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 <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <array>
#include <set>
#include <queue>

using namespace std;

int main() {
	int t;
	cin >> t;
	for (int _ {}; _ < t; _++) {
		int n, m;
		cin >> n;
		vector<int> r (n);
		for (int i {}; i < n; i++)
			cin >> r[i];
		vector<int> gr {0};
		bool ok {false};
		for (int i {}; i < n; i++) {
			if (r[i] >= 0 || !ok) {
				gr[gr.size()-1] += r[i];
				ok = true;
			} else {
				gr.push_back(r[i]);
				ok = false;
			}
		}
		if (gr[gr.size()-1] <= 0)
			gr.erase(gr.begin()+gr.size()-1);
		cin >> m;
		vector<int> b (m);
		for (int i {}; i < m; i++)
			cin >> b[i];
		vector<int> gb {0};
		ok = false;
		for (int i {}; i < m; i++) {
			if (b[i] >= 0 || !ok) {
				gb[gb.size()-1] += b[i];
				if (b[i] >= 0)
					ok = true;
			} else {
				gb.push_back(b[i]);
				ok = false;
			}
		}
		if (gb[gb.size()-1] <= 0)
			gb.erase(gb.begin()+gb.size()-1);
		vector<int> ggr {gr}, ggb {gb};
		for (int i {}; i < (int)ggr.size(); i++) {
			while (ggr[i] <= 0 && i+1 < (int)ggr.size()) {
				ggr[i] += ggr[i+1];
				ggr.erase(ggr.begin()+i+1);
			}
		}
		for (int i {}; i < (int)ggb.size(); i++) {
			while (ggb[i] <= 0 && i+1 < (int)ggb.size()) {
				ggb[i] += ggb[i+1];
				ggb.erase(ggb.begin()+i+1);
			}
		}
		int it1 {}, it2 {}, res {}, cnt {};
		while (it1 < (int)ggr.size() || it2 < (int)ggb.size()) {
			if (it1 < (int)ggr.size() && it2 < (int)ggb.size()) {
				if (ggr[it1] > ggb[it2]) {
					cnt += ggr[it1];
					it1++;
				} else {
					cnt += ggb[it2];
					it2++;
				}
			} else if (it1 < (int)ggr.size()) {
				cnt += ggr[it1];
				it1++;
			} else {
				cnt += ggb[it2];
				it2++;
			}
			res = max(res, cnt);
		}
		cout << res << "\n";
	}
}
