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 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
using namespace std;
int main()
{
    int n,n1,n2,n3;
    scanf("%d",&n);


    n1 = n % 100;
    n2 = n /100;
    n3 = n2 % 100;
    if(n3<=12 && n1<=12)
        printf("AMBIGUOUS");
    else
    {
        if(n3!=0 && n1!=0)
        {
            if(n3>12 && n1<=12)
                printf("YYMM");
            else if(n3<=12 && n>12)
                printf("MMYY");


        }
        else if(n3!=0 && n1==0)
        {
            if(n3>12)
                printf("NA");
            else if(n3<=12)
                printf("MMYY");

        }
        else if(n3==0 && n1!=0)
        {
            if(n1>12)
                printf("NA");
            else if(n1<=12)
                printf("YYMM");

        }
















    }
}