Monocarp had a sequence a consisting of n + m integers a_1, a_2, ..., a_{n + m}. He painted the elements into two colors, red and blue; n elements were painted red, all other m elements were painted blue.

After painting the elements, he has written two sequences r_1, r_2, ..., r_n and b_1, b_2, ..., b_m. The sequence r consisted of all red elements of a in the order they appeared in a; similarly, the sequence b consisted of all blue elements of a in the order they appeared in a as well.

Unfortunately, the original sequence was lost, and Monocarp only has the sequences r and b. He wants to restore the original sequence. In case there are multiple ways to restore it, he wants to choose a way to restore that maximizes the value of 

$$$f(a) = max(0, a_1, (a_1 + a_2), (a_1 + a_2 + a_3), ..., (a_1 + a_2 + a_3 + ... + a_{n + m}))$$$

Help Monocarp to calculate the maximum possible value of f(a).

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the test cases follow. Each test case consists of four lines.

The first line of each test case contains one integer n (1 ≤ n ≤ 100).

The second line contains n integers r_1, r_2, ..., r_n (-100 ≤ r_i ≤ 100).

The third line contains one integer m (1 ≤ m ≤ 100).

The fourth line contains m integers b_1, b_2, ..., b_m (-100 ≤ b_i ≤ 100).

Output

For each test case, print one integer — the maximum possible value of f(a).

Example

Input


4
4
6 -5 7 -3
3
2 3 -4
2
1 1
4
10 -3 2 2
5
-1 -2 -3 -4 -5
5
-1 -2 -3 -4 -5
1
0
1
0


Output


13
13
0
0

Note

In the explanations for the sample test cases, red elements are marked as bold.

In the first test case, one of the possible sequences a is [6, 2, -5, 3, 7, -3, -4].

In the second test case, one of the possible sequences a is [10, 1, -3, 1, 2, 2].

In the third test case, one of the possible sequences a is [-1, -1, -2, -3, -2, -4, -5, -3, -4, -5].

In the fourth test case, one of the possible sequences a is [0, 0].
/*
少加头文件，爆零两行泪
DP后效性，爆零两行泪
文件读入错，爆零两行泪
文件名写错，爆零两行泪
调完不编译，爆零两行泪
结论不证明，爆零两行泪
数组没开够，爆零两行泪
ll开 int，爆零两行泪
 dp 没初值，爆零两行泪
深搜没边界，爆零两行泪
广搜忘出队，爆零两行泪
模数没看清，爆零两行泪
越界不特判，爆零两行泪
多测不清空，爆零两行泪
空间胡乱开，爆零两行泪
无向变有向，爆零两行泪
调试忘删除，爆零两行泪
文件不保存，爆零两行泪
题目提交错，爆零两行泪
*/
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define ll long long
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB emplace_back
#define PF emplace_front
#define FILL(a,b) memset(a,b,sizeof(a))
#define lwb lower_bound
#define upb upper_bound
using namespace std;
int dp[105][105],a[105],b[105];
void solve(){
	int n,m;
	cin>>n;
	FOR(i,1,n)cin>>a[i],a[i]+=a[i-1];
	cin>>m;
	FOR(i,1,m)cin>>b[i],b[i]+=b[i-1];
	FOR(i,0,n){
		FOR(j,0,m){
			dp[i][j]=-1e9;
			if(i){
				gmax(dp[i][j],max(dp[i-1][j],(a[i]+b[j])));
			}
			if(j){
				gmax(dp[i][j],max(dp[i][j-1],(a[i]+b[j])));
			}
			if(!i&&!j)dp[i][j]=0;
		}
	}
	cout<<max(0,dp[n][m])<<'\n';
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin>>t;
	while(t--)solve();
	RE 0;
}




