<p>You are given an array <code>nums</code> consisting of non-negative integers. You are also given a <code>queries</code> array, where <code>queries[i] = [x<sub>i</sub>, m<sub>i</sub>]</code>.</p>

<p>The answer to the <code>i<sup>th</sup></code> query is the maximum bitwise <code>XOR</code> value of <code>x<sub>i</sub></code> and any element of <code>nums</code> that does not exceed <code>m<sub>i</sub></code>. In other words, the answer is <code>max(nums[j] XOR x<sub>i</sub>)</code> for all <code>j</code> such that <code>nums[j] &lt;= m<sub>i</sub></code>. If all elements in <code>nums</code> are larger than <code>m<sub>i</sub></code>, then the answer is <code>-1</code>.</p>

<p>Return <em>an integer array </em><code>answer</code><em> where </em><code>answer.length == queries.length</code><em> and </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>

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

<pre>
<strong>Input:</strong> nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]
<strong>Output:</strong> [3,3,7]
<strong>Explanation:</strong>
1) 0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.
2) 1 XOR 2 = 3.
3) 5 XOR 2 = 7.
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]
<strong>Output:</strong> [15,-1,5]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length, queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= nums[j], x<sub>i</sub>, m<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>

class Trie:
    L = 30

    def __init__(self):
        self.left = None
        self.right = None

    def insert(self, val: int):
        node = self
        for i in range(Trie.L, -1, -1):
            bit = (val >> i) & 1
            if bit == 0:
                if not node.left:
                    node.left = Trie()
                node = node.left
            else:
                if not node.right:
                    node.right = Trie()
                node = node.right
    
    def getMaxXor(self, val: int) -> int:
        ans, node = 0, self
        for i in range(Trie.L, -1, -1):
            bit = (val >> i) & 1
            check = False
            if bit == 0:
                if node.right:
                    node = node.right
                    check = True
                else:
                    node = node.left
            else:
                if node.left:
                    node = node.left
                    check = True
                else:
                    node = node.right
            if check:
                ans |= 1 << i
        return ans


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        n, q = len(nums), len(queries)
        nums.sort()
        queries = sorted([(x, m, i) for i, (x, m) in enumerate(queries)], key=lambda query: query[1])
        
        ans = [0] * q
        t = Trie()
        idx = 0
        for x, m, qid in queries:
            while idx < n and nums[idx] <= m:
                t.insert(nums[idx])
                idx += 1
            if idx == 0:
                # 字典树为空
                ans[qid] = -1
            else:
                ans[qid] = t.getMaxXor(x)
        
        return ans
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