#include <bits/stdc++.h>
using namespace std;
long long i, j, k, n, m, t, p[1005000], f[1005000], g[1005000], x, y,
    no[1005000], res;
map<long long, long long> mp;
void show() {
  cout << res << '\n';
  for (i = 1; i <= n; i++)
    if (!no[i]) cout << i << ' ';
  exit(0);
}
int main() {
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count() + 1);
  uniform_int_distribution<long long> lim(1, 1e18);
  ios::sync_with_stdio(0);
  for (i = 1; i <= 1000000; i++) f[i] = lim(rng);
  for (i = 2; i <= 1000000; i++)
    if (!p[i])
      for (j = i; j <= 1000000; j += i) p[j] = i;
  cin >> n;
  if (p[n] == n) n--;
  res = n;
  for (i = 1; i <= n; i++) {
    for (j = i; j - 1; j /= p[j]) {
      x ^= f[p[j]];
      if ((n + 1 - i) & 1) y ^= f[p[j]];
    }
    mp[x] = i;
    g[i] = x;
  }
  if (!y) show();
  if (mp.count(y)) {
    res--;
    no[mp[y]] = 1;
    show();
  }
  for (i = 1; i <= n; i++) {
    if (mp.count(g[i] ^ y)) {
      res -= 2;
      no[mp[g[i] ^ y]] = no[i] = 1;
      show();
    }
  };
  y ^= g[n];
  n--;
  res = n;
  for (i = 1; i <= n; i++) {
    if (mp.count(g[i] ^ y)) {
      res -= 2;
      no[mp[g[i] ^ y]] = no[i] = 1;
      show();
    }
  };
}

### Prompt

Your task is to create 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 namespace std;
long long i, j, k, n, m, t, p[1005000], f[1005000], g[1005000], x, y,
    no[1005000], res;
map<long long, long long> mp;
void show() {
  cout << res << '\n';
  for (i = 1; i <= n; i++)
    if (!no[i]) cout << i << ' ';
  exit(0);
}
int main() {
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count() + 1);
  uniform_int_distribution<long long> lim(1, 1e18);
  ios::sync_with_stdio(0);
  for (i = 1; i <= 1000000; i++) f[i] = lim(rng);
  for (i = 2; i <= 1000000; i++)
    if (!p[i])
      for (j = i; j <= 1000000; j += i) p[j] = i;
  cin >> n;
  if (p[n] == n) n--;
  res = n;
  for (i = 1; i <= n; i++) {
    for (j = i; j - 1; j /= p[j]) {
      x ^= f[p[j]];
      if ((n + 1 - i) & 1) y ^= f[p[j]];
    }
    mp[x] = i;
    g[i] = x;
  }
  if (!y) show();
  if (mp.count(y)) {
    res--;
    no[mp[y]] = 1;
    show();
  }
  for (i = 1; i <= n; i++) {
    if (mp.count(g[i] ^ y)) {
      res -= 2;
      no[mp[g[i] ^ y]] = no[i] = 1;
      show();
    }
  };
  y ^= g[n];
  n--;
  res = n;
  for (i = 1; i <= n; i++) {
    if (mp.count(g[i] ^ y)) {
      res -= 2;
      no[mp[g[i] ^ y]] = no[i] = 1;
      show();
    }
  };
}

```