#include<bits/stdc++.h>
#define int long long
using namespace std;
void ts(){cout<<"IAKIOI\n";}
inline int read(){
	int n=0,f=1,ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		n=n*10+ch-'0';
		ch=getchar();
	}
	return n*f;
}
const int mod=998244353;
struct mint{
    int x;mint(int o=0){x=o;}mint&operator+=(mint a){return(x+=a.x)%=mod,*this;}mint&operator-=(mint a){return(x+=mod-a.x)%=mod,*this;}
    mint&operator*=(mint a){return(x=1ll*x*a.x%mod),*this;}mint&operator^=( int b){mint a=*this;x=1;while(b)(b&1)&&(*this*=a,1),a*=a,b>>=1;return*this;}
    mint&operator/=(mint a){return*this*=(a^=mod-2);}friend mint operator+(mint a,mint b){return a+=b;}friend mint operator-(mint a,mint b){return a-=b;}
    friend mint operator*(mint a,mint b){return a*=b;}friend mint operator/(mint a,mint b){return a/=b;}friend mint operator^(mint a,int b){return a^=b;}
    mint operator-(){return 0-*this;}bool operator==(const mint b)const{return x==b.x;}
};
int n,l,e;
int c[305],p[10];
int dp[305][241][305][2];
int ne[241][305][2];
//前 i 次，目前能量 j, 发动了 k 次厉害的攻击，0/1 次不太厉害的攻击的情况数.
signed main(){
	n=read();
	for(int i=1;i<=n;i++)c[i]=read();
	l=read();
	for(int i=1;i<=l;i++)p[i]=read();
	sort(p+1,p+l+1);
	e=read();
	dp[0][e][0][0]=1;
	for(int i=1;i<=n;i++){
		memcpy(dp[i],dp[i-1],sizeof(dp[i]));
		for(int j=1;j<=c[i];j++){
			for(int k=0;k<=240;k++)for(int x=0;x<=300;x++)for(int y=0;y<2;y++)ne[k][x][y]=0;
			for(int k=1;k<=l;k++){
				int v=p[k];
				for(int x=0;x<=240;x++){
					for(int y=0;y<=300;y++){
						for(int z=0;z<2;z++){
							if(!dp[i][x][y][z])continue;
							(ne[min(x+v,240ll)][y][z]+=dp[i][x][y][z])%=mod;
							if(x>=120&&v)(ne[min(240ll,x-115+v)][y+1][z]+=dp[i][x][y][z])%=mod;
							else if(x>=120)(ne[x-115][y+(z+1)/2][(z+1)%2]+=dp[i][x][y][z])%=mod;
						}
					}
				}
			}
			memcpy(dp[i],ne,sizeof(dp[i]));
		}
		for(int k=0;k<=240;k++)for(int x=0;x<=300;x++)for(int y=0;y<2;y++)ne[k][x][y]=0;
		for(int x=0;x<=240;x++){
			for(int y=0;y<=300;y++){
				for(int z=0;z<2;z++){
					if(!dp[i][x][y][z])continue;
					(ne[min(x+20,240ll)][y][z]+=dp[i][x][y][z])%=mod;
				}
			}
		}
		memcpy(dp[i],ne,sizeof(dp[i]));
	}
	int res=0;
	for(int i=0;i<=240;i++){
		for(int j=0;j<=300;j++){
			for(int k=0;k<2;k++)res+=j*dp[n][i][j][k]%mod,res%=mod;
		}
	}
	printf("%lld\n",res);
	return 0;
}