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].
// Just a writer
/*
Don't stop when you are tired
.... Stop when you done.
*/
//make it sense
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef double dbl;
ll n,x,y,m,s,t,k,a,b,f,u;
const int mod=1e9+7;
const int mxn=2e5+5;
//const double by=3.141592653;
ll mx=0,mn=1e18;
//map<char ,int >mp,pm;
//ll a,b;
//char ch[1001];
ll arr[2000000],wan[2000000];
//ll dp[210][500];
//map<ll,ll>mp;
//pair<ll,ll>arrj[1000001];
//char war[1001][1001];
//pair<ll,ll>pa[1001];
//char ch;
//map<string,ll >mp,pm,sa;
string str,str2;
//string str4="abcdefghijklmnopqrstuvwxyz";

//const long long  N=100001;
//ll dp[205][205][205];

//double totle;
/*
bool isprime(int f)
{
    if(f%2==0&&f!=2||f<2)return 0;
    for(int i=3;i*i<=f;i++)
    {
        if(f%i==0)return 0;
    }
    return 1;
}
ll fsp(ll fr, ll se) {
    if (se == 0)
        return 1;
    ll res = fsp(fr, se / 2);
    if (se & 1)
        return (res * res* fr);
        return (res * res);
}
bool prime[(ll)1e7];
set<ll>st;
void seive()
{
   prime[0]=prime[1]=true;
   for(int i=2;i*i<=1e7;i++)
   {
       if(!prime[i])
       {
           for(int j=i*i;j<1e7;j+=i)
            prime[j]=true;
       }
   }

}
*/
map<ll,ll>mp;

int main()
{
    ios_base::sync_with_stdio(0);
    // freopen("input.txt","r", stdin);
	//freopen("output.txt","w", stdout);
       cin>>t;
       while(t--)
       {
          cin>>n;
          ll mx1=0;
          ll mx2=0,sum=0;
          for(int i=0;i<n;i++)
         {
             cin>>arr[i];
             sum+=arr[i];
             mx1=max(mx1,sum);
         }
         sum=0;
          cin>>m;
          for(int i=0;i<m;i++)
          {
              cin>>wan[i];
              sum+=wan[i];
              mx2=max(sum,mx2);
          }
          cout<<mx1+mx2<<endl;
       }
    return 0;
}
