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 ctoi(const char c)
{
    switch (c)
    {
    case '0':
        return 0;
    case '1':
        return 1;
    case '2':
        return 2;
    case '3':
        return 3;
    case '4':
        return 4;
    case '5':
        return 5;
    case '6':
        return 6;
    case '7':
        return 7;
    case '8':
        return 8;
    case '9':
        return 9;
    default:
        return -1;
    }
}

int main()
{
    char S[5];

    scanf("%s", S);
    // printf("%c", S[0]);

    int mae, usi = 0;
    // 先頭の二文字
    mae = ctoi(S[0]) * 10 + ctoi(S[1]);
    // 後ろの二文字
    usi = ctoi(S[2]) * 10 + ctoi(S[3]);

    // printf("%d%d", mae, usi);

    // NA どれもない
    if ((mae == 0 || usi == 0) || (12 < mae && 12 < usi) )  {
        printf("NA");
        return 0;
    }
    // AMBIGUOUS(どちらもあり得る)
    else if(0 < mae && mae <= 12 && 0 < usi && usi <= 12) {
        printf("AMBIGUOUS");
        return 0;
    } 
    // YYMM
    else if(12 < mae) {
        printf("YYMM");
        return 0;
    } else {
        printf("MMYY");
        return 0;
    }
    return 0;
}