Firstly, let's notice that the first $k+1$ elements of the xorbonacci sequence are cyclically repeated throughout the entire sequence. In other words, it holds $X_A=X_{A+k+1}$ for every $A$. The proof of this statement is left as an exercise to the reader.

Now we can recall some properties of the xor operation (labeled with $\oplus$). More precisely, we are interested in the following properties:

$${\displaystyle \begin{array}{rl}A\oplus 0& =A\\ A\oplus A& =0\\ A\oplus B& =B\oplus A\end{array}}$$

Using these properties, we deduce that it holds:

$${\displaystyle X_L\oplus X_{L+1}\oplus \cdots \oplus X_R=(X_1\oplus X_2\oplus \cdots \oplus X_R)\oplus (X_1\oplus X_2\oplus \cdots \oplus X_{L-1})}$$

Using the sequence's cyclicality from the first paragraph and the aforementioned properties, it is clear that the xor of the first $N$ elements of the xorbonacci sequence is equal to the xor of the first $N^{\prime }$ elements of the xorbonacci sequence where $0\le N^{\prime }\le 2(K+1)$. Therefore, it is enough to preprocess the xors of the first $2(K+1)$ elements and, using them, answer all given queries in $\mathcal{O}(1)$.

Unveiling the explicit relation between $N$ and $N^{\prime }$ is left to you as an exercise.