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.*;
import java.lang.*;
public class forces{
	static Scanner sc = new Scanner(System.in);
	static PrintStream op = new PrintStream(System.out);
	public static void main(String[] args) {
		int N = sc.nextInt();
		int[] A = new int[N];
		Map<Integer, Integer> mp = new HashMap<Integer, Integer>();
		for(int i=0; i<N; i++){
			A[i]= sc.nextInt();
			if(mp.containsKey(A[i]))
				mp.put(A[i], mp.get(A[i])+1);
			else
				mp.put(A[i], 1);
		}
		int nn = 0;
		for(int k : mp.keySet())
			if(mp.get(k)==1)
				nn++;
		if(nn%2==0){
			op.println("YES");
			int ok = 0;
			for(int i=0; i<N; i++){
				if(mp.get(A[i]) == 1){
					if(ok==0){
						op.print("A");
						ok = 1;
					}
					else{
						op.print("B");
						ok = 0;
					}
				}
				else
					op.print("A");
			}
			op.println();
		}
		else{
			if(nn==N)
				op.println("NO");
			else{
				int wo = 0;
				for(int i=0; i<N; i++)
					if(mp.get(A[i])>2)
						wo = 1;
				if (wo ==1){
					op.println("YES");
					int c = 0, bj = 0;
					for(int i=0; i<N; i++){
						if(mp.get(A[i]) == 1){
							if(c<=nn/2){
								op.print("A");
								c++;
							}
							else
								op.print("B");
						}
						else if(mp.get(A[i])>2){
							if(bj==0){
								op.print("B");
								bj = 1 ;
							}
							else
								op.print("A");
						}
						else
							op.print("A");
					}
					op.println();
				}
				else
					op.println("NO");
			}
		}

	}
}