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;
import java.util.Arrays;

public class VasyaMultisets{
	public static void main(String args[]){
		Scanner s = new Scanner(System.in);
		int n, i, j, flag;
		n = s.nextInt();
		int arr[] = new int[n];
		char str[] = new char[n];
		int countarr[] = new int[101];

		for(i=0; i<n; i++){
			arr[i] = s.nextInt();
			countarr[arr[i]]++;
		}
		j=0;
		for(i=0; i<n; i++)
			if(countarr[arr[i]] == 1)
				if(j%2 == 0){
					str[i] = 'A';
					j++;
				}
				else{
					str[i] = 'B';
					j++;
				}
		for(i=0; i<n; i++)
			if(str[i] != 'B')
				str[i] = 'A';
		if(j%2 == 0){
			System.out.println("YES");
			for(i=0; i<n; i++)
				System.out.print(str[i]);
			System.out.println("");
		}
		else{
			flag = 0;
			for(i=0; i<n; i++)
				if(countarr[arr[i]] >= 3){
					str[i] = 'B';
					flag = 1;
					break;
				}
			if(flag == 0)
				System.out.println("NO");
			else{
				System.out.println("YES");
				for(i=0; i<n; i++)
					System.out.print(str[i]);
				System.out.println("");
			}
		}
	}
}