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 bin_count1(int n) {
  int count = 0;
  while (n > 0) {
    if (n % 2 == 1) count++;
    n /= 2;
  }
  return count;
}
int f_a[100005];
int main() {
  for (int i = 0; i < 100005; ++i) f_a[i] = -1;
  int n;
  cin >> n;
  int input;
  for (int i = 1; i <= n; ++i) {
    cin >> input;
    f_a[i] = bin_count1(input);
  }
  sort(f_a + 1, f_a + n + 1);
  long long pair = 0;
  for (int i = 1; i <= n; ++i) {
    long long count = 1;
    while (f_a[i] == f_a[i + 1]) {
      count++;
      i++;
    }
    pair = pair + count * (count - 1) / 2;
  }
  cout << pair;
  return 0;
}
