<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 Node{
public:
    Node* child[2];
    Node(){
        child[0]=NULL;
        child[1]=NULL;
    }
};

void insert(Node* root, int num){
    bitset<33>no=num;
    Node* node=root;
    for(int i=32;i>=0;i--){
        if(node->child[no[i]]==NULL){
            Node* children= new Node();
            node->child[no[i]]=children;
        }
        node=node->child[no[i]];
    }
}

int find(Node* root, int num){
    bitset<33>no=num;
    Node* node=root;
    int sum=0;
    for(int i=32;i>=0;i--){
        if(node->child[no[i]^1]==NULL){
            node=node->child[no[i]];
        }else{
            // if(no[i]^1)
            sum+=(1<<i);
            node=node->child[no[i]^1];
        }
    }
    return sum;
}
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Node* root= new Node();
        map<pair<int,int>,int>mp;
        vector<int>ans(queries.size());
        for(int i=0;i<queries.size();i++){
           mp[{queries[i][1],queries[i][0]}]=i;
        }
        sort(nums.begin(),nums.end());
        int index=0,no;
        for(auto it:mp){
            while(index<nums.size() && nums[index]<=it.first.first)insert(root,nums[index++]);
            if(index==0)ans[it.second]=-1;
            else{
                no=find(root,it.first.second);
                ans[it.second]=no;
            }
        }
        for(int i=0;i<queries.size();i++){
           ans[i]=ans[mp[{queries[i][1],queries[i][0]}]];
        }
        return ans;
    }
};
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