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;
//dengyaotriangle!

constexpr int mdn=1000000007;
struct mint{
    int v;
    inline operator int()const{return v;}
    inline mint(int v=0):v(v){}
    inline bool operator==(const mint&a)const{return v==a.v;}
    inline bool operator!=(const mint&a)const{return v!=a.v;}
    inline mint operator~ ()const{int r=1,b=v;for(int i=1;i<=(mdn-2);i<<=1){if((mdn-2)&i)r=static_cast<long long>(r)*static_cast<long long>(b)%mdn;b=static_cast<long long>(b)*static_cast<long long>(b)%mdn;}return mint(r);}
    inline mint operator- ()const{return mint(v==0?0:mdn-v);}
    inline mint operator+ (const mint&a)const{mint c(v+a.v);if(c.v>=mdn)c.v-=mdn;return c;}
    inline mint operator- (const mint&a)const{mint c(v-a.v);if(c.v<0)c.v+=mdn;return c;}
    inline mint operator* (const mint&a)const{long long c=static_cast<long long>(v)*static_cast<long long>(a.v);return mint(c%mdn);}
    inline mint operator/ (const mint&a)const{return (*this)*~a;}
    inline mint&operator++(){v++;if(v==mdn)v=0;return*this;}
    inline mint&operator--(){v--;if(v==-1)v=mdn-1;return*this;}
    inline mint&operator+=(const mint&a){v+=a.v;if(v>=mdn)v-=mdn;return*this;}
    inline mint&operator-=(const mint&a){v-=a.v;if(v<0)v+=mdn;return*this;}
    inline mint&operator*=(const mint&a){long long c=static_cast<long long>(v)*static_cast<long long>(a.v);v=c%mdn;return*this;}
    inline mint&operator/=(const mint&a){long long c=static_cast<long long>(v)*static_cast<long long>((~a).v);v=c%mdn;return*this;}
    friend istream&operator>>(istream&s,mint&a){s>>a.v;return s;}
    friend ostream&operator<<(ostream&s,const mint&a){s<<a.v;return s;}
};

const int maxn=1005;
const int maxk=22;

int n,k;
int a[maxn];
mint dp[maxk][maxn];
mint binom[maxn][maxn];

int main(){
    cin>>n>>k;
    for(int i=0;i<=n;i++){
        binom[i][0]=1;
        for(int j=1;j<=i;j++)binom[i][j]=binom[i-1][j-1]+binom[i-1][j];
    }
    for(int i=1;i<=k;i++)cin>>a[i];
    dp[0][0]=1;
    for(int i=1;i<=k;i++){
        for(int j=0;j<=n;j++){
            for(int k=0;k<=min(j,a[i]);k++){
                mint w=binom[n-j+k][k]*dp[i-1][j-k]*binom[n-k][a[i]-k];
                dp[i][j]+=w;
            }
        }
    }
    cout<<dp[k][n];
    return 0;
}