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
#include <bits/stdc++.h>
using namespace std;
#define F first
#define S second
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;

ll MOD=1e9+7;

ll modinv(ll x)
{
    ll res=1;
    ll pw=MOD-2;
    while(pw>0)
    {
        if(pw%2) res=(res*x)%MOD;
        pw/=2;
        x=(x*x)%MOD;
    }
    return res;
}
ll n, k;

ll ch[1009][1009];
ll a[1009];
ll dp[29][1009];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k;
    for(int i=1; i<=k; i++) cin>>a[i];
    for(int i=0; i<=1000; i++)
    {
        ch[i][0]=1;
        ch[i][i]=1;
    }
    for(int i=1; i<=1000; i++)
    {
        for(int j=1; j<i; j++)
        {
            ch[i][j]=(ch[i-1][j-1]+ch[i-1][j])%MOD;
        }
    }
    for(int i=0; i<=n; i++)
    {
        if(i<=a[1]) dp[1][i]=(ch[a[1]][i]*ch[n][a[1]])%MOD;
        else dp[1][i]=0;
        //cout<<"dp i="<<1<<" db="<<i<<" ::: "<<dp[1][i]<<endl;
    }
    for(int i=2; i<=k; i++)
    {
        for(ll db=0; db<=n; db++)
        {
            for(ll akt=0; akt<=min(db, a[i]); akt++)
            {
                //ll pl=ch[a[i]][akt];
                ll pl=1LL;
                if(n-(db-akt)<akt) continue;
                pl=(pl*ch[n-(db-akt)][akt])%MOD;
                pl=(pl*ch[n-akt][a[i]-akt])%MOD;
                pl=(pl*dp[i-1][db-akt])%MOD;
                //cout<<"plusz ("<<akt<<") "<<pl<<endl;
                dp[i][db]=(dp[i][db]+pl)%MOD;
            }
            //cout<<"dp i="<<i<<" db="<<db<<" ::: "<<dp[i][db]<<endl;
        }
    }
    ll ans=dp[k][n];
    /*for(int i=1; i<=n; i++)
    {
        ans=(ans*ch[n][a[i]])%MOD;
    }*/
    cout<<ans<<endl;
    return 0;
}
