<p>You are given an array <code>nums</code> of <code>n</code> integers, and a 2D integer array <code>queries</code> of size <code>q</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>

<p>For each query, you must find the <strong>maximum XOR score</strong> of any <span data-keyword="subarray">subarray</span> of <code>nums[l<sub>i</sub>..r<sub>i</sub>]</code>.</p>

<p>The <strong>XOR score</strong> of an array <code>a</code> is found by repeatedly applying the following operations on <code>a</code> so that only one element remains, that is the <strong>score</strong>:</p>

<ul>
	<li>Simultaneously replace <code>a[i]</code> with <code>a[i] XOR a[i + 1]</code> for all indices <code>i</code> except the last one.</li>
	<li>Remove the last element of <code>a</code>.</li>
</ul>

<p>Return an array <code>answer</code> of size <code>q</code> where <code>answer[i]</code> is the answer to query <code>i</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,8,4,32,16,1], queries = [[0,2],[1,4],[0,5]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[12,60,60]</span></p>

<p><strong>Explanation:</strong></p>

<p>In the first query, <code>nums[0..2]</code> has 6 subarrays <code>[2]</code>, <code>[8]</code>, <code>[4]</code>, <code>[2, 8]</code>, <code>[8, 4]</code>, and <code>[2, 8, 4]</code> 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.</p>

<p>In the second query, the subarray of <code>nums[1..4]</code> with the largest XOR score is <code>nums[1..4]</code> with a score of 60.</p>

<p>In the third query, the subarray of <code>nums[0..5]</code> with the largest XOR score is <code>nums[1..4]</code> with a score of 60.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [0,7,3,2,8,5,1], queries = [[0,3],[1,5],[2,4],[2,6],[5,6]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[7,14,11,14,5]</span></p>

<p><strong>Explanation:</strong></p>

<table height="70" width="472">
	<thead>
		<tr>
			<th>Index</th>
			<th>nums[l<sub>i</sub>..r<sub>i</sub>]</th>
			<th>Maximum XOR Score Subarray</th>
			<th>Maximum Subarray XOR Score</th>
		</tr>
	</thead>
	<tbody>
		<tr>
			<td>0</td>
			<td>[0, 7, 3, 2]</td>
			<td>[7]</td>
			<td>7</td>
		</tr>
		<tr>
			<td>1</td>
			<td>[7, 3, 2, 8, 5]</td>
			<td>[7, 3, 2, 8]</td>
			<td>14</td>
		</tr>
		<tr>
			<td>2</td>
			<td>[3, 2, 8]</td>
			<td>[3, 2, 8]</td>
			<td>11</td>
		</tr>
		<tr>
			<td>3</td>
			<td>[3, 2, 8, 5, 1]</td>
			<td>[2, 8, 5, 1]</td>
			<td>14</td>
		</tr>
		<tr>
			<td>4</td>
			<td>[5, 1]</td>
			<td>[5]</td>
			<td>5</td>
		</tr>
	</tbody>
</table>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n == nums.length &lt;= 2000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>1 &lt;= q == queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length == 2 </code></li>
	<li><code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code></li>
	<li><code>0 &lt;= l<sub>i</sub> &lt;= r<sub>i</sub> &lt;= n - 1</code></li>
</ul>

class Solution:
    def maximumSubarrayXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        @cache
        def dp(i,j):
            if i==j: return nums[i],nums[i]
            (a,sa),(b,sb) = dp(i+1,j),dp(i,j-1)
            return max(a,b,sa^sb),sa^sb
        res = [dp(a,b)[0] for a,b in queries]
        dp.cache_clear()
        return res
            

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