#include <bits/stdc++.h>
using namespace std;
void FAST_IO(string filein = "", string fileout = "", string fileerr = "") {
  if (fopen(filein.c_str(), "r")) {
    freopen(filein.c_str(), "r", stdin);
    freopen(fileout.c_str(), "w", stdout);
  }
  cin.tie(0), cout.tie(0)->sync_with_stdio(0);
}
void Hollwo_Pelw();
signed main() {
  FAST_IO("hollwo_pelw.inp", "hollwo_pelw.out");
  int testcases = 1;
  for (int test = 1; test <= testcases; test++) {
    Hollwo_Pelw();
  }
  return 0;
}
const int N = 1e6 + 5;
mt19937_64 rng(
    (uint32_t)chrono::steady_clock::now().time_since_epoch().count());
int n, p[N];
uint64_t val[N], hsh[N], tot;
inline void print(int n, vector<int> del) {
  cout << n - del.size() << '\n';
  for (int i = 1; i <= n; i++) {
    bool ok = 1;
    for (auto v : del)
      if (v == i) ok = 0;
    if (ok) cout << i << ' ';
  }
}
void Hollwo_Pelw() {
  cin >> n;
  iota(p + 1, p + n + 1, 1);
  for (int i = 2; i <= n; i++)
    if (p[i] == i)
      for (int j = 2 * i; j <= n; j += i) p[j] = i;
  for (int i = 2; i <= n; i++)
    if (p[i] == i) val[i] = rng();
  for (int i = 2; i <= n; i++) {
    hsh[i] = hsh[i / p[i]] ^ val[p[i]];
  }
  for (int i = 2; i <= n; i++) {
    hsh[i] ^= hsh[i - 1];
    tot ^= hsh[i];
  }
  if (tot == 0) return print(n, {});
  for (int i = 1; i <= n; i++)
    if (tot == hsh[i]) return print(n, {i});
  if (n % 2 == 0) return print(n, {2, n / 2});
  map<uint64_t, int> mp;
  for (int i = 1; i <= n; i++) {
    if (mp.count(hsh[i])) return print(n, {i, mp[hsh[i]]});
    mp[tot ^ hsh[i]] = i;
  }
  --n;
  print(n, {2, n / 2});
}

### Prompt

In CPP, your task is to solve 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;
void FAST_IO(string filein = "", string fileout = "", string fileerr = "") {
  if (fopen(filein.c_str(), "r")) {
    freopen(filein.c_str(), "r", stdin);
    freopen(fileout.c_str(), "w", stdout);
  }
  cin.tie(0), cout.tie(0)->sync_with_stdio(0);
}
void Hollwo_Pelw();
signed main() {
  FAST_IO("hollwo_pelw.inp", "hollwo_pelw.out");
  int testcases = 1;
  for (int test = 1; test <= testcases; test++) {
    Hollwo_Pelw();
  }
  return 0;
}
const int N = 1e6 + 5;
mt19937_64 rng(
    (uint32_t)chrono::steady_clock::now().time_since_epoch().count());
int n, p[N];
uint64_t val[N], hsh[N], tot;
inline void print(int n, vector<int> del) {
  cout << n - del.size() << '\n';
  for (int i = 1; i <= n; i++) {
    bool ok = 1;
    for (auto v : del)
      if (v == i) ok = 0;
    if (ok) cout << i << ' ';
  }
}
void Hollwo_Pelw() {
  cin >> n;
  iota(p + 1, p + n + 1, 1);
  for (int i = 2; i <= n; i++)
    if (p[i] == i)
      for (int j = 2 * i; j <= n; j += i) p[j] = i;
  for (int i = 2; i <= n; i++)
    if (p[i] == i) val[i] = rng();
  for (int i = 2; i <= n; i++) {
    hsh[i] = hsh[i / p[i]] ^ val[p[i]];
  }
  for (int i = 2; i <= n; i++) {
    hsh[i] ^= hsh[i - 1];
    tot ^= hsh[i];
  }
  if (tot == 0) return print(n, {});
  for (int i = 1; i <= n; i++)
    if (tot == hsh[i]) return print(n, {i});
  if (n % 2 == 0) return print(n, {2, n / 2});
  map<uint64_t, int> mp;
  for (int i = 1; i <= n; i++) {
    if (mp.count(hsh[i])) return print(n, {i, mp[hsh[i]]});
    mp[tot ^ hsh[i]] = i;
  }
  --n;
  print(n, {2, n / 2});
}

```