<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 Box:
    def __init__(self):
        self.arr = [None for _ in range(2)]

class Trie:
    def __init__(self):
        self.box = Box()

    def insert(self,val):
        curr = self.box
        #Loop through every bit of val
        #Add it to the trie
        for i in range(31, -1 , -1):
            if val & (1 << i):
                if curr.arr[1]:
                    curr = curr.arr[1]
                else:
                    new_box = Box()
                    curr.arr[1] = new_box
                    curr = new_box
            else:
                if curr.arr[0]:
                    curr = curr.arr[0]
                else:
                    new_box = Box()
                    curr.arr[0] = new_box
                    curr = new_box
    
    def get_max(self,val):
        result = 0
        curr = self.box
        for i in range(31, -1 , -1):
            if val & (1 << i):
                if curr.arr[0]:
                    result |= (1 << i)
                    curr = curr.arr[0]
                else:
                    curr = curr.arr[1]
            else:
                if curr.arr[1]:
                    result |= (1 << i)
                    curr = curr.arr[1]
                else:
                    curr = curr.arr[0]
        return result

class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        nums.sort()
        trie = Trie()
        ans = [-1 for _ in range(len(queries))]
        i = 0
        q = []
        for index in range(len(queries)):
            q.append([queries[index][0], queries[index][1], index])
        q.sort(key=lambda x: x[1])
        for query in q:
            value = query[0]
            upper_bound = query[1]
            index = query[2]
            while i < len(nums) and nums[i] <= upper_bound:
                trie.insert(nums[i])
                i += 1
            if i == 0:
                continue
            ans[index] = trie.get_max(value)
        return ans
            
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