<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 BinaryTrie:
    def __init__(self, max_bit: int = 30):
        self.to = [[-1], [-1]]
        self.cnt = [0]
        self.mn = [1 << 63]
        self.max_bit = max_bit

    def add(self, num: int) -> None:
        cur = 0
        self.cnt[cur] += 1
        self.mn[cur] = min(self.mn[cur], num)
        for k in range(self.max_bit, -1, -1):
            bit = (num >> k) & 1
            if self.to[bit][cur] == -1:
                self.to[bit][cur] = len(self.cnt)
                self.to[0].append(-1)
                self.to[1].append(-1)
                self.cnt.append(0)
                self.mn.append(1 << 63)
            cur = self.to[bit][cur]
            self.cnt[cur] += 1
            self.mn[cur] = min(self.mn[cur], num)

    def remove(self, num: int) -> bool:
        if self.cnt[0] == 0: return False
        cur = 0
        rm = [0]
        for k in range(self.max_bit, -1, -1):
            bit = (num >> k) & 1
            cur = self.to[bit][cur]
            if cur == -1 or self.cnt[cur] == 0: return False
            rm.append(cur)
        for cur in rm: self.cnt[cur] -= 1
        return True

    def count(self, num: int):
        cur = 0
        for k in range(self.max_bit, -1, -1):
            bit = (num >> k) & 1
            cur = self.to[bit][cur]
            if cur == -1 or self.cnt[cur] == 0: return 0
        return self.cnt[cur]

    def max_xor(self, x: int, m: int) -> int:
        if self.mn[0] > m: return -1
        res = cur = 0
        for k in range(self.max_bit, -1, -1):
            bit = (x >> k) & 1
            nxt = self.to[bit ^ 1][cur]
            if nxt == -1 or self.cnt[nxt] == 0 or self.mn[nxt] > m:
                cur = self.to[bit][cur]
            else:
                cur = nxt
                res |= 1 << k
        return res

class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        bt = BinaryTrie()
        for x in nums:
            bt.add(x)
        ans =  []
        for x, m in queries:
            ans.append(bt.max_xor(x, m))
        return ans
71568