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\textit{Problem author and developer: Alexey Mikhnenko}
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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:
\begin{itemize}
\item 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.
\end{itemize}

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.