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 double pi = 3.141592653689793238460;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
const int pr = 31;
int mod = 998244353;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long a[n];
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    long long cnt = 1;
    while (a[i] > 1) {
      if (a[i] % 2 == 0) {
        a[i] /= 2;
      } else {
        cnt++;
        a[i] /= 2;
      }
    }
    m[cnt]++;
  }
  long long res = 0;
  for (auto it = m.begin(); it != m.end(); it++) {
    res += ((it->second - 1) * it->second / 2);
  }
  cout << res << endl;
}
