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.
#include <bits/stdc++.h>
using namespace std;
vector<long long> occurences[100];
int findRecursion(int number) {
  if (number == 0) return 0;
  if (number % 2 == 0)
    return findRecursion(number / 2);
  else
    return findRecursion(number / 2) + 1;
}
int main() {
  int n;
  cin >> n;
  int number;
  int functionResult;
  for (int i = 1; i <= n; i++) {
    cin >> number;
    functionResult = findRecursion(number);
    occurences[functionResult].push_back(i);
  }
  long long count = 0;
  for (int i = 0; i < 100; i++) {
    if (occurences[i].size() > 1)
      count += ((occurences[i].size() - 1) * occurences[i].size()) / 2;
  }
  cout << count << endl;
  return 0;
}
