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 deque

n = int(input())
arr = list(map(int, input().split(" ")))

dic = {}
for i in arr:
	if i in dic:
		dic[i] += 1
	else:
		dic[i] = 1

seta = set()
setb = set()
a = 0
b = 0
dic_ans = {}
ones = set()
twos = set()
more = set()

for i in dic:
	if dic[i] == 1:
		ones.add(i)
	elif dic[i] == 2:
		twos.add(i)
	else:
		more.add(i)

for i in ones:
	if a < b:
		a += 1
		seta.add(i)
		dic_ans[i] = deque(["A"])
	elif b < a:
		b += 1
		setb.add(i)
		dic_ans[i] = deque(["B"])
	else:
		a += 1
		seta.add(i)
		dic_ans[i] = deque(["A"])

for i in twos:
	# if a == b:
	setb.add(i)
	dic_ans[i] = deque(["B", "A"])
	seta.add(i)
	a+=1
	b += 1
	# else:

for i in more:
	if a < b:
		a += 1
		seta.add(i)
		dic_ans[i] = deque(["A"])
		for k in range(dic[i]-1):
			dic_ans[i].append("B")
	elif b < a:
		b += 1
		setb.add(i)
		dic_ans[i] = deque(["B"])
		for k in range(dic[i]-1):
			dic_ans[i].append("A")
	else:
		seta.add(i)
		dic_ans[i] = deque()
		for k in range(dic[i]):
			dic_ans[i].append("A")
# print(*dic_ans)
# print(a,b)
if a == b:
	ans = ""
	for i in arr:
		ans += dic_ans[i].pop()
	print("YES")
	print(ans)
else:
	print("NO")

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