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;
void ins(long long int a[], long long int n, long long int i = 0) {
  for (long long int i = 0; i < n; i++) a[i] = i;
}
template <typename t>
void ins(t a[], long long int n, t b = 0) {
  for (long long int i = 0; i < n; i++) a[i] = b;
}
void print(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  ;
  cout << endl;
}
template <typename t>
void print(t a[], long long int n) {
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  ;
  cout << endl;
}
void po(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
}
template <typename t>
void po(t a[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
}
long long int doit(long long int x) {
  long long int count = 0;
  while (x != 0) {
    if (x % 2 == 1) {
      count++;
    }
    x = x / 2;
  }
  return count;
}
int main() {
  long long int n;
  cin >> n;
  long long int h;
  long long int a[31];
  for (long long int i = 0; i < 31; i++) {
    a[i] = 0;
  }
  for (long long int i = 0; i < n; i++) {
    cin >> h;
    long long int v = doit(h);
    a[v]++;
  }
  long long int sum = 0;
  for (long long int i = 0; i < 31; i++) {
    sum += a[i] * (a[i] - 1) / 2;
  }
  cout << sum << endl;
  return 0;
}
