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 javafx.util.Pair;
import org.omg.Messaging.SYNC_WITH_TRANSPORT;

import java.util.*;
import java.io.*;
import java.util.Collections;

public class Main {
    public static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0)
        {
            n &= (n - 1) ;
            count++;
        }
        return count;
    }

    public static void main(String[] args) throws IOException {
            // TODO Auto-generated method stub
              Scanner S = new Scanner(System.in);
              BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
              PrintWriter pw=new PrintWriter(System.out);
              int n=Integer.parseInt(br.readLine());
              String numst[]=br.readLine().split(" ");
              int []nums=new int[n];
              int dp[]=new int[1000];
              for (int i = 0; i < n; i++) {
                   nums[i]=Integer.parseInt(numst[i]);
                   ++dp[countSetBits(nums[i])];
              }
              long ans=0;
        for (int i = 0; i < 1000; i++) {
            ans+=(((long)((long)dp[i])*((long)dp[i]-1))/(long)2);
        }
        pw.println(ans);pw.flush();


    }



}



