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[1000001];
int getf(int n) {
  if (n <= 1000000) return f[n];
  if (n % 2)
    return getf((n / 2) - 1) + 1;
  else
    return getf(n / 2);
}
long long arth(int n) {
  n--;
  long long ret = (n * (n + 1)) / 2;
  return ret;
}
int main() {
  f[0] = 0;
  for (int i = 0; i <= 1000000; i++) {
    if (i % 2)
      f[i] = f[i / 2] + 1;
    else
      f[i] = f[i / 2];
  }
  int n, a[100001];
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int arr[100001];
  for (int i = 0; i < n; i++) {
    arr[i] = getf(a[i]);
  }
  sort(arr, arr + n);
  int ind = 0;
  long long count = 0;
  for (int i = 1; i < n; i++) {
    if (arr[i] != arr[i - 1]) {
      count += arth(i - ind);
      ind = i;
    }
  }
  if (arr[n - 1] == arr[n - 2]) count += arth(n - ind);
  cout << count << endl;
}
