<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. A pair of integers <code>x</code> and <code>y</code> is called a <strong>strong</strong> pair if it satisfies the condition:</p>

<ul>
	<li><code>|x - y| &lt;= min(x, y)</code></li>
</ul>

<p>You need to select two integers from <code>nums</code> such that they form a strong pair and their bitwise <code>XOR</code> is the <strong>maximum</strong> among all strong pairs in the array.</p>

<p>Return <em>the <strong>maximum</strong> </em><code>XOR</code><em> value out of all possible strong pairs in the array</em> <code>nums</code>.</p>

<p><strong>Note</strong> that you can pick the same integer twice to form a pair.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 7
<strong>Explanation:</strong> There are 11 strong pairs in the array <code>nums</code>: (1, 1), (1, 2), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5) and (5, 5).
The maximum XOR possible from these pairs is 3 XOR 4 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,100]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are 2 strong pairs in the array nums: (10, 10) and (100, 100).
The maximum XOR possible from these pairs is 10 XOR 10 = 0 since the pair (100, 100) also gives 100 XOR 100 = 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [500,520,2500,3000]
<strong>Output:</strong> 1020
<strong>Explanation:</strong> There are 6 strong pairs in the array nums: (500, 500), (500, 520), (520, 520), (2500, 2500), (2500, 3000) and (3000, 3000).
The maximum XOR possible from these pairs is 500 XOR 520 = 1020 since the only other non-zero XOR value is 2500 XOR 3000 = 636.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 2<sup>20</sup> - 1</code></li>
</ul>

class Tril:
    class Node:
        def __init__(self, p: "Node" = None):
            self.nexts = [None] * 2
            self.par = p
            self.val = None
            self.counts = 0
        def add_val(self, v: int) -> None:
            self.val = v
            while self:
                self.counts += 1
                self = self.par
        def del_val(self):
            assert self.val and self.counts
            self.val = None
            while self:
                self.counts -= 1
                self = self.par

    def __init__(self, bits: int):
        self.root = self.Node()
        self.bits = bits
    
    def get(self, n: int) -> Node:
        b = 1 << self.bits
        assert b > n
        b >>= 1
        p = self.root

        while b:
            i = bool(b & n)
            if not p.nexts[i]:
                p.nexts[i] = self.Node(p)
            p = p.nexts[i]
            b >>= 1
        return p
    
    def find_max(self, n: int) -> Node:
        b = 1 << self.bits
        assert b > n
        n = ~n & (b - 1)

        p = self.root
        b >>= 1

        while b:
            i = bool(b & n)
            if not p.nexts[i] or p.nexts[i].counts == 0:
                i = 1 - i
            p = p.nexts[i]
            assert p and p.counts > 0, f"b == {b}, p = {p}"
            b >>= 1
        return p

class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        state = Tril(20)
        nums = list(sorted(set(nums)))
        b = best = 0
        for n in nums:
            print(n)
            state.get(n).add_val(n)
            while nums[b] * 2 < n:
                state.get(nums[b]).del_val()
                b += 1
            best = max(best, state.find_max(n).val ^ n)
        return best
85700