<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 = 1e5 + 5, M = 31 * N;
int son[M][2], idx;
void insert(int x) {
    int p = 0;
    for (int i = 30; i >= 0; i--) {
        int& s = son[p][x >> i & 1];
        if (!s)s = ++idx;
        p = s;
    }
}
int query(int x) {
    int p = 0, res = 0;
    for (int i = 30; i >= 0; i--) {
        int s = x >> i & 1;
        if (son[p][!s]) {
            res += 1 << i;
            p = son[p][!s];
        }
        else p = son[p][s];
    }
    return res;
}

int x = 0;
class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        memset(son, 0, sizeof(son));
        for (int i = 0; i < queries.size(); i++) {
            queries[i].push_back(i);
            swap(queries[i][0], queries[i][1]);
        }
        vector<int> ret(queries.size(), -1);
        for (auto it : nums) {
            vector<int> tmp = { it, -1,-1 };
            queries.push_back(tmp);
        }
        sort(queries.begin(), queries.end(), [](vector<int>&v1, vector<int>&v2) {
            if (v1[0] == v2[0])
                return v1[2] < v2[2];
            return v1[0] < v2[0];
            });
        idx=0;
        int addf = 1;
        for (auto it : queries) {
            if (it[2] == -1) {
                addf = 0;
                insert(it[0]);
            } else {
                if (addf) continue;
                ret[it[2]] = query(it[1]);
            }
        }
        return ret;
    }
};
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