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 int long long
#define ull unsigned long long
#define fio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define trace(x) clog<<#x<<": "<<x<<endl
#define Mod 1000000007
const double PI=acos(-1);
#define lb lower_bound
#define ub upper_bound
#define mkp make_pair
#define pob pop_back
#define pof pop_front
#define pb push_back
#define pf push_front

#define pi pair<int,int>
#define pll pair<ll,ll>
#define pdb pair<db,db>
#define pldb pair<ldb,ldb>

#define vi vector<int>
#define vll vector<ll>
#define vdb vector<db>
#define vldb vector<ldb>
#define vpi vector< pi >
#define vpll vector< pll >
#define yn cout<<"YES\n"
#define nn cout<<"NO\n"

#define si set<int>
#define li list<int>
#define fi first
#define in insert
#define se second
#define fr(i,n) for(i = 0 ; i < n ; i++)
#define  test int t; cin>>t; while(t--)
///NOTE- log2(1e18) = 60
///NOTE- log2(1e9) = 30
/// Last (least significant) set bit of n => n&(-n)
/// 1s Complement of a number in O(1) = ((1<<30)-1)^number //^ = XOR
/// Directly 1's complement is computed using ~a
/// To convert string numbers into integer : string s = "1234" ; int n = stoi(s); (stoi = string to integer)
/// To convert integer to string numbers :  string s = to_string(n);
/// For topological sorting use stack - Push root, the go on pushing its adjacent like in DFS, then when no more neighbours left to discover, pop and print.Printed ans will be in reverse order
/// transform(s.begin(), s.end(), s.begin(), ::tolower); => convert into lower-case
/// Catalan no. (1/n+1)*(2n C n)
/// to check whether a no. is power of 2 => (n&(n-1))==0
template<typename T, typename T1>
T mod(T x, T1 p) {
    x %= p;
    if (x < 0)
        x += p;
    return x;
}
template<typename T>
T inverse(T x, T p) {
    x = mod(x, p);
    if (x == 1)
        return x;
    return mod(1LL * (-p / x) * (inverse(p % x, p)) , p);
}
//NCR mod calculation
/*int fact[1000001];
int ifact[1000001];
int ncr(int n, int r) {
    int res = (1ll * ifact[n - r]*ifact[r]) % Mod ;
	res = (1ll*res*fact[n]) % Mod;
    return res;
}*/


template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p ) {
    return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v ) {
    os << "{";
    typename vector< T > :: const_iterator it;
    for( it = v.begin(); it != v.end(); it++ ) {
        if( it != v.begin() ) os << ", ";
        os << *it;
    }
    return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v ) {
    os << "[";
    typename set< T > :: const_iterator it;
    for ( it = v.begin(); it != v.end(); it++ ) {
        if( it != v.begin() ) os << ", ";
        os << *it;
    }
    return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v ) {
    os << "[";
    typename map< F , S >::const_iterator it;
    for( it = v.begin(); it != v.end(); it++ ) {
        if( it != v.begin() ) os << ", ";
        os << it -> first << " = " << it -> second ;
    }
    return os << "]";
}

#define deb(x) cerr << #x << " = " << x << endl;
#define all(x) x.begin(),x.end()

int32_t main()
{
    fio
    int q;
    cin>>q;
    while(q--)
    {
        int n,m;
        cin>>n;
        int mx1 = 0, mx2 = 0, sum1 = 0, sum2 = 0;
        vector<int> a(n);
        for(int i = 0 ; i < n ; i++)
        {
        cin>>a[i];
        sum1+=a[i];
        mx1 = max(mx1,sum1);
        }
        cin>>m;
        vector<int> b(m);
        for(int i = 0 ; i < m ; i++)
            {cin>>b[i];
            sum2 +=b[i];
        mx2 = max(mx2,sum2);
            }
            cout<<mx1+mx2<<"\n";
    }
    return 0;
}
