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
n = int(input())
a = [*map(int, input().split())]
b = Counter(a)
c = dict()
for i, j in b.items():
	c.setdefault(j, i)
d = list(b.values()).count(1)
t = d//2
ans = str()
if not d & 1:
    print('YES')
    for i in a:
    	if b[i] == 1 and t > 0:
    		ans += 'A'
    		t -= 1
    	else:
    		ans += 'B'
    print(ans)
else:
    if max(c.keys()) < 3:
    	print('NO')
    	exit(0)
    print('YES')
    tt = list(filter(lambda x: x >= 3, c))[0]
    done = True
    for i in a:
    	if b[i] == 1 and t > 0:
    		ans += 'A'
    		t -= 1
    	elif b[i] >= 3 and i == c[tt] and done:
    		ans += 'A'
    		done = False
    	else:
    		ans += 'B'
    print(ans)