<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{
        vector<int> next;
        int cnt = 0;
        Node() {
            next.resize(2, -1);
        }
    };

    class Trie{
        public:
        vector<Node> trie;
        Trie() {
            trie.resize(1);
        }

        void insert(int n) {
            int v = 0;
            for (int i = 29; i >= 0; i--) {
                int idx = (n >> i) & 1;
                if (trie[v].next[idx] == -1) {
                    trie[v].next[idx] = trie.size();
                    trie.emplace_back();
                }
                v = trie[v].next[idx];
            }
            trie[v].cnt++;
        }

        bool search(int n) {
            int v = 0;
            for (int i = 29; i >= 0; i--) {
                int idx = (n >> i) & 1;
                if (trie[v].next[idx] == -1) {
                    return false;
                }
                v = trie[v].next[idx];
            }
            return trie[v].cnt > 0;
        }

        int max_xor(int n) {
            int ans = 0, v = 0;
            for (int i = 29; i >= 0; i--) {
                if ((n >> i) & 1) {
                    if (trie[v].next[0] != -1) {
                        v = trie[v].next[0];
                        ans |= 1 << i;
                    }
                    else {
                        v = trie[v].next[1];
                    }
                }
                else {
                    if (trie[v].next[1] != -1) {
                        v = trie[v].next[1];
                        ans |= 1 << i;
                    }
                    else {
                        v = trie[v].next[0];
                    }
                }
            }
            return ans;
        }
    };

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        vector<vector<int>> q;
        for (int i = 0; i < (int) queries.size(); i++) {
            q.push_back({queries[i][1], queries[i][0], i});
        }
        sort(q.begin(), q.end());
        vector<int> v = nums;
        sort(v.begin(), v.end());
        int idx = 0;
        Trie t;
        map<int, int> ans;
        for (int i = 0; i < (int) q.size(); i++) {
            while (idx < (int) v.size() && v[idx] <= q[i][0]) {
                t.insert(v[idx]);
                idx++;
            }
            ans[q[i][2]] = (idx == 0) ? -1 : t.max_xor(q[i][1]);
        }
        vector<int> a;
        for (auto [x, y] : ans) {
            a.push_back(y);
        }
        return a;
    }
};
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