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.
/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{	
	public static void main (String[] args) throws java.lang.Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
		int  n = Integer.parseInt(br.readLine());
		long[] arr = new long[n];
		String[] z = br.readLine().split(" ");
		for(int i = 0;i<n;i++){
			arr[i] = Long.parseLong(z[i]);
		}
		int pow=0,twot=0,three=0;
		for(int i = 0;i<n;i++){
			if(ans(arr[i])==1){
				pow +=1;
			}
			else if(ans(arr[i])==2){
				twot+=1;
			}
			else if(ans(arr[i])==3){
				three +=1;
			}
		}
		int s = 0;
		if(pow>1){
			s = s + pow*(pow-1)/2;
		}
		if(twot>1){
			s = s + twot*(twot-1)/2;
		}
		if(three>1){
			s = s + three*(three-1)/2;
		}
		pw.println(s);
		pw.close();
	}
	public static int ans(long l){
		while(l%2==0){
			l = l/2;
		}
		if(l==1){
			return 1;
		}
		l = l-1;
		while(l%2==0){
			l = l/2;
		}
		if(l==1){
			return 2;
		}
		else{
			return 3;
		}
	}
}