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 math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)
n = int(input())
a = list(map(int,input().split()))
l = [[] for i in range(105)]
for i in range(n):
	l[a[i]].append(i)
ans = [0 for i in range(n)]
ct = 0
f = 'A'
for i in range(105):
	if(len(l[i]) == 1):
		ans[l[i][0]] = f
		if(f == 'A'):
			f = 'B'
			ct += 1
		else:
			f = 'A'
			ct -= 1
flag = 0
if(ct == 1):
	for i in range(105):
		if(len(l[i]) > 2):
			ans[l[i][0]] = f
			break
	else:
		flag = 1
if(flag == 1):
	print("NO")
else:
	print("YES")
	for i in range(n):
		if(ans[i] == 0):
			ans[i] = 'A'
	print("".join(ans[:n]))