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
from collections import Counter

n = input()

ll = map(int, raw_input().split(" "))

result = []

dd = Counter(ll)

ones = set()
for key in ll:
    if dd[key] == 1:
        ones.add(key)

if len(ones) % 2 == 0:
    count = 0
    for elem in ll:
        if count < len(ones)/2:
            if elem in ones:
                count += 1
                result.append("A")
            else:
                result.append("A")
        else:
            if elem in ones:
                count += 1
                result.append("B")
            else:
                result.append("A")

    print "YES"
    print "".join(result)

else:
    if len(ones) ==  len(dd):
        print "NO"
    else:
        others = None
        for key in dd:
            if dd[key] > 2:
                others = key
                break

        if others is None:
            print "NO"
        else:
            count = 0
            for elem in ll:
                if count < len(ones)/2:
                    if elem in ones:
                        count += 1
                        result.append("A")
                    else:
                        result.append("B")
                else:
                    if elem in ones:
                        count += 1
                        result.append("B")
                    else:
                        result.append("B")

            for i in xrange(n):
                if ll[i] == others:
                    result[i] = "A"
                    break

            print "YES"
            print "".join(result)
