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;
long long m, n;
char a[10];
int b[10];
int ha[100001];
int main() {
  while (scanf("%I64d", &m) != EOF) {
    memset(ha, 0, sizeof(ha));
    long long c = 0;
    for (int i = 0; i < m; i++) {
      scanf("%I64d", &n);
      int sum = 0;
      while (1) {
        if (n == 0)
          break;
        else if (n == 1) {
          sum++;
          break;
        }
        if (n % 2 == 0) {
          n = n / 2;
        } else {
          sum++;
          n = n / 2;
        }
      }
      if (ha[sum] != 0) {
        c += ha[sum];
        ha[sum]++;
      } else {
        ha[sum]++;
      }
    }
    printf("%I64d\n", c);
  }
  return 0;
}
