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 = acos(-1.0);
const double eps = 1e-11;
const int INFINITE = 0x3f3f3f3f;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline T lowbit(T n) {
  return (n ^ (n - 1)) & n;
}
template <class T>
inline int countbit(T n) {
  return (n == 0) ? 0 : (1 + countbit(n & (n - 1)));
}
typedef vector<int> VI;
typedef vector<VI> VII;
typedef vector<string> VS;
#pragma comment(linker, "/STACK:200000000")
const int MAXN = 1e5 + 1;
long long a[MAXN];
bool ispow2(long long x) { return (x & (x - 1)) == 0; }
void solve(void) {
  int n;
  cin >> n;
  long long pow2 = 0ll, rest = 0ll;
  for (int i = (int)(0); i < (int)(n); i++) {
    cin >> a[i];
    if (ispow2(a[i]))
      ++pow2;
    else
      ++rest;
  }
  long long total = pow2 * (pow2 - 1) / 2 + rest * (rest - 1) / 2;
  cout << total << endl;
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}
