<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>

struct trie {
    trie* left;
    trie* right;
};
trie t[3100001];
int tCount;

void insert(int num){
    trie* now = t;
    for(int i=30; i>=0; --i){
        if((1<<i)&num){
            if(now->right==nullptr){
                t[++tCount] = {nullptr, nullptr};
                now->right = t+tCount;
            }
            now = now->right;
        }else{
            if(now->left==nullptr){
                t[++tCount] = {nullptr, nullptr};
                now->left = t+tCount;
            }
            now = now->left;
        }
    }
}

int findBest(int num){
    trie* now = t;
    int best = 0;
    for(int i=30; i>=0; --i){
        if(now->left==nullptr){
            now = now->right;
            best |= (1<<i);
            continue;
        }
        if(now->right==nullptr){
            now = now->left;
            continue;
        }
        if((1<<i)&num){
            now = now->left;
        }else{
            now = now->right;
            best |= (1<<i);
        }
    }
    return best ^ num;
}

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        tCount = 0;
        t[0] = {nullptr, nullptr};
        priority_queue<vector<int>> queryPq;
        priority_queue<int> numPq;
        vector<int> ans(queries.size());
        for(int num:nums) numPq.push(-num);
        for(int i=0; i<queries.size(); ++i){
            queryPq.push({-queries[i][1], i, queries[i][0]});
        }
        while(!queryPq.empty()){
            int m = -queryPq.top()[0];
            int queryIndex = queryPq.top()[1];
            int x = queryPq.top()[2];
            queryPq.pop();
            while(!numPq.empty()){
                if(-numPq.top()>m) break;
                insert(-numPq.top());
                numPq.pop();
            }
            if(tCount==0) ans[queryIndex] = -1;
            else ans[queryIndex] = findBest(x);
        }
        return ans;
    }
};

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