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;
const int u = 1e3;
int adjm[u + 1][u + 1];
vector<int> adjl[(int)(1e6 + 1)];
vector<bool> vis((int)1e6 + 1, 0);
void init_adjm(int n) {
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) adjm[i][j] = 0;
  }
}
void bfs(int s, int n, int m) {
  queue<int> q;
  q.push(s);
  q.push(-1);
  vis[s] = 1;
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    if (p == -1) {
      if (!q.empty()) {
        q.push(-1);
      }
    } else {
      for (int i = 0; i < adjl[p].size(); i++) {
        int curr{adjl[p][i]};
        if (!vis[curr]) {
          q.push(curr);
          vis[curr] = 1;
        }
      }
    }
  }
}
void dfs(int s) {
  vis[s] = 1;
  for (int i{0}; i < adjl[s].size(); i++) {
    if (!vis[adjl[s][i]]) {
      dfs(adjl[s][i]);
    }
  }
}
vector<bool> is_prime(1e6 + 1, true);
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
void seivePrime(long long n) {
  for (long long p{2}; p * p <= n; p++) {
    if (is_prime[p]) {
      for (long long i{p * p}; i <= n; i += p) {
        is_prime[i] = false;
      }
    }
  }
  for (long long p{2}; p <= n; p++) {
    if (is_prime[p]) cout << p << " ";
  }
}
class my_compare {
 public:
  bool operator()(pair<int, int> &p1, pair<int, int> &p2) {
    return p1.second < p2.second;
  }
};
int count_bits(int n) {
  int count{0};
  while (n) {
    count += n & 1;
    n >>= 1;
  }
  return count;
}
void solve() {
  int n;
  long long ans{0};
  cin >> n;
  vector<int> a(n), b(30, 0);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[count_bits(a[i])]++;
  }
  for (int i = 0; i < 30; i++) {
    ans += (1LL * b[i] * (b[i] - 1)) / 2;
  }
  cout << ans;
}
int main() {
  bool test_case{0};
  int t;
  if (test_case) {
    cin >> t;
    while (t--) {
      solve();
    }
  } else
    solve();
}
