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;
    HashMap<Long, Long> saver = new HashMap<Long, Long>();
    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;
            long oddCount = 0;
            HashMap<Long, Long> oddV = new HashMap<Long, Long>();
            while(count<N-1) {
                long tempers = Long.parseLong(temp[count]);
                long temper = getValue(tempers);
                //System.out.println( tempers+ " "+temper);
                if(oddV.containsKey(temper))
                    oddV.put(temper,oddV.get(temper)+1);
                else
                    oddV.put(temper,(long) 1);
                count++;
            }

        
            long ans2 = 0;
            for(long key: oddV.keySet()){
                oddCount = oddV.get(key);
                ans2+= (oddCount)*(oddCount-1)/2;
            }

            ans = ans2;

        }
        catch (Exception e) {
            e.printStackTrace();

        }
        return ans;

    }

    long getValue(long temper){
        if(temper==0)
            return 0;
        while(temper%2==0)
            temper/=2;
        if(temper==1)
            return 1;
        if(temper%2!=0){
            return getValue((temper-1)/2)+1;
        }
        return temper;
    }


    public static void main(String args[]){

        Sorter s = new Sorter();



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



    }



}