Eccoci.

$6174$ is a remarkable number; if we sort its digits in increasing order and subtract that number from the number you get when you sort the digits in decreasing order, we get $7641-1467=6174$.
Even more remarkable is that if we start from any $4$ digit number and repeat this process of sorting and subtracting, we'll eventually end up with $6174$ or immediately with $0$ if all digits are equal.
This also works with numbers that have less than $4$ digits if we pad the number with leading zeroes until we have $4$ digits.
E.g. let's start with the number $0837$:
$8730-0378=8352$
$8532-2358=6174$

$6174$ is called the Kaprekar constant. The process of sorting and subtracting and repeating this until either $0$ or the Kaprekar constant is reached is called the Kaprekar routine.

We can consider the Kaprekar routine for other bases and number of digits.
Unfortunately, it is not guaranteed a Kaprekar constant exists in all cases; either the routine can end up in a cycle for some input numbers or the constant the routine arrives at can be different for different input numbers.
However, it can be shown that for $5$ digits and a base $b = 6t+3\neq 9$, a Kaprekar constant exists.
E.g. base $15$: $(10,4,14,9,5)_{15}$
base $21$: $(14,6,20,13,7)_{21}$

Define $C_b$ to be the Kaprekar constant in base $b$ for $5$ digits. Define the function $sb(i)$ to be

• $0$ if $i = C_b$ or if $i$ written in base $b$ consists of $5$ identical digits
• the number of iterations it takes the Kaprekar routine in base $b$ to arrive at $C_b$, otherwise

Note that we can define $sb(i)$ for all integers $i \lt b^5$. If $i$ written in base $b$ takes less than $5$ digits, the number is padded with leading zero digits until we have $5$ digits before applying the Kaprekar routine.

Define $S(b)$ as the sum of $sb(i)$ for $0 \lt i \lt b^5$.
E.g. $S(15) = 5274369$
$S(111) = 400668930299$

Find the sum of $S(6k+3)$ for $2 \leq k \leq 300$.
Give the last $18$ digits as your answer.