<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* ch[2];
    Node(){
        this->ch[0] = this->ch[1] = NULL;
    }
};
class Trie{
public:
    Node*root;
    Trie(){
        root = new Node();
    }
    void insert(string&s){
        Node*ptr = root;
        for(int i=0;i<30;i++){
            if(!ptr->ch[s[i]-'0']){
                ptr->ch[s[i]-'0'] = new Node();
            }
            ptr= ptr->ch[s[i]-'0'];
        }
    }
    int XOR(int a){
        bitset<30>b(a);
        string s = b.to_string();
        Node*ptr = root;
        bitset<30>ans(0);
        for(int i=0;i<30;i++){
            if(s[i]=='0'){
                if(ptr->ch[1]){
                    ans.set(29-i);
                    ptr = ptr->ch[1];
                }
                else{
                    ptr = ptr->ch[0];
                }
            }
            else{
                if(ptr->ch[0]){
                    ans.set(29-i);
                    ptr = ptr->ch[0];
                }
                else{
                    ptr = ptr->ch[1];
                }
            }
        }
        return (int)ans.to_ullong();
    }
};
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& q) {
        for(int i=0;i<q.size();i++)q[i].push_back(i);
        sort(q.begin(),q.end(),[](const vector<int>&a,const vector<int>&b)->bool{
            if(a[1]!=b[1])return a[1]<b[1];
            return a[2]<b[2];
        });
        sort(nums.begin(),nums.end());
        int pt = 0;
        int pt2 = 0;
        auto t = new Trie();
        vector<int>ans(q.size(),0);
        while(1){
            while(nums.size()>pt2&&nums[pt2]<=q[pt][1]){
                bitset<30>b(nums[pt2]);
                string s = b.to_string();   
                t->insert(s);
                pt2++;
            }
            if(pt2)
                ans[q[pt][2]] = t->XOR(q[pt][0]);
            else ans[q[pt][2]] = -1;
            pt++;
            if(pt>=q.size())break;
        }
        return ans;
    }
};
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