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;
typedef  long long   ll;
typedef  long double ld;
typedef __int128 lx;
#define ceil(n, m) ((n / m) + ( n % m ? 1 : 0))
#define mod(n, m) (((n % m) + m) % m)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
//    freopen("chess.in", "r", stdin);

    int t;
    cin >> t;
    while(t--){
        int n, m;
        cin >> n;
        vector<int> a(n + 1);
        for(int i = 1; i <= n; ++i)
            cin >> a[i];
        cin >> m;
        vector<int> b(m + 1);
        for(int i = 1; i <= m; ++i)
            cin >> b[i];

        for(int i = 1; i <= n; ++i)
            a[i] += a[i - 1];
        for(int i = 1; i <= m; ++i)
            b[i] += b[i - 1];
        int ans = 0;
        for(int i = 0; i <= n; ++i)
            for(int j = 0; j <= m; ++j)
                ans = max(ans, a[i] + b[j]);
        cout << ans << '\n';

    }





}