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

public class Task {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		int[] arr = new int[n];
		ArrayList<Integer> unique = new ArrayList<Integer>();
		int[] ht = new int[100];
		for (int i = 0; i < 100; i++)
			ht[i] = 0;
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();			
			ht[arr[i] - 1]++;
		}
		for(int i = 0; i < 100; i++)
		{
			if(ht[i]==1)
				unique.add(i+1);
		}
		int multiple = 0;
		for (int i = 0; i < 100; i++) {
			if (ht[i] > 2) {
				multiple = i + 1;
				i = 100;
			}
		}
		
		
		if (multiple == 0 && unique.size() % 2 == 1) {
			pw.println("NO");
		} else {
			pw.println("YES");
			
			if (unique.size() % 2 == 0) {
				boolean un = false;
				for (int i = 0; i < arr.length; i++) {
					if (unique.contains(arr[i])) {
						if (un)
							pw.print("A");
						else
							pw.print("B");
						un = !un;
					} else
						pw.print("B");

				}
			}
			else
			{
				boolean un = false;
				boolean mul = true;
				for (int i = 0; i < arr.length; i++) {
					if (unique.contains(arr[i])) {
						if (un)
							pw.print("A");
						else
							pw.print("B");
						un = !un;
					} 
					else if(arr[i]==multiple)
					{
						if(mul)
						{
							pw.print("A");
							mul=false;
						}
						else
							pw.print("B");
					} else pw.print("B");

				}
			}
			pw.println();
		}
		pw.flush();
	}

}