Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

#define MAX 1002
#define MOD 1000000007
template<typename ty1,typename ty2>
inline int add(ty1 x, ty2 y) {
	if(y>=MOD)y%=MOD;
	if(x>=MOD)x%=MOD;
    x += y; return x < MOD ? x : x - MOD;
}
template<typename ty1,typename ty2>
inline void addto(ty1 &x, ty2 y) {
		  if(y>=MOD)y%=MOD;
		  if(x>=MOD)x%=MOD;
	      x += y; if (x >= MOD) x -= MOD;
}
template<typename ty1,typename ty2>
inline int sub(ty1 x, ty2 y) {
		if(y>=MOD)y%=MOD;
		 if(x>=MOD)x%=MOD;
        x -= y; return x < 0 ? x + MOD : x;
}
template<typename ty1,typename ty2>
inline void subto(ty1 &x, ty2 y) {
	if(y>=MOD)y%=MOD;
	if(x>=MOD)x%=MOD;
    x -= y; if (x < 0) x += MOD;
}
template<typename ty1,typename ty2>
inline int mul(ty1 x, ty2 y) {
        if(y>=MOD)y%=MOD;
		if(x>=MOD)x%=MOD;
        return 1ll * x * y % MOD;
}
template<typename ty1,typename ty2>
void multo(ty1 &x, ty2 y) {
        if(y>=MOD)y%=MOD;
		if(x>=MOD)x%=MOD;
        x=1ll * x * y % MOD;
}
 
int n;
int k;

vector<int> v;

int dp[21][MAX];
#define MOD 1000000007
class Combination{
	long long int ppow(long long int i, long long int j){
		long long int res = 1LL;
		while (j){
			if ((j & 1LL)){
				res *= i;
				if (res >= MOD){
					res %= MOD;
				}
			}
			j >>= 1;
			i *= i;
			if (i >= MOD){
				i %= MOD;
			}
		}
		return res;
	}
public:
	vector<long long int> k;
	vector<long long int> r;
	void resize(int N){
		k.resize(N + 2);
		r.resize(N + 2);
		k[0] = 1;
		for (int i = 1; i < N+2; i++){
			k[i] = k[i - 1];
			k[i] *= i;
			if (k[i] >= MOD)k[i] %= MOD;
		}
		long long int al = k[k.size() - 1];
		long long int iv = ppow(k[k.size() - 1],MOD-2);
		r[k.size() - 1] = iv;
		for (int i = (int)(r.size()) - 2; i >= 0; i--){
			r[i] = r[i + 1] * (i + 1);
			if (r[i] >= MOD){
				r[i] %= MOD;
			}
		}
	}
	long long int C(int a, int b){
		if (a < b)return 0;
		long long int up = k[a];
		long long int dw = r[b] * r[a - b];
		dw %= MOD;
		up *= dw;
		up %= MOD;
		return up;
	}
	long long int H(int a, int b){
		return C(a + b - 1, b);
	}
	long long int catalan_number(int n){
		return (C(2 * n, n) + MOD - C(2 * n, n - 1)) % MOD;
	}
};


//Lucas theorem とか
Combination C;
int main(){
	cin>>n>>k;
	C.resize(n+2);
	for(int i=0;i<k;i++){
		int a;
		scanf("%d",&a);
		v.push_back(a);
	}
	dp[0][0]=1;
	for(int i=0;i<k;i++){
		for(int j=0;j<=n;j++){
			if(dp[i][j]==0)continue;
			for(int jj=j;jj<=n;jj++){
				if(jj-j<=v[i]){
					addto(dp[i+1][jj],mul(dp[i][j],mul(C.C(n-j,jj-j),C.C(n-(jj-j),v[i]-(jj-j)))));
				}
			}
		}
	}
	printf("%d\n",dp[k][n]);
	return 0;
}