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())
multiset = list(map(int, input().split()))

counts = [0 for _ in range(101)]

for elem in multiset:
	counts[elem] += 1

nice = 0
gt2elem = -1

for i, count in enumerate(counts):
	if i>0:
		if count==1:
			nice += 1
		elif count > 2:
			gt2elem = i

if nice%2==0 or gt2elem != -1:
	x = 0
	s=[]
	for elem in multiset:
		if counts[elem]==1:
			s.append("AB"[x])
			x = 1-x
		elif nice%2==1 and elem==gt2elem:
			s.append('B')
			gt2elem = -1
		else:
			s.append('A')
	print('YES')
	print(''.join(s))
else:
	print('NO')