There are $K^N$ sequences that will be evaluated; i.e. “a sequence of length $N$ consisting of integers between $1$ and $K$ (inclusive).”
Thus, there are $M^{(K^N)}$ ways to “give each of the evaluated sequences a score between $1$ and $M$.”
Therefore, this problem asks to find the remainder of $M^{(K^N)}$ divided by $P := 998244353$.

If $M$ is a multiple of $P$, then $M^{(K^N)}$ is a multiple of $P$, so we should print $0$.
Hereinafter we assume that $M$ is not a multiple of $P$.

Then, since $P$ is a prime, by Fermat’s little theorem

$M^{P-1} \equiv 1 \pmod{P}$

holds. Therefore, let $q$ and $r$ be the quotient and remainder when dividing $K^N$ by $(P-1)$, respectively, then

$M^{(K^N)} = M^{(q(P-1)+r)} = (M^{P-1})^q \times M^r \equiv 1^q \times M^r = M^r \pmod{P}$,

so the remainder of $M^{(K^N)}$ divided by $P$ is equal to the remainder of $M^r$ by $P$.
Hence, in order to find the remainder of $M^{(K^N)}$ by $P$, it is sufficient to find the following two values:

• The remainder $r$ when dividing $K^N$ by $(P-1)$, and
• the remainder when dividing $M^r$ by $P$.

These can be computed fast enough by means of fast exponentiation.
Be sure to avoid overflows in the process of computation.