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 main(void){
  int S,YY,YYMM=0,MM,MMYY=0;
  
  scanf("%d",&S);
  
  MM = S%100;
  YY = (S -MM)/100;

  if((0<=YY&&YY<=99)&&(1<=MM&&MM<=12)){
    YYMM++;
  }
     
  if((0<=MM&&MM<=99)&&(1<=YY&&YY<=12)){
	MMYY++;
  }

  
  if(YYMM==1&&MMYY==1){
     printf("AMBIGOUS\n");
  }else if(YYMM==1){
     printf("YYMM\n");
  }else if(MMYY ==1){
    printf("MMYY\n");
  }else{
    printf("NA\n");
  }

  
  return 0;
  
}