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.Scanner;

public class VasyaandMultisets {

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int[] a = new int[n];
        int[] f = new int[101];
        for (int i = 0; i < n; i++) {
            a[i] = scan.nextInt();
            ++f[a[i]];
        }
        int c = 0;
        int k = -1;
        for (int i = 0; i < 101; i++) {
            if (f[i] == 1) {
                ++c;
            }
            if (f[i] >= 3) {
                k = i;
            }
        }
        int[] given = new int[n];
        int p = c / 2;
        for (int i = 0; i < n; i++) {
            if (f[a[i]] == 1 && p > 0) {
                given[i] = 1;
                --p;
            }
        }
        p = c / 2;
        for (int i = 0; i < n; i++) {
            if (f[a[i]] == 1 && p > 0) {
                given[i] = 2;
                --p;
            }
        }
        if (c % 2 == 0) {
            for (int i = 0; i < n; i++) {
                if (given[i] == 0) {
                    given[i] = 1;
                }
            }
        } else {
            if (k == -1) {
                System.out.println("NO");
                return;
            } else {

                for (int i = 0; i < n; i++) {
                    if (f[a[i]] == 1 && given[i] == 0) {
                        given[i] = 1;
                    }
                }
                for (int i = 0; i < n; i++) {
                    if (a[i] == k) {
                        given[i] = 2;
                        break;
                    }
                }
                for (int i = 0; i < n; i++) {
                    if (given[i] == 0) {
                        given[i] = 1;
                    }
                }
            }

        }
        System.out.println("YES");
        for (int i = 0; i < n; i++) {
            if(given[i]==1){
                System.out.print("A");
            }else{
                System.out.print("B");
            }
            
        }
        System.out.println();

    }
}