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 = int(input())
a = [int(_) for _ in input().split()]

cnt = {}
for x in set(a):
    cnt[x] = len([y for y in a if y == x])

one = []
more = []
for x, c in cnt.items():
    if c == 1:
        one.append(x)
    elif c >= 3:
        more.append(x)

oneset = set(one)
last = 'B'
first = True
ans = []
if more or len(one) % 2 == 0:
    print('YES')
    for x in a:
        if x in oneset:
            last = 'A' if last == 'B' else 'B'
            ans.append(last)
        elif more and x == more[0]:
            if first:
                if len(one) % 2 == 0:
                    ans.append('A')
                else:
                    ans.append('B')
                first = False
            else:
                ans.append('A')
        else:
            ans.append('A')
    print(''.join(ans))
else:
    print('NO')
