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:
from collections import Counter
inp = lambda: map(int, input().rstrip().split())
n = int(input())
s = list(inp())
c = Counter(s)
t = 0
for x in c:
    if c[x] == 1:
        t += 1
# print(c)
if t % 2 == 1 and t == n:
    print("NO")
else:
    ans = ""
    i = 0
    if t % 2 == 1:
        flag = 1
    else:
        flag = 0
    for x in s:
        if c[x] == 1:
            if i < (t+1) // 2:
                ans += "A"
                i += 1
            else:
                ans += "B"
        elif flag == 1 and c[x] > 2:
            ans += "B"
            flag = 0
        else:
            ans += "A"
    if flag == 1:
        print("NO")
    else:
        print("YES")
        print(ans)