<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 Trie {
    public:
    int tot, n, digit;
    vector<array<int, 2>> son;
    unordered_map<int, int> Map;
    Trie(int n, int d = 31) : n(n), digit(d) {
        son.resize((n + 1) * digit);
    }
    
    void insert(int x) {
        int cur = 0;
        for (int d = digit; d >= 0; d--) {
            int bit = (x >> d) & 1;
            if (!son[cur][bit]) {
                son[cur][bit] = ++tot;
            }
            cur = son[cur][bit];
        }
        Map[cur] = x;
    }
    
    int get(int x) {
        int cur = 0;
        for (int d = digit; d >= 0; d--) {
            int bit = (x >> d) & 1;
            if (son[cur][!bit]) {
                cur = son[cur][!bit];
            } else {
                cur = son[cur][bit];
            }
        }
        return (Map.find(cur) == Map.end()) ? -1 : x ^ Map[cur];
    }
};

class Solution {    
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();
        for (int i = 0; i < queries.size(); i++) {
            queries[i].push_back(i);
        }
        sort(queries.begin(), queries.end(), [&](vector<int>& a, vector<int>& b) {
            return a[1] < b[1];
        });
        
        Trie t(n);
        int m = queries.size();
        vector<int> ans(m);
        int i = 0;
        sort(nums.begin(), nums.end());
        for (auto& q : queries) {
            while (i < n && nums[i] <= q[1]) {
                t.insert(nums[i]);
                i++;
            }
            ans[q[2]] = t.get(q[0]);
        }
        return ans;
    }
};
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