<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>


//leetcode submit region begin(Prohibit modification and deletion)
#include <bits/stdc++.h>

using namespace std;


using ll = long long;
#define pb push_back
#define ep emplace_back
#define fi x
#define se y
#define mp make_pair
#define all(x) x.begin(), x.end()

template <typename T> void chkMax(T &x, T y) { if (y > x) x = y; }
template <typename T> void chkMin(T &x, T y) { if (y < x) x = y; }
//读入
template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}
//打印
template <typename T> void print(T x) {
    if (x < 0) { putchar('-'); print(-x); return ; }
    if (x >= 10) print(x / 10);
    putchar((x % 10) + '0');
}

//树状数组
template<typename T, int N>
struct FenWick {
    vector<T> arr;
    FenWick(): arr(N + 1, 0) {}
    void update(int pos, T val) {
        for (; pos <= N;pos |= (pos + 1)) {
            arr[pos] += val;
        }
    }
    // 获取 [1, pos] 的和
    T get(int pos) {
        T ret = 0;
        for (; pos > 0; --pos) {
            ret += arr[pos];
            pos &= (pos + 1);
        }
        return ret;
    }
    // 获取 [l, r] 的和
    T query(int l, int r) {
        return get(r) - get(l - 1);
    }
};
bool isPrime(int x){
    if(x <= 1) return false;
    int cur = 2;
    while(cur * cur <= x){
        if(x % cur != 0){
            return false;
        }
        ++cur;
    }
    return true;
}

int son[100010 * 31][2], idx;
void add(int num){
    int p = 0;
    for(int i = 30; i >= 0; --i){
        int u = (num >> i) & 1;
        if(!son[p][u]) son[p][u] = ++idx;
        p = son[p][u];
    }
}
int query(int x){
    int p = 0, ret = 0;
    for(int i = 30; i >= 0; --i){
        int u = (x >> i) & 1;
        if(son[p][!u])  ret = ret * 2 + !u, p = son[p][!u];
        else ret = ret * 2 + u, p = son[p][u];  
    }
    return ret ^ x;
}
class Solution {
public:
    Solution(){
        idx = 0;
        memset(son, 0, sizeof son);
    }
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& q) {
        for(int i = 0; i < q.size(); i++) q[i].ep(i);
        sort(all(nums));
        int cnt = 0;
        sort(all(q), [](vector <int> & a, vector<int> & b){return a[1] < b[1];});
        vector <int> ans(q.size());
        for(int i = 0; i < q.size(); i++){
            int k = q[i][1];
            while(cnt < nums.size()){
                if(nums[cnt] <= k) add(nums[cnt++]);
                else break;
            }
            if(cnt) ans[q[i][2]] = query(q[i][0]);
            else ans[q[i][2]] = -1;
        }
        return ans;
    }
};

//leetcode submit region end(Prohibit modification and deletion)

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