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 int num[100005];
long long int bits[50];
int bitcnt(long int x) {
  int ans = 0;
  while (x != 0) {
    if (x % 2 != 0) ans++;
    x /= 2;
  }
  return ans;
}
int main() {
  int N, i, B, mx;
  long long int ans;
  while (scanf("%d", &N) != EOF) {
    for (i = 0; i < N; i++) scanf("%ld", &num[i]);
    for (i = 0; i < 50; i++) bits[i] = 0;
    mx = 0;
    for (i = 0; i < N; i++) {
      B = bitcnt(num[i]);
      bits[B]++;
      mx = (B > mx ? B : mx);
    }
    ans = 0;
    for (i = 0; i <= mx; i++) {
      ans += ((bits[i] * (bits[i] - 1)) / 2);
    }
    cout << ans << endl;
  }
  return 0;
}
