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 dd[] = { 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

int main() {
	static char s[8];
	int a, b, my, ym;

	scanf("%s", s);
	a = (s[0] - '0') * 10 + (s[1] - '0');
	b = (s[2] - '0') * 10 + (s[3] - '0');
	my = a > 0 && a <= 12 && b > 0 && b <= dd[a];
	ym = b > 0 && b <= 12 && a > 0 && a <= dd[b];
	if (my && ym)
		printf("AMBIGUOUS\n");
	else if (my)
		printf("MMYY\n");
	else if (ym)
		printf("YYMM\n");
	else
		printf("NA\n");
	return 0;
}