顯然可以容斥去掉每人都不為空的限制。每種物品分配方式獨立，各自算一個可重組合乘起來即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 1000000007
#define N 1010 
char getc(){char c=getchar();while ((c<'
 
A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1
 
)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N],C[N<<1][N<<1],f[N],ans;
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4710.in","r",stdin);
    freopen("bzoj4710.out","w",stdout);
    const char LL[]="%I64d\n";
#else
 

    const char LL[]="%lld\n";
#endif
    m=read(),n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    C[0][0]=1;
    for (int i=1;i<=2000;i++)
    {
        C[i][0]=C[i][i]=1;
        for (int j=1;j<i;j++) C[i][j]=(C[i-1][j-1]+C[i-1][j])%P;
    }
    for (int i=1;i<=m;i++)
    {
        f[i]=1;
        for (int j=1;j<=n;j++)
        f[i]=1ll*f[i]*C[a[j]+i-1][i-1]%P;
    }
    for (int i=m;i>=1;i--)
    if (m-i&1) inc(ans,P-1ll*C[m][i]*f[i]%P);
    else inc(ans,1ll*C[m][i]*f[i]%P);
    cout<<ans;
    return 0;
}