MO's with Updates - Analysis

With block size $B = n^{2/3}$: $l$-pointer: moves $O(B)$ per query = $O(qn^{2/3})$ $r$-pointer: for each $l$-block, $r$ sweeps the full $r$-range once, so $O(n)$ per $l$-block. With $n^{1/3}$ $l$-blocks, total = $O(n \cdot n^{1/3}) = O(n^{4/3})$ $t$-pointer: moves $O(n)$ per $(l,r)$-block pair, $(n/B)^2$ pairs = $O(n \cdot n^{2/3}) = O(n^{5/3})$

Total: $O(n^{5/3} + qn^{2/3})$. For $q \approx n$, the complexity is $O(n^{5/3})$.