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 long fac(long n)
    {
        if (n == 0)
            return 1;
        else
            return n * fac(n-1);
    }
    /*
    public static long fac(long n) {
        long result = 1;
        for (long i = n; i > 0; i--) {

            result= result*i ;

        }
        return result;

    }
*/
    public static long recurence(long 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) {
        //System.out.prlong(recurence(207643327));

    
        Scanner in = new Scanner(System.in);
        long n = in.nextLong();
        long temp;
        long x;
        long array[] = new long[31];
        for (long j = 1; j <= n; j++) {
            x = in.nextLong();
            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){
                //System.out.prlongln("**"+array[i]);

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

        //System.out.prlong(max);

        


        System.out.print(max);

    }

}
