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:
//package Codef;

import java.util.Scanner;

public class C_1051 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s = new Scanner(System.in);
		//System.out.println((char)('A'+ 1));
		int n = s.nextInt();
		int[] a = new int[n];
		int[] c = new int[101];
		for (int i = 0; i < n; i++) {
			a[i] = s.nextInt();
			c[a[i]]++;
		}
		StringBuilder sb = new StringBuilder();
		int ct = 0, c1 = 0;
		for (int i = 0; i < n; i++) {
			if (c[a[i]] == 1) {
				ct++;
			}
			else if (c[a[i]] > 2) {
				c1 = 1;
			}
		}

		if (ct % 2 == 1 && c1 == 0) {
			System.out.println("NO");
              return;
		}
		System.out.println("YES");
		int j=0;
		for (int i = 0; i < n; i++) {
			if(c[a[i]] == 1) {
				System.out.print((char)('A'+j%2));
				j++;
//				System.out.println(i);
				continue;
			}
			if(c1 == 1 && c[a[i]] > 2 && ct%2 == 1 ) {
				j=1;
				System.out.print((char)('A'+j%2));
				j++;
				c1 = 0;
				continue;
			}
			System.out.print('A');
		}
	}

}