<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 {
public:
struct Node{
    Node* left;
    Node* right;
    Node(){
        left = nullptr;
        right = nullptr;
    }
};

void insert(long long int n, Node* head){
    Node* cur = head;
    for(int i = 31; i >= 0; i--){
        long long int cur_bit = ((1LL) << i) & n;
        if(cur_bit > 0){
            if(cur -> right == nullptr){
                cur -> right = new Node();
            }
            cur = cur -> right;
        } else {
            if(cur -> left == nullptr){
                cur -> left = new Node();
            }
            cur = cur -> left;
        }
    }
}

long long int find_max(long long int n, Node* head){
    Node* cur = head;
    long long int ans = 0;
    for(int i = 31; i >= 0; i--){
        long long int cur_bit = ((1LL) << i)&n;
        if(cur_bit > 0){
            if(cur -> left != nullptr){
                ans |= (1 << i);
                cur = cur -> left;
            } else {
                cur = cur -> right;
            }
        } else {
            if(cur -> right != nullptr){
                ans |= (1 << i);
                cur = cur -> right;
            } else {
                cur = cur -> left;
            }
        }
    }
    return ans;
}
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> q;
        int n = queries.size();
        for(int i = 0; i < n; i++){
            vector<int> v = {queries[i][1], queries[i][0], i};
            q.push_back(v);
        }
        sort(q.begin(), q.end());
        vector<int> result(n, -1);
        int k = 0;
        Node* head = new Node();
        for(auto& it : q){
            int lim = it[0];
            long long int n = it[1];
            int index = it[2];
            while(k < nums.size() && nums[k] <= lim){
                insert(nums[k++], head);
            }
            if(head -> left != nullptr || head -> right != nullptr){
                long long int ans = find_max(n, head);
                result[index] = ans;
            }
        }
        return result;
    }
};
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