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 com.rishabh.math;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class VasyaAndMultisets {

    private static int lim=100;
    public static void main(String[] args) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(br.readLine());
        String s[] = br.readLine().split(" ");
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(s[i]);
        }

        int hash[] = new int[lim+1];
        boolean numGreaterThan2 = false;
        int niceNumbers = 0;
        for (int i = 0; i < n; i++) {
            hash[a[i]]++;
        }

        for (int i = 1; i <= lim; i++) {
            if (hash[i] == 1) {
                niceNumbers += 1;
            }

            if (hash[i] > 2) {
                numGreaterThan2 = true;
            }
        }

        StringBuilder ans = new StringBuilder();

        if (niceNumbers % 2 == 0 || numGreaterThan2) {
            int cnt = 0;
            boolean numGreaterThan2Visited = false;

            ans.append("YES").append("\n");

            for (int i = 0; i < n; i++) {
                if (hash[a[i]] == 1) {
                    if (cnt >= (niceNumbers + 1) / 2) {
                        ans.append("B");
                    } else {
                        ans.append("A");
                        cnt += 1;
                    }
                } else if (niceNumbers%2!=0 && hash[a[i]] > 2 && !numGreaterThan2Visited) {
                    ans.append("B");
                    numGreaterThan2Visited=true;
                } else {
                    ans.append("A");
                }
            }
        } else {
            ans.append("NO").append("\n");
        }

        System.out.println(ans);

    }

}