There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)

Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).

Constraints

* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N

Input

Input is given from Standard Input in the following format:


N K
a_1 a_2 \ldots a_K


Output

Print the answer.

Examples

Input

3 2
3 2


Output

12


Input

856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63


Output

337587117
#include<cstring>
#include<algorithm>
#include<cstdio>
#define rep(a,b,c) for (int a=b;a<=c;a++)
#define per(a,b,c) for (int a=b;a>=c;a--)
typedef long long LL;
typedef double dob;
using namespace std;
const int P=1000000007;
const int N=1010;
int c[N][N],f[N],g[N],n,k,a[N];
void upd(int &x,int y){x=(x+y)%P;}
int C(int x,int y){return c[x][y];}
int main(){
	scanf("%d%d",&n,&k);
	rep(i,1,k) scanf("%d",a+i);
	f[0]=1;
	rep(i,0,1000) c[i][i]=c[i][0]=1;
	rep(i,1,1000) rep(j,1,i-1) c[i][j]=(c[i-1][j-1]+c[i-1][j])%P;
	rep(i,1,k){
		memset(g,0,sizeof(g));
		rep(j,0,n) if (f[j]){
			rep(t,0,min(a[i],n-j)){
				upd(g[j+t],(LL)f[j]*C(n-j,t)%P*C(n-t,a[i]-t)%P);
			}
		}
		swap(f,g);
	}
	f[n]=(f[n]%P+P)%P;
	printf("%d\n",f[n]); 
	return 0;
}
