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 <iostream>
using namespace std;
#define ll long long
#define ld long double
#define mp make_pair
#define pb push_back
#define fo(i,n) for(ll i=0;i<n;i++)
#define fo1(i,n) for(ll i=1;i<=n;i++)
#define loop(i,a,b)for(ll i=a;i<=b;i++)
#define loopr(i,a,b)for(ll i=b;i>=a;i--)
#define all(x) x.begin(), x.end()
#define sz(x) (ll)(x).size()
#define vll vector<ll> 
#define vvl vector<vll>
#define vpll vector<pll>
#define pll pair<ll,ll>
#define F first
#define S second
#define MOD 1000000007
ll max(ll a,ll b){if (a>b) return a; else return b;}
ll gcd(ll a, ll b){if(b==0)return a;return gcd(b, a%b);}
ll lcm(ll a, ll b){return a*b/gcd(a, b);}
ll fexp(ll a, ll b){ll ans = 1;while(b){if(b&1) ans = ans*a%MOD; b/=2;a=a*a%MOD;}return ans;}
ll inverse(ll a, ll p){return fexp(a, p-2);}
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	//freopen("input.txt","r",stdin);
	ll T;
	cin >> T;
	ll t=0;
	while(t++<T){
		ll n;
		cin>>n;
		vll r(n);
		fo(i,n)cin>>r[i];
		ll m;
		cin>>m;
		vll b(m);
		fo(i,m)cin>>b[i];
		ll mx=0,su=0;
		ll ans=0;
		fo(i,n){su+=r[i];mx=max(mx,su);}
		ans+=mx;
		mx=0;su=0;
		fo(i,m){su+=b[i];mx=max(mx,su);}
		ans+=mx;
		cout<<ans<<endl;
	}
return 0;
}
