<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>

struct Node{
    Node *arr[2];
    int val;
    Node(){
        val = -1;
        arr[0] = NULL;
        arr[1] = NULL;
    }
};
class Solution {
public:
    void insert(Node *root, int num){
        Node *temp=root;
        for(int i=31;i>=0;i--){
            bool el = ((1<<i) & num);
            if(temp->arr[el]==NULL)temp->arr[el] = new Node();
            temp = temp->arr[el];
        }
        temp->val = num;
    }
    int find(Node *root, int x){
        Node *temp=root;
        for(int i=31;i>=0;i--){
            bool el = ((1<<i) & x);
            int req = 1-el;
            if(temp->arr[req]!=NULL){
                temp = temp->arr[req];
            }
            else {
                temp = temp->arr[1-req];
            }
           if(temp==NULL) {
                return -1;
            }
            
        }
        return temp->val;
    }
    static bool cmp(pair<vector<int>,int>&a,pair<vector<int>,int>&b){
        return a.first[1] < b.first[1];
    }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Node *root=new Node();
        sort(nums.begin(),nums.end());
        vector<pair<vector<int>,int>> arr;
        for(int i=0;i<queries.size();i++){
            arr.push_back({queries[i],i});
        }
        sort(arr.begin(),arr.end(),cmp);
        vector<int>ans(arr.size(),-1);
        int i=0;
        int n = nums.size();
        for(auto it:arr){
            int el = it.first[1];
            int x = it.first[0];
            int ind = it.second;
            while(i<n and nums[i]<=el){
                insert(root,nums[i]);
                i++;
            }
            int el1 = find(root,x);
            if(el1==-1)ans[ind] = -1;
            else ans[ind] = x^el1;
        }
        return ans;     
    }
};
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