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>
int main() {
  long long n, s1 = 0, a, s2 = 0, sum[100005];
  while (scanf("%I64d", &n) != EOF) {
    s1 = 0;
    s2 = 0;
    for (int i = 0; i < n; i++) sum[i] = 0;
    for (int i = 0; i < n; i++) {
      scanf("%I64d", &a);
      if (a % 2 && a != 1) {
        int x = 0;
        while (a) {
          x += a % 2;
          a = a / 2;
        }
        sum[x]++;
      } else
        s2++;
    }
    long long ans = 0;
    if (s2 >= 2) ans += (s2) * (s2 - 1) / 2;
    for (int i = 1; i <= n; i++)
      if (sum[i] >= 2) ans += (sum[i]) * (sum[i] - 1) / 2;
    printf("%I64d\n", ans);
  }
  return 0;
}
