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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.StringTokenizer;
public class VasyaandMultiSet {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		MyScanner input = new MyScanner();
		int n = input.nextInt();
		HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
		int[] arr = new int[n];
		for(int i=0; i<n;i++){
			int x = input.nextInt();
			arr[i] = x;
			if(map.containsKey(x)){
				map.put(x,map.get(x)+1);
			}
			else
				map.put(x,1);
			
		}
		int count=0;
		int val=0;
		boolean odd = false;
		for(int i : map.keySet()){
			if(map.get(i)==1)count++;
			else if(map.get(i)>2 ){
				odd = true;
				val = i;
			}
		}
		boolean condition = false;
		if(count%2==0){
			System.out.println("YES");
			for(int i=0;i<n;i++){
				if(map.get(arr[i])>1){
					System.out.print('A');
				}
				else{
					if(condition)System.out.print('A');
					else
						System.out.print('B');
					condition = !condition;
				}
				
			}
		}
		else {
			
		   if(!odd){
			   System.out.println("NO");
			   return;
		   }
		   System.out.println("YES");
			for(int i=0; i<n;i++){
				if(map.get(arr[i])>1){
			         if(arr[i]==val && odd){
			        	 System.out.print('B');
			        	 odd = false;
			         }
			         else
			        	 System.out.print('A');
				}
				else{
					if(condition)System.out.print('B');
					else
						System.out.print('A');
				   condition = !condition;
				}
			}		
		}
	}

	static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {while (st == null || !st.hasMoreElements()) {
			try {st = new StringTokenizer(br.readLine());}
			catch (IOException e) {e.printStackTrace();}}
		return st.nextToken();}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() {return Double.parseDouble(next());}
		String nextLine(){String str = "";
		try {str = br.readLine();}
		catch (IOException e) {e.printStackTrace();}
		return str;}
	}
}