<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>

class Trie{
    public:
vector<vector<int>> node = vector<vector<int>>(1, vector<int>(2, -1));
    int c=1;
    void insert(int val){
        int x=0;
        for(int i=31;i>=0;i--){
            int bit=((val>>i)&1);
            if(node[x][bit]==-1){
                node[x][bit]=c++;
                node.push_back(vector<int>(2,-1));
            }
            x=node[x][bit];
        }
    }
    int getmax(int val){
        int ans=0;
        int x=0;
        for(int i=31;i>=0;i--){
            int bit=((val>>i)&1);
            if(node[x][1-bit]==-1){
                x=node[x][bit];
            }
            else{
                ans|=(1<<i);
                x=node[x][1-bit];
            }
        }
        return ans;
    }
};
class Solution {
public:
    static bool cmp(const vector<int>& a,const vector<int>& b){
        return a[1]<b[1];
    }
    vector<int> maximizeXor(vector<int>& v, vector<vector<int>>& q) {
        int n=v.size();
        int l=q.size();
        sort(v.begin(),v.end());
        Trie obj;
        for(int i=0;i<l;i++){
            vector<int> v1=q[i];
            v1.push_back(i);
            q[i]=v1;
        }
        sort(q.begin(),q.end(),cmp);
        vector<int> ans(l);
        int j=0;
        for(int i=0;i<l;i++){
            int x=q[i][0],m=q[i][1];
            int in=q[i][2];
            while(j<n && v[j]<=m){
                obj.insert(v[j]);
                j++;
            }
            if(j==0){
                ans[in]=-1;
                continue;
            }
            ans[in]=obj.getmax(x);
        }
        return ans;
    }
};
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