<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 Node{
    int min0,min1;
    vector<Node*> child;
    Node(){
        min0 = min1 = INT_MAX;
        child.resize(2, nullptr);
    }
    void set0(int x){
        min0 = min(min0, x);
    }
    void set1(int x){
        min1 = min(min1, x);
    }
    int getVal(int x){
        if(x == 0) return min0;
        return min1;
    }
};
void insert(Node* root, int x){
    for(int i = 30; i >= 0; i--){
        int bit = (x&(1<<i)) ? 1 : 0;
        if(bit == 0) root->set0(x);
        else root->set1(x);
        if(root->child[bit] == nullptr) root->child[bit] = new Node();
        root = root->child[bit];
    }
}
int findSol(Node *root, int x, int m){
    int ans = 0;
    for(int i = 30; i >= 0; i--){
        int xBit = (x&(1<<i)) ? 1 : 0;
        int op = 1^xBit;
        if(root->child[op] && root->getVal(op) <= m){
            ans += 1<<i;
            root = root->child[op];
        } else if(root->child[xBit] && root->getVal(xBit) <= m){
            root = root->child[xBit];
        } else return -1;
    }
    return ans;
}
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        Node *ROOT = new Node();
        for(auto i: nums) insert(ROOT, i);
        vector<int> ans;
        for(auto q: queries){
            int res = findSol(ROOT, q[0], q[1]);
            ans.push_back(res);
        }
        return ans;
    }
};
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