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
from collections import Counter


def main():
    _ = int(input())
    multiset = [int(c) for c in input().split()]

    cnt = Counter(multiset)
    appear_once = [k for k, v in cnt.items() if v == 1]
    appear_twice = [k for k, v in cnt.items() if v == 2]
    appear_morethan2 = [k for k, v in cnt.items() if v > 2]

    res_a = []
    res_b = []
    if len(appear_once) % 2 == 0:
        res_a += appear_once[:len(appear_once)//2]
        res_b += appear_once[len(appear_once)//2:]
        for e in appear_twice:
            res_a += [e for _ in range(cnt[e])]
        for e in appear_morethan2:
            res_a += [e for _ in range(cnt[e])]
    else:
        if not appear_morethan2:
            print('NO')
            return

        res_a += appear_once[:len(appear_once)//2+1]
        res_b += appear_once[len(appear_once)//2+1:]
        for e in appear_twice:
            res_a += [e for _ in range(cnt[e])]

        res_b.append(appear_morethan2[0])
        cnt[appear_morethan2[0]] -= 1
        for e in appear_morethan2:
            res_a += [e for _ in range(cnt[e])]

    res = []
    for e in multiset:
        if e in res_a:
            res.append('A')
            res_a.remove(e)
        else:
            res.append('B')
            res_b.remove(e)

    print('YES')
    print(''.join(res))


if __name__ == '__main__':
    main()
