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.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Hashtable;
import java.util.Map;
import java.util.Map.Entry;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class taskC {
	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[] arr = new int[n];
		char[] ans = new char[n];
		for (int i = 0; i < n; i++)
			arr[i] = sc.nextInt();
		TreeMap<Integer, Integer> map = new TreeMap();
		for (int i = 0; i < n; i++) {
			if (!map.containsKey(arr[i]))
				map.put(arr[i], 0);
			map.put(arr[i], map.get(arr[i]) + 1);

		}

		int c = 0, a = 0, b = 0;
		for (int i = 0; i < n; i++) {
			if (map.get(arr[i]) == 1) {
				if (c % 2 == 0) {
					ans[i] = 'A';
					a++;
				} else {
					ans[i] = 'B';
					b++;
				}
				c++;
			}
		}

		for (int i = 0; i < n; i++) {
			if (ans[i] != 'A' && ans[i] != 'B')
				if (map.get(arr[i]) > 2 & a != b) {
					ans[i] = 'B';
					b++;
				} else
					ans[i] = 'A';

		}

		if (a == b) {
			out.println("YES");
			out.print(String.valueOf(ans));
		} else
			out.println("NO");
		out.flush();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}