Hello, Codeforces! 74TrAkToR and I were solving problems and have noticed one super interesting fact.

So, check this out. 801E - Возмутительный Контраст is a very nice problem we have solved. There is interesting an interesting solution with DP on a DAG. I won't go into detail, but in the end we had to solve this problem:

You are given an array (a) with n different numbers from 0 to m — 1. You need to create an array (b) (with same length) with (not necessary different) numbers, so if you multiply first i numbers of b and get in by modulo m, you will receive a[i]. It's guaranteed, that we can create such array.

Ok, it's easy to understand, that we must solve some equations like $$$a \cdot x \equiv b \pmod m $$$.

How to solve it? We wrote 2 similar solutions, but we had one similar mistake, but.... we got AC. I've checked jury solution and.... there is this idea toox.

Well, this solution is based on the following fact. We can get $$$b \pmod m$$$ by multiplying $$$a$$$ on some $$$x$$$ if and only if $$$gcd(b, m) \vdots gcd(a, m)$$$. Let's find x. Let $$$a' = gcd(a, m)$$$, $$$b' = gcd(b, m)$$$, then $$$p_a = \frac a {a'} $$$, $$$p_b = \frac b {b'} $$$. Easy to realize that $$$p_a$$$ and $$$p_b$$$ are coprime with m, we can get $$${p_a}^{-1} = {p_a}^{\phi(m) - 1} $$$, so our $$$x = \frac {b} {a} = \frac {b'} {a'} \cdot \frac {p_b} {p_a} = \frac {b'} {a'} \cdot p_b \cdot {p_a}^{\phi(m) - 1} $$$. So, wasn't so hard, really? Mmmaybe, but that's not true, because $$$p_b$$$ can be not comprime with $$$m$$$. $$$m = 24$$$, $$$b = 16$$$. $$$b' = 4$$$, $$$p_b=2$$$. Holy... Idk, i can't found some information about it. It's getting AC and the author of this problem uses this formula too, I think, it's right. But why??? If we use formula $$$x = b \cdot {a}^{\phi(m) - 1} $$$, it's getting WA7. Maybe are weak tests.

I also came up the right solution: let's solve diophantine equation $$$ax+my=c$$$. ex_gcd can help you, but we must use $$$(x = 0, y = c / gcd)$$$ at the end.

I have observed one thing: all of numbers $$$a$$$ with the same $$$gcd(a, m)$$$ have the same $$${a}^{\phi(m)}$$$ that have partly the same $$$gcd$$$ with $$$m$$$, but all of the prime divisors (p) in maximum power (it power that $$$m \vdots p^x$$$, x — maximal), so, $$${a}^{\phi(m)}$$$ may have other prime divisors.

Can you help us? Thank you in advance. It's really shocking, that we pass a similar mistake.