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:
n=int(input())
arr=list(map(int,input().split()))
d={}
ans=['A']*n
for i in range(n):
  if arr[i] in d:
    d[arr[i]].append(i)
  else:d[arr[i]]=[i]
flag=False
s='A'
cnt=0
sec_arr=[]
for i in d.keys():
  if len(d[i])==1:
    cnt+=1
    a=d[i]
    if s=='A':
      s='B'
      continue
    else:
      ans[a[0]]=s
      s='A'
  elif len(d[i])>2:
     sec_arr=d[i]
     
  
    
if cnt%2==0:
  print('YES')
  print(''.join(ans))
elif cnt%2==1 and len(sec_arr)>0:
   print('YES')
   ans[sec_arr[0]]='B'
   print(''.join(ans))
else:
  print('NO')