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 <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <cmath>
    #include <vector>
    #include <queue>
    #include <map>
    #include <set>
    using namespace std;
    #define ll long long
    #define ull unsigned long long
    #define fi first
    #define se second
    #define dingyi int mid = l + r >> 1, ls = p << 1, rs = p << 1 | 1
    #define y0 y_csyakioi_0
    #define y1 y_csyakioi_1
    #define rep(i, x, y) for(int i = x; i <= y; ++i)
    #define per(i, x, y) for(int i = x; i >= y; --i)
    #define repn(i, x, y, z) for(int i = x; i <= y; i += z)
    #define pern(i, x, y, z) for(int i = x; i >= y; i -= z)
    inline int read(){
    	int x = 0, f = 1; char ch = getchar();
    	while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); }
    	while(ch >= '0' && ch <= '9'){ x = x * 10 + (ch ^ 48); ch = getchar(); }
    	return x * f;
    }
    const int N = 110;
    int n, m, a[N], b[N], s[N], t[N];
    inline void mian(){
    	n = read(); rep(i, 1, n) a[i] = read();
    	m = read(); rep(i, 1, m) b[i] = read();
    	int max1 = 0, max2 = 0;
    	rep(i, 1, n) s[i] = s[i - 1] + a[i];
    	rep(i, 1, m) t[i] = t[i - 1] + b[i];
    	rep(i, 1, n) max1 = max(max1, s[i]);
    	rep(i, 1, m) max2 = max(max2, t[i]);
    	printf("%d\n", max1 + max2);
    }
    signed main(){ int qwq = read(); while(qwq--) mian(); return 0; }