<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 Solution {
    struct Node{
        vector<Node*> link{2, NULL};
    };
public:
    void insert(int n, Node* curr)
    {
        int ex=1<<31;
        for(int i=0; i<32; i++)
        {
            int bit=(n & ex)==0?0:1;
            if(curr->link[bit]==NULL)
            curr->link[bit]=new Node();
            curr=curr->link[bit];
            n=n<<1;
        }
    }
    bool isempty(Node* root){
        return !(root->link[0] || root->link[1]);
    }
    int getmaxxor(int n, Node* curr)
    {
        int output=0, ex=1<<31;
        for(int i=0; i<32; i++)
        {
            output=output<<1;
            int bit=(n & ex)==0?0:1;
            if(curr->link[1-bit]==NULL)
            curr=curr->link[bit];
            else{
                output++;
                curr=curr->link[1-bit];
            }
            n=n<<1;
        }
        return output;
    }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Node* root=new Node();
        vector<int> output(queries.size());
        multiset<pair<int, pair<int,int>>> hsh;
        sort(nums.begin(), nums.end());
        for(int i=0; i<queries.size(); i++)
        hsh.insert({queries[i][1], {queries[i][0], i}});
        int i=0;
        for(auto it: hsh)
        {
            int x=it.second.first;
            int m=it.first;
            int ind=it.second.second;
            while(i<nums.size() && nums[i]<=m)
            {
                insert(nums[i], root);
                i++;
            }
            if(isempty(root))
            output[ind]=-1;
            else
            output[ind]=getmaxxor(x, root);
        }
        return output;
    }
};
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