The problem is asking to find all the vertices to a non-degenerate convex polygon. We can try to find all the vertices by asking the judge a point ~(x, y)~ in which we get any vertex of the polygon that is closest to ~(x, y)~.

For the first subtask all the range of points that could be on the polygon is small (~(10 - (-10) + 1)^2 = 441~ possible candidates). A sufficient algorithm for the first subtask is to query all points in the range and maintain a counter of all the times the queried point is equal to the point that has been returned.

For the other subtask, we note that a convex polygon must have two points ~(x_1, y_1)~ and ~(x_2, y_2)~ such that ~y_1 \ne y_2~. This means that the smallest ~y~-coordinate is not the same as the largest ~y~-coordinate. We can get those two points by querying points ~(0, -10^{18})~ and ~(0, 10^{18})~ for the vertex with the smallest and largest ~y~-coordinate of the polygon.

Now consider, this method of constructing a polygon, knowing that we have at least ~2~ vertices ~u~ and ~v~. We will now check if ~u~ and ~v~ are adjacent. For ~u~ and ~v~ to be adjacent, then all the vertices must strictly lie on one side of the line ~u~ and ~v~. So, if there exist a point on the other side of the line, then ~u~ and ~v~ are not adjacent. We can try to find this point by using the following method.

1. Let ~m~ be the midpoint of the line segment ~u~ and ~v~.
2. Draw a ray ~r~ that is perpendicular to the line segment ~u~ and ~v~ from ~m~ and extends towards the plane where a new point might potentially exist (the half-plane that does not contain the rest of the known points of the convex polygon).
3. Let ~q~ be some point on the ray where either the absolute value of its ~x~ or ~y~ value is large and query that point.

Note that point that is being returned as ~p~. If ~p~ is equal to either ~u~ or ~v~, then we have not found any point between ~x~ and ~y~. Otherwise, ~u~ and ~v~ are not adjacent as ~p~ is between, so we will repeat this process for the pairs ~x~ and ~p~ and ~p~ and ~v~.

Since every edge and every vertex of the polygon will use up a ~1~ query, then we need to use ~2N~ queries to find all the points.

A few tips when implementing. We can work with integer coordinates all the time by mapping all ~(x, y) \to (2x, 2y)~. This way the midpoints will all have integer coordinates. Also, by repeating the process, we can think of it as divide-and-conquer on a polygon, so we can write it recursively or iteratively.