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.*;
import java.io.*;
public class vasyaAndMultisets {
	static int n, max = 101;
	static int[] arr, freq = new int[max];
	public static void main(String[] args) {
		FS sc = new FS();
		n = sc.nextInt();
		arr = new int[n];
		for(int i = 0; i < n; ++i) ++freq[arr[i] = sc.nextInt()];
		
		//put nums with freq 1 in alternating sets
		boolean[] putA = new boolean[max];
		boolean currA = false;
		for(int i = 1; i < max; ++i) {
			if(freq[i] == 1) {
				putA[i] = currA;
				currA = !currA;
			}
		}
		
		//now we may have exactly one more in B than we do in A, find a number with freq > 2 and put one in A and the rest in B
		if(currA) {
			boolean weGood = false;
			for(int i = 1; i < max; ++i) {
				if(freq[i] > 2) {
					weGood = true;
					putA[i] = true;
					break;
				}
			}
			if(!weGood) {
				System.out.println("NO");
				return;
			}
		}
		
		System.out.println("YES");
		boolean[] done = new boolean[max];
		for(int i = 0; i < n; ++i) {
			int curr = arr[i];
			if(!done[curr] && putA[curr]) {
				done[curr] = true;
				System.out.print('A');
			}
			else System.out.print('B');
		}
		System.out.println();
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}