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: Accepted)

#include <stdio.h>
int main(void)
{
    int a,b;
    a = 0;
    b = 0;
    char s[5000];
    scanf("%s", s);
    if(s[0] == '0' && s[1] != '0'){
        a ++;
    }
    else if(s[0] == '1' && s[1] <= '2'){
        a ++;
    }
    if(s[2] == '0' && s[3] != '0'){
        b ++;
    }
    else if(s[2] == '1' && s[3] <= '2'){
        b ++;
    }
    if(a < b){
        printf("YYMM");
    }
    else if(a > b){
        printf("MMYY");
    }
    else if(a == 1 && b == 1){
        printf("AMBIGUOUS");
    }
    else if(a == 0 && b == 0){
        printf("NA");
    }
    
    return 0;
}