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.util.*;
public class Multi
{
    public static void main(String arg[])
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int a[]=new int[n];
        int b[]=new int[101];
        for(int i=0;i<n;i++)
        {
            b[a[i]=sc.nextInt()]++;
        }
        int x=0;
        int y=0;
        int p=0;
        HashSet<Integer> hs=new HashSet<>();
        for(int i=0;i<101;i++)
        {
            if(b[i]==1)
            {
             x++;
            hs.add(i);
            }
            if(b[i]>=3){
                y++;
                p=i;
            }
        }
        if(x%2==1 && y==0)
        {
            System.out.println("NO");
            return;
        }
        StringBuilder sb=new StringBuilder();
        if(x%2==1 && y>0)
        {
            int k=x/2;
            int s=0;
            for(int i=0;i<n;i++)
            {
                if(p==a[i] && s==0)
                {
                    sb.append("A");
                    s=1;
                }
                else if(k>0 && hs.contains(a[i]))
                {
                    sb.append("A");
                    k--;
                }
                else
                sb.append("B");
            }
            
        }
        else
        {
            int k=x/2;
           
            for(int i=0;i<n;i++)
            {
                
                if(k>0 && hs.contains(a[i]))
                {
                    sb.append("A");
                    k--;
                }
                else
                sb.append("B");
            }
            
        }
        System.out.println("YES");
        System.out.println(sb);
    }
}