Let $$$h_n(r) = \displaystyle \sum_{i=1}^{n} i^r$$$.

$$$h_n(r), h_m(r)$$$ can be computed for all $$$1 \leq r \leq 2k$$$ for both $$$n, m$$$ in $$$O(k^2)$$$ using the recursion:

Let $$$P_{ij}$$$ be the probability that $$$(i, j)$$$ has a $$$1$$$ in it after all the operations.

By linearity of expectation, we have to find $$$\displaystyle \sum_{i, j} P_{i,j}$$$.

Let $$$p_{i,j}$$$ be the probability that cell $$$(i, j)$$$ is flipped in one such operation.

The total number of submatrices is $$$\dfrac{n(n+1)}{2} \dfrac{m(m+1)}{2}$$$.

The number of submatrices containing $$$(i, j)$$$ is $$$i(n-i+1)j(m-j+1)$$$

So,

Now, $$$P_{i, j} = $$$ probability that this cell is flipped in odd number of operations:

So, we have to find:

Let $$$t = - \dfrac{8}{n(n+1)m(m+1)}$$$. Also, let $$$f_n(i) = i (n + 1 - i)$$$ and expand using binomial theorem:

We have :

Now,

Hence, $$$\displaystyle \sum_{i=1}^{n} f_n(i)^r$$$ can be computed in $$$O(r) = O(k)$$$.

Thus the original formula can be computed in $$$O(k^2)$$$

We convert the problem into:

Given a value $$$r$$$, find the number of lines with distance $$$\leq r$$$ from point $$$q$$$.

For this, consider a circle $$$C$$$ with radius $$$r$$$ centered at $$$q$$$. Consider two points $$$A$$$ and $$$B$$$, both outside the circle $$$C$$$. The line passing through $$$A$$$ and $$$B$$$ has distance $$$\leq r$$$ from $$$q$$$ iff it intersects the circle $$$C$$$. Let $$$F_A$$$ and $$$G_A$$$ be the points of contacts of tangents drawn from $$$A$$$ to the circle. Similarly define $$$F_B$$$ and $$$G_B$$$.

We can prove that the line passing through points $$$A$$$ and $$$B$$$ intersects the circle $$$C$$$ if and only if the line segments $$$F_{A} G_{A}$$$ and $$$F_{B}G_{B}$$$ do NOT intersect.

So, we need to draw $$$2 n$$$ tangents, and then draw $$$n$$$ chords passing through the respective points of contacts, and count the number of intersections of these chords.

For this, sort the points by polar angles in $$$O(n \log{n}) $$$. Now we can count the number of intersections of line segments in $$$O(n \log{n})$$$, by iterating on the points.

Therefore, this question can be answered in $$$O(n \log{n})$$$.

Then we can just binary search to find the answer in complexity $$$O(n \log{n} \log{\frac{R}{\epsilon}})$$$