What is the remainder of : $\sum\_{i=0}^{2019} 3^i$ divided by $3^4$ ?

I know that $$\sum\_{i=0}^{2019} 3^i = \frac{3^{2020}}{3-1} = \frac{3^{2020}}{2} $$

so $$\frac{\sum\_{i=0}^{2019} 3^i}{3^4} = \frac{3^{2016}}{2}$$

how do I proceed from here? I tried running a program in python on it but I think the number is too big