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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

n,k = map(int,input().split())
a = tuple(map(int,input().split()))
mod = 10**9+7
rng = 1001
fctr = [1]
finv = [1]
for i in range(1,rng):
  fctr.append(fctr[-1]*i%mod)
for i in range(1,rng):
  finv.append(pow(fctr[i],mod-2,mod))
def cmb(n,k):
  if n<0 or k<0:
    return 0
  else:
    return fctr[n]*finv[n-k]*finv[k]%mod
dp = [[0 for i in range(n+1)] for j in range(k+1)]
dp[0][0] = 1
for i in range(1,k+1):
  x = a[i-1]
  for j in range(n+1):
    for t in range(min(n-j,x)+1):
      dp[i][j+t]=(dp[i][j+t]+dp[i-1][j]*cmb(n-j,t)*cmb(n-t,x-t))%mod
print(dp[k][n])