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:

$$\displaystyle \begin{align*} A \oplus 0 &= A \\ A \oplus A &= 0 \\ A \oplus B &= B \oplus A \end{align*}$$

Using these properties, we deduce that it holds:

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

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'~ elements of the xorbonacci sequence where ~0 \le N' \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'~ is left to you as an exercise.