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

from collections import Counter
cnt = Counter(a)

solo = set()
big = set()
for x, v in cnt.items():
	if v == 1:
		solo.add(x)
	if v >= 3:
		big.add(x)
if len(solo) % 2 == 0:
	print("YES")
	cur = 0
	for x in a:
		if x in solo:
			print("AB"[cur], end="")
			cur ^= 1
		else:
			print("A", end="")
	print()
else:
	if not big:
		print("NO")
	else:
		print("YES")
		did_big = 0
		cur = 0
		for x in a:
			if x in solo:
				print("AB"[cur], end="")
				cur ^= 1
			elif x in big and not did_big:
				print("B", end="")
				did_big = 1
			else:
				print("A", end="")
		print()