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.Scanner;
import java.util.TreeSet;

/**
 *
 * @author meskill
 */
public class JavaApplication57 {

    static class f implements Comparable<f> {

        int a;
        int count;

        f(int a) {
            this.a = a;
            count = 0;
        }

        @Override
        public int compareTo(f o) {
            if (this.a == o.a) {
                o.count++;
            }
            return this.a - o.a;
            //throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
        }

        @Override
        public String toString() {
            return a + " - " + count; //To change body of generated methods, choose Tools | Templates.
        }
    }

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        TreeSet<f> t = new TreeSet<>();
        int i = 0;
        while (i < n) {
            int buf = sc.nextInt();
            int next = 0;
            while (buf > 0) {
                next += buf % 2;
                buf /= 2;
            }
            t.add(new f(next));
            i++;
        }
        i = 0;
        long result = 0;
        while (!t.isEmpty()) {
            long buf = t.pollLast().count;
            result += buf * (buf - 1) / 2;
        }
        System.out.println(result);
    }
}