<p>There is a rooted tree consisting of <code>n</code> nodes numbered <code>0</code> to <code>n - 1</code>. Each node&#39;s number denotes its <strong>unique genetic value</strong> (i.e. the genetic value of node <code>x</code> is <code>x</code>). The <strong>genetic difference</strong> between two genetic values is defined as the <strong>bitwise-</strong><strong>XOR</strong> of their values. You are given the integer array <code>parents</code>, where <code>parents[i]</code> is the parent for node <code>i</code>. If node <code>x</code> is the <strong>root</strong> of the tree, then <code>parents[x] == -1</code>.</p>

<p>You are also given the array <code>queries</code> where <code>queries[i] = [node<sub>i</sub>, val<sub>i</sub>]</code>. For each query <code>i</code>, find the <strong>maximum genetic difference</strong> between <code>val<sub>i</sub></code> and <code>p<sub>i</sub></code>, where <code>p<sub>i</sub></code> is the genetic value of any node that is on the path between <code>node<sub>i</sub></code> and the root (including <code>node<sub>i</sub></code> and the root). More formally, you want to maximize <code>val<sub>i</sub> XOR p<sub>i</sub></code>.</p>

<p>Return <em>an array </em><code>ans</code><em> where </em><code>ans[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>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/c1.png" style="width: 118px; height: 163px;" />
<pre>
<strong>Input:</strong> parents = [-1,0,1,1], queries = [[0,2],[3,2],[2,5]]
<strong>Output:</strong> [2,3,7]
<strong>Explanation: </strong>The queries are processed as follows:
- [0,2]: The node with the maximum genetic difference is 0, with a difference of 2 XOR 0 = 2.
- [3,2]: The node with the maximum genetic difference is 1, with a difference of 2 XOR 1 = 3.
- [2,5]: The node with the maximum genetic difference is 2, with a difference of 5 XOR 2 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/c2.png" style="width: 256px; height: 221px;" />
<pre>
<strong>Input:</strong> parents = [3,7,-1,2,0,7,0,2], queries = [[4,6],[1,15],[0,5]]
<strong>Output:</strong> [6,14,7]
<strong>Explanation: </strong>The queries are processed as follows:
- [4,6]: The node with the maximum genetic difference is 0, with a difference of 6 XOR 0 = 6.
- [1,15]: The node with the maximum genetic difference is 1, with a difference of 15 XOR 1 = 14.
- [0,5]: The node with the maximum genetic difference is 2, with a difference of 5 XOR 2 = 7.
</pre>

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

<ul>
	<li><code>2 &lt;= parents.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= parents[i] &lt;= parents.length - 1</code> for every node <code>i</code> that is <strong>not</strong> the root.</li>
	<li><code>parents[root] == -1</code></li>
	<li><code>1 &lt;= queries.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= node<sub>i</sub> &lt;= parents.length - 1</code></li>
	<li><code>0 &lt;= val<sub>i</sub> &lt;= 2 * 10<sup>5</sup></code></li>
</ul>

class Trie:
    def __init__(self):
        self.root = {}

    def insert(self, x):
        node = self.root
        for i in range(18, -1, -1):
            bit = (x >> i) & 1
            node = node.setdefault(bit, {})
            node["mult"] = 1 + node.get("mult", 0)
        node["#"] = x

    def search(self, x):
        node = self.root
        for i in range(18, -1, -1):
            bit = (x >> i) & 1
            if 1 ^ bit in node: node = node[1 ^ bit]
            else: node = node[bit]
        return x ^ node["#"]

    def remove(self, x):
        node = self.root
        for i in range(18, -1, -1):
            bit = (x >> i) & 1
            node[bit]["mult"] -= 1
            if node[bit]["mult"] == 0:
                node.pop(bit)
                break
            node = node[bit]

class Solution:
    def maxGeneticDifference(self, parents: List[int], queries: List[List[int]]) -> List[int]:
        nodeQueries = {}
        for i, (node, val) in enumerate(queries):
            nodeQueries.setdefault(node, []).append([val, i])

        tree, root = {}, -1
        for node, parent in enumerate(parents):
            if parent == -1: root = node
            else: tree.setdefault(parent, []).append(node)
        
        res = [0] * len(queries)
        trie = Trie()

        def fn(node):
            trie.insert(node)
            for val, i in nodeQueries.get(node, []):
                res[i] = trie.search(val)
            for child in tree.get(node, []): fn(child)
            trie.remove(node)

        fn(root)
        return res
        
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