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;
void f(int b[]) {
  int i;
  for (i = 0; i < 100000; i++) {
    if (i == 0 || i == 1)
      b[i] = i;
    else {
      if (i % 2 == 0)
        b[i] = b[i / 2];
      else
        b[i] = b[i / 2] + 1;
    }
  }
}
int main() {
  int n, i, j, z, y, ci;
  long long count = 0;
  cin >> n;
  int a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  int b[100000], c[100000] = {0};
  f(b);
  for (i = 0; i < n; i++) {
    z = a[i];
    ci = 0;
    while (z >= 100000) {
      if (z % 2 == 0)
        z = z / 2;
      else
        z = z / 2, ci++;
    }
    y = b[z] + ci;
    c[y]++;
  }
  for (i = 0; i < 100000; i++)
    if (c[i] > 0) count += (c[i] * (c[i] - 1) / 2);
  cout << count;
  return 0;
}
