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;
#define pb push_back
#define fi first
#define se second
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define repp(i, a, b) for (int i = a; i <= (b); ++i)
#define repr(i, a, b) for (int i = a; i >= (b); --i)
#define bit(n) (1LL << (n))
#define sz(x) ((ll)(x).size())
typedef long long ll;
const int INF = 1001001001;
const ll LINF = 1001001001001001001ll;
const int MOD = 1000000007;

template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

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;
  }
  /* 変数名.pow(n) */
  mint pow(ll t) const {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // 以降はMODが素数のみ
  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 = 2000;
long long fac[MAX], finv[MAX], inv[MAX];

/* 前処理 O(n) */
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; 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;
  }
}

/* 計算 O(1) */
long long COM(int n, int k) {
  if (n < k)
    return 0;
  if (n < 0 || k < 0)
    return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main() {
  int N, K;
  cin >> N >> K;
  vector<int> a(K);
  rep(i, K) {
    cin >> a[i];
  }

  COMinit();
  vector<vector<mint>> dp(K + 1, vector<mint>(N + 1));
  dp[0][0] = 1;
  rep(k, K) {
    rep(n, N + 1) {
      repp(nn, 0, N - n) {
        dp[k + 1][n + nn] += dp[k][n] * COM(N - n, nn) * COM(N - nn, a[k] - nn);
      }
    }
  }

  cout << dp[K][N].x << endl;
}