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
import sys
cout = sys.stdout

n = int(raw_input())
ar = map(int, raw_input().split())

hs = {}
for i in ar:
    if i in hs:
        hs[i] += 1
    else:
        hs[i] = 1

single = trump = 0
for i in hs:
    if hs[i] == 1:
        single += 1
    elif hs[i] > 2:
        trump += 1

if trump == 0 and single % 2 == 1:
    print "NO"
    exit(0)

print 'YES'

req = False
if single % 2 == 1:
    req = True

double = set()
last = True
for i in ar:
    if hs[i] == 1:
        if last:
            cout.write('B')
            last = False
        else:
            cout.write('A')
            last = True
    elif hs[i] == 2:
        if i in double:
            cout.write('B')
        else:
            cout.write('A')
            double.add(i)
    else:
        if req:
            cout.write('A')
            req = False
        else:
            cout.write('B')