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())

xs = [int(x) for x in input().split()]

from collections import Counter

c = Counter(xs)

nice = []
pair = []
other = []
for i in c:
    if c[i] == 1:
        nice.append(i)
    elif c[i] == 2:
        pair.append(i)
    else:
        other.append(i)


bs = set()
for i in range(len(nice)):
    if i%2==1:
        bs.add(nice[i])
if len(nice)%2 != 0:
    if len(other) == 0:
        print("NO")
        exit()
    else:
        bs.add(other[0])

print("YES")
l = []
for i in xs:
    if i in bs:
        l.append("B")
        bs.remove(i)
    else:
        l.append("A")
print(''.join(l))