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<bits/stdc++.h>


using namespace std;
using ll = long long;
const ll OO = 1e18;
#define ed '\n'

#define sz(a)a.size()
#define vi vector<int>

#define all(c)c.begin(),c.end()
#define RT(v)return cout<<v<<ed,0


#define reset(mem,val)memset(mem,val,sizeof mem)

string ys = "YES", no = "NO"; const int N = 1 << 20, M = 1 << 10;
const ll mod = 1e9 + 7;

int n, m; 
vi v, v2;
ll mx ;
ll dp[105][105];
ll solve(int i, int j,ll sum){
	mx = max(mx, sum);
	if (i == n&&j == m)return 0;
	ll &rt = dp[i][j];
	if (~rt)return dp[i][j];
	rt = 0;
	if (i>n&&j<m)rt = max(solve(i + 1, j, sum + v[i]) + v[i], solve(i, j + 1, sum + v2[j]) + v2[j]);
	if (i < n)rt = max(rt,solve(i + 1, j, sum + v[i]) + v[i]);
	if(j<m) rt=max(rt,solve(i, j + 1, sum + v2[j])+v2[j]);
	return dp[i][j]=rt;
}
int main()
{

	int t; cin >> t;
	while (t--){
		cin >> n;
		v.clear();
		v.resize(n);
		for (auto &i : v) cin >> i;
		cin >> m;
		v2.clear();
		v2.resize(m);
		for (auto &i : v2) cin >> i;
		mx = 0;
		reset(dp, -1);
		solve(0, 0, 0);
		cout << mx << ed;
	}
}
