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:
//I AM THE CREED
/* //I AM THE CREED
/* package codechef; // don't place package name! */
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
import java.awt.Point;
public class Main{
    public static void main(String[] args) throws IOException 
    { 
  
        Scanner input = new Scanner(System.in);
        while(input.hasNext()){
            int n=input.nextInt();
            HashSet<Integer> s=new HashSet<>();
            int[] l=new int[n];
            HashMap<Integer, Integer> freq=new HashMap<>();
            for(int i=0;i<n;i++){
                l[i]=input.nextInt();
                freq.put(l[i], freq.getOrDefault(l[i], 0)+1);
            }
            for(int k:freq.keySet()){
                if(freq.get(k)==1){
                    s.add(k);
                }
            }
            if(s.size()%2==1 && n-s.size()==0){
                System.out.println("NO");
                continue;
            }
            if(s.size()%2==1){
                boolean added=false;
                int ct=0;
                StringBuilder sb=new StringBuilder();
                for(int i=0;i<n;i++){
                    if(s.contains(l[i])){
                        if(ct%2==0){
                            sb.append("A");
                            ct++;
                            continue;
                        }
                        sb.append("B");
                        ct++;
                        continue;
                    }
                    if(added || freq.get(l[i])==2){
                        sb.append("A");
                        continue;
                    }
                    sb.append("B");
                    added=true;
                }
                if(added==false){
                    System.out.println("NO");
                    continue;
                }
                System.out.println("YES");
                System.out.println(sb.toString());
                continue;
            }
            int ct=0;
            StringBuilder sb=new StringBuilder();
            for(int i=0;i<n;i++){
                if(s.contains(l[i])){
                    if(ct%2==0){
                        sb.append("A");
                        ct++;
                        continue;
                    }
                    sb.append("B");
                    ct++;
                    continue;
                }
                sb.append("A");
            }
            System.out.println("YES");
            System.out.println(sb.toString());
        }
        
    }
    


}