Last year we set a problem on bachelor arithmetic which made some bachelors really unhappy. So to even things up, we are making a problem on the tough schedule of a married man.

 

Our dashing hero Danny has recently got married and that has created a lot of problems for him, at least that is what his friends think. So many broken promises, so many missed appointments and dinners. Err! Danny, now is losing tracks of even simplest of calculations, so you must help him to decide whether he can attend his meeting or not. Danny is busy with his wife for a large portion of the day. This large portion is denoted by a starting time and an ending time. Then Danny has an important meeting in a day, he misses that if it overlaps or touches (For example, if Danny�s time span with his wife finishes at 18:00 and the meeting starts at 18:00 then the two schedules conflict and Danny misses the meeting) the time scheduled for his wife. Given the time span Danny has allotted for his wife and the time span of the meeting you, will have to find whether Danny misses that meeting or not.

 

Input

First line of the input file contains an integer N (0<N<2001) which denotes how many sets of inputs are there. The input for each set is given in two lines. The description for each set is given below:

 

First line of each set contains two strings separated by a single space. These two strings denote the time span Danny is busy with his wife. The second line also contains two strings which denotes the time when Danny has to attend a meeting. All the strings that denote time are of the format hh:mm (two digit for hour and two digit for minute). For example �forty five past eight� (Morning) is denoted as 08:45, �forty five past 9� (night) is denoted as 21:45.� You can assume that all times are valid 24-hour clock time, starting time strictly precedes ending time and all times are within a single day.

 

Output

For each set of input produce one line of output. This line contains the serial of output followed by a string which denotes Danny�s decision. If Danny can attend the meeting then print �Hits Meeting� and if Danny misses (Mrs) the meeting as it conflicts with the time allotted for his wife print �Mrs Meeting� instead.

 

Sample Input��������������������������� Output for Sample Input

3 17:47 22:40 06:18 17:04 10:44 17:05 01:11 01:27 03:36 19:02 14:33 15:24

Case 1: Hits Meeting Case 2: Hits Meeting Case 3: Mrs Meeting

Problem setter: S.Manzoor, M. M. Rahman, Special Thanks: S. Hafiz, M. A. Arif

題目描述問兩個時段有沒有衝突。

沒有衝突的話輸出 "Hits Meeting", 反之。

所有時段都在一天內，不會跨越午時 (0:00)

#include <stdio.h>

int main() {
    int testcase, cases = 0;
    scanf("%d", &testcase);
    while(testcase--) {
        int sh, sm, eh, em;
        scanf("%d:%d %d:%d", &sh, &sm, &eh, &em);
        sh = sh*60 + sm, eh = eh*60 + em;
        int ah, am, bh, bm;
        scanf("%d:%d %d:%d", &ah, &am, &bh, &bm);
        ah = ah*60 + am, bh = bh*60 + bm;
        printf("Case %d: ", ++cases);
        if(eh < ah || sh > bh)
            puts("Hits Meeting");
        else
            puts("Mrs Meeting");
    }
    return 0;
}