<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* l, *r;
    int cnt;
    trie(){
        l=r=NULL;
        cnt=0;
    }
};
class Solution {
public:
    trie* head=NULL;
    void add(int val){ 
        trie* cur=head;
        for (int i=29; i>=0; i--){
            if (val&(1<<i)){
                if (!(cur->r)){
                    trie *temp=new trie();
                    cur->r=temp;
                }
                cur=cur->r;
            }else{
                if (!(cur->l)){
                    trie* temp=new trie();
                    cur->l=temp;
                }
                cur=cur->l;
            }
        }
    }
    int get(int val){
        trie* cur=head;
        int ans=0;
        for (int i=29; i>=0; i--){
            if (val&(1<<i)){
                if ((cur->l)){cur=cur->l;ans|=(1<<i);}
                else if ((cur->r))cur=cur->r;
                else return ans;
            }else{
                if ((cur->r)){cur=cur->r;ans|=(1<<i);}
                else if ((cur->l))cur=cur->l;
                else return ans;
            }
        }
        return ans;
    }

    bool static comp(vector<int>& v1, vector<int>& v2){
        if (v1[1]<v2[1])return true;
        else return false;
    }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& q) {
        int n=nums.size();
        for (int i=0; i<q.size(); i++)q[i].push_back(i);
        sort(q.begin(), q.end(), comp);
        sort(nums.begin(), nums.end());
        vector<int> ans(q.size());
        int j=0;
        head = new trie();
        for (auto x: q){
            while (j<n && nums[j]<=x[1]){
                add(nums[j]);
                j++;
            }
            if (j==0)ans[x[2]]=-1;
            else ans[x[2]]=get(x[0]);
        }
        return ans;
    }
};
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