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 fst ios_base::sync_with_stdio(false);cin.tie(0), cout.tie(0);
#define f(i,a,b) for(int i=a;i<b;i++)
#define fr(i,a,b) for(int i=a;i>=b;i--)
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define gcd __gcd
#define vi vector<int>
#define vll vector<long long>
#define lb lower_bound
#define ub upper_bound

typedef long long ll;
#define mod 1000000007
int main()
{
    fst;
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int a[n];
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
        }
        int m;
        cin>>m;
        ll sum=0;
        int b[m];
        for(int i=0;i<m;i++)
        {
            cin>>b[i];
            
        }
        //ll sum=0;
        ll ans1=0;
       
        for(int i=0;i<n;i++)
        {
            sum=0;
           for(int j=0;j<=i;j++)
           {
               sum+=a[j];
           }
           ans1=max(ans1,sum);
        }
        ll ans2=0;
        for(int i=0;i<m;i++)
        {
            sum=0;
           for(int j=0;j<=i;j++)
           {
               sum+=b[j];
           }
           ans2=max(ans2,sum);
        }
        cout<<ans1+ans2<<endl;
    }
}
