<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>

#define ll long long

const ll W = 2e6 + 6; // why was so much W reqd?
ll trie[2][W];
ll node_cnt = 0;
// int cnt[W];
ll MX = 35; // max bits
void add(ll k){
    ll x = 0; // root
    for(ll i=MX;i>=0;i--){
        ll bit = 0;
        if(k & (1LL << i)){
            bit = 1;
        }
        if(trie[bit][x] == 0){
            node_cnt++;
            trie[bit][x] = node_cnt; // create a new node
        }
        x = trie[bit][x];
        // cnt[x]++;
    }
}

ll maxXor(ll k){
    ll x = 0;
    ll ans = 0;
    for(ll i = MX ; i >= 0; i--){
        ll bit = 0;
        if(k & (1LL << i)){
            bit = 1;
        }
        if(trie[1-bit][x] != 0){
            // maximised
            x = trie[1-bit][x];
            ans += (1LL << i);
        }
        else{
            // no choice
            x = trie[bit][x];
        }
    }
    return ans;
}

class Solution {
public:
#define fr(i, n) for (long long i = 0; i < (n); i++)
#define vi vector<long long>
#define dbg(x) cout << #x << " : " << x << endl

    vector<int> maximizeXor(vector<int>& a, vector<vector<int>>& qry) {
        // max xor -> think trie
        // idea 1 : sort queries w.r.t. m[i] and get max xor from trie -> 0(32n)
        // idea 2 : binary search and 2 pointers -> 0(32n logn)

        int n = a.size();
        int q = qry.size();
        
        sort(a.begin(), a.end());
        node_cnt = 0;

        memset(trie , 0 , sizeof(trie));

        vector<int> ans(q);
        vector<vector<ll>> qry1;
        fr(i_, q) {
            ll x = qry[i_][0];
            ll m = qry[i_][1];
            qry1.push_back({m , x , i_});
        }

        sort(qry1.begin() , qry1.end()); // sorted wrt max value

        ll prev = 0;
        fr(i_,q){
            ll m = qry1[i_][0];
            ll x = qry1[i_][1];
            ll ind = qry1[i_][2];

            ll idx = upper_bound(a.begin() , a.end() , m) - a.begin();

            if(idx == 0){
                ans[ind] = -1;
                continue;
            }

            for(int i=prev;i<idx;i++){
                add(a[i]);
            }

            ans[ind] = maxXor(x);
            prev = idx;
        }
        return ans;
    }
};
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