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() {
  long long int i, j, k = 1, l, m, n, sum = 0, c = 0, t, q = 2, d, sum1 = 0,
                      flag = 0;
  vector<int> v1, v2(100, 0), v3;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> m;
    v1.push_back(m);
  }
  for (i = 0; i < n; i++) {
    m = v1[i];
    if (m < 2) {
      if (m == 0)
        v2[0]++;
      else
        v2[1]++;
    } else {
      while (m > 1) {
        c++;
        k = k * 2;
        if (k == v1[i]) {
          flag = 1;
          v2[1]++;
          break;
        }
        m = m / 2;
        v3.push_back(k);
      }
      if (flag != 1) {
        sum1 = sum1 + v3[c - 1];
        t = k / 2;
        j = v2[i] - t;
        d = c - 1;
        if (sum1 > j) {
          v2[q]++;
        } else {
          q = 1;
          while (sum < j) {
            sum1 = sum1 + v3[d];
            q++;
            d--;
          }
          v2[q]++;
          q = 2;
        }
      }
      c = 0;
      q = 2;
      flag = 0;
      sum1 = 0;
      v3.clear();
    }
    k = 1;
    c = 0;
    q = 2;
    flag = 0;
    v3.clear();
  }
  for (i = 0; i < v2.size(); i++) {
    if (v2[i] != 0) {
      j = (v2[i] * (v2[i] - 1)) / 2;
      sum = sum + j;
    }
  }
  cout << sum << endl;
  return 0;
}
