<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 TrieNode(object):
    def __init__(self):
        self.children = {}
class Solution:
    def __init__(self):
        self.root = TrieNode()
    def insert(self,curs):
        cur = self.root
        for i in range (len(curs)):
            if curs[i] not in cur.children:
                cur.children[curs[i]] = TrieNode()
            cur = cur.children[curs[i]]
    def query(self,curs,vals):
        cur = self.root
        res = 0
        for i in range (len(curs)):
            if curs[i] == '1':
                if '0' in cur.children:
                    res+=vals[i]
                    cur = cur.children['0']
                else:
                    if '1' not in cur.children:
                        return -1
                    cur = cur.children['1']
            else: #curs[i] == '0'
                if '1' in cur.children:
                    res+=vals[i]
                    cur = cur.children['1']
                else:
                    if '0' not in cur.children:
                        return -1
                    cur = cur.children['0']
        return res
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        vals = [0]*32
        vals[-1] = 1
        for i in range (len(vals)-2,-1,-1):
            vals[i] = vals[i+1]*2
        temp = []
        for i in range (len(queries)):
            temp.append((queries[i][1],queries[i][0],i))
        temp.sort()
        nums = sorted(list(set(nums)))
        res = [-1]*len(temp)
        index = 0
        for i in range (len(temp)):
            while index < len(nums) and nums[index] <= temp[i][0]:
                t = bin(nums[index])[2:]
                t = '0'*(len(vals)-len(t)) + t
                self.insert(t)
                index+=1
            tt = bin(temp[i][1])[2:]
            tt = '0'*(len(vals)-len(tt)) + tt
            curres = self.query(tt,vals)
            res[temp[i][2]] = curres
            # break
        return res
        
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