Cats and a Mouse

Two cats named and are standing at integral points on the x-axis. Cat is standing at point and cat is standing at point . Both cats run at the same speed, and they want to catch a mouse named that's hiding at integral point on the x-axis. Can you determine who will catch the mouse?
You are given queries in the form of , , and . For each query, print the appropriate answer on a new line:

• If cat catches the mouse first, print Cat A.
• If cat catches the mouse first, print Cat B.
• If both cats reach the mouse at the same time, print Mouse C as the two cats fight and mouse escapes.

Input Format

The first line contains a single integer, , denoting the number of queries.
Each of the subsequent lines contains three space-separated integers describing the respective values of (cat 's location), (cat 's location), and (mouse 's location).

Constraints

Output Format

On a new line for each query, print Cat A if cat catches the mouse first, Cat B if cat catches the mouse first, or Mouse C if the mouse escapes.

Sample Input 0

3
1 2 3
1 3 2
2 1 3

Sample Output 0

Cat B
Mouse C
Cat A

Explanation 0

Query 0: The positions of the cats and mouse are shown below:
Cat will catch the mouse first, so we print Cat B on a new line.
Query 1: In this query, cats and reach mouse at the exact same time:
Because the mouse escapes, we print Mouse C on a new line.

C-code:

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    int q;
    scanf("%d",&q);
    for(int a0 = 0; a0 < q; a0++){
        int x,y,z;
        scanf("%d %d %d",&x,&y,&z);
        if(abs(z-x)<abs(z-y))
            printf("Cat A\n");
        else if(abs(z-x)>abs(z-y))
             printf("Cat B\n");
        else
             printf("Mouse C\n");
    }
    return 0;
}