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 C1051 
{
    public static void main(String args[])throws IOException
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int arr[]=new int[100+1];
        int a[]=new int[n];
        boolean f=false;
        for(int i=0;i<n;i++)
        {
            a[i]=sc.nextInt();
            arr[a[i]]++;
        }
        int c=0;
        for(int i=1;i<=100;i++)
        {
            if(arr[i]>2)
                f=true;
            if(arr[i]==1)
                c++;
        }
        if(!f && c%2==1)
            System.out.println("NO");
        else
        {
            System.out.println("YES");
           
            int j=c/2;
            if(c%2==0)
                f=false;
            for(int i=0;i<n;i++)
            {
                if(arr[a[i]]==1)
                {
                    if(j>0)
                    {
                        System.out.print("A");
                        j--;
                    }
                    else
                    {
                        System.out.print("B");
                    }
                }
                else if(f && arr[a[i]]>2)
                {
                    System.out.print("A");
                    f=false;
                }
                else
                {
                    System.out.print("B");
                }
                //System.out.print(ch[i]);
            }
        }
    }
}