<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>


from dataclasses import dataclass


class TrieNode:
  def __init__(self):
    self.children: list[TrieNode | None] = [None] * 2


class BitTrie:
  def __init__(self, maxBit: int):
    self.maxBit = maxBit
    self.root = TrieNode()

  def insert(self, num: int) -> None:
    node = self.root
    for i in range(self.maxBit, -1, -1):
      bit = num >> i & 1
      if not node.children[bit]:
        node.children[bit] = TrieNode()
      node = node.children[bit]

  def getMaxXor(self, num: int) -> int:
    maxXor = 0
    node = self.root
    for i in range(self.maxBit, -1, -1):
      bit = num >> i & 1
      toggleBit = bit ^ 1
      if node.children[toggleBit]:
        maxXor = maxXor | 1 << i
        node = node.children[toggleBit]
      elif node.children[bit]:
        node = node.children[bit]
      else:  # There's nothing in the Bit Trie.
        return 0
    return maxXor


@dataclass(frozen=True)
class IndexedQuery:
  queryIndex: int
  x: int
  m: int

  def __iter__(self):
    yield self.queryIndex
    yield self.x
    yield self.m


class Solution:
  def maximizeXor(self, nums: list[int], queries: list[list[int]]) -> list[int]:
    ans = [-1] * len(queries)
    maxBit = int(math.log2(max(max(nums), max(x for x, _ in queries))))
    bitTrie = BitTrie(maxBit)

    nums.sort()

    i = 0  # nums' index
    for queryIndex, x, m in sorted([IndexedQuery(i, x, m)
                                    for i, (x, m) in enumerate(queries)],
                                   key=lambda x: x.m):
      while i < len(nums) and nums[i] <= m:
        bitTrie.insert(nums[i])
        i += 1
      if i > 0 and nums[i - 1] <= m:
        ans[queryIndex] = bitTrie.getMaxXor(x)

    return ans

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