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.
/**
 * Created with IntelliJ IDEA.
 * User: akashm
 * Date: 6/19/13
 * Time: 4:56 PM
 * To change this template use File | Settings | File Templates.
 */

import java.util.*;

public class B272 {
    static HashMap<Integer, Integer> dp = new HashMap<Integer, Integer>();

    public static int f(int x) {
        if (!dp.containsKey(x))
            dp.put(x, f(x / 2) + (x % 2));
        return dp.get(x);
    }

    public static void main(String[] args) {
        dp.put(0, 0);
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int a[] = new int[n];
        int ff[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
            ff[i] = f(a[i]);
        }
        Arrays.sort(ff);
        long sameVal = 1;
        long count = 0;
        for (int i = 1; i < n; i++) {
            if (ff[i] == ff[i - 1])
                ++sameVal;
            else {
                count += (sameVal * (sameVal - 1)) / 2;
                sameVal = 1;
            }
        }
        count += (sameVal * (sameVal - 1)) / 2;
        System.out.println(count);
    }
}
