Why does skipping occurs when using std::remove() to remove more than one value from a vector?

I am doing a challenge that I stumbled upon Reddit the other night where you do one programming challenge a day. It seems really nice, and provides a lot of practice for myself as I learn.

I have hit this one problem where I try to remove something from a vector using std::remove() and it removes the desired numbers, and here is the issue, the next number after it. I have tried doing other functions and various ways of doing this, but this seems the most practical, and feels so close to what I need.

My code:

#include <iostream>
#include <algorithm>  // std::sort(), std::remove()
#include <vector>


void print_array(std::vector<int> array)
{
  std::cout << "Array: ";
  for(int i = 0; i < array.size(); i++)
  {
    std::cout << array.at(i) << " ";
  }
  std::cout << std::endl;
}

int main()
{
  std::vector<int> num_array = {2, 3, 4, 2, 3, 5, 4, 6, 4, 6, 9, 10, 9, 8, 7, 8, 10, 7};

  // Sort ant then print the array.
  std::sort(num_array.begin(), num_array.end());
  print_array(num_array);

  for(int i = 0; i < num_array.size(); i++)
  {
    if(num_array.at(i) == num_array.at(i + 1))
    {
      std::cout << num_array.at(i) << " is duped\n";
      num_array.erase(std::remove(num_array.begin(), num_array.end(), num_array.at(i)), num_array.end());
      print_array(num_array);
    }
    else
    {
      std::cout << num_array.at(i) << " is not duped.\n";
    }
  }


  print_array(num_array);
  return 0;
}

Output