I hope my solution and explanation would help you!

Consider a rectangle,we have know that rectangles cannot intersect,and all sides of the rectangles have odd length.As we all know,the Four color theorem,so it must have a solution.What we need to do is to find a solution to give a color to every rectangle.If the lower left quarter of the rectangle,(x1,y1),so that we can know the situation of (x1+odd,x2+odd),we can confirm that we just need to color the rectangle according to its x1 and y1's parity.

Hello!I have replied the question in the first page,I hope you can have a look at it please.

No,No,No,our school training program always contains many problems in codeforces,so we ususally solve the same problem.We share our solutions,because our teacher encourage us to have discussions together,it's a part of our study program,too.

If you say we are fake id of a red coder,that's interesting.And I really hope my id can become a red id one day.:)

First,I have to say that this problem has been done by all of my classmates in our teams.As you know,we come from the same school and the same city,and I want to tell you that div2 E has appeared in our school test we attended in the past.

Also,that problem has been published by our fellow students in China before,our teacher can prove that we all solved the problem ourselves,because she has told us the solution to the problem a month ago when we first read the problem which is similar to the div2 E(maybe a little different,but the solution is similar,too).

If you say we write code in same style,what I want to say is that we have studied together for two years,and all of my classmates' styles are the same.:) I have to say if the code has been written and published before the start of the round,if I need to write it again,and our classmates are taught by the same teacher,and we all have the solution by our fellow students,it's not difficult to explain this situation.

div2 E is a difficult but intersting problem,I hope you can solve this problem soon,too.:)

UPD:div2 D is also a interesting problem ,maybe everyone who has solved this problem used the same solution.:)