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 REP(i,a,n) for(int i=(a);i<(n);++i)
#define BREP(i,a,n) for(int i=(a);i>(n);--i)
#define TRAV(i, v) for(auto &i: v)
#define fi first
#define se second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()
#define INDEX(arr,ind)	(upper_bound(ALL(arr),ind)-arr.begin())
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define pow10(n) stoll("1"+string(n,'0'))
#define EPS 1e-9
// #define LOCAL 1

typedef long long ll; 
typedef long double ld;
typedef pair<int, int> ii; 
typedef pair<ll, ll> pll; 
typedef vector < int > vi;
typedef vector < ll > vll;
typedef vector<ii> vii; 
typedef map<int, int> mii;

const long long INF = 1LL << 60;



void solve(){
	int n, m;
	cin >> n;
	vi r(n);
	REP(i,0,n) cin >> r[i];

	cin >> m;
	vi b(m);
	REP(j,0,m) cin >> b[j];

	int mxr = max(0, r[0]), mxb=max(0, b[0]);
	REP(i,1,n){
		r[i]=r[i]+r[i-1];
		mxr=max(r[i], mxr);
	}
	REP(i,1,m){
		b[i]=b[i]+b[i-1];
		mxb=max(b[i], mxb);
	}
	cout << mxr+mxb << endl;	
}

int main(){
	
	#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);		
	#endif

	int T;
	cin >> T;
	for (int tc = 1; tc <= T; ++tc)
	{
		// cout << "Case #" << tc << ": ";
		solve();
	}
	return 0;
}