Problem: 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

Solution (in PYTHON3):

Solution:
from collections import Counter

c = Counter()
N = int(input())
s = list(map(int, input().split()))

one = 0
more = 0

for x in s:
    c[x] += 1
for x in c:
    if c[x] == 1:
        one += 1
    if c[x] > 2:
        more += 1
if one%2==1 and more==0:
    print("NO")
    exit(0)
now = 'A'
ans = ''
if one%2 == 0:
    more = -1
ch = ''
sel = -1
for x in s:
    if c[x] == 1:
        ans = ans+now
        now = ('A' if now=='B' else 'B')
    elif x == sel:
        ans = ans+ch
    elif c[x] > 2 and more > 0:
        more = -1
        sel = x
        ans = ans+now
        now = ('A' if now=='B' else 'B')
        ch = now
    else:
        ans = ans+'A'
print("YES")
print(ans)