<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:
    __slots__ = ("left", "right")
    def __init__(self, left=None, right=None):
        self.left = left
        self.right = right


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:

        trie = Trie()

        def add(num):
            cur = trie
            for k in range(max_k, -1, -1):
                if num >> k & 1:
                    if not cur.right:
                        cur.right = Trie()
                    cur = cur.right
                else:
                    if not cur.left:
                        cur.left = Trie()
                    cur = cur.left
        
        def check(num):
            cur = trie
            x = 0
            for k in range(max_k, -1, -1):
                xk = 1 << k
                if num & xk:
                    if cur.left:
                        cur = cur.left
                        x |= xk
                    elif cur.right:
                        cur = cur.right
                    else:
                        break  # 稍微剪点枝
                else:
                    if cur.right:
                        cur = cur.right
                        x |= xk
                    elif cur.left:
                        cur = cur.left
                    else:
                        break
            return x

        nums = sorted(set(nums))
        queries = sorted(enumerate(queries), key=lambda x: x[1][1])
        max_k = max(nums[-1], queries[-1][1][1]).bit_length()

        n, q = len(nums), len(queries)
        res = [-1] * q

        # 双指针
        n0 = q0 = 0
        while n0 < n and q0 < q:
            idx, [xi, mi] = queries[q0]
            if nums[n0] <= mi:  # 防止有比mi大的numi过早被加进树中
                add(nums[n0])
                n0 += 1
            else:
                if n0:  # 没有比mi小的numi，查询结果自然是-1
                    res[idx] = check(xi)
                q0 += 1
        if q0 < q:
            for q1 in range(q0, q):
                idx, [xi, _] = queries[q1]
                res[idx] = check(xi)
        return res
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