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 main() {
  int n;
  long long a[100005], f[100005];
  int num;
  while (~scanf("%d", &n)) {
    num = 0;
    memset(f, 0, sizeof(f));
    for (int i = 0; i < n; i++) {
      scanf("%I64d", &a[i]);
      long long temp = a[i];
      num = 0;
      while (1) {
        if (temp == 0 || temp == 1 || temp == 2) {
          num += 1;
          break;
        } else {
          if (temp % 2) {
            temp /= 2;
            num += 1;
          } else {
            temp /= 2;
          }
        }
      }
      f[i] = num;
    }
    num = 0;
    sort(f, f + n);
    int j = 0;
    int sum[100005];
    int k = 0;
    memset(sum, 0, sizeof(sum));
    for (int i = 1; i < n; i++) {
      if (f[i] == f[j])
        sum[k]++;
      else {
        j = i;
        k++;
      }
    }
    for (int i = 0; i <= k; i++) {
      for (int j = 0; j <= sum[i]; j++) {
        num += j;
      }
    }
    printf("%d\n", num);
  }
  return 0;
}
