<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{
    vector<Node*> list;

    Node(){
        list.resize(2,NULL);
    }

    bool exists(int i){
        return (list[i]!=NULL);
    }

    void put(int i){
        list[i]=new Node();
    }

    Node* get(int i){
        return list[i];
    }
};

class Trie{
    Node* root;
public:
    Trie(){
        root=new Node();
    }

    void add(int i){
        Node* node=root;
        for(int j=31;j>=0;j--){
            int k=((1<<j)&i)>>j;
            if(node->exists(k)){
                node=node->get(k);
            }else{
                node->put(k);
                node=node->get(k);
            }
        }
        return;
    }

    int xorMax(int num){
        Node* node=root;
        int ans=0;
        for(int j=31;j>=0;j--){
            int k=((1<<j)&num)>>j;
            if(node->exists(!k)){
                ans+=(1<<j);
                node=node->get(!k);
            }else if(node->exists(k)){
                node=node->get(k);
            }else{
                return -1;
            }
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Trie* trie=new Trie();
        sort(nums.begin(),nums.end());
        vector<pair<int,pair<int,int>>> a;
        for(int i=0;i<queries.size();i++){
            a.push_back({queries[i][1],{queries[i][0],i}});
        }
        sort(a.begin(),a.end());
        vector<int> ans(queries.size());
        int i=0;
        for(auto p:a){
            while(i<nums.size() && nums[i]<=p.first){
                trie->add(nums[i]);
                i++;
            }
            ans[p.second.second]=trie->xorMax(p.second.first);
        }
        return ans;
    }
};
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