<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 {
    int ini;
    vector<Node*> addre;
    Node(int c) : ini(c) {};
};

Node* root;

void getin(int lol) {
    Node* temp = root;
    for (int j = 31; j >= 0; j--) {
        bool founded = false;
        int bit = (lol >> j) & 1;
        for (auto& c : temp->addre) {
            if (c->ini == bit) {
                founded = true;
                temp = c;
                break;
            }
        }
        if (!founded) {
            Node* newnode = new Node(bit);
            temp->addre.push_back(newnode);
            temp = newnode;
        }
    }
}

int sabse_bada(int bol) {
    Node* temp = root;
    int ans = 0;
    for (int i = 31; i >= 0; i--) {
        bool founded = false;
        int bit = 1 - ((bol >> i) & 1);
        for (auto& c : temp->addre) {
            if (c->ini == bit) {
                founded = true;
                temp = c;
                ans |= (1 << i);
                break;
            }
        }
        if (!founded && !temp->addre.empty()) {
            temp = temp->addre[0];
        }
    }
    return ans;
}

vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
    root = new Node(0);

    // Create a vector of pairs: each pair consists of a query and its original index
    vector<pair<vector<int>, int>> indexedQueries;
    for (int i = 0; i < queries.size(); i++) {
        indexedQueries.push_back({ queries[i], i });
    }

    // Sort nums and indexedQueries
    sort(nums.begin(), nums.end());
    sort(indexedQueries.begin(), indexedQueries.end(), [](pair<vector<int>, int>& a, pair<vector<int>, int>& b) {
        return a.first[1] < b.first[1];
    });

    vector<int> ans(queries.size(), -1);
    int numIndex = 0;

    for (const auto& queryPair : indexedQueries) {
        int x = queryPair.first[0];
        int m = queryPair.first[1];
        int queryIndex = queryPair.second;

        // Insert all numbers <= m into the Trie
        while (numIndex < nums.size() && nums[numIndex] <= m) {
            getin(nums[numIndex]);
            numIndex++;
        }

        // If Trie has elements, find the max XOR
        if (numIndex > 0) {
            ans[queryIndex] = sabse_bada(x);
        }
    }

    return ans;
}

};
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