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())
s = list(map(int, input().split()))
res = ['A']*n

type1, type3 = [], []
w = list(set(s))

for v in w:
	indices = [i for i, x in enumerate(s) if x == v]
	if len(indices) == 1:
		type1.append(indices[0])
	elif len(indices) > 2:
		for i in range(len(indices)):
			type3.append((v, indices[i]))

count_a, count_b = 0, 0

for i in range(len(type1)):
	if count_a > count_b:
		res[type1[i]] = 'B'
		count_b += 1
	else:
		count_a += 1

if count_a > count_b and len(type3) != 0:
	res[type3[0][1]] = 'B'
	count_b += 1

if count_a == count_b:
	print('YES')
	print(''.join(res))
else:
	print('NO')