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.io.PrintStream;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

/**
 * @author artyom
 */
public class DimaAndSequence implements Runnable {
	private BufferedReader in;
	private PrintStream out;
	private StringTokenizer tok;

	private void solve() throws IOException {
		Map<Integer, Integer> remainderCounts = new HashMap<>();
		long pairsCount = 0;
		for (int i = 0, n = nextInt(); i < n; i++) {
			int remainder = getRemainder(nextInt());
			if (remainderCounts.containsKey(remainder)) {
				int count = remainderCounts.get(remainder);
				pairsCount += count;
				remainderCounts.put(remainder, count + 1);
			} else {
				remainderCounts.put(remainder, 1);
			}
		}
		out.println(pairsCount);
	}

	private static int getRemainder(int nextInt) {
		int rem = 0;
		while (nextInt > 1) {
			rem += nextInt % 2;
			nextInt /= 2;
		}
		return rem;
	}

	//--------------------------------------------------------------
	public static void main(String[] args) {
		new DimaAndSequence().run();
	}

	@Override
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = System.out;
			tok = null;
			solve();
			in.close();
		} catch (IOException e) {
			System.exit(0);
		}
	}

	private String nextToken() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	private int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
}