<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 TrieNode {
public:
    int val;
    TrieNode* child[2];
    bool isTerminal;
    TrieNode(int val) {
        this->val = val;
        child[0] = child[1] = NULL;
        isTerminal = false;
    }
};
class Trie {
public:
    TrieNode* root;
    Trie() { root = new TrieNode(-1); }
    void insert(int num) {
        TrieNode* t = root;
        string s = "";
        for (int j = 29; j >= 0; j--) {
            if ((1LL << j) & num) {
                s += '1';
            } else {
                s += '0';
            }
        }
        for (int i = 0; i < s.length(); i++) {
            if (t->child[s[i] - '0'] != NULL) {
                t = t->child[s[i] - '0'];
            } else {
                t->child[s[i] - '0'] = new TrieNode(s[i] - '0');
                t = t->child[s[i] - '0'];
            }
        }
    }
    int getmax(int num) {
        TrieNode* t = root;
        string s = "";
        for (int j = 29; j >= 0; j--) {
            if ((1LL << j) & num) {
                s += '1';
            } else {
                s += '0';
            }
        }
        int ans = 0, curr = 29;
        bool f1 = false;
        for (int i = 0; i < s.length(); i++) {
            int f = 0;
            if (s[i] == '0') {
                f = 1;
            }
            if (t->child[f] != NULL) {
                ans |= (1 << curr);
                t = t->child[f];
            } else {
                t = t->child[1 - f];
            }
            curr--;
        }
        return ans;
    }
};
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& q) {
        Trie* t = new Trie();
        vector<vector<int>> q1;
        int idx = 0;
        for (auto i : q) {
            q1.push_back({i[1], i[0], idx});
            idx++;
        }
        sort(q1.begin(), q1.end());
        vector<int> ans(q.size());
        int n = q.size();
        sort(nums.begin(), nums.end());
        int k = 0;
        for (auto i : q1) {
            cout << i[0];
            while (k < nums.size() && i[0]>=nums[k]) {
                t->insert(nums[k]);
                k++;
            }
            if(k == 0){
                ans[i[2]] = -1;
            }else{
                ans[i[2]] = (t->getmax(i[1]));
            }

        }
        return ans;
    }
};
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