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 f first
#define s second
#define pb push_back
#define ar array
#define all(x) x.begin(), x.end()
#define siz(x) (int)x.size()

#define FOR(x, y, z) for(int x = (y); x < (z); x++)
#define ROF(x, z, y) for(int x = (y-1); x >= (z); x--)
#define F0R(x, z) FOR(x, 0, z)
#define R0F(x, z) ROF(x, 0, z)
#define trav(x, y) for(auto&x:y)

using ll = long long;
using vi = vector<int>;
using vl = vector<long long>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;

template<class T> inline bool ckmin(T&a, T b) {return b < a ? a = b, 1 : 0;}
template<class T> inline bool ckmax(T&a, T b) {return b > a ? a = b, 1 : 0;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const char nl = '\n';
const int mxN = 2e5 + 10;
const int MOD = 1e9 + 7;
const long long infLL = 1e18;

void solve(){
    int n; cin >> n;
    vi a(n);
    F0R(i, n) cin >> a[i];
    int m; cin >> m;
    vi b(m);
    F0R(i, m) cin >> b[i];
    int ma = 0, mb = 0;
    int sum = 0;
    F0R(i, n) sum += a[i], ckmax(ma, sum);
    sum = 0;
    F0R(i, m) sum += b[i], ckmax(mb, sum);
    cout << ma + mb << nl;
}

int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    int t; cin >> t;
    while(t--){
        solve();
    }

    return 0;
}