#include<iostream>
using namespace std;
int n,m;
int a[5005],b[5005];
int dp[5005][5005][2];//取到a[i],b[j]时的最小最大值,最后是0a  1b
const int inf=0x3f3f3f3f3f3f3f3f;
int next(int x,int jiou){
	if((x+1)%2==jiou)	return x+1;
	else	return x+2;
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)	cin>>a[i];
	cin>>m;
	for(int i=1;i<=m;i++)	cin>>b[i];
	for(int i=0;i<=n;i++){
		for(int j=0;j<=m;j++){
			if(i==0&&j==0)	continue;
			//取a[i]过来
			if(i>=1)	dp[i][j][0]=min(next(dp[i-1][j][0],a[i]),next(dp[i-1][j][1],a[i]));
			else	dp[i][j][0]=inf;
			//取b[j]过来
			if(j>=1)	dp[i][j][1]=min(next(dp[i][j-1][0],b[j]),next(dp[i][j-1][1],b[j]));
			else	dp[i][j][1]=inf;
		}
	}
	/*for(int i=0;i<=n;i++){
		for(int j=0;j<=m;j++){
			cout<<dp[i][j][0]<<","<<dp[i][j][1]<<" ";
		}
		cout<<endl;
	}*/
	cout<<min(dp[n][m][0],dp[n][m][1]);
	return 0;
}