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>
int abs(int a) {
  if (a < 0) return -1 * a;
  return a;
}
int big(int a, int b) {
  if (a > b) return a;
  return b;
}
int small(int a, int b) {
  if (a < b) return a;
  return b;
}
int compare(const void *a, const void *b) {
  long long int *p1 = (long long *)a;
  long long int *p2 = (long long *)b;
  return *p1 - *p2;
}
long long f(long long n) {
  if (n == 0) return 0;
  return f(n / 2) + (n & 1);
}
long long nc2(long long a) { return (a * (a - 1)) >> 1; }
long long a[100004] = {};
int main() {
  long long ans = 0, x;
  int n, i, j, k, y, t;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &x);
    a[i] = f(x);
  }
  qsort(a, n, sizeof(long long), compare);
  long long count = 0;
  for (i = 0; i < n; i++) {
    count++;
    if (i != n - 1 && a[i] == a[i + 1]) continue;
    ans += nc2(count);
    count = 0;
  }
  printf("%lld", ans);
  return 0;
}
