You are given an event that runs from time $0$ to time eventTime. There are n non overlapping meetings scheduled during the event, where the ith meeting occupies the half open interval [startTime[i], endTime[i]).

You may reschedule at most k meetings. Rescheduling a meeting means choosing a new half open interval of the same duration that lies entirely within $[0,$ eventTime $]$ and does not overlap any other meeting after all reschedules are applied.

Return the maximum possible length of a continuous time interval during the event that contains no meetings.

Note: Meetings are given in non overlapping order, meaning each meeting ends no later than the next meeting starts.

Constraints:

• $1 \leq$ eventTime $\leq 10^9$

• n $==$ startTime.length $==$ endTime.length

• $2 \leq$ n $\leq 10^5$

• $1 \leq$ k $\leq$ n

• $0 \leq$ startTime[i] $<$ endTime[i] $\leq$ eventTime

• endTime[i] $\leq$ startTime[i + 1] where i lies in the range $[0,$ n $- 2]$

Input: eventTime = 12-int, k = 1-int, startTime = [1,6,9]-List[int], endTime = [3,7,10]-List[int]
Output: 5-int
Explanation: The meetings are [1,3), [6,7), and [9,10). By rescheduling at most one meeting (for example, moving one meeting to merge gaps and leave a larger uninterrupted interval), you can create a longest continuous free block of length 5.

Input: eventTime = 20-int, k = 2-int, startTime = [2,5,11,16]-List[int], endTime = [4,6,12,18]-List[int]
Output: 11-int
Explanation: The meetings are [2,4), [5,6), [11,12), and [16,18). With up to two reschedules, you can “pack” meetings closer together (without changing durations or overlapping), producing a maximum free interval of length 11.

Input: eventTime = 8-int, k = 3-int, startTime = [0,2,4,6]-List[int], endTime = [1,3,5,7]-List[int]
Output: 4-int
Explanation: The meetings are [0,1), [2,3), [4,5), and [6,7). Since you may reschedule up to 3 meetings, you can cluster them into a tighter region, leaving a continuous free interval of length 4 somewhere within [0,8).

Input: eventTime = 1000000000-int, k = 1-int, startTime = [100,200000000,700000000,900000000]-List[int], endTime = [1000,200000100,700000050,900000010]-List[int]
Output: 699999850-int
Explanation: Even with a huge timeline, the same rules apply: reschedule at most one meeting (keeping its duration) to avoid overlaps and maximize a single continuous free block. The optimal rearrangement yields a longest free interval of 699999850.

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