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 Main{
    public static void main(String[] args)
    {
        Scanner s = new Scanner (System.in);
        int n = s.nextInt();
        int[] arr = new int[n];
        int[] freq = new int[101];
        int cnt1 = 0;
        StringBuilder res = new StringBuilder();
        for (int i=0;i<n;i++)
        {
            arr[i] = s.nextInt();
            freq[arr[i]]++;
        }
        for (int i=0;i<101;i++)
        if (freq[i]==1) cnt1++;
        HashMap<Integer , Integer> m1 = new HashMap<>();
        HashMap<Integer , Integer> m2 = new HashMap<>();
        boolean eq = true;
        for (int i=0;i<101;i++)
        {
            if (freq[i]==0) continue;
            if (freq[i]==1)
            {
                if (eq)
                {
                    m1.put(i , 1);
                }
                else
                {
                    m2.put(i,1);
                }
                eq=!eq;
                cnt1--;
            }
            else if (freq[i]==2)
            {
                m1.put(i,freq[i]);
            }
            else if ((eq &&cnt1%2==1)||(!eq && cnt1%2==0))
            {
                if (eq)
                {
                    m1.put(i , 1);
                    m2.put(i , freq[i]-1);
                }
                else
                {
                    m2.put(i,1);
                    m1.put(i , freq[i]-1);
                }
                eq=!eq;
            }
            else
            {
                m1.put(i,freq[i]);
            }
        }
        if (!eq)
        {
            System.out.println("NO");
            return;
        }
        else
        {
            for(int i=0;i<n;i++)
            {
                if (m1.containsKey(arr[i]))
                {
                    res.append('A');
                    if (m1.get(arr[i])==1) m1.remove(arr[i]);
                    else m1.put(arr[i] , m1.get(arr[i])-1);
                }
                else 
                {
                    res.append('B');
                    if (m2.get(arr[i])==1) m2.remove(arr[i]);
                    else m2.put(arr[i] , m2.get(arr[i])-1);
                }
            }
        }
        System.out.println("YES");
        System.out.println(res);
    }
}