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.*;
import java.util.regex.*;
public class vk18
{
    public static void main(String[]stp) throws Exception
    {
    	Scanner scan=new Scanner(System.in);
    	PrintWriter pw = new PrintWriter(System.out);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        //String []s;
        int n=Integer.parseInt(st.nextToken()),i,j;
        st=new StringTokenizer(br.readLine());
        HashMap<Integer,TreeSet<Integer>> hm=new HashMap<>();
        TreeSet<Integer> ts;
        for(i=1;i<=100;i++) {ts=new TreeSet<>(); hm.put(i,ts); }
        Integer a[]=new Integer[n];
    	Character c[]=new Character[n];
    	Arrays.fill(c,'P');
        Integer fre[]=new Integer[101];
        Arrays.fill(fre,0);
        int max=0,k=0,one=0,two=0;
        Integer b[]=new Integer[n];
        for(i=0;i<n;i++) 
        {
        	a[i]=Integer.parseInt(st.nextToken());
        	fre[a[i]]++; 
        	(hm.get(a[i])).add(i);
        	max=Math.max(a[i],max);
        }
        boolean virat=true;
        int index=0;
        for(i=0;i<=max;i++)
        {
        	if(fre[i]==1) { one++; b[k++]=i; }
        	else if(fre[i]>2 && virat) { two++; index=i; virat=false; };
        }
        if(one %2 ==0)
        {
        	boolean flag=true;
        	for(i=0;i < k ;i++)
        	{
        		if(flag) { c[(hm.get(b[i])).first()]='A'; flag=false; }
        		else { c[(hm.get(b[i])).first()]='B'; flag=true; }
        	}
        }
        else if(two==1)
        {
        	boolean flag=true;
        	for(i=0;i < k ;i++)
        	{
        		if(flag) { c[(hm.get(b[i])).first()]='A'; flag=false; }
        		else { c[(hm.get(b[i])).first()]='B'; flag=true; }
        	}
        	c[hm.get(index).first()]='B';
        }
        else { pw.write("NO"); pw.flush(); System.exit(0); }
        pw.println("YES");
        for(i=0;i<n;i++)
        {
        	if(c[i]=='P') pw.print("A");
        	else pw.print(c[i]);

        }
        pw.flush();

        
    }
	
}