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.HashMap;
import java.util.Map;
import java.util.Scanner;

/**
 * Created by shereifhawary on 1/12/16.
 */
public class DimaAndSequence {

    private static Map<Long, Long> count = new HashMap<Long, Long>();

    public static void main(String... args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();

        for (int i = 0; i < n; i++) {
            long x = scanner.nextLong();
            long res = solve(x);
            if (count.containsKey(res)) {
                count.put(res, count.get(res) + 1);
            } else {
                count.put(res, 1l);
            }
        }

        long c = 0;

        for (Long key : count.keySet()) {
            c += com(count.get(key));
        }
        System.out.println(c);

    }

    private static long solve(long x) {

        if(x ==0) {
            return 0;
        }
        if (x == 1) {
            return 1;
        }
        if (x % 2 == 1) {
            return solve(x-1) + 1;
        }
        while (x % 2 == 0) {
            x = x / 2;
        }
        return solve(x);

    }


    private static long com(Long total) {
        return total * (total - 1) / 2;
    }


}
