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 $2$, ${111}_2\otimes {11}_2={1001}_2$:

$$\begin{array}{r}\phantom{\otimes 111}{111}_2\\ \otimes \phantom{1111}{11}_2\\ \phantom{\otimes 111}{111}_2\\ \oplus \phantom{11}{111}_2\phantom{9}\\ \phantom{\otimes 11}{1001}_2\end{array}$$ We consider the equation: $$\begin{array}{r}(a\otimes a)\oplus (2\otimes a\otimes b)\oplus (b\otimes b)=5\end{array}$$ 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 $X({10}^{18})$.