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;
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return 1;
    }
    return 0;
}
#define DEBUG
#ifdef DEBUG
template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
    os << '(' << p.first << ',' << p.second << ')';
    return os;
}
template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
    os << '{';
    for(int i = 0; i < (int)v.size(); i++) {
        if(i) { os << ','; }
        os << v[i];
    }
    os << '}';
    return os;
}
void debugg() { cerr << endl; }
template <class T, class... Args>
void debugg(const T &x, const Args &... args) {
    cerr << " " << x;
    debugg(args...);
}
#define debug(...)                                                             \
    cerr << __LINE__ << " [" << #__VA_ARGS__ << "]: ", debugg(__VA_ARGS__)
#define dump(x) cerr << __LINE__ << " " << #x << " = " << (x) << endl
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif

struct Setup {
    Setup() {
        cin.tie(0);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(15);
    }
} __Setup;

using ll = long long;
#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
#define repl(i, a, b) for(int i = a; i < int(b); i++)
#define rep(i, n) repl(i, 0, n)
const int INF = 1 << 30;
const ll LLINF = 1LL << 60;
constexpr int MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

//-------------------------------------

int main() {
    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        vector<int> r(n), sumr(n + 1, 0);
        rep(i, n) {
            cin >> r[i];
            sumr[i + 1] = sumr[i] + r[i];
        }
        int m;
        cin >> m;
        vector<int> b(m), sumb(m + 1, 0);
        rep(i, m) {
            cin >> b[i];
            sumb[i + 1] = sumb[i] + b[i];
        }
        int ans = *max_element(ALL(sumr)) + *max_element(ALL(sumb));
        cout << ans << "\n";
    }
}