<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>

using Ti3 = std::tuple<int, int, int>;

constexpr int MAXN = 100010;
constexpr int MAXM = 32;

struct Trie {
  int nxt[MAXN * MAXM][2], cnt = 1;

  void Clear() {
    std::memset(nxt, -1, sizeof(nxt));
    cnt = 1;
  }

  void Insert(int v) {
    int x = 0;
    for (int i = 30; i >= 0; --i) {
      int d = ((v >> i) & 1);
      if (nxt[x][d] == -1) {
        nxt[x][d] = cnt++;
      }
      x = nxt[x][d];
    }
  }

  int Query(int v) const {
    int x = 0, ret = 0;
    for (int i = 30; i >= 0; --i) {
      int d = ((v >> i) & 1);
      if (nxt[x][d ^ 1] == -1) {
        x = nxt[x][d];
      } else {
        ret |= (1 << i);
        x = nxt[x][d ^ 1];
      }
    }
    return ret;
  }
} trie;

Ti3 q[MAXN];

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
      int n = nums.size(), m = queries.size();
      std::sort(nums.begin(), nums.end());
      for (int i = 0; i < m; ++i) {
        int x = queries[i][0], v = queries[i][1];
        q[i] = std::make_tuple(v, x, i);
      }
      std::sort(q, q + m);
      std::vector<int> ret(m);
      trie.Clear();
      int p = 0;
      for (int i = 0; i < m; ++i) {
        auto [v, x, k] = q[i];
        for (; p < n && nums[p] <= v; ++p) {
          trie.Insert(nums[p]);
        }
        ret[k] = p ? trie.Query(x) : -1;
      }
      return ret;
    }
};
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