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
num_numbers = int(input())
numbers = list(input().split())
appear_once = set()
appear_twice = set()
appear_multiple = set()
for number in numbers:
    if number not in appear_multiple:
        if number in appear_twice:
            appear_multiple.add(number)
            appear_twice.remove(number)
        elif number in appear_once:
            appear_twice.add(number)
            appear_once.remove(number)
        else:
            appear_once.add(number)

if len(appear_once) % 2 == 0 or len(appear_multiple) > 0:
    print("YES")
    nice_to_a = True
    solution = list()
    split_multiple = len(appear_once) % 2 == 1
    for number in numbers:
        if number in appear_once:
            if nice_to_a:
                solution.append('A')
            else:
                solution.append('B')
            nice_to_a = not nice_to_a
        elif split_multiple and number in appear_multiple:
            solution.append('B')
            split_multiple = False
        else:
            solution.append('A')
    print(''.join(solution))
else:
    print("NO")
