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

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

rv = ['A'] * len(a)
flag = 'A'
for i in range(len(a)):
    if c[a[i]] == 1:
        rv[i] = flag
        if flag == 'A':
            flag = 'B'
        else:
            flag = 'A'

if flag == 'B':
    new_c = dict(filter(lambda it: it[1] > 2, list(c.items())))
    if new_c:
        key = list(new_c.keys())[0]
        for i in range(len(a)):
            if a[i] == key:
                rv[i] = 'B'
                break
    else:
        print('NO')
        exit(0)

print('YES')
print(''.join(rv))