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, b, c;
using namespace std;
int fu(int r) {
  int t = 0;
  while (r) {
    if (r & 1) t++, r--;
    r = r >> 1;
  }
  return t;
}
int main() {
  pair<int, int> ch[100000];
  cin >> a;
  for (int i = 0; i < a; ++i) {
    cin >> ch[i].second;
    ch[i].first = fu(ch[i].second);
  }
  sort(ch, ch + a);
  int p = ch[0].first, e = 1;
  long long ot = 0;
  for (int i = 1; i < a; ++i) {
    if (p == ch[i].first)
      e++;
    else {
      p = ch[i].first;
      if (e == 1) continue;
      ot += (e * (e - 1)) >> 1;
      e = 1;
    }
  }
  if (e != 1) {
    ot += (e * (e - 1)) >> 1;
    e = 1;
  }
  cout << ot;
}
