You have `k` servers numbered from `0` to `k-1` that are being used to handle multiple requests simultaneously. Each server has infinite computational capacity but cannot handle more than one request at a time. The requests are assigned to servers according to a specific algorithm:

- The `ith` (0-indexed) request arrives.
- If all servers are busy, the request is dropped (not handled at all).
- If the `(i % k)th` server is available, assign the request to that server.
- Otherwise, assign the request to the next available server (wrapping around the list of servers and starting from 0 if necessary). For example, if the `ith` server is busy, try to assign the request to the `(i+1)th` server, then the `(i+2)th` server, and so on.

You are given a strictly increasing array `arrival` of positive integers, where `arrival[i]` represents the arrival time of the `ith` request, and another array `load`, where `load[i]` represents the load of the `ith` request (the time it takes to complete). Your goal is to find the busiest server(s). A server is considered busiest if it handled the most number of requests successfully among all the servers.

Return a list containing the IDs (0-indexed) of the busiest server(s). You may return the IDs in any order.

Example 1:

    Input: k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3]
    Output: [1]
    Explanation:
    All of the servers start out available.
    The first 3 requests are handled by the first 3 servers in order.
    Request 3 comes in. Server 0 is busy, so it's assigned to the next available server, which is 1.
    Request 4 comes in. It cannot be handled since all servers are busy, so it is dropped.
    Servers 0 and 2 handled one request each, while server 1 handled two requests. Hence server 1 is the busiest server.

Example 2:

    Input: k = 3, arrival = [1,2,3,4], load = [1,2,1,2]
    Output: [0]
    Explanation:
    The first 3 requests are handled by first 3 servers.
    Request 3 comes in. It is handled by server 0 since the server is available.
    Server 0 handled two requests, while servers 1 and 2 handled one request each. Hence server 0 is the busiest server.

Example 3:

    Input: k = 3, arrival = [1,2,3], load = [10,12,11]
    Output: [0,1,2]
    Explanation: Each server handles a single request, so they are all considered the busiest.

Constraints:

- `1 <= k <= 105`
- `1 <= arrival.length, load.length <= 105`
- `arrival.length == load.length`
- `1 <= arrival[i], load[i] <= 109`
- `arrival` is strictly increasing.