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


public class sequences {

    /**
     * @param args
     */
    public static int functionValue(int i){
        if (i == 0){
            return 0;
        } else {
            if ((i % 2) == 0){
                return functionValue(i / 2);
            } else{
                return 1 + functionValue((i - 1) / 2);
            }
        }
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int sequenceLength = in.nextInt();
        int[] sequence = new int[sequenceLength];
        int[] functionValues = new int[sequenceLength];
        for (int i = 0; i < sequenceLength; i++){
            sequence[i] = in.nextInt();
            functionValues[i] = functionValue(sequence[i]);
        }
        Arrays.sort(functionValues);
        int numberOfTheSameValues = 1;
        int counter = 0;
        for (int i = 1; i < functionValues.length; i++){
            if(functionValues[i] == functionValues[i - 1]){
               numberOfTheSameValues = numberOfTheSameValues + 1;
            } else{
                counter = counter + (numberOfTheSameValues * (numberOfTheSameValues - 1) / 2);
                numberOfTheSameValues = 1;
            }
        }
        counter = counter + (numberOfTheSameValues * (numberOfTheSameValues - 1) / 2);
        System.out.println(counter);

    }

}
