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
from collections import Counter
n = list(map(int, input().split()))
t = list(map(int, input().split()))
c = Counter(t)
#print(c)
ones = [i for i in c if c[i]==1]
twos = [i for i in c if c[i]==2]
mores = [i for i in c if c[i]>2]

if len(ones)%2 == 0:
	print('YES')
	oc = 'A'
	for i in range(len(t)):
		if t[i] in ones:
			print(oc, end='')
			oc = ('B' if oc=='A' else 'A')
		else:
			print('A', end='')

elif len(mores)==0:
	print('NO')
	
else:
	print('YES')
	oc = 'A'
	volt = False
	for i in range(len(t)):
		if t[i] in ones:
			print(oc, end='')
			oc = ('B' if oc=='A' else 'A')
		elif t[i] in mores and not volt:
			print('B', end='')
			volt = True
		else:
			print('A', end='')
		
	
	
	
