#include <bits/stdc++.h>
const int N = 1000005;
int n, lp[N], in[N], u[N];
unsigned long long pr[N], ex[N];
std::set<std::pair<long long, int> > st;
std::vector<int> p;
std::mt19937_64 rng(
    std::chrono::steady_clock::now().time_since_epoch().count());
int main() {
  scanf("%d", &n);
  memset(lp, -1, N << 2);
  for (int i = 2; i <= n; ++i) {
    if (lp[i] == -1) {
      pr[p.size()] = rng();
      lp[i] = p.size();
      p.push_back(i);
    }
    for (int j = 0; j < (int)p.size() && j <= lp[i] && p[j] * i < N; ++j)
      lp[p[j] * i] = j;
  }
  unsigned long long cur = 0, ov = 0;
  for (int i = 1; i <= n; ++i) {
    int val = i;
    while (val > 1) {
      int t = lp[val];
      do {
        cur ^= pr[t];
        val /= p[t];
      } while (!(val % p[t]));
    }
    st.insert(std::make_pair(cur, i));
    ov ^= cur;
    ex[i] = cur;
  }
  std::set<std::pair<long long, int> >::iterator it, it1;
  if (ov) {
    if ((it = st.lower_bound(std::make_pair(ov, 0))) != st.end() &&
        it->first == ov)
      u[it->second] = 1;
    else {
      bool ok = 0, fi = 0;
      int i = n;
      while (!ok) {
        if (fi) st.erase(std::make_pair(ex[i], i));
        for (it = st.begin(); it != st.end() && !ok; ++it) {
          it1 = st.lower_bound(
              std::make_pair(ov ^ it->first ^ (fi ? ex[i] : 0), 0));
          if (it1 == it) ++it1;
          if (it1 != st.end() &&
              it1->first == (ov ^ it->first ^ (fi ? ex[i] : 0))) {
            ok = 1;
            u[it->second] = 1;
            u[it1->second] = 1;
            if (fi) u[i] = 1;
          }
        }
        if (fi) st.insert(std::make_pair(ex[i], i));
        if (fi) --i;
        fi = 1;
      }
    }
  }
  int ans = n;
  for (int i = 1; i <= n; ++i) ans -= u[i];
  printf("%d\n", ans);
  for (int i = 1; i <= n; ++i)
    if (!u[i]) printf("%d ", i);
  printf("\n");
  return 0;
}

### Prompt

Develop a solution in cpp 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>
const int N = 1000005;
int n, lp[N], in[N], u[N];
unsigned long long pr[N], ex[N];
std::set<std::pair<long long, int> > st;
std::vector<int> p;
std::mt19937_64 rng(
    std::chrono::steady_clock::now().time_since_epoch().count());
int main() {
  scanf("%d", &n);
  memset(lp, -1, N << 2);
  for (int i = 2; i <= n; ++i) {
    if (lp[i] == -1) {
      pr[p.size()] = rng();
      lp[i] = p.size();
      p.push_back(i);
    }
    for (int j = 0; j < (int)p.size() && j <= lp[i] && p[j] * i < N; ++j)
      lp[p[j] * i] = j;
  }
  unsigned long long cur = 0, ov = 0;
  for (int i = 1; i <= n; ++i) {
    int val = i;
    while (val > 1) {
      int t = lp[val];
      do {
        cur ^= pr[t];
        val /= p[t];
      } while (!(val % p[t]));
    }
    st.insert(std::make_pair(cur, i));
    ov ^= cur;
    ex[i] = cur;
  }
  std::set<std::pair<long long, int> >::iterator it, it1;
  if (ov) {
    if ((it = st.lower_bound(std::make_pair(ov, 0))) != st.end() &&
        it->first == ov)
      u[it->second] = 1;
    else {
      bool ok = 0, fi = 0;
      int i = n;
      while (!ok) {
        if (fi) st.erase(std::make_pair(ex[i], i));
        for (it = st.begin(); it != st.end() && !ok; ++it) {
          it1 = st.lower_bound(
              std::make_pair(ov ^ it->first ^ (fi ? ex[i] : 0), 0));
          if (it1 == it) ++it1;
          if (it1 != st.end() &&
              it1->first == (ov ^ it->first ^ (fi ? ex[i] : 0))) {
            ok = 1;
            u[it->second] = 1;
            u[it1->second] = 1;
            if (fi) u[i] = 1;
          }
        }
        if (fi) st.insert(std::make_pair(ex[i], i));
        if (fi) --i;
        fi = 1;
      }
    }
  }
  int ans = n;
  for (int i = 1; i <= n; ++i) ans -= u[i];
  printf("%d\n", ans);
  for (int i = 1; i <= n; ++i)
    if (!u[i]) printf("%d ", i);
  printf("\n");
  return 0;
}

```