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;
void solve() {
  long long int n, ans = 0;
  cin >> n;
  vector<long long int> A(n);
  map<long long int, long long int> fr;
  for (long long int i = 0; i < (n); ++i) cin >> A[i], fr[A[i]]++;
  sort(A.begin(), A.end());
  map<long long int, long long int>::reverse_iterator it;
  for (it = fr.rbegin(); it != fr.rend(); it++) {
    long long int a = it->first;
    auto k = fr.begin();
    if (a & 1)
      k = fr.find((floor(a / 2.0)));
    else
      k = fr.find(a / 2);
    if (k != fr.end()) {
      ans += k->second * it->second;
    }
  }
  cout << ans << "\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
