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;

template <int MOD_> struct modnum {
	static constexpr int MOD = MOD_;
	static_assert(MOD_ > 0, "MOD must be positive");

private:
	using ll = long long;

	int v;

	static int minv(int a, int m) {
		a %= m;
		assert(a);
		return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
	}

public:

	modnum() : v(0) {}
	modnum(ll v_) : v(int(v_ % MOD)) { if (v < 0) v += MOD; }
	explicit operator int() const { return v; }
	friend std::ostream& operator << (std::ostream& out, const modnum& n) { return out << int(n); }
	friend std::istream& operator >> (std::istream& in, modnum& n) { ll v_; in >> v_; n = modnum(v_); return in; }

	friend bool operator == (const modnum& a, const modnum& b) { return a.v == b.v; }
	friend bool operator != (const modnum& a, const modnum& b) { return a.v != b.v; }

	modnum inv() const {
		modnum res;
		res.v = minv(v, MOD);
		return res;
	}
	friend modnum inv(const modnum& m) { return m.inv(); }
	modnum neg() const {
		modnum res;
		res.v = v ? MOD-v : 0;
		return res;
	}
	friend modnum neg(const modnum& m) { return m.neg(); }

	modnum operator- () const {
		return neg();
	}
	modnum operator+ () const {
		return modnum(*this);
	}

	modnum& operator ++ () {
		v ++;
		if (v == MOD) v = 0;
		return *this;
	}
	modnum& operator -- () {
		if (v == 0) v = MOD;
		v --;
		return *this;
	}
	modnum& operator += (const modnum& o) {
		v += o.v;
		if (v >= MOD) v -= MOD;
		return *this;
	}
	modnum& operator -= (const modnum& o) {
		v -= o.v;
		if (v < 0) v += MOD;
		return *this;
	}
	modnum& operator *= (const modnum& o) {
		v = int(ll(v) * ll(o.v) % MOD);
		return *this;
	}
	modnum& operator /= (const modnum& o) {
		return *this *= o.inv();
	}

	friend modnum operator ++ (modnum& a, int) { modnum r = a; ++a; return r; }
	friend modnum operator -- (modnum& a, int) { modnum r = a; --a; return r; }
	friend modnum operator + (const modnum& a, const modnum& b) { return modnum(a) += b; }
	friend modnum operator - (const modnum& a, const modnum& b) { return modnum(a) -= b; }
	friend modnum operator * (const modnum& a, const modnum& b) { return modnum(a) *= b; }
	friend modnum operator / (const modnum& a, const modnum& b) { return modnum(a) /= b; }
};

template <typename T> T pow(T a, long long b) {
	assert(b >= 0);
	T r = 1; while (b) { if (b & 1) r *= a; b >>= 1; a *= a; } return r;
}

using num = modnum<int(1e9) + 7>;

int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	int n, k;
	cin >> n >> k;
	vector<num> fact(n+1), ifact(n+1);
	fact[0] = 1;
	for (int i = 1; i <= n; i++) fact[i] = fact[i-1] * i;
	for (int i = 0; i <= n; i++) ifact[i] = inv(fact[i]);
	vector<num> dp(n+1);
	vector<num> ndp(n+1);
	dp[0] = 1;
	for (int i = 0; i < k; i++) {
		int a;
		cin >> a;
		fill(ndp.begin(), ndp.end(), 0);
		for (int s = 0; s <= n; s++) {
			for (int v = 0; v <= a && s+v <= n; v++) {
				ndp[s+v] += dp[s] * fact[n-v] * ifact[n-a] * ifact[a-v] * ifact[v];
			}
		}
		swap(dp, ndp);
	}
	num ans = dp[n] * fact[n];
	cout << ans << '\n';
}
