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())
q=0
a = [int(i) for i in input().split()]
cnt =cnt3 = 0
B = False
kek = True

for i in a:
    if a.count(i)==1:
        cnt+=1
    if a.count(i)>=3:
        cnt3+=1

if cnt%2==0:
    print('YES')
    for i in a:
        if a.count(i)==1:
            if B:
                print('B',end='')
            else:
                print('A',end='')
            B = not B
        else:
            print('A',end='')
else:
    if cnt3>0:
        print('YES')
        for i in a:
            if a.count(i)==1:
                if B:
                    print('B',end='')
                else:
                    print('A',end='')
                B = not B
            #else:
            #    print('A',end='')

            elif (a.count(i)>=3 )& (kek):
                print('B',end='')
                kek = False
                continue
            else:
                print('A',end='')
    else:
        print('NO')