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 <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <tuple>
#include <iomanip>
#include <climits>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
//typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n)-1; i >= 0; i--)
#define pb push_back
#define f first
#define s second
#define chmin(x, y) x = min(x, y);
#define chmax(x, y) x = max(x, y);

//const ll INFL = LLONG_MAX;//10^18 = 2^60
const ll INFL = 1LL<<60;
//const int INF = INT_MAX;
const ll INF = 1 << 30;//10^9
ll MOD = 1000000007;
//ll MOD = 998244353;

vector<ll> dy = {0, 1, 0, -1, 1, 1, -1, -1, 0};
vector<ll> dx = {1, 0, -1, 0, 1, -1, 1, -1, 0};

void pres(double A, ll x = 10){cout << fixed << setprecision(x) << A << endl;}
void BinarySay(ll x, ll y = 60){rep(i, y) cout << (x>>(y-1-i) & 1); cout << endl;}
ll cnt_bit(ll x){return __builtin_popcountll(x);}

ll pow_long(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x; k /= 2;
  }
  return res;
}

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

ll inverse(ll x){return pow_mod(x, MOD - 2);};

ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll x, ll y){
  ll res = x / gcd(x, y);
  res *= y;
  return res;
};

//コンビネーション
const int MAXcomb = 200010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
//facはn!,finvは1/n!
//invは逆元
void COMinit(){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(int i = 2; i < MAXcomb; i++){
        fac[i] = fac[i-1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
ll comb(int n, int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * finv[k] % MOD * finv[n-k] % MOD;
}

const int MAXkai = 200010;
ll kai_memo[MAXkai];
ll kai(ll N){
  if(kai_memo[N] != 0) return kai_memo[N];
  if(N <= 1) return 1;
  return kai_memo[N] = N * kai(N-1) % MOD;
}

ll N, K;
vector<ll> A;
ll dp[22][1010];
void solve(){
  COMinit();
  cin >> N >> K;
  A.resize(K);
  rep(i, K) cin >> A[i];
  dp[0][0] = 1;
  rep(i, K)rep(j, N+1)rep(k, N+1){
    if(j+k>N) continue;
    dp[i+1][j+k] += dp[i][j] * comb(N-j, k) % MOD * comb(N-k, A[i]-k) % MOD;
    dp[i+1][j+k] %= MOD;
  }
  cout << dp[K][N] << endl;
}


int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}