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.*;
import java.io.*;
public class VasyaandMultisets{
    public static void main(String[] args){
        Scanner scan=new Scanner(System.in);
        PrintWriter pr=new PrintWriter(System.out);
        int n=scan.nextInt();
        int[] a=new int[n];
        int[] b=new int[101];
        for(int i=0;i<n;i++){
            a[i]=scan.nextInt();
            b[a[i]]++;
        }
        int multi=0;int current=1;
        for(int i=1;i<101;i++){
            if(b[i]==1){
                b[i]=current;
                current*=-1;
            }
            if(b[i]>2){
                multi=i;
                b[i]=3;
            }
        }
        if(current==-1 && multi==0){
            System.out.println("NO");
            return;
        }
        if(current==-1)
            b[multi]=2;
        for(int  i=0;i<n;i++){
            if(b[a[i]]>2)
                pr.print("A");
            if(b[a[i]]==1)
                pr.print("A");
            if(b[a[i]]==-1)
                pr.print("B");
            if(b[a[i]]==2)
            {pr.print("B");b[a[i]]=3;}
        }
        System.out.println("YES");
        pr.close();
    }
}