#include <bits/stdc++.h>
using namespace std;
int n;
int vis[1000010], prime[1000010], tot;
unsigned long long val[1000010];
map<unsigned long long, int> all;
void GetPrime(int N) {
  for (int i = 2; i <= N; i++) {
    if (!vis[i])
      prime[tot++] = i, val[i] = ((long long)(rand() << 15 | rand()) << 30 |
                                  (rand() << 15 | rand()));
    for (int j = 0; j < tot && i * prime[j] <= N; j++) {
      vis[i * prime[j]] = 1, val[i * prime[j]] = val[i] ^ val[prime[j]];
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 1; i <= N; i++) {
    val[i] ^= val[i - 1];
  }
}
int main() {
  scanf("%d", &n);
  GetPrime(n);
  unsigned long long S = 0;
  for (int i = 1; i <= n; i++) {
    S ^= val[i];
  }
  if (S == 0) {
    printf("%d\n", n);
    for (int i = 1; i <= n; i++) {
      printf("%d ", i);
    }
    printf("\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (val[i] == S) {
      printf("%d\n", n - 1);
      for (int j = 1; j <= n; j++) {
        if (i == j) continue;
        printf("%d ", j);
      }
      printf("\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (all.count(val[i] ^ S)) {
      int x = all[val[i] ^ S], y = i;
      printf("%d\n", n - 2);
      for (int j = 1; j <= n; j++) {
        if (j == x || j == y) continue;
        printf("%d ", j);
      }
      printf("\n");
      return 0;
    }
    all[val[i]] = i;
  }
  if (n == 1) {
    printf("1\n1\n");
    return 0;
  }
  if (n & 1) n--;
  if (n % 2 == 0) {
    int k = n / 2;
    if (k % 2 == 0) {
      printf("%d\n", n - 1);
      for (int i = 1; i <= n; i++) {
        if (i == k) continue;
        printf("%d ", i);
      }
      printf("\n");
    } else {
      printf("%d\n", n - 2);
      for (int i = 1; i <= n; i++) {
        if (i == 2 || i == k) continue;
        printf("%d ", i);
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 n;
int vis[1000010], prime[1000010], tot;
unsigned long long val[1000010];
map<unsigned long long, int> all;
void GetPrime(int N) {
  for (int i = 2; i <= N; i++) {
    if (!vis[i])
      prime[tot++] = i, val[i] = ((long long)(rand() << 15 | rand()) << 30 |
                                  (rand() << 15 | rand()));
    for (int j = 0; j < tot && i * prime[j] <= N; j++) {
      vis[i * prime[j]] = 1, val[i * prime[j]] = val[i] ^ val[prime[j]];
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 1; i <= N; i++) {
    val[i] ^= val[i - 1];
  }
}
int main() {
  scanf("%d", &n);
  GetPrime(n);
  unsigned long long S = 0;
  for (int i = 1; i <= n; i++) {
    S ^= val[i];
  }
  if (S == 0) {
    printf("%d\n", n);
    for (int i = 1; i <= n; i++) {
      printf("%d ", i);
    }
    printf("\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (val[i] == S) {
      printf("%d\n", n - 1);
      for (int j = 1; j <= n; j++) {
        if (i == j) continue;
        printf("%d ", j);
      }
      printf("\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (all.count(val[i] ^ S)) {
      int x = all[val[i] ^ S], y = i;
      printf("%d\n", n - 2);
      for (int j = 1; j <= n; j++) {
        if (j == x || j == y) continue;
        printf("%d ", j);
      }
      printf("\n");
      return 0;
    }
    all[val[i]] = i;
  }
  if (n == 1) {
    printf("1\n1\n");
    return 0;
  }
  if (n & 1) n--;
  if (n % 2 == 0) {
    int k = n / 2;
    if (k % 2 == 0) {
      printf("%d\n", n - 1);
      for (int i = 1; i <= n; i++) {
        if (i == k) continue;
        printf("%d ", i);
      }
      printf("\n");
    } else {
      printf("%d\n", n - 2);
      for (int i = 1; i <= n; i++) {
        if (i == 2 || i == k) continue;
        printf("%d ", i);
      }
      printf("\n");
    }
  }
  return 0;
}

```