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:
//package codeforces.educational_r_51_1051;

import java.util.Scanner;

public class C {
    static int[] numbers;
    static int[] counts;
    static int n;
    static Digit[] digits;

    static class Digit {
        DigitCount count;
        int number;
        boolean included;

        Digit(int number) {
            this.number = number;
            this.included = false;
        }
    }

    enum DigitCount {
        SINGLE, DOUBLE, MULTIPLE
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        counts = new int[101];
        n = scanner.nextInt();
        numbers = new int[n];
        for (int i = 0; i < n; i++) {
            int number = scanner.nextInt();
            numbers[i] = number;
            counts[number]++;
        }
        int alreadyNice = 0;
        int multipleDigits = 0;
        digits = new Digit[101];
        for (int i = 1; i <= 100; i++) {
            digits[i] = new Digit(i);
        }
        for (int i = 1; i <= 100; i++) {
            if (counts[i] != 0) {
                if (counts[i]==1) {
                    alreadyNice++;
                    digits[i].count = DigitCount.SINGLE;
                } else if (counts[i]==2){
                    digits[i].count = DigitCount.DOUBLE;
                } else {
                    digits[i].count = DigitCount.MULTIPLE;
                    multipleDigits++;
                }
            }
        }
        boolean needToAddNice = false;
        if (alreadyNice % 2 == 0) {
            System.out.println("YES");
        } else {
            if (multipleDigits==0) {
                System.out.println("NO");
                return;
            } else {
                needToAddNice = true;
                System.out.println("YES");
            }
        }

        String setRepresentation = "";
        int niceCount = 0;
        for (int i = 0; i < n; i++) {
            int number = numbers[i];
            Digit digit = digits[number];
            if (digit.count == DigitCount.SINGLE) {
                if (niceCount % 2 == 0) {
                    setRepresentation += "A";
                } else {
                    setRepresentation += "B";
                }
                niceCount++;
            } else if (digit.count != DigitCount.DOUBLE && needToAddNice) {
                setRepresentation += "B";
                needToAddNice = false;
            } else {
                setRepresentation += "A";
            }
        }
        System.out.println(setRepresentation);
    }


}