<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 {
    Node* arr[2];
};
bool com(long long a ,long long b){
    return a<=b;
}
class Solution {
public:
    // static bool comp(pair<long long, pair<long long, long long>>& a, pair<long long, pair<long long, long long>>& b) {
    //     return a.second.first <= b.second.first;
    // }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& q) {
        Node* root = new Node();
        long long n = q.size();
        vector<int> ans(n);
        vector<pair<long long, pair<long long, long long>>> v;
        for (long long i = 0; i < n; i++) {
            v.push_back({q[i][1], {q[i][0], i}});
        }
        sort(v.begin(), v.end());
        sort(nums.begin(), nums.end());
        long long i = 0;
        for (auto x : v) {
            while (i < nums.size() && nums[i] <= x.first) {
                Node* node = root;
                for (long long j = 31; j >= 0; j--) {
                    if (nums[i] & (1 << j)) {
                        if (node->arr[1] == NULL) {
                            node->arr[1] = new Node();
                        }
                        node = node->arr[1];
                    } else {
                        if (node->arr[0] == NULL) {
                            node->arr[0] = new Node();
                        }
                        node = node->arr[0];
                    }
                }
                i++;
            }
            Node* node = root;
            bool ok = true;
            long long num = 0;
            for (long long j = 31; j >= 0; j--) {
                if ((x.second.first & (1 << j))) {
                    if (node->arr[0] != NULL) {
                        num |= (1 << j);
                        node = node->arr[0];
                    } else if (node->arr[1] != NULL) {
                        node = node->arr[1];
                    } else {
                        ok = false;
                        break;
                    }
                } else {
                    if (node->arr[1] != NULL) {
                        num |= (1 << j);
                        node = node->arr[1];
                    } else if (node->arr[0] != NULL) {
                        node = node->arr[0];
                    } else {
                        ok = false;
                        break;
                    }
                }
            }
            if (ok) {
                ans[x.second.second] = num;
            } else {
                ans[x.second.second] = -1;
            }
        }
        return ans;
    }
};
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