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())
dict1={}
arr=list(map(int,input().split()))
for i in range(n):
	try:
		dict1[arr[i]]+=1
	except:
		KeyError
		dict1[arr[i]]=1
if(1 not in dict1.values()):
	print("YES")
	print("A"*n)
else:
	count1=0
	count2=0
	arr1=[]
	arr2=[]
	arr3=[]
	for i in dict1:
		if(dict1[i]==1):
			count1+=1
			arr1.append(i)
		elif(dict1[i]>2):
			count2+=1
			arr3.append(i)
		else:
			arr2.append(i)
	s=""
	arrx1=arr1[:count1//2]
	arrx2=arr1[count1//2:]
	for i in range(n):
		if(arr[i] in arrx1):
			s+="A"
		else:
			s+="B"
	if(count1%2==0):
		print("YES")
		print(s)
	else:
		if(count2==0):
			print("NO")
		else:
			for i in range(n):
				if(arr[i]==arr3[0]):
					s=s[:i]+"A"+s[i+1:]
					break
			print('YES')
			print(s)
