Hello, Codeforces!
I am very excited to invite you to my second-ever contest, Codeforces Round 1065 (Div. 3), which will start on Nov/20/2025 17:35 (Moscow time)! In this contest, you will be given 2 hours and 30 minutes to solve 7 problems, two of which have been split into subtasks. The subtasks for a given problem are not necessarily placed adjacently in the problemset.
The round will be hosted by the rules of educational rounds (extended ICPC). Thus, all solutions will be judged on preliminary tests during the round, and after the round, there will be a 12-hour phase of open hacks. After the open hack phase, all accepted solutions will be rejudged on successful hacks. Also, note that there is no score distribution — rank will be determined by number of problems solved, followed by penalty; wrong submissions will incur the usual penalty of 10 minutes, following the rules of educational rounds.
As a reminder, only trusted participants of the third division will be included in the official standings table. This is a compulsory measure for combating unsporting behavior. To qualify as a trusted participant of the third division, you must:
- take part in (and solve at least one problem in) at least five rated rounds
- and not have had a rating of 1900 or higher at any moment in time.
Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you (unless you register unrated).
Also, note the rule restricting the use of AI. If you are caught breaking this rule, you will be condemned to life spent in prison the basement of some unspecified Codeforces user. Said person's basement is a rather unenjoyable place to live in, so I would advise adhering to the rules.
I would like to thank the following people for helping make this round possible!
- cry for his excellent coordination;
- Vladosiya for translating the statements into Russian;
- satyam343 and Lilypad for discussing problem ideas with me;
- djm03178, satyam343, Dominater069, and Intellegent for being very important people who have contributed significantly to the round;
- Dominater069, nifeshe, Intellegent, catgirl, temporary1, Friedrich, djm03178, wuhudsm, -firefly-, efishel, Edeeva, yse, SpyrosAliv, ritam1234, chromate00, Ermiooo159, kevlu8, Argentum47, expertaq, simplelife, ByteRaider, and RJ_Parin_fr for testing;
- macaquedev for pretending to test;
- MikeMirzayanov for the incredible Codeforces and Polygon platforms;
- and most importantly of all, You for participating!
Good luck, and have fun!
UPDATE: Due to issues with Cloudflare, the round has been rescheduled for Nov/20/2025 17:35 (Moscow time).
UPDATE: Editorial
As a tester, yuri is peak
Yuri (from Angel Beats!) is peak.
Totally agree.
K-On and Kobayashi could be great, I think. Though you said Angel Beats is the peak.
Well, I think Toaru Kagaku no Railgun is also great, according to your avatar.:)
It must be. :) And even the blogger is Misaka ;)
the gif is peak
that is true
a @reirugan round ? i wish it will be as good as your previous div3 round !!
anime kiss jumpscare
peak div.3 content
unfortunately i was forced to remove the previous gif... but have an adachi instead!
unfortunately my comment got removed too, I thought it was funny.
No anime kiss ): Thats cool too
What the heck? How dare you post something that contains LGBTQ content?”
Why my rating was not improved even though I solved 5 problems?
Because ratings haven't been updated yet
I saw your submission to problem D. It's clear the answer is AI generated. I hope you can see the error of your ways and stop cheating in the future. It'll only hinder your progress.
what's today's codeforces topic?
yuri
thanks
Could you please remove the gif?
Are you homophobic
No comment. ( Though, I commented :) )
Yeah I am now remove it (and give source thanks).
what the f-
Bruh why LGBTQ in codeforces?????? I am homophobic
ru sociable, then??
More than you?
yes pls
codeforces when my mom enters the room
LOL
literally loading page in GTA San Andreas when mum enters
as a tester, this gif actually spoils the entire contest
I got jumpscared so hard by that gif.
Game's fucking gone
'condemned to life spent in the basement of some unspecified Codeforces user' when are we specifying this lad
we internally run an auction for each subject to determine whose basement theyll be assigned to
I can confirm that we have increasing bids on the amount of distress given to punish the cheaters, and then the basement with the lowest bid is chosen (We don't want all of our basements occupied by cheaters)
I hope its not wakanda-forever
As a tester, unspecified Codeforces user is yuri
As a tester, reirugan tested my yuri knowledge.
Hope I can solve a question or two
What is the anime name so I can watch it with my family later?
I believe the GIF is from Samurai Flamenco.
Edit: the GIF was changed today, current one is from Adachi and Shimamura.
Tester here — had fun with this round! Cool problems (gif got changed)
source? for cf practice ofc
wtf is that gif
we got gay codeforces before GTA VI?
Who is gay?
you
sauce ?
add some problems with yuri :-)
I am now curious about the problem statement for this contest.
can't wait for the contest's problem statements
is this a hint that there will be at most 18 problems?
$$$7 \lt 18$$$ you know...
No way this was an codeforces notification!!
opening codeforces is never disappointing
This announcement just relieved all the fatigue from the latest contest in a blink! thanks for the GIF of Adachi!
Previous one was better (ToT)
I will give it a try; I hope it's as good as my previous Div-3 round!
waiting to become pupil!
As a participant always love to taking part in Div. 3 contests
Why do some problems feel trivial only in hindsight? Is the idea inherently simple, or do we just overthink when the timer is running?
Wth! is going on in the comment section. Someone please exolain in alice and bob terms
All my favorite NYU students have helped in this contest. Wow!
who is your favorite? :v
you r my og <3
The picture is beautiful, feeling happy when opening codeforces.
Btw, who is the charactor in the picture?
sakura adachi!
Having heard about it for a long time, I haven't seen it. I will do that after the contest, thanks!
As a reirugan fan good luck guys
I hope this contest will be much easier-
安達としまむら round！
Who is in u avatar
sui chat bro
I am a Newbie.
cool
Related ? $$$67$$$
THIS GENERATION IS DOOMED.
Adachi Sakura!!!
How lucky I am to see a Yuri picture on CodeForces! It really makes me curious.
"macaquedev for pretending to test;" lol
yeah, I got bored and fell asleep.
Uhh what do you mean by that??
Usually subtasks are enumerated, for example, G1 and G2, and are therefore adjacent.
In this case, it's possible the subtasks are enumerated with different nonadjacent letters.
My cf is ruined
Excited for battle!!!
Can I try? I'm just a new person
Yes, you can. Div 3 and 4 are intended for beginners.
Llegar, partir y merendar
le sabes
Hope to solve at least 5 problems,thus increase my rating.:)
bro how you predict that?i really dont know, can you say for ex how many problems i need solve every div2 contests to reach speacialist
In common, you should solve 3 problems or solve 2 problems fast(that depends on the difficulty of the contest) in div.2 to reach specialist. In div.3, the number of problems you solve is usually 5.
i find one extension named CF predictor and today i wanna check it
Thank God...No more interactive problems
Today is my birthday:)
Happy Birthday!
Thanks!
Happy birthday:)
Thanks!
happy birthday! :)
Thanks!
Happy Birthday
Thanks!
how can i participate as a tester in a competition?
To qualify as a trusted participant of the third division, you must:
take part in (and solve at least one problem in) at least five rated rounds
what does it mean ?
You need to have completed 5 CodeForces rated rounds, simply registering doesn't suffice, you also need to have solved at least one problem in those 5 rounds (actual participate). For example this div 3 is also a rated round.
hope everybody get a positive Delta
Unfortunately, that’s impossible… unless, of course, a miracle happens!
All the best to everyone hope i get closer to specialist
If I didn't participate in 5 rated rounds, Why I ain't trusted?
Looking for a "Demon Slayer" based contest if we are entering the zone of anime in cf!
Thank you cloudflare for forcing this round to be delayed
I blame the vibecoder that broke cloudflare's code
me too
was the contest originally now and they postponed it or am I tripping
they delayed it
The contest was originally now, but was postponed due to Cloudflare's incident I think.
I though i misread the day :v
looks like due to codeforces being down for some time the round will be delayed for two days?
What just even happened XD
Cloudflare exploded
what is wrong with cloudflare ：》
I was looking forward to the yuri. At least it didn't happen during the actual round.
Why is the contest delayed?
cloudflare (was or is) down for like an hour.
I don't think it was just an hour, more like 3 hours
cloudflare became wrong
cloudflare crashed so everything crashed
because of the potato server of cloudflare:)
hello?
Wtf???Rescheduled , was looking forward for the contest.
Yes but why???
Now Contest is Delayed!
Actually it went well just before the contest begin...lol
I hate Cloudflare:(
I hate Cloudflare(
I hate Cloudflare:(
reCAPTCHA better
just wanted to solve some Div3 for me
It's amazing that Cloudflare is fixed a minute before the original contest start time.
It has delayed.
I'm seeing that Shayan's and Arpa's streams are scheduled to release within the next three hours. Please get rid of them!
Arpa's video has already been delayed
I believe Shayan hasn't prerecorded his video; my understanding is that he usually screenrecords live during the contest
Thanks for pointing it out, though!
Cloudflare be trippin :cry:
Let's go, the contest got rescheduled, and now I'll get my well deserved normal sleep time.
As Cloudfare is back in service, are there any chances that the contest's date returns to being today?
Why the contest has delayed but the stream has not delayed?
I am looking looking forward to the contest.
Was praying that it gets rescheduled. Thanks:) . As a contestant , I hope to get plus delta!
Why 48 hours delay :(
There's nothing we can do bro
Just after rescheduling the contest ,the cloudflare came back ! noice:)
The issue with Cloudflare is fixed, you should reschedule it back to today
So, Cloudflare has always been the villain?
I came home in turbo mode for the contest and codeforces said, 'Relax bro, we are delayed!' Perfect, I rushed for maximizing my frustration!
cloudflare disappointed us all
ChatGPT as well)
thanks i have my exams tomorrow. 2.5 hrs saved. I'll utilise this in sleeping
there are two long days left to wait
W organizers!!! I thought I would miss todays round (╥﹏╥)
What was the earlier gif :) reirugan
I hate Cloudflare...
Dead internet theory has a new definition now
how?
Damn it. We just lost an opportunity to witness a fairest CF contest ever!
Non-hackers getting a crawling ban is sweet justice given all the pain hackers have experienced over the past year.
Try CF Submitter : https://marketplace.visualstudio.com/items?itemName=DevXSayan.cf-submitter — Fetch all the problems of a live contest inside vscode, view questions without reloading hustle, run test cases, and submit in one click, all without leaving vscode
Aim to solve A to E
This is going to be my first rated contest, im very excited, hoping get some rating :D
Adashima is the best!
CONTEST ABOUT TO START
C2 > DEFG
is c2 not exactly the same as c1? you just do the same logic for all the bits.
Really?
Yah . just check which player can get which bit .
D looked so simple but its soo hard. I am curious to know its greedy strategy
there's a greedy construction, traverse from the back searching for 1 and simultaneously keep a note of the minimum excluded value, when you reach 1, you can jump to the mex value position and iterate like this ...
Really ? C2 is just a little bit harder than C1, we only need more 1 for loop to check from the Most significant bit. Could you show how to solve E ?
loop 2,1,4,6,5,3,9,11,12,10,7,8
Div. 3>>1
For problem E,my code 349964120 construct a permutation with atmost 1 bad index.I wonder whether 6 was there just to confuse.
master for a reason!
Your solution is really great!
but 6 is so that there is more freedom for more solutions and no need to do small and annoying tweaks
for example my code might even have 3 or 4 bad indexes 349936459
mine with 0 :)
what about n = 3?
N>8 :(
you can't change the range of n to justify your answer.it gives 1 for n=11,for n=3,answer will always be 1.
just saw , I mostly checked for larger ones , my claim is wrong :(
5 2 4 7 6 8 (11 10 3) 9 1
https://codeforces.me/contest/2171/submission/349987169
I have started with taking prime number and then there multiples like 2's multiple, 3's multipe, 5's multiple.
If in any case no multiple's are present then I am adding them in between 2's multiple with position gap of 2.
For larger n, #prime numbers <<< #2's multiple.
My B solution kept failing on some hidden case, I am confident my logic is correct. Also for D, I was trying to use DSU based approach but kept timing out. Any ideas on approach for problem D?
For problem D
My idea was that if we have vertex minimum number, it has to be paired with vertices that are to the right in the array. So, we can "squeeze" these vertexes into one with maximum value among this group. We can continue do it, until we have 1 vertex (so we built a tree) or the minimum element is the last in the remaining part of the array (it means, that last vertex can't be connected and will be separate).
To find minimum we can use set with pair { value, index }. To "crop" the array we can make copy of n, let it call m, and after each operation set m = i + 1. Complexity will be O(n*log(n))
I hope I wrote it clearly)
In Problem E, why multiples of 2, 5 and 10 doesn't work but multiples of 2, 3 and 6 work ?? Is it because we can have at most 6 bad indices and not 10 indices ?
we can make 0 bads as well , the fact that matters is that for any N there are total primes always <=N/2 with N :)
a proof for why 2,3 specifically work:
consider a solution of the form
**_**_**_...
where the * themselves form a sequence s.t. the adjacent * are not coprime. clearly, such a solution has no bad indices. so, it suffices to find a sequence of the * numbers with length >= 2n/3.
consider multiples 2 and 3. it now suffices to show that these cover 2/3rds of 1 .. n. consider partitioning 1 .. n into disjoint groups of 3 starting from 3. let j be the first number in the block (e.g. 3,6,9,etc.). if j%2==0, then j%2==0 and (j+2)%2==0. so we have covered 2/3 numbers in the range j,j+1,j+2. if j%2!=0, then j%3==0 and (j+1)%2==0. so we have also covered 2/3 numbers in this range. so, from any n we can form a valid sequence * using multiples of 2 and 3, but making sure we carefully place 6 to connect the 2 and 3 multiples. so we use 2 4 8 .. 6 3 9 15 .. for the *'s and fill in the rest.
felt like this was hard compared to other div3
I was able to do G in $$$O(N * log(N) ^ 2)$$$
Code link: https://codeforces.me/contest/2171/submission/349941964
Is there a better solution?
I also thought of the same solution, but too lazy to implement
I transformed b into a, and my solution works in O (N log(max(A)).
Basically, if all b[i] >= 2 * a[i] then we make all even and divide by 2 else we subtract the differences 1 by one
Submission link
That was an awesome contest for me! After giving so much effort, i solved the c problem
I solved A, C1 and C2 but not B wtf?
Woah! C2 >> B
B is a f**king problem.
The middle ones cancel out
B is tough if you miss the fact that only a[n]-a[1] matters on the first glance.
Is this div2?
++
Trash round.
Felt difficult than almost all the div3 I participated in till now ...
F is much easier than E, and is too close to D, separating them by E was just evil.
Also, I skipped the first sentence of each problem since it add nothing to the statement, this also caused me to not notice that it has two versions...
E was easy and tough at the same time :)
I got an WA on pD but I don't know why. Can anyone help me? thx
Check this case 4 6 2 5 1 3
Your answer is NO, but the correct answer is YES.
n = 9 consider { {5, 6, 7, 9}, {2, 8}, {1, 3, 4}} All the three components are connected. But min(first) > max(last)
I personally not satisfied by today's div3 :( Maybe I will get downvotes. I just shared my sadness
How tf is this a wrong edge list !?
you have to print the numbers not the indices :)
Please say I didn't waste so much time thinking I had to print indices :)
There can't be an edge between 2 and 3, since 3 appears before 2 in the permutation
I'm so happy to reach such a high place in this round! The thrill just won't go away!
Congrats! Well done.
Your code looks good but the hacking round isn’t over haha. Good luck hope you stay at #1.
C1 & C2 are nice problems! I was looking at C2 and had absolutely no idea how to do it but after a quick peek at C1 I had an observation that got me through C2!
Had it not been for C1 i would have been stuck at C.
Why would you want to make the
mod1e6 + 3 on G?Do you have a solution that does not require precomputing factorials?
I missed this. You're right. Thanks for explanation :)
If all the constraints are kept the same except for the mod (most importantly, $$$1 \leq a_i \leq b_i \leq 10^6$$$ is kept) and we only considered $$$t = 1$$$, I'm aware of a (partial) solution with two steps than can be done in $$$O(n)$$$ for the first step and $$$\tilde{O}(sqrt(mod))$$$ for the second step (per test case). This idea still requires precomputation, but only for factorials up to $$$b_i - a_i$$$, so we need an additional $$$O(max(b_i))$$$ precomputation. With $$$t = 1$$$, this idea can support mods around $$$10^9$$$, however it 1) cannot be used to solve G for significantly larger $$$t$$$ (I suspect $$$t = 10$$$ already causes TLE) and 2) is definitely not suitable for Div 3, for reasons I will explain later.
We start from the part of the solution where we are required to evaluate $$$\frac{(\sum c_{i})!}{\prod(c_i!)}$$$.
The denominator can be solved by simply precomputing all possible factorials up to $$$10^6$$$, and using mod inverse.
For the numerator, as we restricted ourselves to $$$t = 1$$$ test case, it suffices to find a fast solution to $$$n!$$$ $$$(mod$$$ $$$p)$$$. This has already been previously explored. It suffices to find a solution to the case where $$$n$$$ is a square number, and let $$$P(x)=(x+1)(x+2) \dots (x+m)$$$, and evaluate $$$P(x)$$$ at $$$x = 0, m, 2m, ... m(m-1).$$$ A naive evaluation takes way too long, so instead evaluate this function using FFT. This calculates $$$(m^2)!$$$ $$$(mod$$$ $$$p)$$$. To extend to $$$n!$$$ $$$(mod$$$ $$$p)$$$, let $$$m = \lfloor \sqrt{n} \rfloor$$$, and perform the calculation. For the final $$$O(\sqrt{n})$$$ missing multiplications, it suffices to do standard naive multiplication over mod $$$p$$$ starting from the $$$(m^2)!$$$ $$$(mod$$$ $$$p)$$$ result. The bottleneck of this solution will be the evaluation of $$$P(x)$$$. In the worst case, $$$(\sum c_{i})!= mod-1$$$ (otherwise, we immediately know our answer is 0), and so $$$m = \lfloor \sqrt{mod} \rfloor$$$. Thus, this step takes $$$\tilde{O}(sqrt(mod))$$$ time.
https://fredrikj.net/blog/2012/03/factorials-mod-n-and-wilsons-theorem/
Can someone please tell me why my dsu code for D is giving WA? Submission:349981598
Consider the case 3 5 2 1 4
Your answer is NO, but the correct answer is YES.
Step 1, the code performs union(5, 3).
Step 2, the code only performs union(4, 1) and union(4, 3).
Thank you so much. I spent half an hour trying to find the problem.
I gave this contest after registration properly. And it was my first contest. Then why it is appearing in unrated section and not in rated section for me? Help pls
The contest will be rated after the hacking phase is over.
Uh thankyou
I introduce a solution of problem E,but I can't prove it(maybe I just know a little) Firstly, make a number of (n/2)+1 called Rt , and continue ues it ,if it's a prime number,put in the first.And then,from 2 to n,make the number mutiples,every time chose two number,such as (prime number, a,b(a and b have mutiples relation)),continue to put such numbers,until Rt > n ,put other number using mutiples relation, ans you will pass the problem.
Rt is a number chosing from (n/2)+1 to n which is a prime number.
i was registered as unrated by mistake in Codeforces Round 1065 (Div. 3) .please check my submissions my name is Manisha0369
Can anyone tell / give any hints IN C problem in genral?
try playing the game starting from the index 1 , think like u are the player playing the game and what optimal moves(swap) u need to play in order to :
or in more simple words , at the ith move , u can either choose to swap or not to swap. swap only if the operation increase ur score than current or it decreases ur opponents score.
// here the easy version assumes only 0s and 1s hence the solution is a lil bit different , the above explained algo fits best with the hard version
// C1 — easy
// C2 — hard
do reply me if u got the approach and consider upvoting it :)
Why my C1 solution showing in queue ? Is this issue with other's also ?
dont worry....it was actually under system testing phase , it will be reflected into the final results
(In China, CF1065 starts at 22:35)
Yesterday night: OK I passed A B C1 C2 D F
Then I go to bed
This morning: Yes I was not been hacked
(I had to do a stupid contest so I wasn't on CF in the morning)
In the afternoon: Let me see... Oh no FST on C1 C2 F
I wanted to see how I get hacked, then I see it was just in queue
me be like:https://cdn.luogu.com.cn/upload/image_hosting/3bz1srce.png
How to upsolve any problem, when i try to submit problem after contest is over while upsolving it says "Contest is Over" ???
simply revisit the contest page from contest section,select the problem u want to solve, u will get an upload solution button there.
i guess $$$O(m \sqrt m)$$$ memory gets passed in problem H (you just need to use 16 bit integers) 349971224
I’m new to cp and codeforces. I registered for this round and participated but I am seeing that it was unrated for me. I am confused because I didn’t choose to register as unrated and there wasn’t an option for choosing between rated and unrated (at least for me). I registered days before the competition. What might be the cause of me being unrated for the competition?
dont worry! the rating will be reflected into your account soon.
when you register there is a checkbox, which asks if you want to register unrated, maybe you clicked on it by mistake. otherwise your rating will update soon
My rating is 1177 and had never got higher than now. I have solved at least one problem in at least five rated rounds, such as 1858, 1879, 1878, 2166, 2169. And I had never cheated in any round. But I wonder why it shows "Unrated allowed"?
how much more time for ratings to be updated?
Hello
Can anyone please help me to find cases in which mine code will fail (Give Tle) for problem — F It is giving tle on test case — 12 Thanks for helping
how to solve D using DSU?
i cant write a comment here
I have a very easy solution to c1 and c2, you just have to change the checking condition in the c2 version. Now it says that the play should be optimal, that means if it's the first player's turn and he is ahead in score , then he won't swap. Likewise , if he is down by one then on his turn if on swapping his score increases then only he will swap otherwise not. Second player would do the same. Therefore,the solution might be a very simple greedy approach where on the i'th turn, each player checks the score and then swaps.
Now C1 SOLUTION----->
The snippet shows that first we calculate the original xor of all elements,then on each move swap them. If the score increases like it can be 1 or 0 only. Then on swapping the score of both of them will become their negation.
C2 SOLUTION -------->
In the C2 problem, the score now won't remain 0 or 1, but the logic remains the same, first calculate the default xor of all numbers then the greedy approach to check on each move whether the scores will change or not. We just update the scores in different way in C2 because in C1 only 0 or 1 were possible pair of total xor scores. Thank You, Niksy
I missed that. As i am a beginner in here in codeforces.
Cloudflare engineer: Oh no, there's a bug! Quick, get GPT-5 to fix it!
GPT-5: Hey, I need Cloudflare to work properly!
Microsoft engineer: LOL
Codeforce: Hahahahahahahahahahahahahahahahahahahahahahahahahaha, kids, you're going to unrated again!
(Source: Luogu)
Hello, I received a plagiarism warning for solution 349932806 for problem 2171C1. I want to clarify that I did not share my code with anyone, nor did I use anyone else’s code.
I solved the problem by myself and only used my own template / common CP snippets that many users also use. If needed, I can explain my approach or share my local work.
Please review my case. Thank you.
Hello, I received a plagiarism warning for submission 349917059 for problem 2171D. I want to clarify that I did not share my code with anyone, nor did I use anyone else’s code.
I solved the problem myself. However, I used Ideone to test during the contest, and I later realized the code was accidentally left in public mode. This may have exposed my code and caused the similarity.
I did not share it intentionally. Please review my case. Thank you.
Hello — I received the plagiarism notice for my submission 349871347 to problem 2171B.
I did not copy other users’ code. I wrote the solution myself during the contest. To support this, My submission time for that particular question was 20:14 UTC +5.5 20 Nov 2025 and the other person's timing was 21:03 UTC +5.5 20 Nov 2025 I don't know the person and I have written my code myself If you look at the code, it is very simple so it has a high chance to coincide in my opinion
Please let me know if you need any additional logs. I’m happy to cooperate to resolve this. Thank you.
Here is my solution : ~~~~~ t = int(input()) for _ in range(t): n = int(input()) a = list(map(int, input().split())) if a[0] != -1 and a[-1] != -1: ans = abs(a[-1] — a[0]) else: ans = 0 if a[0] == -1 and a[-1] == -1: a[0] = 0 a[-1] = 0 elif a[0] == -1: a[0] = a[-1] elif a[-1] == -1: a[-1] = a[0] for i in range(n): if a[i] == -1: a[i] = 0 print(ans) print(*a) ~~~~~
Hello, I received a similarity warning for my solution in this round. I did not share my code with anyone or copy from anyone. It is possible that I once used an online IDE that unintentionally made my code public. I understand this is against the rules even if unintentional, and I sincerely apologize. I will make sure this never happens again. Thank you for checking.
very ezzz round for div.3
great one to perform but loss the pointtttt..
shoutout to the lesbians for this goated yuri themed contest
Tbh when i see a div 3 i smile it's easy and so amazing to pass , Hope they will be more div 3's by the time and also Good luck To everyone on this kind of contests or else !
That animation is so cool. I just love it