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

def main():
    _ = sys.stdin.readline()
    s = list(map(int, sys.stdin.readline().strip().split(' ')))

    freq = [0] * 101
    for i in s:
        freq[i] += 1

    ans = [0] * len(s)
    currMulti = 0
    for i, v in enumerate(s):
        if freq[v] == 1:
            ans[i] = currMulti
            currMulti ^= 1
    if currMulti == 1:
        ok = False
        for i, v in enumerate(s):
            if freq[v] > 2:
                ans[i] = currMulti
                ok = True
                break
        if ok is not True:
            print("NO")
            return
    print("YES")
    print("".join(["A" if x == 0 else "B" for x in ans]))

if __name__ == '__main__':
    main()
