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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.StringTokenizer;


public class T272B {

	public void solve(int n, long[] a) {
		Arrays.sort(a);
		HashMap<Long, HashSet<Long>> m = new HashMap<Long, HashSet<Long>>();
		HashSet<Long> s = new HashSet<Long>();
		
		for (int i = 0; i < n; i ++) {
			long b = getVal(a[i]);
			s = new HashSet<Long>();
			if (m.containsKey(b)) {
				s = m.get(b);
			}
			s.add(a[i]);
			m.put(b, s);
		}
		
		long ans = 0;
		for (long x : m.keySet()) {
			s = m.get(x);
			if (s.size() >= 2) {
				ans += (s.size() * (s.size() - 1)) / 2;
			}
		}
		System.out.println(ans);
	}
	private long getVal(long a) {
		if (a == 1) return 1;
		if (a == 0) return 0;
		
		if (a % 2 == 0)
			return getVal(a / 2);
		else
			return getVal(a - 1) + 1;
	}
	
	public static void main(String[] args) {
		FastScanner in = new FastScanner();
		int n = in.nextInt();
		long[] a = new long[n];
		for (int i = 0; i < n; i ++) a[i] = in.nextLong();
		new T272B().solve(n, a);
	}
	static class FastScanner {

		BufferedReader br;
		StringTokenizer st;

		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

		long nextLong() {
			return Long.parseLong(nextToken());
		}

		double nextDouble() {
			return Double.parseDouble(nextToken());
		}

	}
}
