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>
#include <chrono>
#define ll long long int 

using namespace std;

class Time{
	chrono::high_resolution_clock::time_point start;
public:
	Time(){
		start = chrono::high_resolution_clock::now();
	}
	~Time(){
		cout << "\nTime: " << chrono::duration_cast<chrono::duration<double>>(chrono::high_resolution_clock::now() - start).count() << "s\n";
	}
};

void fIO(){
	// freopen("output.txt","w",stdout);
	freopen("input.txt","r",stdin);
}

template <typename T>
ostream &operator << (ostream &os, vector<T> &v) {
	for (ll i = 0; i < v.size(); i++)
	{
		os << v[i] << ' ';
	}
	return os;
}

ll powx(ll a, ll b, ll mod){
	ll ans = 1;
	a = a % mod;
	while (b)
	{
		if(b & 1)ans = (ans * a) % mod;
		a = (a * a) % mod;
		b >>= 1;
	}
	return ans;
}

ll powll(ll a, ll b) {
	ll ans = 1;
	while (b)
	{
		if(b & 1)ans = (ans * a);
		a = (a * a);
		b >>= 1;
	}
	return ans;
}

double PI = acos(-1.0);
ll mod = 1000000007;
/*vector<ll> fac(200000);

void precompute(){
	fac[0] = 1;
	for (ll i = 1; i < 200000; i++)
	{
		fac[i] = (fac[i - 1] * i) % mod;
	}
}*/

void testCase()
{
	ll n, m;
	cin >> n;
	vector<ll> r(n);
	for (auto &&i : r)
	{
		cin >> i;
	}
	cin >> m;
	vector<ll> b(m);
	for (auto &&i : b)
	{
		cin >> i;
	}
	ll mxr = 0, sum = 0, zero = 0;
	for (ll i = 0; i < n; i++)
	{
		sum += r[i];
		if (mxr < sum) mxr = sum;
	}
	ll mxb = 0;
	sum = 0;
	for (ll i = 0; i < m; i++)
	{
		sum += b[i];
		if (mxb < sum) mxb = sum;
	}
	cout << mxr + mxb << '\n';
	
}

int main() 
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	// fIO();
	//Time time_this;
	//precompute();
	ll t = 1;
	cin >> t;
	for(; t; --t)
	{
	testCase();
	}
	return 0;
}
// 75 72 18 64 44 95 81 74 -> 58