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;
int n;
int f[100005];
inline int function(int x) {
  int count = 1;
  while (x) {
    if (x % 2 == 1) ++count;
    x >>= 1;
  }
  return count;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    f[i] = x;
  }
  sort(f + 1, f + n + 1);
  int count = 1;
  long long total = 0;
  f[n + 1] = -1;
  for (int i = 1; i <= n; ++i) {
    if (f[i] == f[i + 1])
      ++count;
    else {
      long long add = count * (count - 1);
      add >>= 1;
      total += (long long)add;
      count = 1;
    }
  }
  printf("%I64d\n", total);
}
