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 first[100009], N;
map<int, int> id;
int fun(int num) {
  int ans = 0;
  while (num) {
    if (num % 2 == 0)
      num = num / 2;
    else {
      num--;
      ans++;
      num = num / 2;
    }
  }
  return ans;
}
void Read() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &first[i]);
  }
}
bool cmp(int a, int b) { return a < b; }
long long sum(int p) { return p * (p + 1) / 2; }
void Solve() {
  long long ans = 0;
  for (int i = 1; i <= N; i++) {
    first[i] = fun(first[i]);
    id[first[i]]++;
  }
  for (map<int, int>::iterator it = id.begin(); it != id.end(); ++it) {
    ans += sum(it->second - 1);
  }
  printf("%I64d\n", ans);
}
int main() {
  Read();
  Solve();
}
