Each contiguous portion of a hiring agent's path can be represented as a 2D segment with \(6\) properties (agent's index, segment's initial row, segment's initial column, turns taken by the agent before reaching the segment's initial row/column, direction, and segment length). Each agent's full path corresponds to one such segment (with \(0\) turns taken before reaching it), but we may want to split/trim/remove some of these segments, based on which portions of them "cover" portions of others (visit cells later, thus replacing their flyers).

For each relevant row containing \(a_1\) `'E'` agents and \(a_2\) `'W'` agents (with \(a_1 + a_2 \gt 0\)), our first goal will be to transform all of its `'E'` agents' segments into a disjoint set of segments. For these, if segment \(s_1\) begins in an earlier column than segment \(s_2\) (or also in the same column if \(s_1\)'s agent index is greater than \(s_2\)'s), then \(s_1\) will cover \(s_2\) up to the end of \(s_1\), meaning that \(s_2\) should be trimmed to begin past that column (or removed entirely if it would become empty). Based on this idea, the `'E'` agents can be handled in \(O(a_1 \log(a_1))\) time, involving sorting them by their initial columns. We can repeat much the same process to transform all of the `'W'` agents' segments into a disjoint set of segments.

Our next step will be to combine the `'E'` and `'W'` segments together into a single disjoint set of segments. This can be done in \(O((a_1 + a_2)\log(a_1 + a_2))\) time using a line sweep across the segments' endpoints. For each interval between such endpoints, if it has at most ongoing segment (either `'E'` or a `'W'`), we'll simply keep that segment (trimmed down to that interval). On the other hand, if the interval has both a `'E'` and a `'W'` segment ongoing, we can compute how they interact within that interval, with the `'E'` segment potentially covering some right portion of the `'W'` segment and vice versa (resulting in at most disjoint \(2\) segments for the interval). Note that we'll end up with \(O(a_1 + a_2)\) segments overall.

Finally, all of the above should be repeated for each relevant column, transforming all of the `'S'` and `'N'` agents' segments into a disjoint set of segments. Based on the \(O(N)\) transformed segments, we can then trivially compute a base answer, ignoring interactions of horizontal segments with vertical ones.

What remains is subtracting from the answer based on each cell at which a horizontal segment intersects with a vertical one, with one segment covering the other there. We'll consider each possible pair of directions (`'E'` covered by `'S'`, `'N'` covered by `'W'`, etc.) independently. Through a combination of \(90 \degree\) rotations and axis-aligned flips of the entire grid and its segments, all \(8\) direction pairs can be transformed into just considering `'E'` covered by `'S'`.

We'll perform a line sweep downward through the `'E'` segments' rows and the `'S'` segments top/bottom rows, while maintaining some info about the ongoing `'S'` segments. It's possible to compute a constant "time value" for each `'E'`/`'S'` segment such that, whenever \(2\) segments intersect, the one with the larger time value will cover the other. In particular, a way to compute this time value (properly accounting for row/column offsets and breaking ties based on agent indices) is \(N*(\)(turns previously taken by agent) \(-\) (initial row) \(-\) (initial column)\() + \) (agent index).

We can therefore maintain a 2D segment tree (over the \(O(N)\) coordinate-compressed relevant columns), with each node of the tree storing the set of time values of ongoing `'S'` segments in its interval of columns (most conveniently in an order statistics tree, though a coordinate-compressed Fenwick tree can also work). Whenever a `'S'` segment starts or ends, we'll update the segment tree accordingly, and whenever an `'E'` segment is encountered, we can compute the number of its cells which are covered by querying for the number of time values greater than its own time value present in its interval of columns.

The total time complexity of this solution is \(O(N \log^2(N))\), due to updating and querying the 2D segment tree \(O(N)\) times.