#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;
const int N = 1e6 + 5;
mt19937_64 rnd(time(0));
int n, K, flag[N], pr[N], cnt;
ull H[N], sum;
unordered_map<ull, int> pos;
int main() {
  scanf("%d", &n);
  K = n >> 1;
  flag[0] = flag[1] = 1, H[1] = 0;
  for (int i = 2; i <= n; i++) {
    if (!flag[i]) pr[++cnt] = i, H[i] = rnd();
    for (int j = 1; j <= cnt && i * pr[j] <= n; j++) {
      flag[i * pr[j]] = 1, H[i * pr[j]] = H[i] ^ H[pr[j]];
      if (i % pr[j] == 0) break;
    }
  }
  for (int i = 2; i <= n; i++) H[i] ^= H[i - 1], sum ^= H[i];
  if (sum == 0) {
    printf("%d\n", n);
    for (int i = 1; i <= n; i++) printf("%d ", i);
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (H[i] == sum) {
      printf("%d\n", n - 1);
      for (int j = 1; j <= n; j++)
        if (j != i) printf("%d ", j);
      return 0;
    }
  for (int i = 1; i <= n; i++) {
    if (pos.count(sum ^ H[i])) {
      int p = pos[sum ^ H[i]];
      printf("%d\n", n - 2);
      for (int j = 1; j <= n; j++)
        if (j != i && j != p) printf("%d ", j);
      return 0;
    }
    pos[H[i]] = i;
  }
  printf("%d\n", n - 3);
  for (int i = 1; i <= n; i++)
    if (i != 2 && i != K && i != n) printf("%d ", i);
  return 0;
}

### 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;
using ull = unsigned long long;
const int N = 1e6 + 5;
mt19937_64 rnd(time(0));
int n, K, flag[N], pr[N], cnt;
ull H[N], sum;
unordered_map<ull, int> pos;
int main() {
  scanf("%d", &n);
  K = n >> 1;
  flag[0] = flag[1] = 1, H[1] = 0;
  for (int i = 2; i <= n; i++) {
    if (!flag[i]) pr[++cnt] = i, H[i] = rnd();
    for (int j = 1; j <= cnt && i * pr[j] <= n; j++) {
      flag[i * pr[j]] = 1, H[i * pr[j]] = H[i] ^ H[pr[j]];
      if (i % pr[j] == 0) break;
    }
  }
  for (int i = 2; i <= n; i++) H[i] ^= H[i - 1], sum ^= H[i];
  if (sum == 0) {
    printf("%d\n", n);
    for (int i = 1; i <= n; i++) printf("%d ", i);
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (H[i] == sum) {
      printf("%d\n", n - 1);
      for (int j = 1; j <= n; j++)
        if (j != i) printf("%d ", j);
      return 0;
    }
  for (int i = 1; i <= n; i++) {
    if (pos.count(sum ^ H[i])) {
      int p = pos[sum ^ H[i]];
      printf("%d\n", n - 2);
      for (int j = 1; j <= n; j++)
        if (j != i && j != p) printf("%d ", j);
      return 0;
    }
    pos[H[i]] = i;
  }
  printf("%d\n", n - 3);
  for (int i = 1; i <= n; i++)
    if (i != 2 && i != K && i != n) printf("%d ", i);
  return 0;
}

```