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:
n = int(input())
ar = list(map(int, input().split()))
goodi = set()
has_more = -1
for i, x in enumerate(ar):
    if ar.count(x) == 1:
        goodi.add(i)
    else:
        if ar.count(x) > 2:
            has_more = i
mask = [True for i in range(n)]
if len(goodi) % 2 == 0:
    print("YES")
    for i in list(goodi)[:len(goodi) // 2]:
        mask[i] = False
    for c in mask:
        print({True:'A', False:'B'}[c], end='')
    print()
elif has_more != -1:
    print("YES")
    for i in list(goodi)[:len(goodi) // 2]:
        mask[i] = False
    mask[has_more] = False
    for c in mask:
        print({True:'A', False:'B'}[c], end='')
    print()
else:
    print("NO")