description: Let's call a set of positive integers a_1, a_2, ..., a_k quadratic if the product of the factorials of its elements is a square of an integer, i. e. ∏_{i=1}^{k} a_i! = m^2, for some integer m.
  
  You are given a positive integer n.
  
  Your task is to find a quadratic subset of a set 1, 2, ..., n of maximum size. If there are multiple answers, print any of them.
  
  Input
  
  A single line contains a single integer n (1 ≤ n ≤ 10^6).
  
  Output
  
  In the first line, print a single integer — the size of the maximum subset. In the second line, print the subset itself in an arbitrary order.
  
  Examples
  
  Input
  
  
  1
  
  
  Output
  
  
  1
  1 
  
  
  Input
  
  
  4
  
  
  Output
  
  
  3
  1 3 4 
  
  
  Input
  
  
  7
  
  
  Output
  
  
  4
  1 4 5 6 
  
  
  Input
  
  
  9
  
  
  Output
  
  
  7
  1 2 4 5 6 7 9 
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 1000005;
  int sum() { return 0; }
  template <typename T, typename... Args>
  T sum(T a, Args... args) {
    return a + sum(args...);
  }
  template <typename... T>
  void read(T &...args) {
    ((cin >> args), ...);
  }
  template <typename... T>
  void write(string delimiter, T &&...args) {
    ((cout << args << delimiter), ...);
  }
  template <typename T>
  void readContainer(T &t) {
    for (auto &e : t) {
      read(e);
    }
  }
  template <typename T>
  void writeContainer(string delimiter, T &t) {
    for (const auto &e : t) {
      write(delimiter, e);
    }
    write("\n");
  }
  template <typename... T>
  void printer(T &&...args) {
    ((cout << args << " "), ...);
  }
  int n, t, m;
  vector<int> fac(N);
  vector<unsigned long long> p(N), f(N);
  map<unsigned long long, int> ch;
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    read(n);
    mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
    iota(fac.begin(), fac.end(), 0);
    for (int i = 2; i <= n; i++)
      if (fac[i] == i) {
        for (int j = i; j <= n; j += i) fac[j] = min(fac[j], i);
        p[i] = rnd();
      }
    for (int i = 2; i <= n; i++) {
      f[i] = f[i - 1];
      int x = i;
      while (x > 1) {
        f[i] = f[i] ^ (p[fac[x]]);
        x /= fac[x];
      }
      ch[f[i]] = i;
    }
    auto solve = [&](int n) -> vector<int> {
      unsigned long long ans = 0;
      for (int i = 2; i <= n; i++) ans ^= f[i];
      if (ans == 0) return {};
      if (ch.count(ans)) return {ch[ans]};
      for (int i = 2; i <= n; i++) {
        unsigned long long x = ans ^ f[i];
        if (ch.count(x)) return {ch[x], i};
      }
      return {2, n, n / 2};
    };
    auto ans = solve(n);
    cout << n - ans.size() << endl;
    for (int i = 1; i <= n; i++) {
      if (find(ans.begin(), ans.end(), i) == ans.end()) cout << i << ' ';
    }
    return 0;
  }
  