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.*;
import java.util.*;
import java.math.*;

public class Olimp {

    FastScanner in;
    PrintWriter out;

    public void solve() throws IOException {
        int n = in.nextInt();
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = Long.parseLong(in.next());
        }
        ArrayList<Long> q1 = new ArrayList<Long>();
        ArrayList<Long> q2 = new ArrayList<Long>();
        for (int i = 0; i < n; i++) {
            if (a[i] == 0) {
                if (q1.contains(0)) {
                    q2.set(q1.indexOf(0), q2.get(q1.indexOf(0)) + 1);
                } else {
                    q1.add((long) 0);
                    q2.add((long) 1);
                }
            } else {
                long tmp = 1;
                long x = a[i];
                while (x > 1) {
                    if (x % 2 == 0) {
                        x /= 2;
                    } else {
                        tmp++;
                        x /= 2;
                    }
                }
                if (q1.contains(tmp)) {
                    q2.set(q1.indexOf(tmp), q2.get(q1.indexOf(tmp)) + 1);
                } else {
                    q1.add((long) tmp);
                    q2.add((long) 1);
                }

            }
        }
        long ans = 0;
        for(int i = 0; i < q2.size(); i++)
            ans += q2.get(i) * (q2.get(i) - 1) / 2;
        System.out.println(ans);
    }

    public void run() {
        try {
            in = new FastScanner();
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {

        BufferedReader br;
        StringTokenizer st;

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

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

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

    public static void main(String[] arg) {
        new Olimp().run();
    }
}
