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
import collections
n = int(input())
a = [int(i) for i in input().split()][:n]
cnt = collections.Counter(a)
odd = 0
for i in a:
	if cnt[i] == 1:
		odd += 1
A = 0
ans = ''
if odd % 2 == 0:
	for i in a:
		if cnt[i] == 1:
			if A % 2 == 0:
				ans += 'A'
			else:
				ans += 'B'
			A += 1
		else:
			ans += 'A'
else:
	C = 0
	for i in a:
		if cnt[i] > 2 and C == 0:
			ans += 'B'
			C += 1
		elif cnt[i] == 1:
			if A % 2 == 0:
				ans += 'A'
			else:
				ans += 'B'
			A += 1
		else:
			ans += 'A'
		# print(C,A)
yes = 0
for i in a:
	if cnt[i] > 2:
		yes += 1
if yes > 0 or odd % 2 == 0:
	print('YES\n' + ans)
else:
	print('NO')
