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;
import java.util.HashMap;


public class Main {
	
	
private static HashMap<Long, Long> map = new HashMap<Long, Long>();
	
	public static void main(String[] args) throws IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String line;
		
		Long n = Long.parseLong(br.readLine());
		
		line = br.readLine();
		String[] tokens = line.split("\\s+");
		long[] numbers = new long[tokens.length];
		
		long validPairs = 0;
		
		HashMap<Long, Long> someMap = new HashMap<Long, Long>();
		long functionRes;
		
		for (long i = 0; i < numbers.length; i++) {
			numbers[(int) i] = Long.parseLong(tokens[(int) i]);
			functionRes = seq(numbers[(int) i]);
			if (someMap.containsKey(functionRes)) someMap.put(functionRes, someMap.get(functionRes) + 1);
			else someMap.put(functionRes, (long) 1);
		}
		
		for (Long key: someMap.keySet()) {
			validPairs += choose2(someMap.get(key));
		}
		
		System.out.println(validPairs);
	
	}
	
	public static long seq (long num) {
		
		if (map.containsKey(num)) return map.get(num);
		
		long value;
		if (num == 0) return 0;
		if (num % 2 == 0) value = seq(num / 2);
		else value = seq(num / 2) + 1;
		map.put(num, value);
		return value;
	}
	
	public static long choose2(long n) {
		return n * (n - 1) / 2;
	}
}