<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>

from collections import defaultdict
from sortedcontainers import SortedList
class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:


        #sorting nums
        l=len(nums)
        nums.sort()
        ptr=0

        #Initialinsing Trie
        Trie=defaultdict(defaultdict)

        # answer
        nq=len(queries)
        ans=[0 for i in range(nq)]
        qury=SortedList()

        #sorting of queries
        for i in range(nq):
            qury.add((queries[i][1],queries[i][0],i))
        
        for i in range(nq):
            ref=qury[0]
            qury.discard(ref)
            while ptr<l and nums[ptr]<=ref[0]:
                tt=nums[ptr]
                local_trie=Trie

                for i in range(31,-1,-1):
                    point=0
                    if tt & 1<<i:
                        point=1
                    if not local_trie[point]:
                        local_trie[point]=defaultdict(defaultdict) 
                    local_trie=local_trie[point]
                local_trie['end']=True
                ptr+=1
            local_trie=Trie
            nm=0
            flag=True
            for i in range(31,-1,-1):
                point=0
                if ref[1] & 1<<i:
                    point=1
                if local_trie[point^1]:
                    nm+=2**i
                    local_trie=local_trie[point ^ 1]
                elif local_trie[point]:
                    local_trie=local_trie[point]
                else:
                    flag=False
                    break
            if not flag:
                ans[ref[2]]=-1
            else:
                ans[ref[2]]=nm                
        return ans

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