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 n, ans = 0, len = 0;
long long a[100010], cnt[10000000];
long long lalala(long long tt) {
  long long ok = 0;
  while (tt) {
    if (tt % 2 == 0)
      tt /= 2;
    else {
      tt /= 2;
      ok++;
    }
  }
  return ok;
}
long long hahaha(long long tt, long long s) {
  long long ok = 1;
  for (long long i = 1; i <= s; i++) {
    ok *= tt;
    tt--;
  }
  for (long long i = s; i >= 1; i--) {
    ok /= i;
  }
  return ok;
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i <= n; i++) {
    a[i] = lalala(a[i]);
    cnt[a[i]]++;
    len = max(len, a[i]);
  }
  for (long long i = 1; i <= len; i++) {
    if (cnt[i] >= 2) {
      ans += hahaha(cnt[i], 2);
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
