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:
from collections import Counter
from sys import stdin
n = int(input())
a = [int(x) for x in stdin.readline().split(" ")]
p = dict(Counter(a))
b = [x for (x,y) in p.items() if y==1]
magic = -1
if len(b)%2!=0 :
	c = [x for (x,y) in p.items() if y>2]
	if not len(c):
		print("NO")
		exit(0)
	
	else:
		magic = c[0]
		
print("YES")
b = b[:len(b)//2]
for x in a:
	if x in b or x==magic:
		if x==magic:
			magic=-1
		print('B',end='')
	else:
		print('A',end='')