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

def print_answer(first, only_first_occurance=set()):
    print 'YES'
    answer = ''
    for i in n:
        if i in first:
            answer += 'A'
        elif i in only_first_occurance:
            answer += 'A'
            only_first_occurance.remove(i)
        else:
            answer += 'B'
    print answer


raw_input()
n = map(int, raw_input().split())
d = dict()
for x in n:
    if x in d:
        d[x] += 1
    else:
        d[x] = 1

ones = {x for x in d if d[x] == 1}

if len(ones) % 2 == 0:
    print_answer(first={x for i, x in enumerate(ones) if i < len(ones) / 2})
elif len(d) > len(ones):
    for k, v in d.items():
        if v > 2:
            print_answer(first={x for i, x in enumerate(ones) if i < len(ones) / 2}, only_first_occurance={k})
            exit(0)
    print 'NO'
else:
    print 'NO'
