You are given a sorted array `nums` of `n` non-negative integers and an integer `maximumBit`. You want to perform the following query `n` times:

1. Find a non-negative integer `k < 2maximumBit` such that `nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k` is maximized. `k` is the answer to the `ith` query.
2. Remove the last element from the current array `nums`.

Return an array `answer`, where `answer[i]` is the answer to the `ith` query.

Example 1:

    Input: nums = [0,1,1,3], maximumBit = 2
    Output: [0,3,2,3]
    Explanation: The queries are answered as follows:
    1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
    2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
    3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
    4th query: nums = [0], k = 3 since 0 XOR 3 = 3.

Example 2:

    Input: nums = [2,3,4,7], maximumBit = 3
    Output: [5,2,6,5]
    Explanation: The queries are answered as follows:
    1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
    2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
    3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
    4th query: nums = [2], k = 5 since 2 XOR 5 = 7.

Example 3:

    Input: nums = [0,1,2,2,5,7], maximumBit = 3
    Output: [4,3,6,4,6,7]

Constraints:

- `nums.length == n`
- `1 <= n <= 105`
- `1 <= maximumBit <= 20`
- `0 <= nums[i] < 2maximumBit`
- `nums`​​​ is sorted in ascending order.