设计状态$f_i$为分给i个同学，使得每个同学都可以拿到特产的方案数

$g_i$为分给$i$个同学，至多$i$个同学都拿到的方案数

设特产数量分别为$a_1，a_2，...，a_m$

$g_{n}=\prod_{i=1}^{m} (^{n-1}_{n+a_{i}-1})$

$g_{n}=\sum_{i=1}^{n}f_i\ * \ (^{i}_{n})\\$ 反演得到：$\\$ $f_{n}=\sum_{i=1}^{n} (-1)^{n-i}\ * \ (^{i}_{n})\ * \ \prod_{j=1}^{m} (^{i-1}_{i+a_{j}-1})$

Code:

#include<bits/stdc++.h>
using namespace std;
long long n, m;
long long a[10010];
const long long mod = 1e9 + 7;
long long jc[10010], inv[10010];
long long quickpow(long long x, long long power) {
	long long sum = 1;
	while (power) {
		if (power & 1)sum *= x;
		sum%=mod;x*=x;x%=mod;power/=2;
	}
	return sum;
}
void init() {
	jc[1] = 1;
	for (long long i = 2; i <= n; i++) {
		jc[i] = jc[i - 1] * i;
		jc[i] %= mod;
	}
	inv[n]=quickpow(jc[n],mod-2);
	for(long long i=n-1;i>=1;i--){
		inv[i]=inv[i+1]*(i+1);
		inv[i]%=mod;
	}
	return;
}
long long c(long long big,long long sma){
	if(sma==0||big==sma)return 1;
	return jc[big]*inv[big-sma]%mod*inv[sma]%mod;
}
signed main() {
	cin>>n>>m;
	for(long long i=1;i<=m;i++)cin>>a[i];
	init();
	long long ans=0;
	for(long long i=1;i<=n;i++){
		long long f=((n-i)&1?-1:1);
		long long z=c(n,i);
		long long sum=1;
		for(long long j=1;j<=m;j++){
			sum*=c(i+a[j]-1,i-1);
			sum%=mod;
		}
		ans+=f*(z*sum%mod);
		ans=(ans+mod)%mod;
	}
	cout<<ans;
}