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 x, n, v, a[1000000], j;
long long answ, answ1;
long long f(int k) {
  long long dd = 1, dd1 = 1;
  for (int i = 1; i <= k; i++) dd1 *= i;
  for (int i = 1; i <= k - 2; i++) dd *= i;
  if (dd == 0) dd = 1;
  if (dd1 == 0) dd1 = 1;
  return dd1 / (2 * dd);
}
int main() {
  cin >> n;
  j = 0;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x % 2 == 0 || x == 1)
      v++;
    else {
      a[j] = x;
      j++;
    }
  }
  if (v > 1) answ1 += f(v);
  n = j;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j &&
          ((((a[i] - 1) / 2) == a[j] - 1) || (((a[j] - 1) / 2) == a[i] - 1)))
        answ++;
  answ /= 2;
  cout << answ + answ1;
  return 0;
}
