def cmb(n, r, mod):#コンビネーションの高速計算　
    if ( r<0 or r>n ):
        return 0
    r = min(r, n-r)
    return (g1[n] * g2[r] % mod) * g2[n-r] % mod

mod = 998244353#出力の制限
N = 2*10**3
g1 = [1]*(N+1) # 元テーブル
g2 = [1]*(N+1) #逆元テーブル
inverse = [1]*(N+1) #逆元テーブル計算用テーブル

for i in range( 2, N + 1 ):
    g1[i]=( ( g1[i-1] * i ) % mod )
    inverse[i]=( ( -inverse[mod % i] * (mod//i) ) % mod )
    g2[i]=( (g2[i-1] * inverse[i]) % mod )
inverse[0]=0

import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import log,gcd

input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

def solve(N,M,A):
    A = [0] + A
    dp = [[[0 for j in range(A[m-1]*(m>0)+1)] for i in range(A[m]-A[m-1]*(m>0)+1)] for m in range(M+1)]

    for m in range(1,M+1):
        if m!=1:
            for j in range(A[m-1]+1):
                inv = g2[A[m-1]] * g1[A[m-1]-j] * g1[j] % mod
                for k in range(A[m-1]-A[m-2]+1):
                    tmp = cmb(A[m-1]-A[m-2],k,mod) * cmb(A[m-2],j-k,mod)
                    if not tmp:
                        continue
                    tmp *= inv * dp[m-1][k][j-k]
                    tmp %= mod
                    dp[m][0][j] += tmp
                    dp[m][0][j] %= mod

        for i in range(1,A[m]-A[m-1]+1):
            for j in range(A[m-1]+1):
                dp[m][i][j] = A[m] + i * dp[m][i-1][j] + j * dp[m][i][j-1]
                dp[m][i][j] %= mod
                dp[m][i][j] *= inverse[i+j]
                dp[m][i][j] %= mod

    return dp[M][A[M]-A[M-1]][A[M-1]]

N,M = mi()
A = li()
print(solve(N,M,A))