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

import javafx.util.Pair;
import org.omg.CORBA.INTERNAL;
import sun.awt.image.ImageWatched;
import sun.reflect.generics.tree.Tree;

import java.nio.channels.ScatteringByteChannel;
import java.text.SimpleDateFormat;
import java.util.*;

public class Main {


    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);


        int n = s.nextInt();

        int[] count = new int[101];

        int[] arr = new int[n];
        for (int i =0; i < n; i++) {

            int k = s.nextInt();
            arr[i] = k;
            count[k]++;
        }

        int[] answer = new int[n];

        int Acount = 0;

        int Bcount = 0;
        boolean hasOdd = false;

        for (int i = 0; i < n; i++) {

            int num = arr[i];

            if (count[num] == 1) {

                if (Acount > Bcount) {

                    Bcount++;

                    answer[i] = 2;
                } else {

                    Acount++;

                    answer[i] =1;
                    // 같으면 A 에 넣는다 A 가 항상 더 같거나 크다
                }

            } else if (count[num] > 2) {

                hasOdd = true;
            }
        }

        if (Acount == Bcount) {

            System.out.println("YES");

            for (int i = 0 ; i < n; i++) {

                if (answer[i] == 0) {

                    System.out.print("A");

                } else if (answer[i] == 1) {

                    System.out.print("A");
                } else {

                    System.out.print("B");
                }
            }

            return;
        } else if (hasOdd == false) {

            System.out.print("NO");
        } else {

            System.out.println("YES");

            for (int i = 0; i < n; i++) {

                int num = arr[i];

                if (count[num] >2) {

                    answer[i] = 2;

                    break;
                }
            }

            for (int i = 0; i < n; i++) {

                if (answer[i] == 0) {

                    System.out.print("A");

                } else if (answer[i] == 1) {

                    System.out.print("A");
                } else {

                    System.out.print("B");
                }
            }
        }
    }

}
