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.Scanner;

public class Main {

    public static void main(String[] args) {
	Scanner scar= new Scanner(System.in);
	int n=scar.nextInt();
	int unic=0;
	boolean v;
	boolean govna=false;
	int kek=0;
	int []a=new int[n];
	int []b=new int [101];
	for (int i=0;i<n;i++){
	    a[i]=scar.nextInt();
        b[a[i]]++;
    }
    for(int i=0;i<n;i++){
	    if (b[a[i]]==1){
	       unic++;
        }
        if(b[a[i]]>2){
	     govna=true;
	     kek=a[i];
        }
    }
    if(unic%2!=0&&!govna){
        System.out.println("NO");
        v=true;
        }
        else {
        v = false;
        System.out.println("YES");
    }

    if (!v) {
        if(unic%2==0) {
            int g=unic/2;
            int k = 1;
            for (int i = 0; i < n; i++) {
                switch (b[a[i]]) {
                    case 1:
                        if (k%2==1) {
                            System.out.print("A");
                            k++;
                        } else {
                            System.out.print("B");
                            k++;
                        }
                        break;
                    case 2:
                        System.out.print("A");
                        break;
                    default:
                        System.out.print("B");
                        break;
                }

            }
        }
        else{
            int g=(unic+1)/2;
            int k = 1;
            boolean fef=true;
            for (int i = 0; i < n; i++) {
                switch (b[a[i]]) {
                    case 1:
                        if (k%2==1) {
                            System.out.print("A");
                            k++;
                        } else {
                            System.out.print("B");
                            k++;
                        }
                        break;
                    case 2:
                        System.out.print("A");
                        break;
                    default:
                        if(kek==a[i]&& fef) {
                            System.out.print("B");
                            fef = false;
                        }
                        else {
                            System.out.print("A");
                        }
                        break;

                }

            }
        }
    }
}}