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;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1e9 + 7;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
const double eps = 1e-7;
const double pi = acos(-1.0);
inline void inc(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline void dec(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
inline void chmax(long long &x, long long y) {
  if (y > x) x = y;
}
inline void mulm(long long &x, long long y) {
  x *= y;
  if (x >= mod) x -= mod;
}
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
long long a[100005];
map<long long, long long> help;
int main() {
  long long ans = 0, i, j, k, l, m, n;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = n; i >= 1; i--) {
    long long x = cbit(a[i]);
    ans += help[x];
    help[x]++;
  }
  cout << ans;
}
