Dima got into number sequences. Now he's got sequence a1, a2, ..., an, consisting of n positive integers. Also, Dima has got a function f(x), which can be defined with the following recurrence:

  * f(0) = 0; 
  * f(2·x) = f(x); 
  * f(2·x + 1) = f(x) + 1. 



Dima wonders, how many pairs of indexes (i, j) (1 ≤ i < j ≤ n) are there, such that f(ai) = f(aj). Help him, count the number of such pairs. 

Input

The first line contains integer n (1 ≤ n ≤ 105). The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

The numbers in the lines are separated by single spaces.

Output

In a single line print the answer to the problem.

Please, don't use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

3
1 2 4


Output

3


Input

3
5 3 1


Output

1

Note

In the first sample any pair (i, j) will do, so the answer is 3.

In the second sample only pair (1, 2) will do.
def factorial(n):
	if n == 0 or n == 1:
		return 1
	product = 1
	for i in range(2, n+1):
		product *= i
	return product

def choose2(n):
	return factorial(n) / (2 * factorial(n-2))

comb_table = []
for i in range(3,31):
	comb_table.append(choose2(i))
comb_table.sort()

n = int(raw_input())

line2 = raw_input().split(' ')
x = [int(i) for i in line2]

count_arr = [0 for i in range(30)]
for i in x:
	if i == 0 or i == 1:
		count_arr[i] += 1
	else:
		count = 1
		while i > 1:
			if i%2 == 1:
				count += 1
			i /= 2
		count_arr[count] += 1

i = 1
pairs = 0
for i in count_arr:
	if i >= 2:
		pairs += comb_table[i-2]
print pairs