#include <bits/stdc++.h>
using ll = long long;
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> prime(n + 1, 0);
  for (int i = 2; i <= n; i++) {
    if (!prime[i]) {
      for (int j = i; j <= n; j += i) prime[j] = i;
    }
  }
  mt19937_64 rng(233);
  vector<long long> h(n + 1);
  for (int i = 2; i <= n; i++)
    if (prime[i] == i) h[i] = rng();
  long long P = 0;
  vector<long long> p(n + 1);
  for (int i = 1; i <= n; i++) {
    p[i] = p[i - 1];
    int x = i;
    while (x > 1) {
      p[i] ^= h[prime[x]];
      x /= prime[x];
    }
    P ^= p[i];
  }
  auto print = [&](vector<int> erased) {
    cout << n - erased.size() << '\n';
    for (int i = 1; i <= n; i++)
      if (!count(erased.begin(), erased.end(), i)) cout << i << ' ';
    cout << '\n';
    exit(0);
  };
  if (P == 0) print({});
  for (int i = 1; i <= n; i++)
    if (p[i] == P) print({i});
  map<long long, int> mp;
  for (int i = 1; i <= n; i++) {
    if (mp.count(P ^ p[i])) print({i, mp[P ^ p[i]]});
    mp[p[i]] = i;
  }
  assert(n % 2 == 1);
  print({2, (n - 1) / 2, n});
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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 

### Response

```cpp
#include <bits/stdc++.h>
using ll = long long;
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> prime(n + 1, 0);
  for (int i = 2; i <= n; i++) {
    if (!prime[i]) {
      for (int j = i; j <= n; j += i) prime[j] = i;
    }
  }
  mt19937_64 rng(233);
  vector<long long> h(n + 1);
  for (int i = 2; i <= n; i++)
    if (prime[i] == i) h[i] = rng();
  long long P = 0;
  vector<long long> p(n + 1);
  for (int i = 1; i <= n; i++) {
    p[i] = p[i - 1];
    int x = i;
    while (x > 1) {
      p[i] ^= h[prime[x]];
      x /= prime[x];
    }
    P ^= p[i];
  }
  auto print = [&](vector<int> erased) {
    cout << n - erased.size() << '\n';
    for (int i = 1; i <= n; i++)
      if (!count(erased.begin(), erased.end(), i)) cout << i << ' ';
    cout << '\n';
    exit(0);
  };
  if (P == 0) print({});
  for (int i = 1; i <= n; i++)
    if (p[i] == P) print({i});
  map<long long, int> mp;
  for (int i = 1; i <= n; i++) {
    if (mp.count(P ^ p[i])) print({i, mp[P ^ p[i]]});
    mp[p[i]] = i;
  }
  assert(n % 2 == 1);
  print({2, (n - 1) / 2, n});
  return 0;
}

```