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 T top()
#define E end()
#define PS push
#define Q queue
#define PP pop()
#define L0 (LL)0
#define V vector
#define ST stack
#define FI first
#define MAX 4e18
#define MIN -MAX
#define DQ deque
#define SZ size()
#define C clear()
#define B begin()
#define F front()
#define SE second
#define DL double
#define G getline
#define IN insert
#define endl "\n"
#define EM empty()
#define cnn continue
#define MD 1000000007
#define PSB push_back
#define PI acos(-1.00)
#define PPB pop_back()
#define PSF push_front
#define ub upper_bound
#define lb lower_bound
#define PPF pop_front()
#define line cout<<endl
#define LL long long int
#define one cout<<1<<endl
#define zer cout<<0<<endl
#define PQ priority_queue
#define nil cout<<-1<<endl
#define N cout<<"NO"<<endl
#define Y cout<<"YES"<<endl
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#define MAU cout<<"Case "<<mau<<": "
#define SP(pre) fixed<<setprecision(pre)
#pragma GCC optimization ("unroll-loops")
#define __lcm(A1,A2) (A1/(__gcd(A1,A2)))*A2
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define file freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
#define UOSMOY v.C;w.C;ww.C;uu.C;uo.C;vo.C;vu.C;ab.C;bc.C;cc.C;uuu.C;a.C;b.C;c.C;u.C
V<LL>v;
V<LL>w,ab,bc;
set<LL>uo;
map<LL,LL>ww,cc,u;
V<pair<LL,LL> >uu,su;
map<LL,V<LL> >uuu;
map<pair<LL,LL>,LL> vo;
map<pair<LL,LL>,LL> vu;
map<LL,pair<LL,pair<LL,LL > > > vs;
priority_queue<LL,V<LL>,greater<LL> > moy;
LL dx[]= {-1,0,1,0,1,1,-1,-1};
LL dy[]= {0,1,0,-1,-1,1,1,-1};
LL dxx[]= {2,1,-1,-2,-2,-1,1,2};
LL dyy[]= {1,2,2,1,-1,-2,-2,-1};
LL dp[101][10001];
LL ar[10][10];
int main()
{
//  BRR;
//  file;
    FAST;
    char ch;
    string a,b,c;
    DL x,y,d,e,f,g,h;
    LL n,i,j,k,p,q,o,l,s,tt=1,m,t,z,aa,r=0;
    cin>>tt;
    for(LL mau=1; mau<=tt; mau++)
    {
        UOSMOY;
        cin>>n;
        s=0;
        t=0;
        for(i=0;i<n;i++)
        {
            cin>>o;
            s+=o;
            t=max(t,s);
        }
        p=0;
        q=0;
        cin>>m;
        for(i=0;i<m;i++)
        {
            cin>>o;
            p+=o;
            q=max(q,p);
        }
        k=0;
        k=max(k,t);
        k=max(k,q);
        k=max(k,q+t);
        cout<<k<<endl;
    }
    return 0;
    //IN
    //AL
}
