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.Arrays;
import java.util.Scanner;

public class Main2 {
	
    public static void main(String args[]){
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int[] a = new int[n];
        for(int i = 0 ; i < n ; i++){
        	a[i] = input.nextInt();
        }
        System.out.println(count(n,a));
    }
    
    public static long count(int n,int[] a){
    	long count = 0;
    	int[] bit = new int[n];
    	for(int i = 0 ; i < a.length ; i++){
    		bit[i] = Integer.bitCount(a[i]);
    	}
    	int same = 1;
    	Arrays.sort(bit);
    	int number = bit[0];
    	for(int i = 1 ; i < a.length ; i++){
    		if(bit[i] == number){
    			same++;
    		}else{
    			count+=(same*(same-1))/2;
    			same = 1;
    			number = bit[i];
    		}
    	}
    	count+=(same*(same-1))/2;
    	return count;
    }

}