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.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.ArrayList;
public class Main {

    static int d[][];
    static int N;
    static boolean used[];
    public static void main(String[] args) 
    {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        long N[] = new long[n];
        long pr[] = new long[64];
        for(int i = 0;i<n;i++)
        {
            N[i] = scan.nextInt();
        }
        for(int i = 0;i<n;i++)
        {
            pr[f(N[i])]++;
        }
        long pairs = 0;
        for(int i=0;i<64;i++)
        {
            pairs+=pr[i]*(pr[i]-1)/2;
        }
        System.out.print(pairs);
    }
    private static int f(long g)
    {
        if(g==0)return 0;
        if(g%2==0) return f(g/2);
        else return f(g/2)+1;
    }
    
}
