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 v[50001];
int N;
int f(int v) {
  if (v <= 1) return v == 0 ? 0 : 1;
  return f(v / 2) + (v % 2);
}
void solve() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    int t;
    scanf("%d", &t);
    v[i] = f(t);
  }
  long long ans(0);
  int cnt = 1;
  for (int i = 0; i < N - 1; i++) {
    if (v[i] == v[i + 1]) {
      cnt++;
    } else {
      if (cnt > 1) {
        ans += (long long)cnt * (cnt - 1) / 2;
      }
      cnt = 1;
    }
  }
  if (cnt > 1) {
    ans += (long long)cnt * (cnt - 1) / 2;
  }
  printf("%I64d\n", ans);
}
int main() {
  solve();
  return 0;
}
