#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  scanf("%d", &n);
  if (n == 1) {
    printf("1\n1\n");
    return 0;
  }
  if (n & 1) {
    long long prod = 1ll * n * ((n >> 1) - 1);
    long long s = sqrt(prod);
    if (s * s == prod) {
      printf("%d\n", n - 2);
      int s1 = (n >> 1) - 2, s2 = n - 2;
      for (int i = 1; i <= n; i++)
        if (i != s1 && i != s2) printf("%d ", i);
      putchar('\n');
      return 0;
    }
  }
  if (n & 1) n--;
  int m = n >> 1;
  if (m & 1) {
    int x = 0, k = (m + 1) >> 1, s = sqrt(k);
    if (s * s == k)
      x = m + 1;
    else if (m == 9)
      x = 7;
    if (x) {
      printf("%d\n", n - 1);
      for (int i = 1; i <= n; i++)
        if (i != x) printf("%d ", i);
      putchar('\n');
    } else {
      printf("%d\n", n - 2);
      for (int i = 1; i <= n; i++)
        if (i != 2 && i != m) printf("%d ", i);
      putchar('\n');
    }
  } else {
    printf("%d\n", n - 1);
    for (int i = 1; i <= n; i++)
      if (i != m) printf("%d ", i);
    putchar('\n');
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int main() {
  int n;
  scanf("%d", &n);
  if (n == 1) {
    printf("1\n1\n");
    return 0;
  }
  if (n & 1) {
    long long prod = 1ll * n * ((n >> 1) - 1);
    long long s = sqrt(prod);
    if (s * s == prod) {
      printf("%d\n", n - 2);
      int s1 = (n >> 1) - 2, s2 = n - 2;
      for (int i = 1; i <= n; i++)
        if (i != s1 && i != s2) printf("%d ", i);
      putchar('\n');
      return 0;
    }
  }
  if (n & 1) n--;
  int m = n >> 1;
  if (m & 1) {
    int x = 0, k = (m + 1) >> 1, s = sqrt(k);
    if (s * s == k)
      x = m + 1;
    else if (m == 9)
      x = 7;
    if (x) {
      printf("%d\n", n - 1);
      for (int i = 1; i <= n; i++)
        if (i != x) printf("%d ", i);
      putchar('\n');
    } else {
      printf("%d\n", n - 2);
      for (int i = 1; i <= n; i++)
        if (i != 2 && i != m) printf("%d ", i);
      putchar('\n');
    }
  } else {
    printf("%d\n", n - 1);
    for (int i = 1; i <= n; i++)
      if (i != m) printf("%d ", i);
    putchar('\n');
  }
  return 0;
}

```