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
n = int(input())
A = list(map(int, input().split()))
D = dict((A[i], A.count(A[i])) for i in range(n))
S = [""] * n

q = True
if list(D.values()).count(1) % 2:
  q = False
else:
  q = True

p = True
for k in D:
  if D[k] == 1:
    if p:
      S[A.index(k)] = 'A'
    else:
      S[A.index(k)] = 'B'

    p = not p

  if D[k] == 2:
    j = A.index(k)
    S[j] = 'A'

    j += 1
    j = j + A[j:].index(k)
    S[j] = 'B'

if not q:
  for k in D:
    if D[k] > 2:
      j = A.index(k)
      S[j] = 'B'
      for i in range(j+1, n):
        if A[i] == k:
          S[i] = 'A'
      q = q = True
      break

if not q:
  print("NO")
  exit()

for i in range(n):
  if S[i] == "":
    S[i] = 'A'

print("YES")
print(''.join(S))
  
