<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 TrieNode:
    def __init__(self):
        self.children = {}

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, num):
        node = self.root
        for bit in range(31, -1, -1):
            bit_val = (num >> bit) & 1
            if bit_val not in node.children:
                node.children[bit_val] = TrieNode()
            node = node.children[bit_val]

    def findMaxXOR(self, num):
        if not self.root.children:
            return -1
        node = self.root
        max_xor = 0
        for bit in range(31, -1, -1):
            bit_val = (num >> bit) & 1
            opposite_bit = 1 - bit_val
            if opposite_bit in node.children:
                max_xor |= (1 << bit)
                node = node.children[opposite_bit]
            else:
                node = node.children[bit_val]
        return max_xor

class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        nums.sort()
        queries = sorted((m, x, i) for i, (x, m) in enumerate(queries))
        trie = Trie()
        answer = [-1] * len(queries)
        j = 0
        
        for m, x, i in queries:
            while j < len(nums) and nums[j] <= m:
                trie.insert(nums[j])
                j += 1
            answer[i] = trie.findMaxXOR(x)
        
        return answer

# Example usage:
solution = Solution()
nums = [0, 1, 2, 3, 4]
queries = [[3, 1], [1, 3], [5, 6]]
print(solution.maximizeXor(nums, queries))  # Output: [3, 3, 7]

nums = [5, 2, 4, 6, 6, 3]
queries = [[12, 4], [8, 1], [6, 3]]
print(solution.maximizeXor(nums, queries))  # Output: [15, -1, 5]
451241