1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 const int N = 22;
 7 int _, n, dp[N][N][N][N];
 8 int a[N], b[N], suma[N], sumb[N];
 9 
10 int DP(int al, int ar, int bl, int br, int flag)//flag为1表示先挑， flag为0表示后挑
11 {
12     if(al>ar&&bl>br) return 0;
13     if(dp[al][ar][bl][br]){
14         if(flag) return dp[al][ar][bl][br];
15         else return suma[ar] - suma[al-1] + sumb[br] - sumb[bl-1]-dp[al][ar][bl][br];
16     }
17     int ans = 0;
18     if(al<=ar){
19         ans = max(DP(al+1, ar, bl, br, 0) + a[al], ans);
20         ans = max(DP(al, ar-1, bl, br, 0) + a[ar], ans);
21     }
22     if(bl<=br){
23         ans = max(DP(al, ar, bl+1, br, 0) + b[bl], ans);
24         ans = max(DP(al, ar, bl, br-1, 0) + b[br], ans);
25     }
26     dp[al][ar][bl][br] = ans;
27     if(!flag) ans = suma[ar] - suma[al-1] + sumb[br] - sumb[bl-1] - ans;//如果是后挑，把剩余卡的综合减去先挑者能得到的最大分数
28     return ans;
29 }
30 
31 void solve()
32 {
33     memset(dp, 0, sizeof(dp));
34     scanf("%d", &n);
35     for(int i=1; i<=n; i++) {
36         scanf("%d", a+i); 
37         suma[i] = suma[i-1] + a[i];
38     }
39     for(int i=1; i<=n; i++) {
40         scanf("%d", b+i); 
41         sumb[i]= sumb[i-1] + b[i];
42     }
43     printf("%d\n", DP(1, n, 1, n, 1));
44 }
45 
46 int main()
47 {
48 //    freopen("in.txt", "r", stdin);
49     scanf("%d", &_);
50     while(_--) solve();
51     return 0;
52 }