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 sys import stdin, stdout

class Solve:
  def __init__(self):
    R = stdin.readline
    W = stdout.write
    n = int(input())
    a = [int(x) for x in R().split()]
    ans = ['A' for i in range(n)]
    check = {}
    auniq = buniq = 0
    for i in range(n):
      if a[i] not in check:
        check[a[i]] = 0
      check[a[i]] += 1
      if check[a[i]] == 1:
        auniq += 1
      elif check[a[i]] == 2:
        auniq -= 1
    for i in range(n):
      if auniq == buniq:
        W('YES\n')
        W(''.join(ans))
        return
      if check[a[i]] == 1 and auniq > buniq+1:
        check[a[i]] = 0 
        ans[i] = 'B'
        auniq -= 1
        buniq += 1
    for i in range(n):
      if auniq == buniq:
        W('YES\n')
        W(''.join(ans))
        return
      if check[a[i]] > 2:
        check[a[i]] = 2 
        ans[i] = 'B'
        buniq += 1
    W('NO\n')

def main():
  s = Solve()

main()