<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 m0, m1;
    Node* child[2];

    Node() {
        m1 = INT_MAX, m0 = INT_MAX;
        child[0] = child[1] = nullptr;
    }
};

void insert(int n, Node* c) {
    Node* root = c;
    for (int i = 31; i >= 0; i--) {
        int x = n >> i & 1;
        if(x) root->m1 = min(root->m1, n);
        else root->m0 = min(root->m0, n);
        if (!root->child[x])
            root->child[x] = new Node();
        root = root->child[x];
    }
}

int query(int n, int K, Node* root) {
    Node* curr = root;
    int ans = 0;
    for(int i = 31; i >= 0; i--){
        int x = n >> i & 1;
        int m1 = root->m1, m0 = root->m0;
        if(x){
            if(root->child[1 - x] && m0 <= K){
                root = root->child[1 - x];
                ans |= (1 << i);
            }else if(root->child[x] && m1 <= K){
                root = root->child[x];
            }else return -1;
        }else{
            if(root->child[1 - x] && m1 <= K){
                root = root->child[1 - x];
                ans |= (1 << i);
            }else if(root->child[x] && m0 <= K){
                root = root->child[x];
            }else return -1;
        }
    }
    return ans;
}

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        int mn = *min_element(nums.begin(), nums.end());
        vector<int> ans(queries.size(), 0);
        Node* root = new Node();
        for (auto& e : nums) {
            insert(e, root);
        }

        for (int i = 0; i < queries.size(); i++) {
            int x = queries[i][0], m = queries[i][1];
            if (m < mn) {
                ans[i] = -1;
                continue;
            }
            ans[i] = query(x, m, root);
        }
        return ans;
    }
};
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