$k$ Markov Numbers (noch nicht übersetzt)

Consider positive integer solutions to

$a^2+b^2+c^2 = 3abc$

For example, $(1,5,13)$ is a solution. We define a 3-Markov number to be any part of a solution, so $1$, $5$ and $13$ are all 3-Markov numbers. Adding distinct 3-Markov numbers $\le 10^3$ would give $2797$.

Now we define a $k$-Markov number to be a positive integer that is part of a solution to:

$\displaystyle \sum_{i=1}^{k}x_i^2=k\prod_{i=1}^{k}x_i,\quad x_i\text{ are positive integers}$

Let $M_k(N)$ be the sum of $k$-Markov numbers $\le N$. Hence $M_3(10^{3})=2797$, also $M_8(10^8) = 131493335$.

Define $\displaystyle S(K,N)=\sum_{k=3}^{K}M_k(N)$. You are given $S(4, 10^2)=229$ and $S(10, 10^8)=2383369980$.

Find $S(10^{18}, 10^{18})$. Give your answer modulo $1\,405\,695\,061$.