<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, queries):
        n = len(nums)
        score = [[0] * (n + 1) for _ in range(n + 1)]
        
        for l in range(n - 1, -1, -1):
            last_pwr = 1
            score[l][1] = nums[l]
            
            for r in range(l + 1, n):
                length = r - l + 1
                score[l][length] = score[l][length - last_pwr] ^ score[l + last_pwr][length - last_pwr]
                
                if last_pwr == length:
                    last_pwr *= 2
        
        suffix = [[0] * (n + 1) for _ in range(n + 1)]
        for r in range(n):
            for c in range(r, n):
                suffix[r][c] = score[r][c - r + 1]
        
        for r in range(n):
            for c in range(1, n):
                suffix[r][c] = max(suffix[r][c], suffix[r][c - 1])
        
        for c in range(n):
            for r in range(n - 2, -1, -1):
                suffix[r][c] = max(suffix[r][c], suffix[r + 1][c])
        
        result = []
        for q in queries:
            result.append(suffix[q[0]][q[1]])
        
        return result

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