Since we have to check which ticks are prime, we can use the Sieve of Eratosthenes to generate a list of primes. We then have to perform interwoven updates and queries, meaning that both operations must be efficient. Using traditional data structures, we can only achieve imbalanced $\mathcal{O}(1)$ and $\mathcal{O}(N)$ operations (with prefix sum/difference arrays), or $\mathcal{O}(\log N)$ and $\mathcal{O}(N\log N)$ operations (with Binary Indexed Trees or BITs). We thus have to use a range update and range query binary indexed tree that would enable $\mathcal{O}(\log N)$ for both operation types.

We notice that, given a single range update of $[a,b]$ by $k$, and then a query to the sum $S_x$ of elements in the range $[0,x]$, there are 3 possibilities for the value of the sum:

$${\displaystyle S_x=\{\begin{array}{ll}0& \text{if }x<a\\ k(x-(a-1))=kx-k(a-1)& \text{if }a\le x\le b\\ k(b-(a-1))=kb-k(a-1)& \text{if }x>b\end{array}}$$

We see that we can express this sum in the form $\delta x-\lambda$, where $\delta$ is the value in the array at index $x$ after update, and $\lambda$ is a "fix" value that we subtract in order to correct the sum. The value of $\lambda$ varies depending on $x$ (draw it out for better understanding):

$${\displaystyle \lambda =\{\begin{array}{ll}0& \text{if }x<a\\ k(a-1)& \text{if }a\le x\le b\\ -kb+k(a-1)& \text{if }x>b\end{array}}$$

Showing that this idea generalizes to multiple updates and queries is left as an exercise for the reader. With this mathematical formulation, we can use 2 BITs to implement this algorithm: one to store the accumulated $\delta$ values (which we maintain like a normal range update point query BIT), and one to store the accumulated $\lambda$ corrections.

Or, you know, use a segment tree with lazy propagation, but the BIT solution is less clunky.

Time complexity: $\mathcal{O}(T\log \log T+T\log N)$