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.math.BigInteger;
import java.util.*;

public class Main {
    
    static HashMap<Integer,Integer> masF = new HashMap<>();
    public static void main(String[] args) throws Exception{
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] mas = new int[n];
        masF.put(0, 0);
        masF.put(1, 1);
        masF.put(2, 1);
        long pairs=0;
        HashMap<Integer,Integer> masT = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int a = sc.nextInt();
            int f = getF(a);
            if(!masT.containsKey(f))
            {
                masT.put(f, 1);
            }else
            {
                pairs-=(masT.get(f) *(long)(masT.get(f) -1));
                masT.put(f, masT.get(f)+1);
                pairs+=(masT.get(f) *(long)(masT.get(f) -1));
            }
        }
        System.out.println(pairs/2);
    }
    static int getF(int a)
    {
        while(a%2==0) a/=2;
        if(a<500000000 && (a==0 || masF.containsKey(a))) return masF.get(a);
        a=(a-1)/2;
        masF.put(a*2+1, getF(a)+1);
        return masF.get(a*2+1);
    }
}
