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.Arrays;
import java.util.Hashtable;
import java.util.Scanner;

public class B167 {

    public static int temp = 0;
    public static Hashtable<Long, Long> htb = new Hashtable<Long, Long>();

    public void solve() throws IOException {
        Scanner sc = new Scanner(System.in);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int i = 0, j = 0, k = 0;
        int ans = 0, sum = 0;
        long count = 0;
        int n = Integer.valueOf(br.readLine());
        long a[] = new long[n];

        String s[] = br.readLine().split(" ");

        long sol[] = new long[n];

        for (i = 0; i < n; ++i) {
            a[i] = Integer.valueOf(s[i]);

            if (htb.containsKey(a[i])) {
                sol[i] = htb.get(a[i]);
            } else {
                //  htb.put(a[i], (ff(a[i]) + temp));
                long f = fff(a[i]);
                htb.put(a[i], (f));
                sol[i] = f;
            }

            temp = 0;
        }

        //System.out.println(Arrays.toString(sol));

        int fg[] = new int[10000];

        for (i = 0; i < n; ++i) {
            fg[(int) sol[i]]++;
        }
        for (i = 0; i < 10000; ++i) {
            if(fg[i]>0){
                count+=C(fg[i],2);
            }
        }


        System.out.println(count);

    }

    static long C(long n, long k) {
        double res = 1;
        for (int i = 1; i <= k; ++i) {
            res = res * (n - k + i) / i;
        }
        return (long) (res + 0.01);
    }

    public static long fff(long num) {
        temp = 0;
        while (num != 1) {
            if (num % 2 != 0 && num != 1) {
                // System.out.println("num = " + num);
                temp++;
            }

            num /= 2;

            if (htb.containsKey(num)) {
                return htb.get(num) + temp;
            }
        }

        return 100 + temp;
    }

    public static void main(String[] args) throws IOException {
        new B167().solve();
        // System.out.println(fff(30));
    }
}