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:
from collections import Counter
n = int(input())
s = list(map(int, input().split()))
cn = Counter(s)
ones = set()
triple = set()
for key in cn.keys():
    if cn[key] == 1:
        ones.add(key)
    if cn[key] >= 3:
        triple.add(key)
l_1 = len(ones)
l_3 = len(triple)
if (l_1 % 2 != 0) and (l_3 == 0):
    print('NO')
else:
    print('YES')
    st = ''
    for x in s:
        if (x in ones) and len(ones) > (l_1 // 2 + l_1 % 2):
            st += 'B'
            ones.remove(x)
        elif (l_1 % 2 != 0) and (x in triple):
            st += 'B'
            triple.clear()
        else:
            st += 'A'
    print(st)