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].
/*
------------------------------------------------------------
--                                                        --
--   cccccc   oooooooo    ddddd      eeeeee    rrrrr      --
--   cc       oo    oo    dd   dd    ee        rr   rr    --
--   cc       oo    oo    dd   dd    eeeeee    rrrrr      --
--   cc       oo    oo    dd   dd    ee        rr  rr     --
--   cccccc   oooooooo    ddddd      eeeeee    rr   rr    --
--                                                        --
------------------------------------------------------------
*/
#include<bits/stdc++.h>
#include<cmath>
#define ll long long int
#define accuracy(n,value) fixed<<setprecision(n)<<double(value)
#define floop(i,a,b) for(ll i=a;i<b;i++)
#define all(v) v.begin(), v.end()
#define pb(a) push_back(a)
#define INF 1e9+7
#define present(container, element) (container.find(element) != container.end())
#define cpresent(container, element) (find(all(container),element) != container.end())
using namespace std;


ll mod=1000000007;
ll getElement(set<int>&st,int index){
    return *(std::next(st.begin(), index));
}


ll power(ll x,ll y){
    ll ans=1;
    while(y--){
        ans*=x;
    }
    return ans;
}

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
  

  int t;
  cin>>t;

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

      
      vector<int>first(n+1,0);
      for(int i=1;i<=n;i++){
          int value;
          cin>>value;
          first[i]=first[i-1]+value;
      }

      int m;
      cin>>m;
      vector<int>second(m+1,0);
      for(int i=1;i<=m;i++){
          int value;
          cin>>value;
          second[i]=second[i-1]+value;
      }

      cout<<*max_element(all(first))+*max_element(all(second))<<'\n';
  }
}


