<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 Solution {
public:
    struct Node {
        Node *C[2] = {0};
    };

    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Node root, *curr;
        for (int n : nums) {
            curr = &root;
            // printf("%i\n", n);
            for (int i = 30; i >= 0; --i) {
                // if (n == 3) {
                //     printf("%i", (n >> i) & 1);
                // }
                if (!curr->C[(n >> i) & 1])
                    curr->C[(n >> i) & 1] = new Node;
                curr = curr->C[(n >> i) & 1];
            }
            // printf("\n");
        }
        vector<int> ans;
        for (auto &Q : queries) {
            int x = Q[0], m = Q[1];
            int n = dfs(&root, 30, 0, x, m);
            ans.push_back(n == -1 ? n : n ^ x);
        }
        return ans;
    }

    int dfs(Node *curr, int i, int n, int x, int m) {
        // printf("%i %i %i\n", x, i, n);
        if (i < 0) return n;
        if ((x & (1 << i)) == 0) {
            int r = -1;
            // if (x == 12 && i < 2)
                // printf("%i %p\n", (n + (1 << i) <= m), curr->C[1]);
            if ((n + (1 << i) <= m) && curr->C[1]) {
                r = dfs(curr->C[1], i-1, n + (1 << i), x, m);
            }
            if (r == -1 && curr->C[0]) {
                r = dfs(curr->C[0], i-1, n, x, m);
            }
            return r;
        }
        else {
            int r = -1;
            if (curr->C[0]) {
                r = dfs(curr->C[0], i-1, n, x, m);
            }
            if (r == -1 && (n + (1 << i) <= m) && curr->C[1]) {
                r = dfs(curr->C[1], i-1, n + (1 << i), x, m);
            }
            return r;
        }
    }
};
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