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 java.io.*;
import java.util.*;

public class C {

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt();

		int[] s = new int[n];
		int[] map = new int[101];

		for (int i = 0; i < n; i++)
			map[s[i] = sc.nextInt()]++;

		int cnt = 0;

		for (int x : map)
			cnt += x == 1 ? 1 : 0;
		int selected = 0;

		if (cnt % 2 != 0) {
			
			for (int i = 1; i <= 100; i++)
				if (map[i] > 2)
					selected = i;

			if (selected == 0) 
			{
				System.out.println("NO");
				return;
			}
		}
		out.println("YES");
		int k = 1;
		
		for (int i = 0; i < n; i++)
			if (map[s[i]] > 2 && selected == s[i])
			{
				out.print("A");
				selected = 0 ;
			}
			else if (map[s[i]] > 1)
				out.print("B");
		
			else if (map[s[i]] == 1) {
				if (2 * k <= cnt) {
					out.print("A");
					k++;
				} else {
					out.print("B");
					k++;
				}
			}

		out.flush();
		out.close();

	}

}