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.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
import java.util.TreeMap;

public class B {
	static final int MAX = 1;//3000000;
	static int[] ary = new int[MAX];

	public static void main(String[] args) {
		doIt();
		//System.out.println(f(469264357));
		//System.out.println(f(996569493));
	}

	static void doIt() {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		long[] num = new long[1000];
		for (int i = 0; i < n; i++) {
			num[f(sc.nextInt())]++;
		}
		long ans = 0;
		for (long i : num) {
			if(2 <= i) ans += (i - 1)*i / 2;
		}
		System.out.println(ans);

	}

	static int f(int x){
		if(x == 0) return 0;
		else{
			if(x < MAX){
				if(ary[x] == 0)
					if(x % 2 == 0) ary[x] = f(x / 2);
					else ary[x] = f((x - 1) / 2) + 1;
				return ary[x];
			}
			else 
				if(x % 2 == 0) return f(x / 2);
				else return f((x - 1) / 2) + 1;
		}

	}
}
