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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <iterator>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <numeric>
#include <iomanip>
#include <limits>
#include <set>
#include <map>
#include <type_traits>
#include <cstdint>
#include <queue>
#define _USE_MATH_DEFINES
#include <math.h>

typedef long long ll;

using namespace std;

//素因数分解
void factoring(ll n, map<ll, ll>& mp) {
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			mp[i]++;
			n /= i;
		}
	}
	if (n != 1) mp[n]++;
}

//絶対値
template<class T, class U = std::make_unsigned_t<T>>
U SafeAbs( T x ) {
    return x < 0 ? -static_cast<uintmax_t>(x) : x;
}

//最大・最小
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

//xのn乗
ll mod_pow(ll x, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}
//xの逆元
ll mod_inv(ll x, ll mod) {//O(log(mod))
    return mod_pow(x, mod - 2, mod);
}

//nCr
ll ncr(ll n, ll r, ll mod){
	ll res=1;
	ll num=1,den=1;
	for(ll i=n;i>=n-r+1;i--){
		(den*=i)%=mod;
	}
	for(ll i=1;i<=r;i++){
		(num*=i)%=mod;
	}
	
	res=den*mod_inv(num,mod);

	return res%mod;
}


//二分探索
bool isOk(ll index, ll key,vector<ll> &a){
	
	if(a[index]>key){
		return true;
	}
	
	return false;
}

ll bin_search(ll key, vector<ll> &a){
	
	ll l=-1;
	ll r=a.size();

	while(abs(r-l)>1){
		ll mid = (l+r)/2;
		if(isOk(mid, key, a)){
			r=mid;
		}
		else{
			l=mid;
		}
	}
	return r;
}

/*
//深さ優先探索
void dfs(vector<vector<ll>> &gra,vector<bool> &seen, ll v){

	seen[v]=true;

	for(auto next : gra[v]){
		if(seen[next]){
			continue;
		}

		dfs(gra,seen,next);
	}
	
}
*/
const ll MOD=1000000007;

void solve(){
	string s;
	cin>>s;
	
	ll a=atoi(s.substr(0,2).c_str());
	ll b=atoi(s.substr(2,2).c_str());

	if((a>12 || a==0)&& (b>12||b==0)){
		cout<<"NA"<<endl;		
	}
	else if(a>12||a==0){
		cout<<"YYMM"<<endl;
	}
	else if(b>12||b==0){
		cout<<"MMYY"<<endl;
	}
	else{
		cout<<"AMBIGUOUS"<<endl;
	}
	
}


int main(){
	
	solve();
	
    return 0;
}