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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>

int main(void){
    int S[4];

    int i=0;
    for(;i<4;i++){
        scanf("%d",&S[i]);
    }

    int ans[2]={0,0};
    int b[2];

    b[0]=S[0]*10+S[1];
    b[1]=S[2]*10+S[3];

    for(i=0;i++;i<2){
        if(b[i]<=12){
            ans[i]=1;
        }
    }

    switch(ans[0]){
        case(0):
        switch(ans[1]){
            case(0):
            printf("NA\n");
            case(1):
            printf("YYMM\n");
        }
        case(1):
        switch(ans[1]){
            case(0):
            printf("MMYY\n");
            case(1):
            printf("AMBIGUOUS\n");
        }
    }
    return 0;
}