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 fa(x,v) for(auto x:v)
#define f(i,a,b) for(ll i=(ll)a;i<=(ll)b;i++)
#define rf(i,a,b) for(ll i=(ll)a;i>=(ll)b;i--)
#define spc " "
#define endl '\n'
#define ll long long
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define len(s) (ll)int(s.length())
#define sz(v) (ll)int(v.size())
#define MOD 1000000007

void Fast_IO() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
    freopen("output.out","w",stdout);
    #endif
} 

// Code Here

int main()
{
    Fast_IO();
    int n_testcases; cin>>n_testcases;
    for(int n_testcase = 1; n_testcase <= n_testcases; n_testcase++)
    {
    	ll n; cin >> n;
    	ll a[n]; f(i,0,n-1) cin >> a[i];
    	ll m; cin >> m;
    	ll b[m]; f(i,0,m-1) cin >> b[i];
    	f(i,1,n-1) a[i] += a[i-1];
    	f(i,1,m-1) b[i] += b[i-1];
    	ll amax = 0, bmax = 0;
    	f(i,0,n-1) amax = max(amax, a[i]);
    	f(i,0,m-1) bmax = max(bmax, b[i]);
    	cout << amax + bmax << endl;
    }
    return 0;
}