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())
ls = list(map(int, input().split()))
nice = []
count = {}
for i in ls:
    if i in count:
        count[i]+=1
    else:
        count[i] = 1
for i in range(len(ls)):
    if count[ls[i]]==1:
        nice.append(ls[i])
candidate = 0
for i in range(len(ls)):
    if count[ls[i]]>2:
        candidate = ls[i]
if len(nice)%2 and candidate==0:
    print('NO')
elif len(nice)%2:
    print('YES')
    result = []
    turn = 'A'
    for i in ls:
        if candidate==i:
            result.append('B')
        elif i in nice:
            if turn=='A':
                turn = 'B'
            else:
                turn = 'A'
            result.append(turn)
        else:
            result.append('B')
    for i in range(len(ls)):
        if candidate==ls[i]:
            result[i] = 'A'
            break
    for i in result:
        print(i, end='')
    print()
else:
    print('YES')
    turn = 'A'
    for i in ls:
        if i in nice:
            if turn=='A':
                turn = 'B'
            else:
                turn = 'A'
            print(turn, end='')
        else:
            print('B', end='')
    print()
