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 Nmax = 1e5 + 100;
template <class T>
int getbit(T x, int pos) {
  return (x >> (pos - 1)) & 1;
}
template <class T>
void turn_on(T &x, int pos) {
  x = x | ((T)1 << (pos - 1));
}
template <class T>
void turn_off(T &x, int pos) {
  x = x & ~((T)1 << (pos - 1));
}
template <class T>
T sqr(T a) {
  return a * a;
}
int a[Nmax], hist[Nmax];
long long ans, n;
int calc(int x) {
  if (x == 0) return 0;
  if (x % 2 == 1) return calc(x / 2) + 1;
  if (x % 2 == 0) return calc(x / 2);
}
int main() {
  scanf("%d", &n);
  for (int i = int(1); i <= int(n); ++i) {
    scanf("%d", &a[i]);
    ++hist[calc(a[i])];
  }
  for (int i = 0; i < int(Nmax); ++i)
    ans += (long long)hist[i] * (hist[i] - 1) / 2;
  cout << ans;
}
