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;
#pragma optimize ("g",on)
#pragma GCC target ("avx2")
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target ("avx,avx2,fma")
#pragma GCC optimize ("03")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#define all(a) a.begin(),a.end()
#define ll long long
#define pb push_back
#define nl '\n'
#define popb pop_back()
#define sz size()
#define ld long double
#define ull unsigned long long
#define F first
#define S second
#define fix fixed<<setprecision
#define pii pair<int,int>
#define E exit (0)
const int inf=1e9;  	

int main() {
    //freopen("planting.in", "r", stdin);
	//freopen("planting.out", "w", stdout);
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t;
	cin>>t;
	while (t--) {
		int n;
		cin>>n;
		int r[n+1];
		for (int i=1;i<=n;i++) {
			cin>>r[i];
		}
		int m;
		cin>>m;
		int b[m+1];
		for (int i=1;i<=m;i++) {
		 	cin>>b[i];
		}
		ll ans=0,sum=0;
		for (int i=0;i<=n;i++) {
			ll sum2=0;
			if (i>0) {
				sum+=r[i];
			}
			for (int j=0;j<=m;j++) {
				if (j>0) {
					sum2+=b[j];
				}
				ans=max(ans,sum+sum2);
			}
		}
		cout<<ans<<nl;
	}
}