#include<cstdio>
using namespace std;
const int ggg=998244353;
int n,m;
int f[111100]={};
int sum[111111]={};
int cheng[511111]={};
int chu[511111]={};
long long shuliang[511111]={};
long long ans[511111]={};
bool heshu[511111]={};
int xiayige[555555]={};//预处理表示下一个因子的值，以便在多次分解质因子时不用重复计算
long long kuaisumi(long long a,long long b)
{
	if(b==0)
		return 1;
	int top=0;
	bool kj[44]={};
	long long te=1;
	long long gg[44]={};
	for(;b;b>>=1)
		if(b&1)
			kj[++top]=1;
		else
			kj[++top]=0;
	gg[1]=a;
	for(int i=2;i<=top;i++)
		gg[i]=gg[i-1]*gg[i-1]%ggg;
	for(int i=1;i<=top;i++)
		if(kj[i])
			te=te*gg[i]%ggg;
	return te;
}
int main()
{
	freopen("asm_formation.in","r",stdin);
	freopen("asm_formation.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&f[i]);
	for(int i=1;i<=n;i++)
		sum[i]=sum[i-1]+f[i];
	for(int i=2;i<=500000;i++)//筛法求质数
		for(int j=i*2;j<=500000;j=j+i)
			heshu[j]=1;
	for(int i=2;i<=500000;i++)
	{
		int j=i;
		if(!heshu[i])
		{
			xiayige[i]=1;
			continue;
		}
		for(int k=2;k*k<=j;k++)
			if(j%k==0)
			{
				j/=k,xiayige[i]=j;break;
			}
	}
	for(int i=2;i<=n;i++)
	{
		for(int j=1;j<f[i];j++)
			cheng[sum[i]-j]++,chu[j]++;
	}
	for(int i=2;i<=500000;i++)
		if(cheng[i]>0)
		{
			int j=i;
			while(j>1)
			{
				int temp=j/xiayige[j];
				shuliang[temp]=shuliang[temp]+cheng[i];
				j=xiayige[j];
			}
		}
	for(int i=2;i<=500000;i++)//计算每个质数乘的个数
		if(chu[i]>0)
		{
			int j=i;
			while(j>1)
			{
				int temp=j/xiayige[j];
				shuliang[temp]=shuliang[temp]-chu[i];
				j=xiayige[j];
			}
		}
	long long daan=1;
	for(int i=2;i<=500000;i++)
		ans[i]=kuaisumi(i,shuliang[i]);//答案用快速幂求解
	for(int i=2;i<=500000;i++)
		daan=daan*ans[i]%ggg;
	printf("%d\n",daan);
}