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.*;
public class test1{
	public static void main(String[] args) {
		
		
		Scanner in =new Scanner(System.in);
	int s[]=new int[101];
	int n=in.nextInt();
	int a[]=new int[n];
	int ca=0,cb=0;
	int k=0;
	HashSet<Integer>hs=new HashSet<>();
char c[]=new char[n];
	
	for(int i=0;i<n;i++)
	{	a[i]=in.nextInt();
	s[a[i]]=s[a[i]]+1;
	
	}
	for(int i=0;i<101;i++)
		if(s[i]==1)
			hs.add(i);
	if(hs.size()%2==0) {
		System.out.println("YES");
		for(int i=0;i<n;i++)
		{
			if(hs.contains(a[i]))
			{
				if(ca==cb) {
					c[i]='A';
					ca++;
				}else
				{c[i]='B';
					cb++;
					
				}
				
				
			}
			else
				c[i]='A';
			
		}
		for(int i=0;i<n;i++)
		System.out.print(c[i]);
		System.out.println();
	}else {int f=0,ch=0;
	int se=-1;
	for(int i=0;i<n;i++)
	{
		if(s[a[i]]!=1&&s[a[i]]!=2&&s[a[i]]!=0)
			{se=a[i];
			//hs.add(a[i]);
			f=1;
			break;
			
			}		
		
	}
	
	if(f==0)	
		System.out.println("NO");
	else {
		System.out.println("YES");
//		System.out.println(se);
		for(int i=0;i<n;i++) {
			if(a[i]==se&&ch==0)
			{c[i]='B';
				ch=1;
			}
			else {
				
				if(hs.contains(a[i]))
				{
					if(ca==cb) {
						c[i]='A';
						ca++;
					}else
					{c[i]='B';
						cb++;
						
					}
					
					
				}
				else
					c[i]='A';
			}	}
		for(int i=0;i<n;i++)
			System.out.print(c[i]);
			
			
			
		
		
		
		
		
	}
	
	
	}
	
	
	}
	
}