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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;

public class Sorter {
    long N,M;
    long[] mat;
    public Sorter() {}



    long reader(){

        BufferedReader br = null;
        long ans = 0;
        try {
            String[] temp;
            br = new BufferedReader(new InputStreamReader(System.in));

            String currLine = br.readLine();
            N = Long.parseLong(currLine)+1;
            currLine = br.readLine();
            temp = currLine.split(" ");
            mat = new long[(int) (N-1)];
            
            int count = 0,evenCount= 0, oddCount = 0;
            while(count<N-1){
                mat[count] = Integer.parseInt(temp[count]);
                if(mat[count]==1 || mat[count]%2==0)
                    evenCount++;
                else
                    oddCount++;
                count++;
            }
            
            long ans1 = (evenCount)*(evenCount-1)/2;
            long ans2 = (oddCount)*(oddCount-1)/2;
            ans = ans1+ans2;
            
        }
        catch (Exception e) {
            e.printStackTrace();

        }
        return ans;

    }




    public static void main(String args[]){

        Sorter s = new Sorter();



        System.out.println(s.reader());



    }



}