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.PrintWriter;
import java.util.ArrayList;
import java.util.Map;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class B {

    static long[] math = new long[100005];
    static Map<Integer, Integer> ans = new TreeMap<Integer, Integer>();

    public static void main(String[] args) throws NumberFormatException,
            IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int n = Integer.parseInt(br.readLine());
        StringTokenizer stk = new StringTokenizer(br.readLine());
        ArrayList<Integer> ar = new ArrayList<Integer>();
        for (int i = 0; i < n; i++) {
            int k = solve(Integer.parseInt(stk.nextToken()));
            if (ans.containsKey(k)) {
                ans.put(k, ans.get(k) + 1);
            } else {
                ans.put(k, 1);
                ar.add(k);
            }
        }
        long sum = 0;
        for (int i : ar) {
            sum += entekhab(ans.get(i), 2);
        }

        out.println(sum);
        br.close();
        out.close();
    }

    static int solve(int a) {
        if (a == 0) {
            return 0;
        }

        if (a % 2 == 0) {
            return solve(a / 2);
        }
        return solve(a / 2) + 1;
    }

    public static long entekhab(int n, int k) {
        if (n < k) {
            return 0;
        }

        if (k > n - k)
            k = n - k;
        if (math[n] > 0) {
            return math[n];
        }
        long t = 1;
        for (int i = 1; i <= k; i++) {
            t = (n - i + 1) * t / i;
        }
        return math[n] = t;
    }

}
