Let $n \geq 2$ be an integer. There are $n$ distinct circles in general position, that is, any two of them meet in two distinct points and there are no three of them meeting at one point. Those circles divide the plane in limited regions by circular edges, that meet at vertices (note that each circle have exactly $2n-2$ vertices). For each circle, temporarily color its vertices alternately black and white (note that, doing this, each vertex is colored twice, one for each circle passing through it). If the two temporarily colouring of a vertex coincide, this vertex is definitely colored with this common color; otherwise, it will be colored with gray. Show that if a circle has more than $n-2 + \sqrt{n-2}$ gray points, all vertices of some region are grey. Observation: In this problem, a region cannot contain vertices or circular edges on its interior. Also, the outer region of all circles also counts as a region.