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 as C
n = int(input())
l = [*map(int, input().split())]
cnt = C(l)
a = 0
bad = set()
reserve = []
for k, v in cnt.items():
    if v == 1:
        if a == 0:
            a = 1
        else:
            a = 0
            bad.add(k)
    if v >= 3 and not reserve:
        reserve = [k]
if a and reserve:
    bad.add(reserve[0])
    a = 0
if a:
    print("NO")
    exit()
print("YES")
res = ['A'] * n
for i in range(n):
    if l[i] in bad:
        res[i] = 'B'
        bad.remove(l[i])
print(''.join(res))