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;
int f(int a) {
  int cnt = 0;
  while (a > 0) {
    if (a % 2 == 0) {
      a /= 2;
    } else {
      cnt++;
      a /= 2;
    }
  }
  return cnt;
}
int main() {
  int N;
  cin >> N;
  vector<int> v;
  for (int i = 0; i < N; i++) {
    int a;
    cin >> a;
    v.push_back(f(a));
  }
  sort(v.begin(), v.end());
  int t = 1;
  long long cnt = 0;
  for (int i = 1; i < v.size(); i++) {
    if (v[i] == v[i - 1])
      t++;
    else if (v[i] != v[i - 1]) {
      if (t == 1) {
        t = 1;
        continue;
      }
      cnt += ((t - 1) + 1) * (t - 1) / 2;
      t = 1;
    }
  }
  if (t > 1) {
    cnt += ((t - 1) + 1) * (t - 1) / 2;
    t = 0;
  }
  cout << cnt;
  return 0;
}
