Problem:
Score : 800 points

Problem Statement

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).

Notes

\binom{n}{k} denotes the number of possible choices of k objects out of given distinct n objects, disregarding order.

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.

Sample Input 1

3 2
3 2

Sample Output 1

12

On the first day, all the children 1, 2, and 3 receive one cookie.

On the second day, two out of the three children 1, 2, and 3 receive one cookie.

Their happiness is 4 in any case, so the expected happiness is 4. Print this value multiplied by \binom{3}{3} \times \binom{3}{2}, which is 12.

Sample Input 2

856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63

Sample Output 2

337587117

Find the expected value multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K}, modulo (10^9+7).

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <numeric>
using i64 = long long;

std::vector<long long> fact, inv;

constexpr auto pow(long long x, long long n, const long long mod) {
    long long ret = 1;
    while(n > 0) {
        if ((n & 1) == 1) ret = (ret * x) % mod;
        x = (x * x) % mod;
        n >>= 1;
    }
    return ret;
}

constexpr auto inverse(const long long x, const long long p) {
    return pow(x, p - 2, p);
}

void init_factorial(const int n, const long long mod) {
    fact.resize(n + 1);
    inv.resize(n + 1);
    fact[0] = 1;
    for (long long i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % mod;
    inv[n] = inverse(fact[n], mod);
    for (long long i = n; i > 0; i--) inv[i - 1] = inv[i] * i % mod;
}

int main() {
    constexpr i64 mod = 1000000007;
    int n, k;
    std::cin >> n >> k;
    init_factorial(n, mod);
    std::vector<int> a(k);
    for(auto &e : a) std::cin >> e;

    auto f = [&](int i, int j) { return fact[n - j] * inv[n - a[i]] % mod * inv[a[i] - j] % mod; };

    const int s = std::accumulate(a.begin(), a.end(), 0);

    std::vector<i64> dp(s + 1);
    dp[0] = 1;
    for (int i = 0; i < k; i++) {
        std::vector<i64> next(s + 1);
        for (int j = 0; j <= s; j++) for (int l = 0; j + l <= s && l <= a[i]; l++) {
            next[j + l] = (next[j + l] + dp[j] * f(i, l) % mod * inv[l] % mod) % mod;
        }
        std::swap(dp, next);
    }

    std::cout << dp[n] * fact[n] % mod << std::endl;

    return 0;
}