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 GFG {
   public static void main (String[] args) {
		Scanner sc=new Scanner(System.in);
		    int n=sc.nextInt();
		    int arr[]=new int[n];
		    int fre[]=new int [101];
	        for(int i=0;i<n;i++)
	        {arr[i]=sc.nextInt();
	        fre[arr[i]]++;}
	        int c=0,d=0,k=0;
	        for(int i=0;i<101;i++)
	        {
	            if(fre[i]==1)
	            c++;
	            if(fre[i]>2)
	            d++;
	        }
	        char ans[]=new char[n];
	       if(c%2==0)
	       {
	           System.out.println("YES");
	           int a=0,i=0;
	           for(i=0;i<n;i++)
	           {
	               if(a==c/2)
	               break;
	               if(fre[arr[i]]==1){
	               ans[i]='A';
	               a++;}
	           }
	           for(int j=i;j<n;j++)
	           {
	               if(fre[arr[j]]==1)
	               ans[j]='B';
	           }
	           for(i=0;i<n;i++)
	           {
	               if(fre[arr[i]]!=1)
	               ans[i]='B';
	           }
	           
	       }
	       if(c%2!=0&&d==0){
	       System.out.println("NO");k=1;}
	       if(c%2!=0&&d>0)
	       {
	            System.out.println("YES");
	           int a=0,i=0;
	           for(i=0;i<n;i++)
	           {
	               if(a==c/2)
	               break;
	               if(fre[arr[i]]==1){
	                ans[i]='A';
	               a++;}
	           }
	           for(int j=i;j<n;j++)
	           {
	               if(fre[arr[j]]==1)
	               ans[j]='B';
	           }
	           int ind=0,pos=0;
	            for(i=0;i<n;i++)
	            {
	                if(fre[arr[i]]>2){ind=1;
	                ans[i]='A';pos=i;break;}
	           }
	           for(i=0;i<n;i++)
	           {
	               if(i!=pos&&fre[arr[i]]!=1)
	               ans[i]='B';
	           }
	       }
	       if(k==0)
	       System.out.println(String.valueOf(ans));
   }
}