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 Vasya_and_Multisets
{
    public static void main(String args[])
    {
        Scanner s1 = new Scanner(System.in);
        int n = s1.nextInt();
        int ct[] = new int[101];
        Arrays.fill(ct,0);
        int s[] = new int[n];
        for(int i=0;i<n;i++)
        {
            s[i] = s1.nextInt();
            ct[s[i]]++;
        }
        int nct = 0;
        for(int i=1;i<=100;i++)
            if(ct[i] == 1)
                nct++;
        int pct;
        StringBuilder sb = new StringBuilder();
        if(nct % 2 == 0)
        {
            pct = 0;
            for(int i=0;i<n;i++)
                if(ct[s[i]] == 1)
                {
                    pct++;
                    sb.append(pct <= nct / 2 ? 'A' : 'B');
                }
                else
                    sb.append('A');
            System.out.println("YES\n"+sb);
        }
        else
        {
            pct = 0;
            boolean flag = true;
            for(int i=0;i<n;i++)
                if(ct[s[i]] == 1)
                {
                    pct++;
                    sb.append(pct <= nct / 2 ? 'A' : 'B');
                }
                else if(ct[s[i]] >= 3 && flag)
                {
                    sb.append('A');
                    flag = false;
                }
                else
                    sb.append('B');
            System.out.println(flag ? "NO" : "YES\n" + sb);
        }
        s1.close();
    }
}