<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 Solution {
public:


    class Trie{

        public:
        vector<Trie*> child;
        int end;

        Trie(){
            child.resize(2);
            for(int i=0;i<2;i++)child[i]=NULL;
            end=0;
        
        }

        

    };

    void insert(Trie* root,int n){

            Trie*curr=root;

            for(int i=31;i>=0;i--){

                int digit=(n&(1<<i))==0?0:1;

                if(root->child[digit]==NULL){
                    root->child[digit]=new Trie();
                }

                root=root->child[digit];

            }
    }


    Trie* root=new Trie();

    int findxor(Trie* root,int val){

        Trie* curr=root;
        int maxval=0;
        for(int i=31;i>=0;i--){

            int digit=(val&(1<<i))==0?0:1;

            if(curr->child[1-digit]!=NULL){
                maxval=maxval|(1<<i);
                curr=curr->child[1-digit];
            }
            else{
                curr=curr->child[digit];
               
            }

        }

        return maxval;
    }

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& q) {


        int n=nums.size();
        int m=q.size();

        for(int i=0;i<m;i++)q[i].push_back(i);

        vector<vector<int>> temp;

        for(int i=0;i<q.size();i++){
            temp.push_back({q[i][1],q[i][0],i});
        }

        sort(nums.begin(),nums.end());
        sort(temp.begin(),temp.end());
        
        int ind=0;
        vector<int> ans(m);
        int curr=-1;

        for(int i=0;i<m;i++){
            
            // cout<<i<<" "<<ind<<" "<<temp[i][0]<<" "<<temp[i][1]<<" "<<temp[i][2]<<endl;
            
            while(ind<n && nums[ind]<=temp[i][0]){
                insert(root,nums[ind]);
                ind++;
            }


            if(ind!=0) curr=findxor(root,temp[i][1]);

           
            ans[temp[i][2]]=curr;


        }

        return ans;
        
    }
};
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