Problem:
Score : 200 points

Problem Statement

You have a digit sequence S of length 4. You are wondering which of the following formats S is in:

YYMM format: the last two digits of the year and the two-digit representation of the month (example: 01 for January), concatenated in this order

MMYY format: the two-digit representation of the month and the last two digits of the year, concatenated in this order

If S is valid in only YYMM format, print YYMM; if S is valid in only MMYY format, print MMYY; if S is valid in both formats, print AMBIGUOUS; if S is valid in neither format, print NA.

Constraints

S is a digit sequence of length 4.

Input

Input is given from Standard Input in the following format:

S

Output

Print the specified string: YYMM, MMYY, AMBIGUOUS or NA.

Sample Input 1

1905

Sample Output 1

YYMM

May XX19 is a valid date, but 19 is not valid as a month. Thus, this string is only valid in YYMM format.

Sample Input 2

0112

Sample Output 2

AMBIGUOUS

Both December XX01 and January XX12 are valid dates. Thus, this string is valid in both formats.

Sample Input 3

1700

Sample Output 3

NA

Neither 0 nor 17 is valid as a month. Thus, this string is valid in neither format.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
    int s;
    int m[4];

    scanf("%d",&s);
    for(int i=3;i>=0;i--){
        m[i]=s%10;
        s=s/10;
    }
    if(m[0]*10+m[1]<=12&&m[2]*10+m[3]<=12&&m[0]*10+m[1]>0&&m[2]*10+m[3]>0){
        printf("AMBIGIOUS\n");
    }else if(m[0]*10+m[1]<=12&&m[0]*10+m[1]>0){
        printf("MMYY\n");
    }else if(m[2]*10+m[3]<=12&&m[2]*10+m[3]>0){
        printf("YYMM\n");
    }else{
        printf("NA");
    }
    return 0;
}