<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_node
{
public:
    vector<Trie_node *> next;
    bool is_num;
    Trie_node()
    {
        next.resize(2, NULL);
        is_num = false;
    }
};

class Trie
{
    Trie_node *head;

public:
    Trie()
    {
        head = new Trie_node();
    }
    void insert(int val)
    {
        Trie_node *cur = head;
        for (int i = 30; i >= 0; --i)
        {
            int cur_val = (val >> i) & 1;
            if (!cur->next[cur_val])
            {
                cur->next[cur_val] = new Trie_node();
            }
            cur = cur->next[cur_val];
            if (i == 0)
            {
                cur->is_num = true;
            }
        }
    }
    int search(int val)
    {
        Trie_node *cur = head;
        int ret = 0;
        for (int i = 30; i >= 0; --i)
        {
            int cur_val = (val >> i) & 1;
            int next_val = !cur_val;
            if (cur->next[next_val])
            {
                ret += (1 << i);
                cur = cur->next[next_val];
            }
            else if (cur->next[cur_val])
            {
                cur = cur->next[cur_val];
            }
            else
            {
                return -1;
            }
        }
        return ret;
    }
};

class Solution
{
public:
    vector<int> maximizeXor(vector<int> &nums, vector<vector<int>> &queries)
    {
        vector<vector<int>> v = queries;
        unordered_map<int, unordered_map<int, int>> m;
        Trie t;
        int j = 0;

        sort(v.begin(), v.end(), [](vector<int> &a, vector<int> &b)
             { return a[1] < b[1]; });
        sort(nums.begin(), nums.end());

        for (int i = 0; i < v.size(); ++i)
        {
            while (j < nums.size())
            {
                if (nums[j] <= v[i][1])
                {
                    t.insert(nums[j]);
                    j++;
                }
                else
                {
                    break;
                }
            }
            m[v[i][0]][v[i][1]] = t.search(v[i][0]);
        }
        vector<int> ret;
        for (const auto &q : queries)
        {
            ret.push_back(m[q[0]][q[1]]);
        }
        return ret;
    }
};
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