Problem:
Score : 800 points

Problem Statement

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).

Notes

\binom{n}{k} denotes the number of possible choices of k objects out of given distinct n objects, disregarding order.

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.

Sample Input 1

3 2
3 2

Sample Output 1

12

On the first day, all the children 1, 2, and 3 receive one cookie.

On the second day, two out of the three children 1, 2, and 3 receive one cookie.

Their happiness is 4 in any case, so the expected happiness is 4. Print this value multiplied by \binom{3}{3} \times \binom{3}{2}, which is 12.

Sample Input 2

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

Sample Output 2

337587117

Find the expected value multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K}, modulo (10^9+7).

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p)pom(a,p-2,p)


//階乗,二項係数
//*
ll fact[100010];
ll invfact[100010];
void makefact(int n,ll m){
	n=min(n,m-1);
	fact[0]=1;
	rep(i,0,n)fact[i+1]=fact[i]*(i+1)%m;
	invfact[n]=invp(fact[n],m);
	for(int i=n;i>0;i--)invfact[i-1]=invfact[i]*i%m;
}
ll choose(int n,int r,int m){return n<0||r<0||n-r<0?0:fact[n]*invfact[r]%m*invfact[n-r]%m;}
//*/

ll dp[30][1010];
int a[1010];
int main(){
	makefact(3000,MOD);
	int n,k;
	scanf("%d%d",&n,&k);
	rep(i,0,k)scanf("%d",a+i);
	dp[0][0]=1;
	rep(i,1,n+1)dp[0][0]=dp[0][0]*i%MOD;

	rep(i,0,k)rep(j,0,n+1){
		rep(k,0,n-j+1)dp[i+1][j+k]=(dp[i+1][j+k]+dp[i][j]*choose(n-k,a[i]-k,MOD)%MOD*invfact[k])%MOD;
	}
//	rep(i,0,k+1){rep(j,0,n+1)printf("%d ",dp[i][j]);puts("");}
	
	printf("%lld",dp[k][n]);
}