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].
/*------U Have To DO It------*/
/*      BY-> RicoProg        */
/*        ___ 
           __|__|
          |  |___           */
#include <bits/stdc++.h>
using namespace std;
 
/**/
 
//---------------------------------------------------MACROS----------------------------------------------------------
#define ll long long
#define ld long double
#define beg(i ,n) for(ll i=0;i<n;i++)
#define beg1(i ,n) for(ll i=1;i<n;i++)
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define pb push_back 
#define eb emplace_back
#define mp make_pair
#define nl "\n" 
#define sp " "
#define vec(v , n) vector<ll> v(n)
#define all(x) x.begin() , x.end()
#define pii pair<int , int>
#define pll pair<ll , ll>
#define mii map<int , int>
#define mll map<ll , ll>
#define mci map<char , int>
#define f1 first
#define f2 second
#define deb(a) cout << #a << "=" << a << nl
#define deb2(a , b) cout << #a << "=" << a << "," << #b << "=" << b << nl
#define deb3(a , b , c) cout << #a << "=" << a << "," << #b << "=" << b << "," << #c << "=" << c << nl
#define deb4(a , b , c , d) cout << #a << "=" << a << "," << #b << "=" << b << "," << #c << "=" << c << "," << #d << "=" << d << nl
#define deb5(a , b , c , d , e) cout << #a << "=" << a << "," << #b << "=" << b << "," << #c << "=" << c << "," << #d << "=" << d << "," << #e << "=" << e << nl

//---------------------------------------------------GLOBAL----------------------------------------------------------
const ll MOD2 = 998244353 ;
const ll MOD = 1e9 + 7 ;
const ll MAX = 1e5 + 7 ;
const ll inf = 1e18L + 5 ;
const double pi = 3.14159265358979323846 ;
ll fpow(ll n , ll m ){ll res=1 ;while(m>0){if(m&1){res*=n ;}m/=2 ;n*=n ;}return res;}
ll modInv(ll n){return fpow(n , MOD-2);}
//---------------------------------------------------FUNCTIONS-------------------------------------------------------

const ll nax = 1e5 + 5 ;//Check constraints from question again
//Global Variable ahead make sure to reinitialise them in case of multiple TC's

ll util(vector<ll> v , ll n){
    ll ans = 0 , temp = 0 ;
    beg(i , n){
        temp += v[i] ;
        ans = max(ans , temp) ;
    }
    return ans ;
}

void solve(){
    ll n , m ;
    cin >> n ;
    vector<ll> red(n) ;
    beg(i , n) cin >> red[i] ;
    cin >> m ;
    vector<ll> blue(m) ;
    beg(i , m) cin >> blue[i] ;
    cout << util(red , n) + util(blue , m) << nl ;
}

//---------------------------------------------------DRIVER----------------------------------------------------------
int main(){
    fast ;
    int t=1 ;
    cin >> t ;
    while(t--){
        solve() ;
    }
    cerr << "Time elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " sec \n";
    return 0;
}  

//if nothing strikes, read question again make sure it is correctly interpreted 