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 PYTHON3):

Solution:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Feb 18 17:19:20 2019

@author: umang
"""

n = int(input())

freq_dict = {}

a = list(map(int, input().split()))

for i in a:
    if i in freq_dict.keys():
        freq_dict[i] += 1
    else:
        freq_dict[i] = 1

freq_freq_dict = {1: 0, 2: 0, -1: 0}

for i in freq_dict.values():
    if i == 1:
        freq_freq_dict[1] += 1
    elif i == 2:
        freq_freq_dict[2] += 1
    else:
        freq_freq_dict[-1] += 1
    
if freq_freq_dict[1] != 0:
    if freq_freq_dict[1] % 2:
        if freq_freq_dict[-1] != 0:
            print("YES")
            flag = True
            other_letter = ''
            for i in a:
                if freq_dict[i] == 1:
                    if flag:
                        print("A", end = '')
                    else:
                        print("B", end = '')
                    flag = not flag
                elif freq_dict[i] != 1 and freq_dict[i] != 2:
                    if other_letter == '':
                        other_letter = i
                        print("B", end = '')
                    else:
                        print("A", end = '')
                else:
                    print("A", end = '')
            
        else:
            print("NO")
    else:
        print("YES")
        flag = True
        for i in a:
            if freq_dict[i] == 1:
                if flag:
                    print("A", end = '')
                else:
                    print("B", end = '')
                flag = not flag
            else:
                print("A", end = '')
        
else:
    print("YES")
    print("A"*len(a))