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, defaultdict

n = int(input())
s = list(map(int, input().split()))
counter = Counter(s)
strict = [i for i, cnt in counter.items() if cnt == 1]
stupid = [i for i, cnt in counter.items() if cnt == 2]
flex = [i for i, cnt in counter.items() if cnt > 2]
if len(strict) % 2 != 0 and not flex:
    print("NO")
    exit(0)
print("YES")
dest = defaultdict(list)
for num in stupid:
    dest[num] = ["A"] * counter[num]
for num in flex:
    dest[num] = ["A"] * counter[num]
for i, num in enumerate(strict):
    dest[num] = ["A" if i % 2 else "B"]
if len(strict) % 2 != 0:
    dest[flex[0]] = ["A"] + ["B"] * (counter[flex[0]] - 1)
for num in s:
    print(dest[num].pop(), end="")
print()
