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.PrintWriter;
import java.util.*;

public class B {

	public static void main(String[] args) {
		PrintWriter out = new PrintWriter(System.out);
		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();
		long[] arr = new long[N];
		for (int i = 0; i < arr.length; i++) {
			arr[i] = sc.nextLong();
			long t = f(arr[i]);
			map.put(arr[i], t);
			if (size.containsKey(t)) {
				long a = size.get(t);
				size.put(t, a + 1);
			} else {
				size.put(t, 1L);
			}
		}
		//System.out.println(size);
		long ans = 0;
		for (Long t : size.keySet()) {
			long a = size.get(t);
			ans += (a * (a - 1)) / 2;
		}
		out.println(ans);
		sc.close();
		out.close();
	}

	private static long getF(long l) {
		if (map.containsKey(l))
			return map.get(l);
		else {
			long t = f(l);
			map.put(l, t);
			return t;
		}
	}

	private static HashMap<Long, Long> map = new HashMap<>();
	private static HashMap<Long, Long> size = new HashMap<>();

	private static long f(long num) {
		if (map.containsKey(num))
			return map.get(num);

		if (num == 0)
			return 0;
		else if (num % 2 == 0) {
			return f(num / 2);
		} else {
			long a = (num - 1) / 2;
			return f(a) + 1;
		}
	}
}
