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.FileReader;
import java.io.PrintWriter;
import java.util.Locale;
import java.util.Scanner;
public class Main implements Runnable { 
    final boolean LOCAL = System.getProperty("ONLINE_JUDGE") == null;
    Scanner sc;
    PrintWriter out;
    
    int freq[] = new int[30];
    private int helper( int x)
    {
        if(x == 0)
            return 0;
        return helper( x >> 1 ) + (x & 1);      
    }
    void solve()
    {       
        long ans = 0;
        int n =  sc.nextInt();
        for (int i = 0; i < n; ++i)
            freq[helper(sc.nextInt())]++;
        for (int i = 0; i < 30; ++i) 
            ans += freq[i] * (freq[i] - 1L) / 2;
        out.println(ans);   
    }
    public void run(){
        try {
            long t1 = 0, t2 = 0, m1 = 0, m2 = 0;
            Locale.setDefault(Locale.US);
            if (LOCAL) {
                sc = new Scanner(new FileReader("input.txt"));
                out = new PrintWriter("output.txt");                
                t1 = System.currentTimeMillis();
                m1 = Runtime.getRuntime().freeMemory();
                System.err.println("Memory = " + (m1/ 1024) + " KB.");
            }
            else
            {
                sc = new Scanner(System.in);
                out = new PrintWriter(System.out);
            }
            solve();
            sc.close();
            out.close();
            if (LOCAL) {
                t2 = System.currentTimeMillis();
                m2 = Runtime.getRuntime().freeMemory();
                System.err.println("Time = " + (t2 - t1) + " ms.");
                final long usedMem = Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory();
                System.err.println("Memory = " + (usedMem/ 1024) + " KB.");
                System.err.println("Memory = " + ((m1 - m2) / 1024) + " KB.");
            }
        } catch (Throwable e) {     
            e.printStackTrace(System.err);
            throw new RuntimeException();
        }
    }
    
    public static void main(String...arg)
    {
        new Thread(null,new Main(),"",256*(1 << 20)).start();
    }
}