Counting multiples with a bounded digit sum

Consider a number of length $N$ whose every position holds a digit from 0 to 9. The leading digit may be 0.

Among those numbers, count the ones that are divisible by $P$ and whose digit sum is at most $M$. Write a program that reports this count modulo 998244353 for every $M$ from 0 to $M_{\max}$.

The first line contains $N$, $P$, and $M_{\max}$, separated by spaces. ($1 \le N \le 10^9$, $1 \le P \le 50$, $1 \le M_{\max} \le 500$)

On the first line, print $M_{\max} + 1$ integers separated by single spaces. They are the answers for $M = 0, 1, \ldots, M_{\max}$, in that order.

For $N = 2$ and $P = 3$, the numbers that satisfy the condition are:

• $M = 0$: 00
• $M = 1$: 00
• $M = 2$: 00
• $M = 3$: 00, 03, 12, 21, 30

For $N = 2$ and $P = 4$:

• $M = 0$: 00
• $M = 1$: 00
• $M = 2$: 00, 20
• $M = 3$: 00, 12, 20
• $M = 4$: 00, 04, 12, 20, 40