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())
s = list(map(int, input().split()))
f = dict()
for si in s:
    if si in f:
        f[si] += 1
    else:
        f[si] = 1

nice = set()
for k in f:
    if f[k] == 1:
        nice.add(k)

found = True
toChange = -1
if len(nice)%2 != 0:
    found = False
    for k in f:
        if f[k] >= 3:
            toChange = k
            found = True
            break

nicelength = len(nice)

if not found:
    print("NO")
else:
    print("YES")
    ans = []
    for si in s:
        if si in nice:
            if len(nice) <= nicelength//2:
                ans.append("A")
            else:
                ans.append("B")
            nice.remove(si)
        elif si == toChange:
            ans.append("A")
            toChange = -1
        else:
            ans.append("B")
    print("".join(ans))