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 <stdlib.h>

int main(void){
	char a[4];
	int b,c,i;
	int d,e,f,g;
	
	scanf("%s",a);
	
	d = a[0] -'0';
	e = a[1] -'0';
	f = a[2] -'0';
	g = a[3] -'0';
	
	b = d*10 + e;
	c = f*10 + g;
	
	if(b>=0 && c>=0){
		if(b>0 && c>0 && c<=12 && b<=12) printf("AMBIGUOUS\n");
		else if(c<=12 && c>0) printf("YYMM\n");
		else if(b<=12 && b>0) printf("MMYY\n");
		else printf("NA\n");
	}
	return 0;
}