The number with digit sum 10

Writing down the numbers which have a digit sum of 10 in ascending order, we get: $19,28,37,46,55,64,73,82,91,109,118,\dots$

Let $f(n)$ be the $n^{\text{th}}$ occurrence of the digit sum $10$. For example, $f(1)=19$, $f(10)=109$ and $f(100)=1423$.

Let ${\displaystyle S(k)=\sum _{n=1}^{k}f(n)}$. For example $S(100)=67582$ and $S({10}^5)=178986850167799$.

Find ${\displaystyle \sum _{i=1}^{100}S({11}^i)}$.

Give your answer modulo $({10}^{18}+3)$