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())
a=list(map(int,input().split()))
d=Counter(a)
c=0
s=['A']*n
p=[]
x=-1
for v in d:
    if d[v]==1:
        c=c+1
        p.append(v)
    elif d[v]>2:
        x=v
if c%2==0:
    print('YES')
    p=p[:c//2]
    for u in p:
        q=a.index(u)
        s[q]='B'
    print(''.join(s))
else:
    if x!=-1:
        print('YES')
        p=p[:c//2]
        p.append(x)
        for u in p:
            q=a.index(u)
            s[q]='B'
        print(''.join(s))
    else:
        print('NO')