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;
map<long long, long> cj;
map<long long, long>::iterator it;
long long dp(long long val) {
  if (val == 0) return 0;
  if (val == 1) return 1;
  if (val == 2) return 1;
  if (val == 3) return 2;
  if (val == 4) return 1;
  if (val == 5) return 2;
  if (val % 2 == 0)
    return dp(val / 2);
  else
    return dp(val - 1) + 1;
}
long long zuhe(long val) { return val * (val - 1) / 2; }
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    cj.clear();
    long long tmp, tmp1;
    for (int i = 0; i != n; i++) {
      scanf("%lld", &tmp);
      tmp1 = dp(tmp);
      if (cj.find(tmp1) == cj.end()) {
        it = cj.begin();
        cj.insert(it, pair<long long, int>(tmp1, 1));
      } else {
        cj[tmp1]++;
      }
    }
    long long ans = 0;
    for (it = cj.begin(); it != cj.end(); it++) {
      long time = it->second;
      ans += zuhe(time);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
