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.*;
import java.io.*;

public class DimaAndSequence {
	public static void main(String[] args)  throws IOException {
		Scanner ins = new Scanner(System.in);//new FileInputStream(new File("./input.txt")));//System.in
		PrintStream outs = System.out;

		int[] num = new int[34];
		for (int i=0; i<34; i++) num[i]=0;
		int n = ins.nextInt();
		for (int i=0; i<n; i++) {
			int tmp = ins.nextInt(); 
			int cnt=0;
			while (tmp != 0) {
				if (tmp % 2 == 1) cnt++;
				tmp /= 2;
			}
			if (cnt != 0) num[cnt]++;
		}

		long ans=0;
		for (int i=0; i<34; i++) {
			if (num[i] >= 2) {
				long x=num[i];
				ans += x*(x-1)/2;
			}
		}

		outs.println(ans);
	}
	
}
