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].
#include<iostream>
#include<bits/stdc++.h>
#define itr(i,n,m) for(int i=n;i<=m;i++)
#define itr1(it,M) for(auto& it : M)
#define pb push_back
#define qc std::ios::sync_with_stdio(0);
using namespace std;
typedef long long ll;
typedef unsigned long long int ul;
typedef double dl;
int main()
{
	int t,n,m;
	int a[105],b[105];
	scanf("%d",&t);
	while(t--)
	{
		int flag=0;
		scanf("%d",&n);
		int suma[105],sumb[105];
		memset(suma,0,sizeof(suma));
		memset(sumb,0,sizeof(sumb));
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			suma[i]+=suma[i-1]+a[i];
		}
		
		scanf("%d",&m);
		for(int i=1;i<=m;i++)
		{
			scanf("%d",&b[i]);
			sumb[i]+=sumb[i-1]+b[i];
		}
		
		int max=-100000000;
		int cnta=0,cntb=0;
		for(int i=0;i<=n;i++)
		for(int j=0;j<=m;j++)
		{
			if(max<suma[i]+sumb[j]) {
				max=suma[i]+sumb[j];cnta=i;cntb=j;
			}
		}
		if(max<=0) printf("0\n");
		else printf("%d \n",max);
	}
}