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>
#include<string.h>
int main(){
    char s[4];
    scanf("%s",s);
    int f = 10*(s[0]-'0') + s[1]-'0';
    //printf("%d\n",f);
    int b = 10*(s[2]-'0') + s[3]-'0';
    if(f>=1 && f<=12 && b >=1 && b<=12){ //AMBIGUOUS
        printf("AMBIGUOUS\n");
    }
    else if(f>=0 && f <100 && b>=1 && b<=12 ){//YYMM
        printf("YYMM\n");
    }
    else if(f>=1 && f<=12 && b >=0 && b<100){//MMYY
        printf("MMYY\n");
    }
    else{
        printf("NA\n");
    }

    return 0;
}