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.IOException;
import java.util.Scanner;

public class TaskB {
	static int[][] a;

	static void qSort(int start, int end) {
		if (start >= end)
			return;
		int i = start, j = end;
		int cur = i - (i - j) / 2;
		while (i < j) {
			while ((i < cur)
					&& ((a[i][0] < a[cur][0]) || ((a[i][0] == a[cur][0]) && (a[i][1] < a[cur][1])))) {
				i++;
			}
			while ((j > cur)
					&& ((a[j][0] > a[cur][0]) || ((a[j][0] == a[cur][0]) && (a[j][1] > a[cur][1])))) {
				j--;
			}
			if (i < j) {
				int temp = a[i][0];
				a[i][0] = a[j][0];
				a[j][0] = temp;
				temp = a[i][1];
				a[i][1] = a[j][1];
				a[j][1] = temp;

				if (i == cur)
					cur = j;
				else if (j == cur)
					cur = i;
			}
		}
		qSort(start, cur);
		qSort(cur + 1, end);
	}

	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long ans = 0;
		a = new int[n][2];
		for (int i = 0; i < n; i++) {
			int x = sc.nextInt();
			while (x > 0) {
				if (x % 2 != 0) {
					a[i][0]++;
				}
				x = x / 2;
			}
			a[i][1] = i;
		}
		qSort(0, n - 1);

		int i = 0;

		while (i < n) {
			int j = i;
			while ((j + 1 < n) && (a[j + 1][0] == a[j][0])) {
				j++;
			}
			long p = j - i + 1;
			ans += p * (p - 1) / 2;
			i = j + 1;
		}
		System.out.println(ans);
	}
}