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

counts = {}
for number in a:
    if number not in counts:
        counts[number] = 0
    counts[number] += 1

left = 0
right = 0
tags = ['A'] * n
for count in counts:
    if counts[count] == 1:
        left += 1

i = 0
for number in a:
    if counts[number] == 1 and left > right+1:
        left -= 1
        right += 1
        tags[i] = 'B'
    i += 1
i = 0
for number in a:
    if counts[number] >= 3 and left > right:
        tags[i] = 'B'
        right += 1
        break
    i += 1

if left == right:
    print ('YES')
    print (''.join(tags))
else:
    print('NO')