<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>

int power(int a, int b) {
    if (b == 0) {
        return 1;
    }
    int res = power(a,b/2);
    if (b%2) {
        return res*res*a;
    }
    else {
        return res*res;
    }
}

class Node{
    public:
    Node* links[2];

    Node() {
        links[0] = NULL;
        links[1] = NULL;
    }

    bool exists(int num) {
        return links[num] != NULL;
    }

    void newReference(int num, Node* node) {
        links[num] = node;
    }

    Node* next(int num) {
        return links[num];
    }
};

class Trie{
    private: Node* root;
    public:
    Trie() {
        root= new Node();
    }
    void insert(vector<int> &num) {
        Node* node = root;
        for(int i = 0; i<num.size(); i++) {
            if (!(node->exists(num[i]))) {
                node->newReference(num[i], new Node());
            }
            node = node->next(num[i]);
        }
    }

    int check(vector<int> &num) {
        Node* node = root;
        int value = 0;
        vector<int> ans;
        for(int i = 0; i<num.size(); i++) {
           if (num[i] == 0) {
            if (!(node->exists(1))) {
                node = node->next(0);
                ans.push_back(0);
            }
            else {
                node = node->next(1);
                //value += power(2, 31-i);
                ans.push_back(1);
            }
           }
           else {
            if (!(node->exists(0))) {
                node = node->next(1);
                //value += power(2, 31-i);
                ans.push_back(1);
            }
            else {
                node = node->next(0);
                ans.push_back(0);
            }
           }     
        }
        for(int i = 0; i<31; i++) {
            if (ans[i] == 1) {
            value += power(2, 30-i);
            }
        }
        return value;
    }
};

 bool func1(vector<int> &pair1, vector<int> &pair2) {
        return pair1[2] < pair2[2];
    }

    bool func2(vector<int> &pair1, vector<int> &pair2) {
        return pair1[0] < pair2[0];
    }

class Solution {
public:

   

    vector<int> tobits(int num) {
        vector<int> bits(31,0);
        int pos = 0;
        while(num > 0) {
            bits[pos] = num%2;
            num = num/2;
            pos++;
        }
        reverse(bits.begin(), bits.end());
        return bits;
    }

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        vector<vector<int> > newqueries;
        
        //sort(queries.begin(), queries.end(), func);
        int nq = queries.size();
        int n = nums.size();
        for(int i = 0; i<nq; i++) {
            vector<int> temp;
            temp.push_back(i);
            temp.push_back(queries[i][0]);
            temp.push_back(queries[i][1]);
            newqueries.push_back(temp);
        }
        sort(nums.begin(), nums.end());
        sort(newqueries.begin(), newqueries.end(), func1);
        Trie* t = new Trie();
        int pos = 0;
        vector<int> ff;
        for(int i = 0; i<nq; i++) {
            int tt = pos;
            while(nums[pos] <= newqueries[i][2] && pos < n) {
                vector<int> temp = tobits(nums[pos]);
                t->insert(temp);
                pos++;
            }
            if (newqueries[i][2] < nums[0]) {
                newqueries[i].push_back(-1);
                continue;
            }
            vector<int> todo = tobits(newqueries[i][1]);
            int ans = t->check(todo);
            newqueries[i].push_back(ans^newqueries[i][1]);
        }
        sort(newqueries.begin(), newqueries.end(), func2);
        for(int i = 0; i<newqueries.size(); i++) {
            ff.push_back(newqueries[i][3]);
        }
        return ff;

    }
};
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