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 CodeForces.C;

import java.util.Scanner;

public class Task1051C {
    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        int n = scn.nextInt();
        int[] a = new int[101];
        int[] numbers = new int[n];
        for(int i = 0; i<n; i++){
            int ind = scn.nextInt();
            numbers[i]=ind;
            a[ind]++;
        }
        int[] b = new int[3];
        for(int i = 0; i<=100; i++){
            if(a[i]==1){
                b[1]++;
            }
            else{
                if(a[i]==2){
                    b[2]++;
                }
                else{
                    if(a[i]>2) {
                        b[0]++;
                    }
                }
            }
        }
        boolean isChet = false;
        if(b[1]%2==0){
            System.out.println("YES");
            isChet=true;
        }
        else{
            if(b[0]>=1){
                System.out.println("YES");
            }
            else{
                System.out.println("NO");
                return;
            }
        }
        int sch = 0;
        for(int i = 0; i<n; i++){
            if(a[numbers[i]]==1){
                if(sch%2==0){
                    System.out.print("B");
                }
                else{
                    System.out.print("A");
                }
                sch++;
                continue;
            }
            if(a[numbers[i]]==2){
                System.out.print("B");
                continue;
            }
            else{
                if(!isChet){
                    System.out.print("A");
                    isChet=true;
                }
                else{
                    System.out.print("B");
                }
            }
        }
    }
}