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].
import java.io.*;
import java.util.*;

public class redAndBlue {

	public static void main(String[] args) throws IOException {
		
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

		StringTokenizer st = new StringTokenizer(f.readLine());
		
		int t = Integer.parseInt(st.nextToken());
		
		for(int x = 0; x < t; x++) {
			
			int N = Integer.parseInt(f.readLine());
			
			ArrayList<Integer> r = new ArrayList<Integer>();
			
			st = new StringTokenizer(f.readLine());
			for(int y = 0; y < N; y++) {
				r.add(Integer.parseInt(st.nextToken()));
			}
			
			int M = Integer.parseInt(f.readLine());
			
			ArrayList<Integer> b = new ArrayList<Integer>();
			
			st = new StringTokenizer(f.readLine());
			for(int y = 0; y < M; y++) {
				b.add(Integer.parseInt(st.nextToken()));
			}
			
			int[] nums = new int[N+M];
			
			for(int y = 0; y < nums.length; y++) {
				if (r.size() != 0 && b.size() != 0) {
					
					if (r.get(0) > b.get(0)) {
						if (function2(b) > function2(r)) {
							nums[y] = b.remove(0);
						}
						else {
							nums[y] = r.remove(0);
						}
					}
					else {
						if (function2(r) > function2(b)) {
							nums[y] = r.remove(0);
						}
						else {
							nums[y] = b.remove(0);
						}
					}
				}
				else {
					
					if (r.size() == 0) {
						nums[y] = b.remove(0);
					}
					else {
						nums[y] = r.remove(0);
					}
				}
				
			}
			
			out.println(function1(nums));
		}
		
		out.close();
		
	}

	public static int function2(ArrayList<Integer> a1) {
		int output = 0;
		int sum = 0;
		
		for(int x = 0; x < a1.size(); x++) {
			
			sum += a1.get(x);
			output = Math.max(output, sum);
			
		}
		
		return output;
	}
	
	public static int function1(int[] a1) {
		int output = 0;
		int sum = 0;
		
		for(int x = 0; x < a1.length; x++) {
			
			sum += a1[x];
			output = Math.max(output, sum);
			
		}
		
		return output;
	}
}
