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;
            long evenCount= 0, oddCount = 0,midCount=0;
            HashMap<Long, Long> oddV = new HashMap<Long, Long>();
            while(count<N-1) {
                long temper = Long.parseLong(temp[count]);

                if(temper == 1 || temper %2 ==0)
                    evenCount++;
                else{
                    temper--;   
                    while(temper%2==0)
                        temper/=2;
                    
                        if(oddV.containsKey(temper))
                            oddV.put(temper,oddV.get(temper)+1);
                        else
                            oddV.put(temper,(long) 1);
                    
                    
                }
            

            count++;
        }

        long ans1 = (evenCount)*(evenCount-1)/2;
        long ans2 = 0;
        for(long key: oddV.keySet()){
            oddCount = oddV.get(key);
            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());



}



}