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 n;
long long int arr[100005], value[100005];
long long int solve(long long int a) {
  long long int b, c;
  if (a == 0) {
    return 0;
  }
  if (a % 2 == 0) {
    return solve(a / 2);
  } else {
    return solve(a / 2) + 1;
  }
}
int main() {
  long long int i, j, a, b, c, ans;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
    value[i] = solve(arr[i]);
  }
  sort(&value[1], &value[n + 1]);
  ans = 0;
  for (i = 1; i <= n; i++) {
    a = 1;
    j = i + 1;
    while (j <= n && value[j] == value[i]) {
      a++;
      j++;
    }
    i = j - 1;
    ans += (a * (a - 1) / 2);
  }
  cout << ans;
  return 0;
}
