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 long int n, m, i, j, k, l, a[1000009], b[1000009], p[1000009], ans, cn, t,
    x, y, z, e, o;
char c[1000009], d[1000009], ch;
void pa() {
  memset(p, 0, sizeof(p));
  for (i = 2; i < 1000000; i++) {
    p[i] = p[i - 1] + i - 1;
  }
}
int main() {
  pa();
  while (scanf("%I64d", &n) != EOF) {
    ans = 0;
    memset(b, 0, sizeof(b));
    for (i = 0; i < n; i++) {
      scanf("%I64d", &x);
      cn = 0;
      while (x != 0) {
        if (x % 2 == 0) {
          x /= 2;
        } else {
          cn++;
          x = (x - 1) / 2;
        }
      }
      b[cn]++;
    }
    for (i = 0; i < 100; i++) {
      ans = ans + p[b[i]];
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
