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.*;
public class CF272B_DimaSequence {
public static void main(String[] args){
    Scanner input =new Scanner(System.in);
    int n=input.nextInt();
    HashMap<Integer,Integer> num=new HashMap<Integer,Integer>();
    int a[]=new int[n];
    long total=0;
    for (int i=0;i<n;i++){
        a[i]=f(input.nextInt());
    }
    for (int i=0;i<n;i++){
        if (num.containsKey(a[i])){
            int r=num.get(a[i]);
            num.put(a[i],r+1);
        }
        else {
            num.put(a[i],1);
        }
    }
    for (int e:num.keySet()){
//      System.out.println(e+" "+num.get(e));
        long y=((long)num.get(e)*((long)num.get(e)-1))/2;
        total+=y;
        
    }
    System.out.print(total);
    
}
public static int f(int x){
    if (x==0){
        return 0;
    }
    if (x%2==0){
        return f(x/2);
    }
    else {
        return f(x/2)+1;
    }
}
}
