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;

typedef long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}

int nextInt() { int x; scanf("%d", &x); return x;}
ll nextLong() { ll x; scanf("%lld", &x); return x;}


const ll MOD = (ll)(1e9 + 7);

const int SZ = 312345;
ll inv[SZ];
ll fact[SZ];
ll fact_inv[SZ];
ll choose(int n, int r) {
  if (n < 0 || r < 0 || n < r) return 0;
  return fact[n] * fact_inv[n-r] % MOD * fact_inv[r] % MOD;
}

struct mint {
  ll x;
  mint(ll x=0):x((x%MOD+MOD)%MOD){}
  mint& operator+=(const mint a) {if ((x += a.x) >= MOD) x -= MOD;return *this;}
  mint& operator-=(const mint a) {if ((x += MOD-a.x) >= MOD) x -= MOD;return *this;}
  mint& operator*=(const mint a) {(x *= a.x) %= MOD;return *this;}
  mint operator+(const mint a) const {mint res(*this);return res+=a;}
  mint operator-(const mint a) const {mint res(*this);return res-=a;}
  mint operator*(const mint a) const {mint res(*this);return res*=a;}
  mint pow(ll b) const {
    mint res(1), a(*this);
    while (b) {
      if (b & 1) res *= a;
      a *= a;
      b >>= 1;
    }
    return res;
  }
  // for prime MOD
  mint inv() const {return pow(MOD-2);}
  mint& operator/=(const mint a) {return (*this) *= a.inv();}
  mint operator/(const mint a) const {mint res(*this);return res/=a;}
};

const int MAX_N = 1005;
const int MAX_K = 22;

mint E[MAX_K][MAX_N];
mint dp[MAX_K][MAX_N];

int main2() {

  REP(i, MAX_K) REP(j, MAX_N) E[i][j] = 0;
  REP(i, MAX_K) REP(j, MAX_N) dp[i][j] = 0;

  int N = nextInt();
  int K = nextInt();
  vector<ll> a(K);
  REP(k, K) a[k] = nextInt();

  REP(k, K) {
    E[k][0] = 1;
    for (int i = 1; i <= N; i++) {
      ll rest = (a[k] - i + 1);
      mint v = (rest <= 0) ? mint(0) : mint(rest) / mint(N-i+1);
      E[k][i] = E[k][i-1] * v;
    }
  }

  dp[0][0] = mint(fact[N]);
  for (int k = 0; k < K; k++) {
    for (int n1 = 0; n1 <= N; n1++) {
      for (int n2 = 0; n1 + n2 <= N; n2++) {
        dp[k+1][n1 + n2] += dp[k][n1] * E[k][n2] * mint(fact_inv[n2]);
      }
    }
  }

  mint ans = dp[K][N];
  REP(k, K) ans *= mint(choose(N, a[k]));
  cout << ans.x << endl;

  return 0;
}

int main() {

  inv[1] = 1;
  for (int i = 2; i < SZ; i++) inv[i] = inv[(int) (MOD % i)] * (MOD - MOD / i) % MOD;
  fact[0] = 1;
  for (int i = 1; i < SZ; i++) fact[i] = fact[i-1] * i % MOD;
  fact_inv[0] = 1;
  for (int i = 1; i < SZ; i++) fact_inv[i] = fact_inv[i-1] * inv[i] % MOD;

#ifdef LOCAL
  for (;!cin.eof();cin>>ws)
#endif
    main2();
  return 0;
}
