#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 2;
int lpf[N];
long long val[N];
long long val1[N];
long long val2[N];
int use[N];
int n;
void finish() {
  cout << accumulate(use + 1, use + n + 1, 0) << '\n';
  for (int i = 1; i <= n; i++) {
    if (use[i]) cout << i << ' ';
  }
  cout << '\n';
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  mt19937_64 rng(
      (long long)chrono::steady_clock::now().time_since_epoch().count());
  for (int i = 1; i < N; i++) {
    lpf[i] = i;
  }
  for (int i = 2; i < N; i++) {
    if (lpf[i] == i) {
      val[i] = rng();
      for (int j = 2 * i; j < N; j += i) {
        lpf[j] = i;
      }
    } else {
      val[i] = val[lpf[i]] ^ val[i / lpf[i]];
    }
  }
  for (int i = 1; i < N; i++) {
    val1[i] = val1[i - 1] ^ val[i];
    val2[i] = val2[i - 1] ^ val1[i];
  }
  cin >> n;
  for (int i = 1; i <= n; i++) use[i] = 1;
  long long x = val2[n];
  if (!x) finish();
  for (int i = 1; i <= n; i++) {
    if (val1[i] == x) {
      use[i] = 0;
      finish();
    }
  }
  unordered_map<long long, int> m;
  for (int i = 1; i <= n; i++) {
    if (m.count(val1[i] ^ x)) {
      use[i] = 0;
      use[m[val1[i] ^ x]] = 0;
      finish();
    }
    m[val1[i]] = i;
  }
  assert(n % 4 == 3);
  use[n] = 0;
  use[n / 2] = 0;
  use[2] = 0;
  finish();
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 namespace std;
const int N = 1e6 + 2;
int lpf[N];
long long val[N];
long long val1[N];
long long val2[N];
int use[N];
int n;
void finish() {
  cout << accumulate(use + 1, use + n + 1, 0) << '\n';
  for (int i = 1; i <= n; i++) {
    if (use[i]) cout << i << ' ';
  }
  cout << '\n';
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  mt19937_64 rng(
      (long long)chrono::steady_clock::now().time_since_epoch().count());
  for (int i = 1; i < N; i++) {
    lpf[i] = i;
  }
  for (int i = 2; i < N; i++) {
    if (lpf[i] == i) {
      val[i] = rng();
      for (int j = 2 * i; j < N; j += i) {
        lpf[j] = i;
      }
    } else {
      val[i] = val[lpf[i]] ^ val[i / lpf[i]];
    }
  }
  for (int i = 1; i < N; i++) {
    val1[i] = val1[i - 1] ^ val[i];
    val2[i] = val2[i - 1] ^ val1[i];
  }
  cin >> n;
  for (int i = 1; i <= n; i++) use[i] = 1;
  long long x = val2[n];
  if (!x) finish();
  for (int i = 1; i <= n; i++) {
    if (val1[i] == x) {
      use[i] = 0;
      finish();
    }
  }
  unordered_map<long long, int> m;
  for (int i = 1; i <= n; i++) {
    if (m.count(val1[i] ^ x)) {
      use[i] = 0;
      use[m[val1[i] ^ x]] = 0;
      finish();
    }
    m[val1[i]] = i;
  }
  assert(n % 4 == 3);
  use[n] = 0;
  use[n / 2] = 0;
  use[2] = 0;
  finish();
}

```