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.HashMap;
import java.util.Scanner;

// Codeforces 272B
public class DimaAndSequence {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int[] arr = new int[in.nextInt()];
		for (int i = 0; i < arr.length; i++) {
			arr[i] = in.nextInt();
		}
		HashMap<Integer, Integer> imageFreq = new HashMap<Integer, Integer>();
		
		for (int x : arr) {
			int y = f(x);
			if (imageFreq.containsKey(y)) {
				imageFreq.put(y, imageFreq.get(y) + 1);
			}
			else {
				imageFreq.put(y, 1);
			}
		}
		int sum = 0;
		for (int v : imageFreq.values()) {
			sum += v * (v - 1) / 2;
		}
		System.out.println(sum);
		
	}
	
	static int f(int x) {
		if (x == 0) return 0;
		if (x % 2 == 0) return f(x / 2);
		return 1 + f((x - 1) / 2);
	}
}
