Fifty-Fifty

Let $p_{4,3}(n)$ be the smallest prime $p$ such that $p>n$ and $p\equiv 3(\mathrm{mod}4)$ and let $$p_{4,3}^k(n):=\{\begin{array}{ll}{p}_{4,3}^{k-1}(p_{4,3}(n))& \text{if }k>1\\ p_{4,3}(n)& \text{if }k=1\end{array}.$$

Let $F(n):=\frac{n-1}{2}!modn$.

For example, $F(3)=1$ and $F(47)=46$.

Let $S(n,m):=\sum _{k=1}^{m}F(p_{4,3}^k(n))$.

You are given $S(1,32)=1856$, $S({10}^7,32)=150008239$ and $S({10}^{12},32)=18000000012930$.

Find $S({10}^{18},32)$.

[My timing: 1.8 seconds (PyPy)]

Note: It would be hard without parigp.