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.*;

public class Solution {
    public static void main (String[] args) throws IOException {
        boolean online = "true".equals(System.getProperty("ONLINE_JUDGE"));
        if (online) {
            in = new Scanner(System.in);
            out = new PrintWriter(System.out);
        }
        else {
            in = new Scanner(new File("input.txt"));
            out = new PrintWriter(new File("output.txt"));
        }
        
        new Solution().run();
        
        out.close();
    }
    
    private void run () {
        int n = in.nextInt();
        int[] d = new int[51];
        for (int i = 0; i < n; ++i)
            ++d[v(in.nextInt())];
        long ans = 0;
        for (int i = 0; i <= 50; ++i)
            ans += c(d[i]);
        out.println(ans);
    }
    
    private int v (int x) {
        if (x == 0) return 0;
        return v(x/2) + (x % 2);
    }
    
    private long c (int x) {
        return (long)x*(x-1)/2;
    }
    
    private static Scanner in;
    private static PrintWriter out;
}