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].
/*
    Anuraj Agarwal
*/
#include<bits/stdc++.h>
#include<chrono>

using namespace std;
using namespace chrono;

#define PI 3.1415926535897932384626
#define sp(x,y)                     fixed<<setprecision(y)<<x
#define mk(arr,n,type)              type *arr=new type[n]; // dynamic allocation of memory
#define del(arr)                    delete []arr;           // deletion of allocated memory
// const int m = 1'000'000'007;
const int inf = 1e9 + 5;


#define fo(i, n) for(i=0; i<n; i++)
#define Fo(i,k,n) for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define ll long long
#define pb push_back
#define F first
#define S second
#define vbe(x) x.begin(), x.end()
#define sortvbe(x) sort(vbe(x))
#define deb(x) cout << #x << " = " << x << endl; //to check the values

inline void swap(int &a, int &b)
{
    int temp = a;
    a = b;
    b = temp;
}
inline int modulo(int a, int b)
{
    return (b + (a % b)) % b;
}
inline ll expo_mod(ll e, ll p, ll m)
{
    if(p == 0)return (1);
    ll ans = expo_mod(e, p / 2, m);
    ans = ((ans % m) * (ans % m)) % m;
    if(p % 2 != 0)ans = ((e % m) * (ans % m)) % m;
    return ans % m;
}
inline ll expo(ll e, ll p)
{
    if(p == 0)return (1);
    ll ans = expo(e, p / 2);
    ans = (ans) * (ans);
    if(p % 2 != 0)ans = (e) * (ans);
    return ans;
}
inline ll com_wd_m(ll n, ll r, ll m)
{
    ll p = 1, d = 1;
    if(r == 0 or r == n )return(1);
    if(r == 1 )return (n % m);
    r = min(r, n - r);
    for(ll i = 0; i < r; i++)
    {
        p = ((p % m) * ((n % m) - (i % m)) % m) % m;
        p = ((p % m) * (expo((i + 1), m - 2) % m)) % m;
    }
    return (p % m);
}
inline ll com(ll n, ll k) 
{ 
    ll res = 1; 
    if (k > n - k) 
        k = n - k; 
    for (ll i = 0; i < k; ++i) { 
        res *= (n - i); 
        res /= (i + 1); 
    } 
    return res; 
} 
inline ll fact(ll n)
{
    return (n == 1 || n == 0) ? 1 : n * fact(n - 1);
}
inline ll fact_mod(ll n, ll x, ll m)
{
    if(n == x) return 1;
    return (n == 1 || n == 0) ? 1 : ((n % m) * (fact((n - 1) % m)) % m) % m;
}
ll MOD=1e9+7;
ll ans[200001];

void A_A()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
    freopen("output.in", "w", stdout);
    freopen("error.in", "w", stderr);
#endif
}
ll num;
void solve()
{
    ll n,m,i;
    cin>>n;
    vector<ll> a(n);
    fo(i,n) cin>>a[i];
    cin>>m;
    vector<ll> b(m);
    fo(i,m) cin>>b[i];
    ll mx=0,sum=0;
    fo(i,n){
        sum+=a[i];
        mx=max(sum,mx);
    }
    ll mx1=0;
    sum=0;
    fo(i,m){
        sum+=b[i];
        mx1=max(sum,mx1);
    }
    cout<<mx+mx1<<endl;

}


int main()
{
    A_A();
    ll t = 1;
    cin>>t;
    while(t--)
    {
        solve();
    }
    cerr << "Itna time laga hai dekh lo : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
    return 0;
}
/*

*/