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 main() {
  int n, seq[100000], result[100000], cnt[100000];
  int i, j, max_num = (1 << 31), ans = 0;
  for (i = 0; i < 100000; i++) cnt[i] = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &seq[i]);
  int add = 0, reg;
  for (i = 0; i < n; i++) {
    reg = seq[i];
    while (reg > 1) {
      if (reg % 2 == 0)
        reg /= 2;
      else if (reg % 2 == 1) {
        reg = (reg - 1) / 2;
        add++;
      }
    }
    if (reg == 0)
      result[i] = add;
    else if (reg == 1)
      result[i] = add + 1;
    cnt[result[i]]++;
    if (max_num < result[i]) max_num = result[i];
  }
  for (i = 1; i <= max_num; i++) {
    if (cnt[i] > 1) ans += (cnt[i]) * (cnt[i] - 1) / 2;
  }
  printf("%d\n", ans);
  return 0;
}
