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 nice;

import java.util.Scanner;

public class TaskC {
public static void main(String []V) {
	int size;
	Scanner in=new Scanner(System.in);
	size=in.nextInt();
	int [] save=new int[101];
	int [] numbers=new int[size];
	int moreThanTwo=0;
	int Two=0;
	int count=0;
	for(int i=0;i<size;i++) {
			numbers[i]=in.nextInt();
			save[numbers[i]]++;
			
		}
	for(int i=0;i<101;i++) {
		if(save[i]==1) {
			count++;
		}
		if(save[i]==2) {
			Two++;
		}
		if(save[i]>2) {
			moreThanTwo++;
		}
	}
	if(count%2!=0 && moreThanTwo==0) {
		System.out.println("NO");
	}
	else if(count%2==0) {
		System.out.println("YES");
		int count_one=0;
		for(int i=0;i<size;i++) {
			if(save[numbers[i]]==1) {
				// save ==1
				if(count_one<count/2) { /// the part  A
					System.out.print("A");
				}else { /// the  part B
					System.out.print("B");
				}
				count_one++;
				
			}else {   /// save > 1
				System.out.print("A");
			}
			
		}
	}else {
		System.out.println("YES");
		boolean flag=true;
		int count_one=0;
for(int i=0;i<size;i++) {
			
			if(save[numbers[i]]==1) { // save ==1
				
				if(count_one<=count/2) { /// the part  A
					System.out.print("A");
				}else { /// the  part B
					System.out.print("B");
				}
				count_one++;
				
			}else {   /// save > 1
				if(save[numbers[i]]>2&&flag==true) { // if first occur of the number more than 3
					flag=false;
					System.out.print("B");
					
				}else {
					System.out.print("A");
				}
				
			}
		
		}
	}
		
	/*for(int i=0;i<101;i++) {
		System.out.println(save[i]);
			}*/
	/*for(int i=0;i<size;i++) {
		
		
	}*/
	//System.out.println(count);
	System.exit(0);
 }
}