You are given an array nums of n integers, and a 2D integer array queries of size q, where queries[i] = [li, ri].
For each query, you must find the maximum XOR score of any subarray of nums[li..ri].
The XOR score of an array a is found by repeatedly applying the following operations on a so that only one element remains, that is the score:
a[i] with a[i] XOR a[i + 1] for all indices i except the last one.a.Return an array answer of size q where answer[i] is the answer to query i.
\n
Example 1:
\n\nInput: nums = [2,8,4,32,16,1], queries = [[0,2],[1,4],[0,5]]
\n\nOutput: [12,60,60]
\n\nExplanation:
\n\nIn the first query, nums[0..2] has 6 subarrays [2], [8], [4], [2, 8], [8, 4], and [2, 8, 4] each with a respective XOR score of 2, 8, 4, 10, 12, and 6. The answer for the query is 12, the largest of all XOR scores.
In the second query, the subarray of nums[1..4] with the largest XOR score is nums[1..4] with a score of 60.
In the third query, the subarray of nums[0..5] with the largest XOR score is nums[1..4] with a score of 60.
Example 2:
\n\nInput: nums = [0,7,3,2,8,5,1], queries = [[0,3],[1,5],[2,4],[2,6],[5,6]]
\n\nOutput: [7,14,11,14,5]
\n\nExplanation:
\n\n| Index | \n\t\t\tnums[li..ri] | \n\t\t\tMaximum XOR Score Subarray | \n\t\t\tMaximum Subarray XOR Score | \n\t\t
|---|---|---|---|
| 0 | \n\t\t\t[0, 7, 3, 2] | \n\t\t\t[7] | \n\t\t\t7 | \n\t\t
| 1 | \n\t\t\t[7, 3, 2, 8, 5] | \n\t\t\t[7, 3, 2, 8] | \n\t\t\t14 | \n\t\t
| 2 | \n\t\t\t[3, 2, 8] | \n\t\t\t[3, 2, 8] | \n\t\t\t11 | \n\t\t
| 3 | \n\t\t\t[3, 2, 8, 5, 1] | \n\t\t\t[2, 8, 5, 1] | \n\t\t\t14 | \n\t\t
| 4 | \n\t\t\t[5, 1] | \n\t\t\t[5] | \n\t\t\t5 | \n\t\t
\n
Constraints:
\n\n1 <= n == nums.length <= 20000 <= nums[i] <= 231 - 11 <= q == queries.length <= 105queries[i].length == 2 queries[i] = [li, ri]0 <= li <= ri <= n - 1给你一个由 n 个整数组成的数组 nums，以及一个大小为 q 的二维整数数组 queries，其中 queries[i] = [li, ri]。
对于每一个查询，你需要找出 nums[li..ri] 中任意 子数组 的 最大异或值。
数组的异或值 需要对数组 a 反复执行以下操作，直到只剩一个元素，剩下的那个元素就是 异或值：
i，同时将 a[i] 替换为 a[i] XOR a[i + 1] 。返回一个大小为 q 的数组 answer，其中 answer[i] 表示查询 i 的答案。
\n\n
示例 1：
\n\n输入： nums = [2,8,4,32,16,1], queries = [[0,2],[1,4],[0,5]]
\n\n输出： [12,60,60]
\n\n解释：
\n\n在第一个查询中，nums[0..2] 的子数组分别是 [2], [8], [4], [2, 8], [8, 4], 和 [2, 8, 4]，它们的异或值分别为 2, 8, 4, 10, 12, 和 6。查询的答案是 12，所有异或值中的最大值。
在第二个查询中，nums[1..4] 的子数组中最大的异或值是子数组 nums[1..4] 的异或值，为 60。
在第三个查询中，nums[0..5] 的子数组中最大的异或值是子数组 nums[1..4] 的异或值，为 60。
示例 2：
\n\n输入： nums = [0,7,3,2,8,5,1], queries = [[0,3],[1,5],[2,4],[2,6],[5,6]]
\n\n输出： [7,14,11,14,5]
\n\n解释：
\n\n| 下标 | \n\t\t\tnums[li..ri] | \n\t\t\t最大异或值子数组 | \n\t\t\t子数组最大异或值 | \n\t\t
|---|---|---|---|
| 0 | \n\t\t\t[0, 7, 3, 2] | \n\t\t\t[7] | \n\t\t\t7 | \n\t\t
| 1 | \n\t\t\t[7, 3, 2, 8, 5] | \n\t\t\t[7, 3, 2, 8] | \n\t\t\t14 | \n\t\t
| 2 | \n\t\t\t[3, 2, 8] | \n\t\t\t[3, 2, 8] | \n\t\t\t11 | \n\t\t
| 3 | \n\t\t\t[3, 2, 8, 5, 1] | \n\t\t\t[2, 8, 5, 1] | \n\t\t\t14 | \n\t\t
| 4 | \n\t\t\t[5, 1] | \n\t\t\t[5] | \n\t\t\t5 | \n\t\t
\n\n
提示：
\n\n1 <= n == nums.length <= 20000 <= nums[i] <= 231 - 11 <= q == queries.length <= 105queries[i].length == 2queries[i] = [li, ri]0 <= li <= ri <= n - 1nums[i..j], nums[i..j + 1], nums[i..j + 2], …. Do you notice any pattern?",
"If dp[i][j] is the XOR score of subarray nums[i..j], dp[i][j] = dp[i - 1][j] XOR dp[i - 1][j + 1]."
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