#include<bits/stdc++.h>
using namespace std;
#define int long long
const int nn=505,mod=998244353,mzhxi=504;
int n,a[nn],dp[nn][nn],sum,fac[nn],inv[nn],ans;
inline int poww(int x,int y){
	if(y==0) return 1;
	int g=poww(x,y/2);
	if(y&1) return g*g%mod*x%mod;
	return g*g%mod;
}
inline int C(int x,int y){
	if(x<0||y<0||x-y<0) return 0;
	return fac[x]*inv[x-y]%mod*inv[y]%mod;
}
signed main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	fac[0]=1;
	for(int i=1;i<=mzhxi;i++) fac[i]=fac[i-1]*i%mod;
	inv[mzhxi]=poww(fac[mzhxi],mod-2);
	for(int i=mzhxi-1;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
dp[0][0] = 1;
	int sum = 0;
	for(int i=0; i<n; i++,sum+=a[i])
	{
		// 全部放前面 
		for(int j=2; j<=sum; j++)
			dp[i+1][j+a[i+1]] = dp[i][j];
		dp[i+1][0] = dp[i][0];
		//枚举前面放多少个(后面至少放一个)
		for(int x=0; x<a[i+1]; x++)
		{
			//枚举剩余部分从哪个位置后面开始放 
			for(int k=1; k<=sum; k++)
			{
				for(int j=k+1; j<=sum; j++)
					dp[i+1][k+1+x] = (1ll*dp[i+1][k+1+x] + 1ll*dp[i][j] * C(j - k - 1 + a[i+1] - x - 1, a[i+1] - x - 1)) % mod;
					
				//j == 0 的情况 
				dp[i+1][k+1+x] = (1ll*dp[i+1][k+1+x] + 1ll*dp[i][0] * C(sum - k + a[i+1] - x - 1, a[i+1] - x - 1)) % mod;
			}
		}
	}

	int ans = 0;
	for(int i=0; i<=sum; i++) ans = (1ll*ans + dp[n][i]) % mod;
	cout << ans << '\n'; 
	return 0;
}