Equal digit sums in an arithmetic sequence

The infinite arithmetic sequence $A(n) = Cn + D$ is defined for every natural number $n$. Find $M$ distinct natural numbers $n_1, n_2, \dots, n_M$, each at most $10^{15}$, such that $A(n_1), A(n_2), \dots, A(n_M)$ all have the same sum of digits in base $B$.

Every positive integer $N$ has exactly one representation in base $B$: the digit string $x_k x_{k-1} \dots x_1 x_0$ with $0 \le x_i < B$ for each $i$ and $x_k B^k + x_{k-1} B^{k-1} + \dots + x_1 B + x_0 = N$. Its sum of digits is $x_k + \dots + x_0$.

Several answers usually satisfy the condition, so exactly one of them is selected by the following rule. For a natural number $n$, let $f(n)$ be the sum of the digits of $A(n)$ in base $B$. Compute $f(n)$ for $n = 1, 2, 3, \dots$ in that order and count how many times each value has appeared. Stop at the moment some value reaches $M$ occurrences and call that value $s$. Exactly $M$ of the numbers $n$ examined so far satisfy $f(n) = s$, and those $M$ numbers are the answer.

The first line contains the integers $C$, $D$, $B$ and $M$, separated by spaces ($1 \le C, D \le 10000$, $2 \le B \le 5000$, $1 \le M \le 250000$).

The input is such that some digit sum reaches $M$ occurrences while $n$ runs from $1$ to $10^7$.

On one line, print the $M$ numbers selected by the rule above in increasing order, separated by single spaces.

Print the numbers $n_i$, not the numbers $A(n_i)$. Every printed number is at most $10^{15}$.

In the first example $A(2) = 5 \times 2 + 3 = 13$ and $A(5) = 5 \times 5 + 3 = 28$. In base 2 the number $13$ is written $1101$ and the number $28$ is written $11100$, so both digit sums are $3$. The value $3$ reaches two occurrences at $n = 5$, and no other value reaches two occurrences earlier.

In the second example $A(2) = 5$, $A(11) = 23$ and $A(20) = 41$. All three have digit sum $5$ in base 10, and no value reaches three occurrences before $n = 20$.