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
import sys
from collections import Counter
input = sys.stdin.readline

n = int(input())
a = list(map(int, input().split(' ')))
c = Counter(a)
ones = 0
twos = 0
more = 0
for i in c.items():
    if i[1] == 1:
        ones += 1
    elif i[1] == 2:
        twos += 1
    else:
        more += 1
if ones % 2 == 1 and more == 0:
    print('NO')
else:
    counter = 0
    flag = ones % 2 == 1
    print('YES')
    for i in a:
        if c[i] == 2:
            print('A', end='')
        elif c[i] == 1:
            print(('A', 'B',)[counter % 2], end='')
            counter += 1
        else:
            if flag:
                print('B', end='')
                flag = False
            else:
                print('A', end='')
