//Achen
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<cstdio>
#include<queue>
#include<cmath>
#define For(i,a,b) for(int i=(a);i<=(b);i++)
#define Rep(i,a,b) for(int i=(a);i>=(b);i--)
const int N=2e5+7,mod=20130427;
typedef long long LL;
typedef double db;
using namespace std;
LL power[N],sum[N],S[N],f[N][2],ans1,ans2,p[N],q[N],a[N],b[N],B,n,m;;template<typename T>void read(T &x)  {char ch=getchar(); x=0; T f=1;while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();if(ch=='-') f=-1,ch=getchar();for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
}void jian(LL &n) {if(a[1]) a[1]--;else {a[1]=B-1;For(i,2,n) {a[i]--;if(a[i]>=0) break;a[i]=B-1;}}while(n&&!a[n]) n--;
}void pre(int len) {power[0]=1; sum[0]=1;For(i,1,len) power[i]=power[i-1]*B%mod;For(i,1,len) sum[i]=(sum[i-1]+power[i])%mod;
}void calc(LL s[],int n,LL& ans) {memset(f,0,sizeof(f));p[0]=q[n+1]=0;For(i,1,n) p[i]=(p[i-1]+s[i]*power[i-1]%mod)%mod;Rep(i,n,1) q[i]=(q[i+1]*B%mod+s[i])%mod;For(i,1,n) {LL tp=B*(B-1)/2%mod;f[i][0]=f[i-1][0]*B%mod+tp*sum[i-1]%mod*power[i-1]%mod;tp=s[i]*(s[i]-1)/2%mod;f[i][1]=(tp*sum[i-1]%mod*power[i-1]%mod+s[i]*f[i-1][0]%mod+f[i-1][1]+s[i]*sum[i-1]%mod*(p[i-1]+1)%mod)%mod;ans=(ans+(q[i+1]>1?(q[i+1]-1):0LL)*f[i][0]%mod+f[i][1])%mod;}
} int main() {
#ifdef DEBUGfreopen(".in","r",stdin);freopen(".out","w",stdout);
#endifread(B);read(n); For(i,1,n) read(a[n-i+1]);read(m); For(i,1,m) read(b[m-i+1]); jian(n);pre(200000);calc(a,n,ans1); calc(b,m,ans2); printf("%lld\n",(ans2-ans1+mod)%mod);return 0;
}
/*
55
5 54 12 6 27 14
7 45 11 25 48 7 45 52
*/