Show that if L is CFL and R is a regular language then {w ∈ Σ^∗ | xw ∈ L for some x ∈ R} is context free

Show that if $$L$$ is CFL and $$R$$ is a regular language such that they both share the same input alphabet $$\Sigma$$, then $$C = \{w \in \Sigma^*\mid xw \in L$$ for some $$x \in R\}$$ is context free.

Hi I've been struggling quite a lot with this question. There's a similar quesion already answered here (Closure of CFL against right-quotient with regular languages)

But the difference here is that we have $$xw \in L$$ and not $$wx \in L$$.

Can anyone give some ideas/help?

The reverse of a language $$L$$ is the language $$L^R = \{ w^R : w \in L \}$$, where $$w^R$$ is obtained from $$w$$ by reversing the orders of the letters.
The reverse of a context-free language is context-free. This can be shown by starting with a grammar and reversing all production rules, that is, replacing $$A \to \alpha$$ by $$A \to \alpha^R$$. Similarly, the reverse of a regular language is regular. This can be shown in many ways: by reversing the arrows in an NFA, by reversing the production rules in a regular grammar, by reversing regular expressions, and so on.
Now define $$L/R = \{ w : wx \in L \text{ for some } x \in R \}$$ and $$R \backslash L = \{ w : xw \in L \text{ for some } x \in R \}$$, and notice that $$R \backslash L = (L^R/R^R)^R$$. Hence closure under regular right quotient implies closure under regular left quotient.