One more one (noch nicht übersetzt)

Consider the following process that can be applied recursively to any positive integer $n$:

• if $n = 1$ do nothing and the process stops,
• if $n$ is divisible by 7 divide it by 7,
• otherwise add 1.

Define $g(n)$ to be the number of 1's that must be added before the process ends. For example:

$125\xrightarrow{\scriptsize{+1}} 126\xrightarrow{\scriptsize{\div 7}} 18\xrightarrow{\scriptsize{+1}} 19\xrightarrow{\scriptsize{+1}} 20\xrightarrow{\scriptsize{+1}} 21\xrightarrow{\scriptsize{\div 7}} 3\xrightarrow{\scriptsize{+1}} 4\xrightarrow{\scriptsize{+1}} 5\xrightarrow{\scriptsize{+1}} 6\xrightarrow{\scriptsize{+1}} 7\xrightarrow{\scriptsize{\div 7}} 1$.

Eight 1's are added so $g(125) = 8$. Similarly $g(1000) = 9$ and $g(10000) = 21$.

Define $S(N) = \sum_{n=1}^{N} g(n)$ and $H(K) = S\left(\frac{7^K-1}{11}\right)$. You are given $H(10) = 690409338$.

Find $H(10^9)$ modulo $1\,117\,117\,717$.