<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>

bool customSort(const vector<int>&v1, const vector<int>&v2){
           return v1[1] < v2[1]; 
}

class Trie{

    public:
    Trie* child[2];
    int count, mini; // mini element that I can get from a node

    Trie(){
        child[0] = child[1] = nullptr;
        count = 0;
        mini = INT_MAX;
    }

    void insert(int x){
        Trie* root = this;
        for(int i = 31; i>=0; i--){
            int nxt = 0;
            if((x >> i)&1){ nxt = 1; }
            if(root->child[nxt]==nullptr){
                root->child[nxt] = new Trie();
            }
            root = root-> child[nxt];
            root -> count++;
            root -> mini = min(root->mini, x);
        }
    }
    
    // findMaxXorEle <= m 

    int findMaxXorEle(int x, int m){
          Trie* root = this;
          int ele = 0;
          for(int i = 31; i>=0; i--){
            int nxt = 1;
            if((x>>i)&1)nxt = 0;
            if((root->child[nxt]==nullptr) || (root->child[nxt]->mini > m)){
                nxt = !nxt;
            }
            if((root->child[nxt]==nullptr) || (root->child[nxt]->mini > m)){
                return -1;
            }
            ele += (1ll << i)*nxt;
            root = root->child[nxt];
          }
          return ele;
    }
};

class Solution {
public:
    
    // Solving Online 

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        
        int m = queries.size();
    
        Trie tr;    
        for(int x: nums)tr.insert(x);

        vector<int>res;

        for(vector<int>&qry: queries){
            int ans = tr.findMaxXorEle(qry[0], qry[1]);
            if(ans==-1)res.push_back(-1);
            else res.push_back(ans^qry[0]);
        } 

        return res; 
        
    }
};
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