<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>

const int N = 3E6 + 10;
int tree[N][2];

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin(), nums.end());
        memset(tree, 0, sizeof(tree));
        int idx = 0, counter = 0;
        vector<array<int,3>> ask;
        for (auto &q : queries){
            ask.push_back({q[1], q[0], idx++});
        }
        sort(ask.begin(), ask.end());
        int sz = queries.size(), n = nums.size();
        vector<int> res(sz, 0);
        for (int i = 0, j = 0; i < sz; i++){
            while(j < n && nums[j] <= ask[i][0]){
                int cur = 0;
                for (int k = 29; k >= 0; k--){
                    int x = 1 & (nums[j] >> k);
                    if (!tree[cur][x]) tree[cur][x] = ++counter;
                    cur = tree[cur][x];
                }
                j++;
            }
            if (j == 0) {
                res[ask[i][2]] = -1;
                continue;
            }
            int val = 0, cur = 0;  
            for (int k = 29; k >= 0; k--){
                int x = 1 & (ask[i][1] >> k);
                if (tree[cur][1 ^ x]){
                    val |= (1 << k);
                    cur = tree[cur][1 ^ x];
                }
                else cur = tree[cur][x];
            }
            res[ask[i][2]] = val;
        }
        return res;  
    }
};
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