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.