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 f[100005];
int n;
int a[100005];
int fac(int x) {
  if (x % 2) return f[x / 2] + 1;
  return f[x / 2];
}
int main() {
  f[0] = 0;
  for (int i = 1; i <= 100000; i++) f[i] = fac(i);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, tmp = 0;
    scanf("%d", &x);
    while (x > 100000) {
      if (x % 2) tmp++;
      x = x >> 1;
    }
    a[i] = tmp + f[x];
  }
  sort(a + 1, a + 1 + n);
  int j = 0;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    while (a[i] == a[i + 1]) i++;
    long long tmp = i - j;
    ans += (1 + tmp - 1) * (tmp - 1) / 2;
    j = i;
  }
  printf("%I64d\n", ans);
  return 0;
}
