You can use sieve type algorithm. First, sort the list, now if an element occurs only once, mark all its multiples and push back to the answer list. If an element occurs multiple times and has not been marked, mark its multiples but don’t push back to the answer list. Here is my code(although I didn’t submit it).

Overall Complexity : O(M*log(log(M))), where M = 10^6.

##
Code

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> vec(n);
int i;
for(i=0;i<n;i++)
cin >> vec[i];
sort(vec.begin(), vec.end());
int mark[1000001] = {0};
int j;
vector<int> ans;
for(i=0;i<n;i++)
{
j = i;
while(j < n && vec[j] == vec[i])
j++;
if(j == i + 1)
{
if(!mark[vec[i]])
{
ans.push_back(vec[i]);
int s = vec[i];
while(s < 1000001)
mark[s] = 1, s+=vec[i];
}
}
else
{
if(!mark[vec[i]])
{
int s = vec[i];
while(s < 1000001)
mark[s] = 1, s+=vec[i];
}
}
i = j - 1;
}
if(ans.size() == 0)
cout << -1 << '\n';
else
{
cout << ans.size() << '\n';
for(auto j : ans)
cout << j << " ";
}
}
```