<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 Solution:
    def maxGeneticDifference(self, parents: List[int], queries: List[List[int]]) -> List[int]:
        adj=defaultdict(list)
        for i,p in enumerate(parents):
            adj[p].append(i)
        
        trie={}

        def add_to_trie(n):
            temp=trie
            for i in range(17,-1,-1):
                msb=n>>i&1
                temp.setdefault(msb,{'cnt':0})
                temp=temp[msb]
                temp['cnt']+=1
        
        def remove_from_trie(n):
            temp=trie
            removed_from_Trie=False
            for i in range(17,-1,-1):
                msb=n>>i&1

                if removed_from_Trie:
                    continue

                if temp[msb]['cnt']==1:
                    temp.pop(msb)
                    removed_from_Trie=True
                else:
                    temp=temp[msb]
                    temp['cnt']-=1
        
        def best_xor(val):
            temp=trie
            res=0
            for i in range(17,-1,-1):
                msb=val>>i&1
                target_msb=1-msb
                if target_msb in temp:
                    temp=temp[target_msb]
                    res|=1<<i
                else:
                    temp=temp[msb]
            return res
        
        res=[0]*len(queries)
        query_map=defaultdict(list)
        for i,(node,val) in enumerate(queries):
            query_map[node].append((i,val))

        
        def dfs(node):
            add_to_trie(node)
            for i,val in query_map[node]:
                res[i]=best_xor(val)
            for child in adj[node]:
                dfs(child)
            remove_from_trie(node)

        dfs(adj[-1][0])
        return res
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