<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:
    Trie(int bit_length)
     : bit_length_(bit_length)
     , nodes_(1) {}
    
    void insert(int num) {
        int idx = 0;
        for (int i = bit_length_; i >= 0; --i) {
            int curr = (num >> i) & 1;
            if (!nodes_[idx][curr]) {
                nodes_.emplace_back();
                nodes_[idx][curr] = size(nodes_) - 1;
            }
            idx = nodes_[idx][curr];
        }
    }

    int query(int num) {
        if (size(nodes_) == 1) {
            return -1;
        }
        int result = 0, idx = 0;
        for (int i = bit_length_; i >= 0; --i) {
            int curr = (num >> i) & 1;
            if (nodes_[idx][1 ^ curr]) {
                idx = nodes_[idx][1 ^ curr];
                result |= 1 << i;
            } else {
                idx = nodes_[idx][curr];
            }
        }
        return result;
    }

private:
    const int bit_length_;
    vector<array<int, 2>> nodes_;
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(begin(nums), end(nums));
        int max_val = max(nums.back(),
                          (*max_element(cbegin(queries), cend(queries),
                                        [](const auto& a, const auto& b) {
                                            return a[0] < b[0];
                                        }))[0]);
        for (int i = 0; i < size(queries); ++i) {
            queries[i].insert(begin(queries[i]), i);
        }
        sort(begin(queries), end(queries),
             [](const auto& a, const auto& b) {
                 return a[2] < b[2];
             });
        vector<int> result(size(queries));
        Trie trie(bit_length(max_val));
        int j = 0;
        for (const auto& q : queries) {
            int i = q[0], x = q[1], m = q[2];
            for (; j < size(nums) && nums[j] <= m; ++j) {
                trie.insert(nums[j]);
            }
            result[i] = trie.query(x);
        }
        return result;
    }

private:
    int bit_length(int x) {
        return x != 0 ? 32 - __builtin_clz(x) : 1;
    }
};
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