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 B {
	static int calc(int a){
		int cnt = 1;
		while(a!=1){
			if(a%2==0)
				a/=2;
			else{
				a--;
				cnt++;
			}				
		}
		return cnt;
	}
	
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out); 
		int n = sc.nextInt();
		int[] a = new int[n];
		int[] num = new int[n];
		for(int i=0;i<n;i++){
			a[i] = sc.nextInt();
			num[i] = calc(a[i]);
		}			
		sc.close();
		
		Arrays.sort(num);
//		for(int i=0;i<a.length;i++)
//			out.print(num[i]);
		
		long cnt = 0;
		long same = 1;
		for(int i=0;i<num.length;i++){
			if(i+1<num.length && num[i]==num[i+1]){
				same++;
			}
			else{
				cnt += same*(same-1)/2;
				same = 1;
			}
		}
		out.print(cnt);
		out.flush();
	}
}
