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 <iomanip>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <complex>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <numeric>
#include <random>
#include <bitset>
#define pb push_back
#define pop pop_back
#define tr(it, vi) for(auto it = vi.begin(); it != vi.end(); it++)
#define endl "\n";
#define IOS ios:: sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define deb(x) cout << #x << " " << x <<endl;
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define INF 2000000000
#define all(a) a.begin(), a.end()
#define fo(i,n) for(int i=0;i<n;i++)
#define Fo(i,k,n) for(int i=k;i<n;i++)
#define eof cout<<"-1"<<endl;
typedef long long ll;
const int mod = 1000000007;
/*//Long Way: 
int max_of_3 = max(a, max(b, c));
int max_of_4 = max(max(a, b), max(c, d));
//Easier Way - Can be extended to any number of variables:
int max_of_3 = max({a, b, c});
int max_of_4 = max({a, b, c, d});*/
using namespace std;

void solve() {
    int n;cin>>n;
    int r[n];
    fo(i,n) {
        cin>>r[i];
    }
    int m; cin>>m;
    int b[m];
    fo(i,m){
        cin>>b[i];
    }
    int rt[n];
    int bt[m];

    int rs = 0;
    int bs = 0;
    fo(i,n){
        rs = rs + r[i];
        rt[i] = rs;
    }
    fo(i,m){
        bs = bs + b[i];
        bt[i] = bs;
    }

    int rmax = 0;
    int bmax = 0;

    fo(i,n){
        if(rt[i]>rmax){
            rmax = rt[i];
        }
    }
    fo(i,m){
        if(bt[i]>bmax){
            bmax = bt[i];
        }
    }

    cout<<rmax+bmax<<endl;
}

int main() {

    #ifndef ONLINE_JUDGE 
    freopen("binput.txt", "r", stdin);
    freopen("boutput.txt", "w", stdout);
    #endif

    IOS;
    int t;
    cin >> t;
    while(t--) {
        solve();
    }
}

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