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:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] freq = new int[101];
		int[] arr = new int[n];
		for(int i=0; i<n; i++)
		{
		    arr[i] = sc.nextInt();
		    freq[arr[i]]++;
		}
		int nice=0, two=0;
		for(int i=1; i<=100; i++)
		{
		    if(freq[i]==1)
		    nice++;
		    if(freq[i]==2)
		    two++;
		}
		if(nice + 2*two==n && nice%2!=0)
		{
		    System.out.println("NO");
		    return;
		}
		System.out.println("YES");
		if(nice%2==0)
		{
		    int count=0;
		    for(int i=0; i<n; i++)
		    {
		        if(freq[arr[i]]==1)
		        {
		            if(count%2==0)
		            {
		                System.out.print("A");
		            }
		            else
		            {
		                System.out.print("B");
		            }
		            count++;
		        }
		        else
		        {
		            System.out.print("A");
		        }
		    }
		    return;
		}
		int ind =0;
		for(int i=1; i<=100; i++)
		{
		    if(freq[i]>=3)
		    {
		        ind=i;
		        break;
		    }
		}
		int count=0;
		boolean flag=false;
		for(int i=0; i<n; i++)
		{
		    if(freq[arr[i]]==1)
	        {
	            if(count%2==0)
	            {
	                System.out.print("A");
	            }
	            else
	            {
	                System.out.print("B");
	            }
	            count++;
	        }
	        else if(arr[i]==ind)
	        {
	            if(flag)
	            {
	                System.out.print("A");
	            }
	            else
	            {
	                System.out.print("B");
	                flag = true;
	            }
	        }
	        else
	        {
	            System.out.print("A");
	        }
		}
	}
}