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;
const int INF = INT_MAX;
const long long INFL = LLONG_MAX;
const int output_precision = 15;
const bool debug = true;
stringstream ss;
int N, cnt[200];
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(output_precision);
  cout << fixed;
  ss.precision(output_precision);
  ss << fixed;
  cin >> N;
  for (int(i) = 1, j1234 = N; (i) <= j1234; (i)++) {
    int foo;
    cin >> foo;
    int acc = 0;
    while (foo > 0) acc += foo & 1, foo >>= 1;
    cnt[acc]++;
  }
  long long ans = 0;
  for (int(i) = 1, j1234 = 100; (i) <= j1234; (i)++)
    if (cnt[i] > 0) ans += 1LL * cnt[i] * (cnt[i] - 1) / 2;
  cout << ans << '\n';
}
