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;
long long int f(long long int x) {
  if (x == 0)
    return 0;
  else if (x % 2 == 0)
    return f(x / 2);
  else
    return (f(x / 2) + 1);
}
int main() {
  long long int n, m = 0, i, j = 0, k = 0, flag = 1;
  cin >> n;
  long long int *a = new long long int[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < n; i++) {
    a[i] = f(a[i]);
  }
  sort(a, a + n);
  for (i = 0; i < n - 1; i++) {
    for (j = i; j < n - 1; j++) {
      if (a[j] == a[j + 1]) {
        m++;
        i++;
        flag = 1;
      }
      if (a[j] != a[j + 1]) {
        k = k + (m * (m + 1)) / 2;
        m = 0;
        flag = 0;
        break;
      }
    }
  }
  if (flag) k = k + (m * (m + 1)) / 2;
  cout << k;
  return 0;
}
