<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:
    trie* c[2];
    trie(){
        c[0]=NULL;
        c[1]=NULL;
    }
};

void ad(int x,trie* root){
    bitset<32> a = x;
    for(int i=31;i>=0;i--){
        if(root->c[a[i]]==NULL){
            root->c[a[i]]= new trie();
        }
        root=root->c[a[i]];
    }
}

int xo(int x,trie* root){
    int ans = 0;
    bitset<32> a=x;
    for(int i=31;i>=0;i--){
        if(root->c[0]!=NULL&&a[i]==1){
            ans += (1<<i);
            root=root->c[0];
        }
        else if(root->c[1]!=NULL&&a[i]==0){
            ans += (1<<i);
            root=root->c[1];
        }else{
            root=root->c[a[i]];
        }
    }
    return ans;
}

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {

        trie * a = new trie();

        sort(nums.begin(),nums.end());
        vector<pair<int,int>> q;
        for(auto i:queries){
            q.push_back({i[1],i[0]});
        }
        sort(q.begin(),q.end());

        map<pair<int,int>,int> mp;
        int j = 0;
        int ans;
        for(auto i:q){
            while(j<nums.size()&&nums[j]<=i.first){
                ad(nums[j],a);
                j++;
            }
            if(j==0){
                mp[i]=-1;
            }else{
                ans = xo(i.second,a);
                mp[i]=ans;
            }
        }
        
        vector<int> an;
        pair<int,int> p;
        for(auto i: queries){
            p={i[1],i[0]};
            an.push_back(mp[p]);
        }

        return an;
    }
};
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