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;
bool sbs(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
long long int xxxx(long long int a, long long int b) {
  while (a > b) {
    a = a / 3;
  }
  return a;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v;
  map<long long int, vector<long long int>> m;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    v.push_back(x);
  }
  for (long long int i = 0; i < n; i++) {
    long long int x2 = v[i];
    long long int temp = 0;
    while (x2 > 0) {
      if (x2 % 2 == 0) {
        x2 = x2 / 2;
      } else {
        x2--;
        temp++;
      }
    }
    m[temp].push_back(i + 1);
  }
  long long int ans = 0;
  for (auto x : m) {
    long long int y = m[x.first].size();
    ans = ans + y * (y - 1) / 2;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
