Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible. 
A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time. 
Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help. 

2
2
20 25
40
1
8

08:00:40 am
08:00:08 am

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream> 
#include <sstream>
using namespace std;
#define MAXN 2003
/*
求解买票的最短时间，可以单独买票可以两个人合买，给出所有人单独买票和两个人合伙所需时间，最后
输出时间注意一下格式即可
一开始想的是区间DP，发现不对。
dp[n] = max(dp[n-1]+a[n],dp[n-2]+a[n-1 到1])
*/
int dp[MAXN],a[MAXN],b[MAXN];//a是单独买，b是两人一起！
void Print(int time)
{
    int hour = time/3600,minute = time/60%60,second = time%60;
    bool am = (hour<4)?true:false;
    if(!am) hour = (8+hour)%12;
    else hour = 8+hour;
    if(hour<10)
        printf("0%d:",hour);
    else
        printf("%d:",hour);
    if(minute<10)
        printf("0%d:",minute);
    else
        printf("%d:",minute);
    if(second<10)
        printf("0%d ",second);
    else
        printf("%d ",second);
    if(am)
        printf("am\n");
    else
        printf("pm\n");
}
int main()
{
    int n,k;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&k);
        for(int i=1;i<=k;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<k;i++)
            scanf("%d",&b[i]);
        dp[1] = a[1];
        for(int i=2;i<=k;i++)
            dp[i] = min(dp[i-1]+a[i],dp[i-2]+b[i-1]);
        Print(dp[k]);
    }
}