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.*;
import java.util.*;
public class div167B
{
	BufferedReader in;
	PrintWriter ob;
	StringTokenizer st;
	public static void main(String[] args) throws IOException {
		new div167B().run();
	}
	void run() throws IOException {
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solve();
		ob.flush();
	}
	void solve() throws IOException {
		int n=ni();
		int a[]=new int[n];
		HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
		for (int i=0; i<n ; i++ ) {
			a[i]=ni();
			int c=Integer.bitCount(a[i])+1;
			if(hm.containsKey(c)==false)
			hm.put(c,1);
			else
			hm.put(c, hm.get(c) + 1 ); 
		}
		long ans=0;
		for (int key : hm.keySet() ) {
			int res=hm.get(key);
			ans+=((res*1L*(res-1))/2);
		}
		ob.println(ans);
	}
	int func(int x) {
		if(x==0) return 1;
		else if(x%2==0)  return func(x/2);	
		else return func(x/2)+1;
	}
	int ni() throws IOException {
		return Integer.parseInt(nextToken());
	}
	String nextToken() throws IOException {
		if(st==null || !st.hasMoreTokens()) 
			st=new StringTokenizer(in.readLine());
		return st.nextToken();
	}
}