You are given a directed acyclic graph of n nodes numbered from 0 to n − 1. This is represented by a 2D array edges of length m, where edges[i] = [ui, vi, costi] indicates a one‑way communication from node ui to node vi with a recovery cost of costi.
Some nodes may be offline. You are given a boolean array online where online[i] = true means node i is online. Nodes 0 and n − 1 are always online.
A path from 0 to n − 1 is valid if:
k.For each valid path, define its score as the minimum edge‑cost along that path.
Return the maximum path score (i.e., the largest minimum-edge cost) among all valid paths. If no valid path exists, return -1.
Example 1:
Input: edges = [[0,1,5],[1,3,10],[0,2,3],[2,3,4]], online = [true,true,true,true], k = 10
Output: 3
Explanation:

The graph has two possible routes from node 0 to node 3:
Path 0 → 1 → 3
Total cost = 5 + 10 = 15, which exceeds k (15 > 10), so this path is invalid.
Path 0 → 2 → 3
Total cost = 3 + 4 = 7 <= k, so this path is valid.
The minimum edge‐cost along this path is min(3, 4) = 3.
There are no other valid paths. Hence, the maximum among all valid path‐scores is 3.
Example 2:
Input: edges = [[0,1,7],[1,4,5],[0,2,6],[2,3,6],[3,4,2],[2,4,6]], online = [true,true,true,false,true], k = 12
Output: 6
Explanation:

Node 3 is offline, so any path passing through 3 is invalid.
Consider the remaining routes from 0 to 4:
Path 0 → 1 → 4
Total cost = 7 + 5 = 12 <= k, so this path is valid.
The minimum edge‐cost along this path is min(7, 5) = 5.
Path 0 → 2 → 3 → 4
Node 3 is offline, so this path is invalid regardless of cost.
Path 0 → 2 → 4
Total cost = 6 + 6 = 12 <= k, so this path is valid.
The minimum edge‐cost along this path is min(6, 6) = 6.
Among the two valid paths, their scores are 5 and 6. Therefore, the answer is 6.
Constraints:
n == online.length2 <= n <= 5 * 1040 <= m == edges.length <= min(105, n * (n - 1) / 2)edges[i] = [ui, vi, costi]0 <= ui, vi < nui != vi0 <= costi <= 1090 <= k <= 5 * 1013online[i] is either true or false, and both online[0] and online[n − 1] are true.