Assume that the contribution of the YYs in \(\mathrm{str}(P)\) to the score are \(1, 3, 5, \ldots\) from left to right. Indeed, if YY occurs \(k\) times in \(\mathrm{str}(P)\), the score is \(k^2\), and we have:

\[1 + 3 + 5 + \cdots + (2k - 1) = \sum_{i = 1}^k (2i - 1) = k^2.\]

If, for every pair of adjacent Ys in the grid, we can compute its total contribution over all possible paths, the answer is the sum of those values.

Assume that both \((i, j)\) and \((i + 1, j)\) have Y written on them, where \(1 \le i < H,\ 1 \le j \le W\). Then, the total contribution of this YY is \(a + 2b\), where:

• \(a\) is the number of paths containing the move \((i, j) \to (i + 1, j)\), and
• \(b\) is the total number of times those paths move between adjacent Y squares before reaching \((i, j)\).

Here, we have:

\[a = \binom{i+j-2}{i-1} \times \binom{H+W-i-j-1}{H-i-1}.\]

Let \(\mathrm{dp}(i, j)\) be the total number of times the paths move between adjacent Y squares before reaching \((i, j)\). After computing these by naive dynamic programming (DP), we can compute:

\[b = \mathrm{dp}(i, j) \times \binom{H+W-i-j-1}{H-i-1}.\]

We can similarly process a horizontal pair of \((i, j)\) and \((i, j + 1)\), both of which have Y written on them, where \(1 \le i \le H,\ 1 \le j < W\). The DP in the pre-computation and the computation of the total contribution both take \(\mathrm{O}(HW)\) time, which is the complexity of this solution.

Sample implementation