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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
string to_str(T str) {
  stringstream stream;
  stream << str;
  return stream.str();
}
template <typename T>
int to_int(T num) {
  int val;
  stringstream stream;
  stream << num;
  stream >> val;
  return val;
}
vector<string> split(const string& s, char delim) {
  vector<string> elems;
  stringstream ss(s);
  string item;
  while (getline(ss, item, delim)) elems.push_back(item);
  return elems;
}
const int dr[] = {-1, -1, 0, 1, 1, 1, 0, -1};
const int dc[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int T2 = 105;
const int T3 = 1005;
const int T4 = 10005;
const int T5 = 100005;
int cnt[35];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    x = __builtin_popcount(x);
    cnt[x]++;
  }
  long long int ans = 0;
  for (int i = 1; i <= 32; i++) {
    ans += cnt[i] * (cnt[i] - 1) / 2;
  }
  cout << ans << "\n";
  return 0;
}
