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.Scanner;
import java.io.IOException;
import java.util.HashMap;
import java.util.Set;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskB solver = new TaskB();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskB {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
	    int n = in.nextInt();
	    int[] a = IOUtils.readIntArray(in, n);
	    Counter<Integer> counter = new Counter<Integer>();
	    for (int num : a) counter.add(calc(num));
	    long res = 0;
	    for (int key : counter.keySet()) {
		    long count = counter.get(key);
		    res += count * (count - 1) / 2;
	    }
	    out.print(res);
    }

	private int calc(int num) {
		if (num == 0) return 0;
		if (num % 2 == 0) return calc(num / 2);
		else return calc(num / 2) + 1;
	}
}

class IOUtils {
    public static int[] readIntArray(Scanner in, int size) {
        int[] array = new int[size];
        for (int i = 0; i < size; i++)
            array[i] = in.nextInt();
        return array;
    }

    }

class Counter<K> extends HashMap<K, Long> {
	public Counter() {
		super();
	}

	public void add(K key) {
		put(key, get(key) + 1);
	}

	public Long get(Object key) {
		if (containsKey(key))
			return super.get(key);
		return 0L;
	}
}

