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 vec vector 
#define PB push_back 

#define MP make_pair 
#define MT make_tuple 
#define F first 
#define S second 

using ll = long long int; // watch out for overflows in your code
using pii = pair<int,int>;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void debug() { cout << "\n"; };
template <typename T, typename... Targs> 
void debug(T t, Targs... args) {
    cout << t << " ";
    debug(args...);
}

template <typename T1, typename T2> 
inline ostream& operator << (ostream& os, const pair<T1,T2>& p) {
    return os << "(" << p.F << ", " << p.S << ")";
}

template <typename T> 
inline ostream& operator << (ostream& os, const vec<T>& v) {
    for(auto& t : v) {
        os << t << " ";
    }
    return os;
}

template <typename T> 
inline ostream& operator << (ostream& os, const vec<vec<T>>& m) {
    for(auto& v : m) {
        for(auto& t : v) {
            os << t << " ";
        }
        os << "\n";
    }
    return os;
}

template <typename T> 
inline ostream& operator << (ostream& os, const set<T>& v) {
    for(auto& t : v) {
        os << t << " ";
    }
    return os;
}

template <typename K, typename V> 
inline ostream& operator << (ostream& os, const map<K,V>& m) {
    for(auto& p : m) {
        os << p.F << " -> " << p.S << "\n";
    }
    return os;
}

const int INF = 2 * 1e9;
int T, N, M;
vec<int> A, B;

int increment(int& i, const vec<int>& v) {
    int S = v.size();
    if(i >= S) {
        return -INF;
    }
    int P = v[i++];
    while(i < S && P < 0) {
        P += v[i];
        i++;
    }
    while(i < S && P <= P + v[i]) {
        P = P + v[i];
        ++i;
    }
    return P;
}

int compute() {
    int j = 0, i = 0;
    int pa, pb;
    int S = 0;
    int ans = 0;

    pa = increment(i, A);
    pb = increment(j, B);

    while(i <= N || j <= M) {
        //debug("pa", pa, "pb", pb);
        if(pa < 0 && pb < 0) {
            break;
        }
        if(pa >= pb) {
            S += pa;
            pa = increment(i, A);
        }
        else {
            S += pb;
            pb = increment(j, B);
        }
        ans = max(ans, S);
    }
    return ans;
}

////////////////////////////////////////////////////////////////////////

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> T;
    while(T--) {
        cin >> N;
        A.assign(N, 0);
        for(int i = 0; i < N; ++i) {
            cin >> A[i];
        }
        cin >> M;
        B.assign(M, 0);
        for(int j = 0; j < M; ++j) {
            cin >> B[j];
        }
        cout << compute() << "\n";
    }

}

