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: Accepted)

#include<bits/stdc++.h>
using namespace std;
// #define LOCAL
#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
    #define eprintf(...) 42
#endif // LOCAL
int const nax = 3e5 + 10, mod = 1e9 + 7;

void plusle(int &a, int b) {
    if ((a += b) >= mod) a -= mod;
}
int add (int a, int b) {
    return ((a += b) >= mod) ? a - mod : a;
}
int mul (int a, int b) {
    return 1LL * a * b % mod;
}

int n, k, a[nax];
int f[22][1050];
int c[1020][1020];


int main(){
    scanf("%d %d", &n, &k);
    c[0][0] = 1;
    for (int i = 1 ; i <= 1000 ; ++ i ) {
        c[i][0] = 1 ;
        for (int j = 1 ;j  <= 1000 ; ++ j)
            c[i][j] = add(c[i - 1][j - 1], c[i - 1][j]);
    }

    for(int i = 1 ;i <= k ;i ++)
        scanf("%d", a  + i);

    for (int i = 0 ; i <= a[1] ; ++ i)
        f[1][i] = mul(c[n][i], c[n - i][a[1] - i]);

    for (int j = 2 ; j <= k ; ++ j)
        for (int i = 0 ; i <= n ; ++ i)
            for (int from = 0 ; from <= n - i && i <= a[j]; ++ from)
                plusle(f[j][from + i] ,mul(f[j - 1][from], mul(c[n - from][ i], c[n - i][a[j] - i])));

    printf("%d\n", f[k][n]);
}
/*
    Good Luck
        -Lucina
*/