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.*;
import java.io.*;

public class DimaSequence272B
{
    public static class MyFasterScanner 
    {
        BufferedReader br;
        StringTokenizer st;

        public MyFasterScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine(){
            String str="";
            try {
                str =  br.readLine().toString();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
    
    public static int onesinbin(int n)
    {
        int returnme = 0;
        while (n>0)
        {
            if (n%2 == 1)
            {
                returnme++;
            }
            n = n/2;
        }
        return returnme;
    }
    
    public static void main(String[] args) 
    {
        // Set up scanner
        MyFasterScanner sc = new MyFasterScanner(); 
        // System.out.println("Enter n");
        int n = sc.nextInt();
        
        long[] onearray = new long[30];

        // For each number, compute the number of 1s in its binary representation
        for (int i=0; i<n; i++)
        {
            // System.out.println("Enter next number");
            int x = sc.nextInt();
            int index = onesinbin(x);
            onearray[index]++;
        }
        
        long answer = 0;
        for (int i=1; i<30; i++)
        {
            long times = onearray[i];
            answer += (times*(times-1))/2;
        }
        System.out.println(answer);
    }
}
