<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 NodeClass:
    def __init__(self):
        self.edges = [None] * 2
class BinaryTrieClass:
    def __init__(self, max_bit):
        self.bit_limit = max_bit
        self.root = NodeClass()
    def add_number(self, value):
        current = self.root
        for bit_position in range(self.bit_limit, -1, -1):
            bit = (value >> bit_position) & 1
            if current.edges[bit] is None:
                current.edges[bit] = NodeClass()
            current = current.edges[bit]
    def compute_max_xor(self, value):
        max_xor_value = 0
        current = self.root
        for bit_position in range(self.bit_limit, -1, -1):
            bit = (value >> bit_position) & 1
            alternate_bit = bit ^ 1
            if current.edges[alternate_bit] is not None:
                max_xor_value |= (1 << bit_position)
                current = current.edges[alternate_bit]
            elif current.edges[bit] is not None:
                current = current.edges[bit]
            else:
                return 0
        return max_xor_value
class Solution:
    def maximizeXor(self, numbers, queries_list):
        results = [-1] * len(queries_list)
        max_number_in_numbers = max(numbers)
        max_number_in_query = max(query[0] for query in queries_list)
        bit_limit = int(max(max_number_in_numbers, max_number_in_query).bit_length()) - 1
        trie = BinaryTrieClass(bit_limit)
        numbers.sort()
        sorted_queries = self.prepare_sorted_queries(queries_list)
        i = 0
        for query_index, x, m in sorted_queries:
            while i < len(numbers) and numbers[i] <= m:
                trie.add_number(numbers[i])
                i += 1
            if i > 0 and numbers[i - 1] <= m:
                results[query_index] = trie.compute_max_xor(x)
        return results
    def prepare_sorted_queries(self, queries_list):
        sorted_queries = [(i, query[0], query[1]) for i, query in enumerate(queries_list)]
        sorted_queries.sort(key=lambda query: query[2])
        return sorted_queries

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