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.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.StringTokenizer;
import static java.lang.Math.*;
import static java.lang.Integer.*;
import static java.lang.Long.*;
import static java.lang.String.*;
import static java.lang.Character.*;

@SuppressWarnings("unused")
public class round167B {

    static BufferedReader br = new BufferedReader(new InputStreamReader(
            System.in));
    static StringTokenizer st = new StringTokenizer("");

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

    static String next() throws Exception {
        while (true) {
            if (st.hasMoreTokens()) {
                return st.nextToken();
            }
            String s = br.readLine();
            if (s == null) {
                return null;
            }
            st = new StringTokenizer(s);
        }
    }

    public static void main(String[] args) throws Exception {
        PrintWriter out = new PrintWriter(System.out);
        int n = nextInt();
        int[] seq = new int[n];
        for (int i = 0; i < n; ++i)
            seq[i] = nextInt();
        Arrays.sort(seq);
        HashMap<Integer, Integer> count = new HashMap<Integer, Integer>();
        long ans = 0;
        for (int i = 0; i < n; ++i) {           
            int d1 = seq[i] % 2 == 0 ? seq[i] / 2 : seq[i] / 2 + 1;         
            if (count.containsKey(d1)){             
                ans += count.get(d1);
            }
            Integer pres = count.get(seq[i]);
            if (pres == null)
                count.put(seq[i], (count.containsKey(d1) ? count.get(d1) : 0) + 1);
            else
                count.put(seq[i], pres + (count.containsKey(d1) ? count.get(d1) : 1));
        }
        out.println(ans);
        out.flush();
    }
}
