Eccoci.

We use $x\oplus y$ for the bitwise XOR of $x$ and $y$.
Define the XOR-product of $x$ and $y$, denoted by $x \otimes y$, similar to a long multiplication in base $2$, except that the intermediate results are XORed instead of the usual integer addition.

For example, $7 \otimes 3 = 9$, or in base $2$, $111_2 \otimes 11_2 = 1001_2$:

\begin{align*} \phantom{\otimes 111} 111_2 \\ \otimes \phantom{1111} 11_2 \\ \hline \phantom{\otimes 111} 111_2 \\ \oplus \phantom{11} 111_2 \phantom{9} \\ \hline \phantom{\otimes 11} 1001_2 \\ \end{align*} We consider the equation: \begin{align} (a \otimes a) \oplus (2 \otimes a \otimes b) \oplus (b \otimes b) = 5 \end{align} For example, $(a, b) = (3, 6)$ is a solution.

Let $X(N)$ be the XOR of the $b$ values for all solutions to this equation satisfying $0 \le a \le b \le N$.
You are given $X(10)=5$.

Find $X(10^{18})$.