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++ (GCC 9.2.1) (judge verdict: Accepted)

/*
	  Author : @evenbao
	  Created : 2020 / 07 / 12
*/

#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

#define pii pair<int , int>
#define mp make_pair
#define fi first
#define se second

template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void read(T &x) {
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}

const int N = 1000 + 10;
const int mod = (int) 1e9 + 7;

int n , K;
int a[N] , binom[N][N] , dp[N][N];

inline void inc(int &x , int y) { x = x + y < mod ? x + y : x + y - mod; }
inline void dec(int &x , int y) { x = x - y >= 0 ? x - y : x - y + mod; }
inline int quickpow(int a , int n) {
		int b = a , res = 1;
		for (; n >>= 1; a = (LL) a * a % mod)
				if (n & 1) res = (LL) res * a % mod;
		return res;
} 

int main() {
		
		read(n); read(K);
		for (int i = 1; i <= K; ++i) read(a[i]);
		for (int i = 0; i <= n; ++i) {	
				binom[i][0] = binom[i][i] = 1;
				for (int j = 1; j < i; ++j)
						binom[i][j] = (binom[i - 1][j] + binom[i - 1][j - 1]) % mod;
		}		
		dp[0][0] = 1;
		for (int i = 1; i <= K; ++i)
		for (int j = 0; j <= n; ++j)
		for (int k = 0; k <= j; ++k)
				if (a[i] >= j - k)
						inc(dp[i][j] , (LL) dp[i - 1][k] * binom[n - k][j - k] % mod * binom[n - j + k][a[i] - j + k] % mod);
		printf("%d\n" , dp[K][n]);
	  return 0;
}