What is the remainder?

${\color{#D61F06}\text{This problem's question:}}$ What is the remainder? $(2^{2^{2^{2}}} \bmod \left(2^{2^2}+3\right))$

As expressed above, this problem can be worked in Wolfram/Alpha as 2^2^2^2 mod (2^2^2+3)

All the numbers are sufficiently small that pencil and paper are sufficient.

Clarification: Δrchish Ray caught a typographical error. It turns out that the answer is the same for both $(2^{2^{2^{2}}} \bmod \left(2^{2^2}+3\right))$ and $(2^{2^{2^{2^2}}} \bmod \left(2^{2^2}+3\right))$ . Therefore, I will leave this problem in Brilliant. I apologize to those that worked the harder version with a five-high tower of twos. Also as a matter of fact, Wolfram/Alpha can handle both versions, four and five high towers.