First, let us assume that the turtle's apartment is infinite and contains no furniture. Consider which cells the turtle will visit if it starts its path from cell $$$(1, 1)$$$. If the turtle eventually shifts to the right by more than $$$m$$$, then the answer to the problem is $$$0$$$. Otherwise, for each $$$y$$$ from $$$1$$$ to $$$m$$$, it is easy to see that the turtle either visits no cells in column $$$y$$$, or visits a segment of cells $$$[l_y, r_y]$$$: that is, all cells $$$(x, y)$$$ where $$$l_y \le x \le r_y$$$.

The values $$$l_y$$$ and $$$r_y$$$ can be precomputed in $$$\mathcal{O}(|s|)$$$. Using these values, it is then easy to check in $$$\mathcal{O}(m)$$$ for a fixed cell $$$(i, j)$$$ whether it satisfies the condition of the problem:

• For each $$$y$$$ from $$$1$$$ to $$$m$$$, if the segment $$$[l_y, r_y]$$$ is non-empty, one must verify that $$$j + y - 1 \le m$$$, and also that all cells in this column on the segment $$$[i + l_y - 1, i + r_y - 1]$$$ exist and are not occupied by furniture.

To perform this check efficiently for each column, it is sufficient to compute prefix sums in every column. This method allows checking a fixed cell in $$$\mathcal{O}(m)$$$, which gives a solution in $$$\mathcal{O}(|s| + nm^2)$$$ if we perform the check independently for every cell. This solution is sufficient to solve the problem for full score.

This problem can also be solved in $$$\mathcal{O}(|s| + nm\log(nm))$$$ using multiplication of two-dimensional polynomials, but this was not required.