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>
int a[1000003];
int f(int i) {
  if (i == 0)
    return 0;
  else
    return f(i >> 1) + (i & 1);
}
void QuickSort(int h, int t) {
  int temp;
  if (h >= t) return;
  int mid = (h + t) / 2, i = h, j = t, x;
  x = a[mid];
  while (i <= j) {
    while (a[i] < x) i++;
    while (a[j] > x) j--;
    if (i <= j) {
      temp = a[i];
      a[i] = a[j];
      a[j] = temp;
      i++;
      j--;
    }
  }
  QuickSort(h, j - 1);
  QuickSort(j + 1, t);
  return;
}
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", a + i);
  for (i = 0; i < n; i++) a[i] = f(a[i]);
  double ans = 0, t = 1;
  QuickSort(0, n - 1);
  for (i = 1; i < n; i++)
    if (a[i] == a[i - 1])
      t += 1;
    else {
      ans += ((t * (t - 1)) / 2);
      t = 1;
    }
  {
    ans += ((t * (t - 1)) / 2);
    t = 1;
  }
  printf("%.0lf\n", ans);
  return 0;
}
