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 x) {
  int c = 0;
  while (x != 0) {
    if (x & 1) {
      ++c;
      x >>= 1;
    } else
      x >>= 1;
  }
  return c;
}
const int maxn = 100500;
long long C[maxn + 1][3];
void zap() {
  for (int n = 1; n <= maxn; ++n) {
    C[n][0] = 1;
    C[1][1] = 1;
    for (int k = 1; k < 3; ++k) C[n][k] = C[n - 1][k - 1] + C[n - 1][k];
  }
}
int main() {
  vector<int> v;
  int a, b, c, i;
  cin >> a;
  for (i = 0; i < a; i++) {
    cin >> b;
    v.push_back(f(b));
  }
  zap();
  sort(v.begin(), v.end());
  long long sum = 0, kol = 0;
  v.push_back(-1);
  for (i = 1; i < v.size(); i++) {
    kol++;
    if (v[i] != v[i - 1]) {
      sum += C[kol][2];
      kol = 0;
    }
  }
  cout << sum;
  return 0;
}
