<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 Tri():
    def __init__(self):
        self.arr=[None,None]
    
    def insert(self,num):
        tmp=self.arr
        a=[]
        tnum=num
        for i in range(64):
            a.append(tnum%2)
            tnum//=2
        a=a[::-1]
        for i in a:
            if tmp[i]==None:
                tmp[i]=Tri().arr
            tmp=tmp[i]    
    def findmx(self,num):
        tmp=self.arr
        #print(tmp)
        a=[]
        tnum=num
        for i in range(64):
            a.append(tnum%2)
            tnum//=2
        a=a[::-1]
        ans=[]
        for i in a:
            if tmp[i]==None and tmp[(i+1)%2]==None:
                return -1
            if tmp[(i+1)%2]!=None :
                ans.append(1)
                tmp=tmp[(i+1)%2]
            else:
                ans.append(0)
                tmp=tmp[i]
        anss=0
        cnt=0
        for i in ans[::-1]:
            anss+=(2**cnt)*i
            cnt+=1
        return anss


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:

        tri=Tri()
        a=[(j,i) for i,j in queries]
        a.sort()
        ans=[]
        d=dict()
        n=len(nums)
        nums.sort()
        i=0
        while i<n:
            if len(a)==0:
                break
            elif a[0][0]>=nums[i]:
                #print(i,a[0][0])
                tri.insert(nums[i])
                i+=1
            else:
                d[(a[0][1],a[0][0])]=tri.findmx(a[0][1])
                a.pop(0)
        
        while len(a)!=0:
            d[(a[0][1],a[0][0])]=tri.findmx(a[0][1])
            a.pop(0)

        for i,j in queries:
            ans.append(d[(i,j)])

        return ans
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