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;
#define int long long

signed main(){

    int t;cin>>t;
    while(t--){

        int n;cin>>n;

        int arr[n];
        int prefix[n];
        
        for(int i=0;i<n;i++){
            cin>>arr[i];
            if(i == 0){
                prefix[0] = arr[0];
            }
            else{
                prefix[i] = prefix[i-1]+arr[i];
            }
        }

        int m;cin>>m;

        int brr[m];
        int suffix[m];

        for(int i=0;i<m;i++){
            cin>>brr[i];
            if(i == 0){
                suffix[0] = brr[0];
            }
            else{
                suffix[i] = suffix[i-1]+brr[i];
            }
        }

        int dp[150][150];
        memset(dp,0,sizeof(dp));
        int bada = 0;

        // int i=0,j=0;
        // int sum = 0;

        // while(i<n && j<m){

        //     int val = max(arr[i],brr[j]);
        //     if(val == arr[i]){
        //         i++;
        //     }
        //     else{
        //         j++;
        //     }
        //     sum += val;
        //     bada = max(bada,sum);
        // }

        // if(i<n){
        //     sum += arr[i];
        //     i++;
        //     bada = max(bada,sum);
        // }

        // if(j<m){
        //     sum += brr[j];
        //     j++;
        //     bada = max(bada,sum);
        // }

        for(int i=0;i<n;i++){
            
            int sum = prefix[i];
            for(int j=0;j<m;j++){
               // cout<<"prefix["<<i<<"] = "<<prefix[i]<<"\n";
               // cout<<"suffix["<<j<<"] = "<<suffix[j]<<"\n";
                dp[i][j] = max(dp[i][j],prefix[i]);
                dp[i][j] = max(dp[i][j],suffix[j]);
                dp[i][j] = max(dp[i][j],prefix[i]+suffix[j]);
               // cout<<"dp["<<i<<"]["<<j<<"] = "<<dp[i][j]<<"\n";
               // cout<<"-------------------------------------"<<"\n";
                bada = max(bada,dp[i][j]);
            }
        }


        cout<<bada<<"\n";
        
    }
}