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.StreamTokenizer;

public class B {
    static StreamTokenizer st = new StreamTokenizer(new BufferedReader(
            new InputStreamReader(System.in)));

    static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    static long nextLong() throws IOException {
        st.nextToken();
        return (long) st.nval;
    }

    static char nextChar() throws IOException {
        st.nextToken();
        return st.sval.charAt(0);
    }

    public static void qSort(long[] A, int low, int high) {
        int i = low;
        int j = high;
        long x = A[(low + high) / 2]; // x - опорный элемент посредине между low
                                        // и high
        do {
            while (A[i] < x)
                ++i; // поиск элемента для переноса в старшую часть
            while (A[j] > x)
                --j; // поиск элемента для переноса в младшую часть
            if (i <= j) {
                // обмен элементов местами:
                long temp = A[i];
                A[i] = A[j];
                A[j] = temp;
                // переход к следующим элементам:
                i++;
                j--;
            }
        } while (i < j);
        if (low < j)
            qSort(A, low, j);
        if (i < high)
            qSort(A, i, high);
    }

    public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub
        int n = nextInt();
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {

            long t = nextLong();
            while (t != 0) {
                if (t % 2 == 0)
                    t /= 2;
                else {
                    t =(t-1)/2;
                    a[i]++;
                }
            }
        }

        qSort(a, 0, n - 1);
        long counter=0;
        for (int i=0; i<n-1; i++){
            long res = 1;
            while (i!=(n-1)&&a[i]==a[i+1]){
                i++;
                res++;
            }
            counter+=(res-1)*res/2;
        }
        System.out.println(counter);
    }
}