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())
lst=[*map(int,input().split())]
d={}
for i,x in enumerate(lst):
    if d.get(x)==None:d[x]=0
    d[x]+=1
one,item,res=0,-1,[]
for i,x in enumerate(d):
    if d[x]==1:one+=1;res.append(x)
    if d[x]>2:item=x
result=['A']*n
if one%2==0:
    for i,x in enumerate(res[:one//2]):
        result[lst.index(x)]='B'
    print('YES');print(''.join(result))
else:
    if item==-1:print('NO')
    else:
        for i,x in enumerate(res[:one//2]):
            result[lst.index(x)]='B'
        result[lst.index(item)]='B'
        print('YES');print(''.join(result))