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.
import java.util.*;

public class B_Dima_and_sequence {

	public static int recurence(int n) {
		if (n == 0)
			return 0;
		// if even
		else if (n % 2 == 0)
			return recurence(n / 2);
		// if oDD
		else
			return recurence((n - 1) / 2) + 1;

	}

	public static void main(String[] args) {

		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int temp;
		int x;
		int array[] = new int[31];
		for (int j = 1; j <= n; j++) {
			x = in.nextInt();
			temp = recurence(x);
			for (int i = 0; i < 31; i++) {
				if (temp == i)
					array[i]++;

			}
		}
		long max = 0;
		long result = 0;

		for (int i = 0; i < 31; i++) {
			if ((array[i]) > 1) {

				result = (array[i] * (array[i] - 1)) / 2;
				max += result;
			}
		}

		System.out.print(max);

	}

}