memset(a,0,sizeof(a));for(int i=0; i<N; i++)scanf("%d",&a[0][i]);for(int i=1; i<N; i++)a[i][i-1]=1;if(n<10) printf("%d\n",n%mod);else {QuickPow(n-9);int ans=0;for(int i=0; i<N; i++)ans+=res[0][i]*(9-i)%mod;printf("%d\n",ans%mod);}}return 0;
}