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 ll long long
#define sd(x) scanf("%d", &(x))
#define pii pair<int, int>
#define F first
#define S second
#define all(c) ((c).begin()), ((c).end())
#define sz(x) ((int)(x).size())
#define ld long double

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<", "<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"{";
	for(int i = 0;i < (int)v.size(); i++){
		if(i)os<<", ";
		os<<v[i];
	}
	os<<"}";
	return os;
}

#ifdef LOCAL
#define cerr cout
#else
#endif

#define TRACE

#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
	cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif

const int N = 1e5 + 10, mod = 1e9 + 7;
int fact[N], invfact[N];
inline int add(int x, int y){ x += y; if(x >= mod) x -= mod; return x;}
inline int sub(int x, int y){ x -= y; if(x < 0) x += mod; return x;}
inline int mul(int x, int y){ return (((ll) x) * y) % mod;}
inline int powr(int a, ll b){
	int x = 1 % mod;
	while(b){
		if(b & 1) x = mul(x, a);
		a = mul(a, a);
		b >>= 1;
	}
	return x;
}
inline int inv(int a){ return powr(a, mod - 2);}
void pre(){
	fact[0] = invfact[0] = 1;
	for(int i = 1;i < N; i++) fact[i] = mul(i, fact[i - 1]);
	invfact[N - 1] = inv(fact[N - 1]);
	for(int i = N - 2; i >= 1; i--) invfact[i] = mul(invfact[i + 1], i + 1);
	assert(invfact[1] == 1);
}

inline int C(int n, int k){
	if(n < k || k < 0) return 0;
	return mul(fact[n], mul(invfact[k], invfact[n - k]));
}

int dp[1005][1005], dp2[1005][1005];
int a[1005];
int main(){
	pre();
	dp[0][0] = 1;
	int n, k; sd(n); sd(k);
	for(int i = 1; i <= k; i++){
		sd(a[i]);
		for(int old = 0; old <= n; old++){
			for(int x = 0; x <= a[i] && x + old <= n; x++){
				dp[i][old + x] = add(dp[i][old + x], mul(dp[i - 1][old], mul(invfact[x], C(n - x, a[i] - x))));
			}
		}
		// trace(dp[i][1], dp[i][2], dp[i][3]);
	}
	printf("%d\n", mul(fact[n], dp[k][n]));
}