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 <cstdio>
#include <algorithm>
#include <map>

using namespace std;

int a[100002];
int b[100002];
int n;
map <int,int> f;
map <int,int> p;

int main()
{
  __int64 ans;
  int i,x;
  f.clear();
  p.clear();
  scanf("%d",&n);
  for (i=1;i<=n;i++)
  {
    scanf("%d",&a[i]);
    b[i]=a[i];
  }
  sort(a+1,a+1+n);
  f[0]=0;
  for (i=1;i<=n;i++)
    if (a[i]&1) f[a[i]]=f[a[i]>>1]+1;
      else f[a[i]]=f[a[i]>>1];
  ans^=ans;
  for (i=1;i<=n;i++)
  {
    x=f[b[i]];
    ans=ans+p[x];
    p[x]=p[x]+1;
  }
  printf("%I64d\n",ans);
  return 0;
}