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>
#define pb push_back
#define eb emplace_back
#define F first
#define S second
#define lp(i,n) for(int i=0;i<n;i++)
#define lp2(i,a,n) for(int i=a;i<n;i++)
#define lli long long int
#define ulli unsigned long long int
#define mod (int)(1e9+7)
#define see(ar)    for(int i=0;i<(sizeof(ar)/sizeof(ar[0]));i++)    cout<<ar[i]<<" "; cout<<endl;
#define seev(v) for( auto it:v) cout<<it<<" "; cout<<endl;
#define vi vector<int>
#define vll vector<lli>
#define pii pair<int,int>
#define pll pair<lli,lli>
#define all(v) v.begin(),v.end()
#define all_rev(v) v.rbegin(),v.rend()
#define lb(v,i) lower_bound(v.begin(),v.end(),i)-v.begin()
#define ub(v,i) upper_bound(v.begin(),v.end(),i)-v.begin()
#define bsr(v,i) binary_search(v.begin(),v.end(),i)
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define umap unordered_map
using namespace std;
lli solve(vi a)
{
    lli cr=0,mxsf=LONG_MIN;
    for(int i=0;i<a.size();i++)
    {
        cr+=a[i];
        // if(cr<0)
        //     cr=0;
        if(cr>mxsf)
            mxsf=cr;
    }
    return max(0LL,mxsf);
}
//lli power(lli a, lli b){ lli r=power(a,b/2); r=(r*r)%mod; if(b&1) r=(r*a)%mod; return r;}
int main()
{
    fast;
    lli t,n,k,v,p,l,m,r,a,b,c;
    string s;
    char ch;
    cin>>t;
    while(t--)
    {
        cin>>n;
        vi a(n,0);
        lp(i,n) cin>>a[i];
        lli sm=solve(a);

        cin>>m;
        vi b(m,0);
        lp(i,m) cin>>b[i];
        sm+=solve(b);
        cout<<sm<<endl;


    }
    
    return 0;
}