<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 Node:
    def __init__(self):
        self.array = [None]*2
    def contain_keys(self, key):
        if self.array[key]:
            return True
        
        return False
    def get(self, key):
        return self.array[key]
    
    def put(self, key,node):
        self.array[key]=node

class trie:
    def __init__(self):
        self.root = Node()
    
    def insert(self, n):
        node= self.root
        for i in range(31, -1, -1):
            temp = n&(1<<i)
            if temp:
                if not node.contain_keys(1):
                    node.put(1,Node())
                node = node.get(1)
            else:
                if not node.contain_keys(0):
                    node.put(0, Node())
                node= node.get(0)
    def find(self,n):
        node=self.root
        if node.get(0) ==None and node.get(1)==None:
            return -1
        ans = 0
        for i in range(31,-1,-1):
            temp = n&(1<<i)
            if temp:
                if node.contain_keys(0):
                    ans = ans|(1<<i)
                    node = node.get(0)
                else:
                    node=node.get(1)
            else:
                if node.contain_keys(1):
                    ans = ans|(1<<i)
                    node = node.get(1)
                else:
                    node=node.get(0)
        return ans


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        nums.sort()
        n = len(nums)
        new_quries = []
        for i,q in enumerate(queries):
            a =q[0]
            b = q[1]
            new_quries.append([a,b,i])
        new_quries.sort(key=lambda x :x[1])

        ans = [None]*len(queries)
        j = 0
        t = trie()
        # print(new_quries)
        for a,b,i in new_quries:
            while j<n and nums[j]<=b:
                t.insert(nums[j])
                j+=1
            
            ans[i]=t.find(a)
        
        return ans




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