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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author Igors
 */
public class CF167B {
    public static void main(String[] args) throws IOException{
        StreamTokenizer s = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        s.nextToken();
        int n = (int)s.nval;
        long nums[]=new long[n];
        for (int i=0;i<n;i++){
            s.nextToken();
            long tmp = (long)s.nval;
            while(tmp>0){
              nums[i]+=tmp%2;
              tmp/=2;  
            }            
        }
        java.util.Arrays.sort(nums);
        long cur=nums[0];
        long curc=0,res=0;
        for (int i=0;i<n;i++){       
            if (cur==nums[i]){
                curc++;
            }
            else {
                res+=(curc*(curc-1))/2;
                cur=nums[i];
                curc=1;
            }
        }
        res+=(curc*(curc-1))/2;
        System.out.println(res);
    }
}
