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.*;
import java.util.*;
import java.math.*;

public class Codeforces
{
  private StreamTokenizer st;
  public PrintWriter pw;
  public static final int modulo = 1000000009; 
  
  void init(String in, String out) 
  {
    if (in.isEmpty())
      st = new StreamTokenizer(System.in);
    else
    {
      try
      {
        st = new StreamTokenizer(new BufferedReader(new FileReader(in)));
      }
      catch(FileNotFoundException e)
      {
        
      }
    }
    
    if (out.isEmpty())
      pw = new PrintWriter(System.out);
    else
    {
      try
      {
        pw = new PrintWriter(new FileWriter(out));
      }
      catch(IOException e)
      {
        
      }
    }
  }
  
  private void close()
  {
    pw.close();
  }
  
  private int nI()
  {
    try{
      st.nextToken();
    }
    catch(IOException e)
    {
      
    }
    return (int)st.nval;
  }
  
  private double nD()
  {
    try{
      st.nextToken();
    }
    catch(IOException e)
    {
      
    }
    return st.nval;
  }
  
  private String nS()
  {
    try{
      st.nextToken();
    }
    catch(IOException e)
    {
      
    }
    return st.sval;
  }
   
  private long nL()
  {
    try{
      st.nextToken();
    }
    catch(IOException e)
    {
      
    }
    return (long)st.nval;
  }
  
  public int calc(long c)
  {
    int cc = 0;
    while (c != 0)
      if (c % 2 == 0)
        c /= 2;
      else
      {
        ++cc;
        c/=2;
      }
    return cc;
  }
  
  public static void main(String[] aslkdjlkgja) throws IOException
  {
    Codeforces z = new Codeforces();
    z.init("", "");
    
    int n = z.nI();
    int a[] = new int[n];
    
    for (int i = 0; i < n; ++i)
      a[i] = z.calc(z.nI());
    
    Arrays.sort(a);
    
    long sum = 0;
    long count = 1;
    int cur = a[0];
    for (int i = 1; i < n; ++i)
      if (a[i] == cur)
        ++count;
      else
      {
        sum += count * (count-1)/2;
        cur = a[i];
        count = 1;
      }
    
    sum += count * (count-1)/2;
    
    
    z.pw.print(sum);
    
   z.close();
  }
}