Prerequisites: you need to be familiar with both modular arithmetic and Fast Fourier Transform / number theoretic transform. The latter can be a rather advanced topic, but I personally found this article helpful. Note that there are some relatively easy optimizations/improvements that can be done, like precalculating the modular inverse of $$$n$$$, the "bit-reversed-indices" (can be done in $$$O(n)$$$ with DP), as well as the powers of $$$\omega_n$$$. Also useful is modifying it so that the input array length can be any power of two $$$\leq n$$$; some problems require multiplying polynomials of many different lengths, and you'd rather the runtime be $$$O(n \log n)$$$ over the sums of lengths, rather than (number of polynomials * maximum capacity).

Also helpful is knowing how to use NTT to solve problems modulo $$$998244353$$$, like 1096G - Счастливые билеты and 1251F - Красно-белый забор. Note that for some problems it's easier to think of FFT not as multiplying polynomials, but of finding multiset $$$C$$$ as the pairwise sums of multisets $$$A$$$ and $$$B$$$, in the form of arrays $$$A[i] =$$$ number of instances of $$$i$$$ in multiset $$$A$$$. This is equivalent to multiplying polynomials of the form $$$\sum _{i=0} ^n A[i]x^i$$$.

Note that $$$\omega_n$$$ can be easily found via the formula $$$g ^ {(m-1) / n} \ \text{ mod } m$$$, provided that:

1. $$$m$$$ is prime
2. $$$g$$$ is any primitive root modulo $$$m$$$. It is easiest to find this before hand and then hardcode it in the submission. You can either implement the algorithm yourself or use Wolfram Alpha to find it via the query PrimitiveRoot[m]. (Spoiler alert, $$$g = 3$$$ works well for $$$998244353$$$)
3. $$$n$$$ divides $$$m-1$$$ evenly. As $$$n$$$ is typically rounded up to the nearest power of 2 for basic NTT implementations, this is easiest when $$$m$$$ is of the form $$${a \cdot 2^{k} + 1}$$$ where $$$2^k \geq n$$$. This is why $$$998244353$$$ is a commonly-appearing modulus; it's $$$119 \cdot 2^{23} + 1$$$. Note that this modulus also appears in many problems that don't require FFT/NTT; this is a deliberate "crying-wolf" strategy employed by puzzle writers, so that you can't recognize immediately that a problem requires FFT/NTT via the given modulus.

Now onto the main topic, the "ultimate" NTT.

Rationale: There are a few problems like 993E - Никита и порядковая статистика that require FFT, however the results aren't output with any modulus, and indeed may exceed the range of a 32-bit integer type. There are several usual solutions for solving this:

1. Do NTT with two moduli and restoring the result via Chinese Remainder Theorem. This has several prominent disadvantages:
   1. Slow, as the NTT routine has to be done twice.
   2. Complicated setup, as several suitable moduli has to be found, and their primitive roots calculated
   3. Restoring the result with CRT requires either brute force or multiplications modulo $$$pq$$$, which may overflow even 64-bit integer types.
2. Do FFT with complex numbers and floating point types. Disadvantages are:
   1. Could be slow due to heavy floating-point arithmetic. Additionally, JVM-based languages (Java, Kotlin, Scala) suffer complications here, as representing complex numbers with object-based tuples adds a significant overhead.
   2. Limited precision due to rounding error. Typically the problems are constructed such that it won't be a problem if care is taken in the implementation, but won't it be nice to just not to have to worry about it?

To solve these problems, I propose the "ultimate" NTT solution — just use one huge modulus. The one I use is $$$m = 9223372036737335297 = 54975513881 \cdot 2^{24} + 1, g = 3$$$. This is just over a hundred million less than $$$2^{63} - 1$$$, the maximum value of a signed 64-bit integer.

However, this obviously raises the issue of how to safely do modular arithmetic with such huge integers. Addition is complicated by possible overflow into the sign bit, thus the usual if(x >= m) x -= m won't work. Instead, first normalize $$$x$$$ into the range $$$[-m, m)$$$; this is easily done with subtracting $$$m$$$ after any addition operation. Then do x += (x >> 63) & m. This has the effect of adding $$$m$$$ to $$$x$$$ if and only if $$$x$$$ is negative.

The elephant in the room however is multiplication. The usual method requires computing a 126-bit product, then doing a modulo operation over a 63-bit integer; this could be slow and error-prone to implement. C++ users could rejoice, as Codeforces recently implemented support for 128-bit integers via this update. However, before you celebrate too early, there are still issues: this may not be available on other platforms, and I can't imagine that straight 128-bit modulo 64-bit is exactly the fastest operation in the world, so the following might still be helpful even to C++ users.

My favored technique for this problem is called Barrett reduction. The explanation is (copied from this website)

1. Choose integer $$$k$$$ such that $$$2^k > n$$$. $$$k = 63$$$ works for our modulus.
2. Precompute $$$\displaystyle r = \left\lfloor \frac {4^k} n \right\rfloor$$$