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())
d=dict()
a=[int(x) for x in input().split()]
for i in range(n):
    d.setdefault(a[i], []).append(i)
di=dict()
for d,s in d.items():
    di.setdefault(len(s), []).append(s)
exist=True
ans=['A']*n
if 1 in di:
    exist = False
    if len(di[1])%2==0:
        exist = True
        for i in range(0, len(di[1]), 2):
            ans[di[1][i][0]] = 'B'
    else:
        for i in range(0, len(di[1])-2, 2):
            ans[di[1][i][0]] = 'B'
        for i in range(3, n+1):
            if i in di:
                ans[di[i][0][0]] = 'B'
                exist = True
                break
print('YES' if exist else 'NO')
if exist:
    print(''.join(ans))