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.InputStreamReader;
import java.util.HashMap;
import java.util.Map.Entry;
import java.util.StringTokenizer;

public class B {

	public static void main(String[] args) throws Exception {
		int count = nextInt();
		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
		int[]parent=new int[count];
		for (int i = 0; i < count; i++) {
			int r=0;
			int x = nextInt();
			while(x>0){
				if((x&1)==1) r++;
				x>>=1;
			}
			if (map.containsKey(r)) {
				map.put(r, map.get(r) + 1);
			} else {
				map.put(r, 1);
			}
		}
		long res = 0;
		for (Entry<Integer, Integer> e : map.entrySet()) {
			res += ((((long) e.getValue()) * ((long) (e.getValue() - 1)))/2);
		}
		System.out.println(res);
	}

	static BufferedReader br = new BufferedReader(new InputStreamReader(
			System.in));
	static StringTokenizer tokenizer = new StringTokenizer("");

	static int nextInt() throws Exception {
		return Integer.parseInt(next());
	}

	static double nextDouble() throws Exception {
		return Double.parseDouble(next());
	}

	static String next() throws Exception {
		while (true) {
			if (tokenizer.hasMoreTokens()) {
				return tokenizer.nextToken();
			}
			String s = br.readLine();
			if (s == null) {
				return null;
			}
			tokenizer = new StringTokenizer(s);
		}
	}
}