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
n = input()
m = raw_input().split()
num = [ int(i) for i in m]

fre = []
ans = ['A' for i in range(n)]
x = 0

other = -1
for i in num:
    if num.count(i) == 1:
        fre.append(x)
    elif num.count(i) == 2:
        pass
    else:
        other = x
    x += 1

if len(fre) % 2 == 1 and other == -1:
    print 'NO'
else:
    print 'YES'
    for i in range(len(fre)/2):
        ans[fre[i]]  = 'B'
    if len(fre) % 2 == 1:
        ans[other] = 'B'
    last = ''
    for i in ans:
        last += i
    print last
