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>
void swap(long long *a, long long *b) {
  long long c = *a;
  *a = *b;
  *b = c;
}
void divide(long long *arr, int l, int r) {
  if (r - l < 1) return;
  int i = l - 1;
  int j = r;
  int razd = (rand() % (r - l)) + l;
  swap(&arr[r], &arr[razd]);
  while (j > i) {
    do i++;
    while (arr[i] < arr[r]);
    do j--;
    while (j != i && arr[j] > arr[r]);
    if (j > i) swap(&arr[i], &arr[j]);
  }
  swap(&arr[i], &arr[r]);
  divide(arr, i + 1, r);
  divide(arr, l, i - 1);
}
void quickSort(long long *arr, int size) {
  srand(500000);
  divide(arr, 0, size - 1);
}
int max1(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
int min1(int a, int b) {
  if (a < b)
    return a;
  else
    return b;
}
long long f(long long x) {
  if (x == 0) return 0;
  if (x % 2 == 0)
    return f(x / 2);
  else
    return f((x - 1) / 2) + 1;
}
int main() {
  static long long n, arr[100005];
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &arr[i]);
    arr[i] = f(arr[i]);
  }
  long long ans = 0;
  long long kol = 1;
  quickSort(arr, n);
  for (int i = 1; i < n; i++)
    if (arr[i] == arr[i - 1]) {
      kol++;
    } else {
      ans += (kol) * (kol - 1) / 2;
      kol = 1;
    }
  if (kol != 1) {
    ans += (kol) * (kol - 1) / 2;
    kol = 1;
  }
  printf("%I64d", ans);
  return 0;
}
