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.
//I AM THE CREED
/* //I AM THE CREED
/* package codechef; // don't place package name! */
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
public class Main{
    public static void main(String[] args) throws IOException 
    { 
  
        Scanner input = new Scanner(System.in);
        while(input.hasNext()){
            int n=input.nextInt();
            HashMap<Integer, Integer> m=new HashMap<>();
            
            for(int i=0;i<n;i++){
                int ele=input.nextInt();
                int val=0;
                while(ele!=0){
                    if(ele%2==0){
                        ele=ele/2;
                        continue;
                    }
                    val++;
                    ele=ele/2;
                }
                
                m.put(val, m.getOrDefault(val, 0)+1);
                
            }
            long ct=0;
            for(int k:m.keySet()){
                long val=(long)(m.get(k));
                ct+=val*(val-1)/2;
            }
            System.out.println(ct);
        }
        
    }
    


}