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;
inline void Read(int &n) {
  char c;
  n = 0;
  do {
    c = getchar();
  } while (!isdigit(c));
  do {
    n = n * 10 + c - 48;
    c = getchar();
  } while (isdigit(c));
}
inline void Write(int n) {
  int a[10], cnt = 0;
  while (n) {
    a[cnt++] = n % 10;
    n /= 10;
  }
  while (cnt--) putchar(a[cnt] + 48);
  putchar('\n');
}
int main() {
  const int maxN = 1e5 + 10;
  int n, a[maxN];
  int64_t cnt[40] = {0};
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    cnt[__builtin_popcount(a[i])]++;
  }
  int64_t ans = 0;
  for (int i = 0; i < 40; ++i) ans += cnt[i] * (cnt[i] - 1) / 2;
  printf("%I64d", ans);
  return 0;
}
