Problem: Vasya has a multiset s consisting of n integer numbers. Vasya calls some number x nice if it appears in the multiset exactly once. For example, multiset \{1, 1, 2, 3, 3, 3, 4\} contains nice numbers 2 and 4.

Vasya wants to split multiset s into two multisets a and b (one of which may be empty) in such a way that the quantity of nice numbers in multiset a would be the same as the quantity of nice numbers in multiset b (the quantity of numbers to appear exactly once in multiset a and the quantity of numbers to appear exactly once in multiset b).

Input

The first line contains a single integer n~(2 ≤ n ≤ 100).

The second line contains n integers s_1, s_2, ... s_n~(1 ≤ s_i ≤ 100) — the multiset s.

Output

If there exists no split of s to satisfy the given requirements, then print "NO" in the first line.

Otherwise print "YES" in the first line.

The second line should contain a string, consisting of n characters. i-th character should be equal to 'A' if the i-th element of multiset s goes to multiset a and 'B' if if the i-th element of multiset s goes to multiset b. Elements are numbered from 1 to n in the order they are given in the input.

If there exist multiple solutions, then print any of them.

Examples

Input

4
3 5 7 1


Output

YES
BABA


Input

3
3 5 1


Output

NO

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;
public class Multiset {

	public static void main(String[] args) throws IOException {
		BufferedReader jk = new BufferedReader(new InputStreamReader(System.in)) ;
		StringTokenizer ana = new StringTokenizer(jk.readLine()) ;
		int n =Integer.parseInt(ana.nextToken());
		int tab[] = new int[100] ;
		Vector<Integer> v = new Vector<Integer>() ;
		ana = new StringTokenizer(jk.readLine()) ;
		for(int i=0 ; i<n ;i++)
		{
			int a = Integer.parseInt(ana.nextToken());
			tab[a-1]++ ;
			v.add(a) ;
		}
		int compt=0 ;
		int in=-2 ;
		int j=1 ;
		for(int h :tab) 
		{
			if(h==1)  compt++ ;
			else if(h>2) in=j ;
			j++ ;
		}

		if(compt%2==1 && in==-2 ) System.out.println("NO");
		else 
		{
			System.out.println("YES");
			int comptaa =0 ; int comptbb =0 ;
			String ta[]= new String[n] ;
			for(int o=0 ; o<n ;o++) ta[o]="0" ;
			String k ="" ;
			for(int i=0 ; i<n ;i++)
			{
				if(compt%2==1 && in==v.elementAt(i))
				{
					if(comptbb<comptaa) 
					{
						ta[i]="B"; comptbb++ ; k="A" ; in=-1 ;
					}
					else 
					{
						ta[i]="A"; comptaa++ ; k="B" ;in=-1 ;
					}
				}
				else if(tab[v.elementAt(i)-1]==1)
				{
					if(comptbb<comptaa) 
					{
						ta[i]="B"; comptbb++ ; 
					}
					else 
					{
						ta[i]="A"; comptaa++ ; 
					}

				}

			}
			for(int o=0 ; o<n ; o++) 
			{

				if(ta[o].equals("0"))
				{
					if(k.equals("")) ta[o]="B" ;
					else ta[o]=k ;
				}
			}
			for(String h : ta) 
			{
				System.out.print(h);
			}

		}


	}

}