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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;

public class B {

	/**
	 * @param args
	 * @throws IOException
	 */
	public static void main(String[] args) throws IOException {
		// Scanner sc = new Scanner(System.in);
		// int n = sc.nextInt();
		BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
		// StringTokenizer t = new StringTokenizer(rd.readLine(), " ");
		int n = Integer.parseInt(rd.readLine());
		StringTokenizer t = new StringTokenizer(rd.readLine(), " ");
		// int x = 1000000000;
		// int[] dp = new int[10000 + 1];
		// for (int i = 1; i < dp.length; i++) {
		// if (i % 2 == 0)
		// dp[i] = dp[i / 2];
		// else
		// dp[i] = dp[i / 2] + 1;
		// }
		// System.out.println(Arrays.toString(dp));
		// int num[] = new int[n];
		// int n_one[] = new int[n];
		int freq[] = new int[33];
		for (int i = 0; i < n; i++) {
			int num = Integer.parseInt(t.nextToken());
			int n_one = 0;
			while (num != 0) {
				if ((num & 1) == 1)
					n_one++;
				num /= 2;
			}
			freq[n_one]++;
		}
		BigInteger out = BigInteger.ZERO;
		for (int i = 0; i < 33; i++) {
			out = out.add((BigInteger.valueOf(freq[i]).multiply(
					BigInteger.valueOf(freq[i] - 1)).divide(BigInteger
					.valueOf(2))));
		}
		System.out.println(out);
	}

}
