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;
using namespace std;
void chekarre(int* arr, int n) {
  cout << "[";
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "]" << endl;
}
int n;
long long m;
int a0 = 0, a1 = 0;
map<int, int> mp, cnt;
set<int> pool;
set<int>::iterator it;
int fun(int x) {
  if (x == 0) return 0;
  if (mp[x] != 0) return mp[x];
  if (x % 2 == 0) return mp[x] = fun(x / 2);
  return mp[x] = fun((x - 1) / 2) + 1;
}
int solve() {
  unsigned long long sum = 0, xx = 0, tot = 0;
  for (it = pool.begin(); it != pool.end(); it++) {
    if (cnt[*it] > 1) {
      if (cnt[*it] % 2 == 0) {
        sum = cnt[*it] / 2;
        sum *= (cnt[*it] - 1);
      } else {
        sum = (cnt[*it] - 1) / 2;
        sum *= cnt[*it];
      }
      tot += sum;
    }
  }
  cout << tot;
  return 1;
}
bool input() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    cin >> m;
    m = fun(m);
    cnt[m]++;
    pool.insert(m);
  }
  return true;
}
int main() {
  int T = 1;
  for (int testnum = 1; testnum <= T; testnum++) {
    if (!input()) break;
    solve();
  }
  return 0;
}
