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.*;
import java.io.*;

public class P166A {

    Scanner in;
    PrintWriter out;
    
    int n, i, l, j;
    long kol=0;
    public void solve() throws IOException {
        n = in.nextInt();
        int[][] a = new int[2][n+1];
        for(i=1; i<=n; i++) {
            a[0][i] = in.nextInt();
            l=0;
            while(a[0][i]!=0){
            	if (a[0][i] % 2 ==1) l++;
            	a[0][i] = a[0][i] / 2;
            }
            a[1][i] = l;
        }
        long t=0;
        Arrays.sort(a[1]);
        for(i=1; i<n; i++) {
        	if (a[1][i]!=a[1][i+1]) {
        		if (t%2==0) kol = kol + (t+1)*(t/2);
        		else kol = kol + (t+1)*(t/2) + (t+1)/2; 
        		t = 0;
        	} else {
        		t++;
        	}
        }
        if (t%2==0) kol = kol + (t+1)*(t/2);
         else kol = kol + (t+1)*(t/2) + (t+1)/2; 
        System.out.print(kol);
    }

    public void run() {
        try {
            in = new Scanner(System.in);
            out = new PrintWriter(System.out);
            
            solve();
            
            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    
    public static void main(String[] args) {
        new P166A().run();
    }

}