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 = list(map(int,input().split()))
count = [0 for _ in range(101)]
multisetA = [0 for _ in range(101)]
multiset = [[],[]]
ans = ['YES','NO']
one = []
two = []
more = []
idex = 0
for v in a:
    count[v] +=1
for i in range(1,101):
    if count[i] == 1:
        one.append(i)
    elif count[i] == 2:
        two.append(i)
    else:
        if count[i]>2:
            more.append(i)
l = len(one)+len(more)
lenOne = len(one)
# print(len(one),print(len(two)),len(more))
if lenOne == 0:
    print(ans[0])
    for i in range(n):
        print('A',end='')
elif len(more) >0 or lenOne%2 ==0:
    if l %2 == 1:
        more.pop(-1)
        l -=1
    amount = l//2
    if amount>lenOne:
        for num in one:
            multisetA[num] = 1
        need = amount - lenOne
        for i in range (need):
            multisetA[more.pop(0)] = 1
        for v in more:
            need = count[v] - 1
            multisetA[v]= need
    else:
        for v in one[:amount]:
            multisetA[v] = 1
        for v in more:
            need = count[v] - 1
            multisetA[v]= need

    for v in two:
        multisetA[v] = 1
    print(ans[0])
    for v in a:
        if multisetA[v]:
            multisetA[v] -=1
            print('A',end='')
        else:
            print('B',end='')
else:
    print(ans[1])


# print(count)
# print('ONE:',one,'Two:',two)
