<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>

namespace details {

    struct trie_node { // 8 byte node 100'000
        trie_node* children[2] {};
    };

    constexpr auto pool_size = 32 * 16 * 100'005;
    struct simple_node_pool {
        
        static trie_node* construct_default() {
            std::size_t old_idx = curr_idx_;
            curr_idx_ += sizeof(trie_node);
            return new (pool_ + old_idx) trie_node{nullptr, nullptr};
        }

        static void reset() {
            curr_idx_ = 0;
        }

    private:
        static unsigned char pool_[pool_size];
        static std::size_t curr_idx_;
    };  

    unsigned char simple_node_pool::pool_[pool_size] {};
    std::size_t simple_node_pool::curr_idx_ {};

    class trie {
    public:
        trie() {
            details::simple_node_pool::reset();
            root_ = simple_node_pool::construct_default();
        }

        void insert(int num) {
            trie_node* curr = root_;
            for (int i {30}; i >= 0; --i) {
                trie_node*& next = curr->children[(num>>i)&1];
                if (!next) next = simple_node_pool::construct_default();
                curr = next;
            }
        }

        int find_max_xor(int num) const {
            trie_node* curr = root_;
            if (!curr->children[0] && !curr->children[1]) return -1;
            int res {}, i {30};
            for (; i >= 0 && curr; --i) {
                int curr_bit = (num>>i)&1;
                trie_node* next = curr->children[1-curr_bit];
                if (next) res += 1 << i;
                else next = curr->children[curr_bit];
                curr = next;
            }
            res |= num & ((1 << (i+1)) - 1);
            return res;
        }

    private:
        trie_node* root_ {nullptr};
    };

}

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        vector<int> res(queries.size());
        using tp = std::tuple<int, int, int>;
        vector<tp> qs; qs.reserve(queries.size());
        for (int i {}; i < queries.size(); ++i) {
            qs.emplace_back(queries[i][0], queries[i][1], i);
        }
        std::sort(nums.begin(), nums.end());
        std::sort(qs.begin(), qs.end(), [](auto const& q1, auto const& q2){
            return std::get<1>(q1) < std::get<1>(q2);
        });
        details::trie t {};
        for (int j {}; auto [x, m, i]: qs) {
            while (j < nums.size() && nums[j] <= m)  t.insert(nums[j++]);
            res[i] = t.find_max_xor(x);
        }
        return res;
    }
};
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