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.Map.Entry;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeMap;
import static java.lang.Integer.*;
public class CF167B{
	static long f(long a){
		if(a==0)return 0;
		if(a<MAX&&mem[(int)a]!=-1)return mem[(int)a];
		if(a%2==0)return f(a/2);
		if(a<MAX)
			return mem[(int)a]=f(a/2)+1;
		return f(a/2)+1;
	}
	static long mem[];
	static int MAX=1000000;
	public static void main(String[] args)throws Throwable{
		BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
		int N=parseInt(in.readLine().trim());
		mem=new long[MAX];
		Arrays.fill(mem,-1);
		TreeMap<Long,Long> mapa=new TreeMap<Long,Long>();
		StringTokenizer st=new StringTokenizer(in.readLine());
		for(int i=0;i<N;i++) {
			long a=f(parseInt(st.nextToken()));
			Long s=mapa.get(a);
			if(s==null)s=0L;
			s++;
			mapa.put(a,s);
		}
		long result=0;
		for(Entry<Long,Long> entry:mapa.entrySet()) 
			result+=((entry.getValue())*(entry.getValue()-1))/2;
		System.out.println(result);
		/*int sol=0;
		for(int i=1;i<=5;i++)if((i+s-1)%(N+1)!=0)sol++;
		System.out.println(sol);*/
	}
}
