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;

public class DimaAndSequence {
	public static int f(int i){
		if(i == 0)
			return 0;
		if(i%2 == 0)
			return f(i/2);
		else
			return f((i-1)/2) + 1;
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		String [] as = br.readLine().split(" ");
		int[]a = new int[n];
		for (int i = 0; i < a.length; i++) {
			a[i] = Integer.parseInt(as[i]);
		}
		long[] counter = new long[1000000];

		for (int i = 0; i < a.length; i++) {
			int val = f(a[i]);
			counter[val]++;
		}
		long c = 0;

		for (int i = 0; i < counter.length; i++) {
			long pair = counter[i]*(counter[i]-1);
			pair/=2;
			c += pair;
		}
		System.out.println(c);
	}
}
