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.*;

public class TaskB {
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new TaskB().solve(in, out);
        out.close();

    }
    public void solve(InputReader in, PrintWriter out) {
        int n = in.nextInt();
        Map<Long, Integer> map = new TreeMap<Long, Integer>();
        for(int i = 0; i < n; i++) {
            long key = f(in.nextLong());
            Integer value = map.get(key);
            if(value == null)
                value = 1;
            else
                ++value;
            map.put(key, value);
        }
        long ans = 0;
        for(Long key : map.keySet()) {
            long value = map.get(key);
            ans += value * (value - 1) / 2;
        }
        out.println(ans);
    }
    private long f(long x) {
        if(x == 0)
            return 0;
        else if(x % 2 == 0)
            return f(x / 2);
        else
            return f(x / 2) + 1;
    }
}

class InputReader {
    BufferedReader buff;
    StringTokenizer tokenizer;

    InputReader(InputStream stream) {
        buff = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }
    boolean hasNext() {
        while (tokenizer == null || !tokenizer.hasMoreTokens())
            try {
                tokenizer = new StringTokenizer(buff.readLine());
            }
        catch (Exception e) {
            return false;
        }
        return true;
    }
    String next() {
        if (!hasNext())
            throw new RuntimeException();
        return tokenizer.nextToken();
    }
    int nextInt() { return Integer.parseInt(next()); }
    long nextLong() { return Long.parseLong(next()); }
}