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

n = int(input())
S = [int(i) for i in input().split()]

c = Counter(S)
V = list(c.values())
cnt1 = V.count(1)
cnt2 = V.count(2)
if cnt1 % 2 == 1 and cnt1 + cnt2 * 2 == n:
    print('NO')
    exit()

d = dict()

s = ['A', 'B']
i = 0
flag = cnt1 % 2 == 1
for k, v in c.items():
    if v == 1:
        d[k] = [s[i]]
        i = (i+1) % 2
    elif flag and v >= 3:
        d[k] = ['B'] + ['A'] * (v - 1)
        flag = False
    else:
        d[k] = ['A'] * v

print('YES')
for s in S:
    print(d[s].pop(), end='')
print()