<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>, and you are allowed to <strong>traverse</strong> between its indices. You can traverse between index <code>i</code> and index <code>j</code>, <code>i != j</code>, if and only if <code>gcd(nums[i], nums[j]) &gt; 1</code>, where <code>gcd</code> is the <strong>greatest common divisor</strong>.</p>

<p>Your task is to determine if for <strong>every pair</strong> of indices <code>i</code> and <code>j</code> in nums, where <code>i &lt; j</code>, there exists a <strong>sequence of traversals</strong> that can take us from <code>i</code> to <code>j</code>.</p>

<p>Return <code>true</code><em> if it is possible to traverse between all such pairs of indices,</em><em> or </em><code>false</code><em> otherwise.</em></p>

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

<pre>
<strong>Input:</strong> nums = [2,3,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).
To go from index 0 to index 1, we can use the sequence of traversals 0 -&gt; 2 -&gt; 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 &gt; 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 &gt; 1.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,9,5]
<strong>Output:</strong> false
<strong>Explanation:</strong> No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.
</pre>

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

<pre>
<strong>Input:</strong> nums = [4,3,12,8]
<strong>Output:</strong> true
<strong>Explanation:</strong> There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.
</pre>

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

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

class UnionFind:
    def __init__(self):
        self.parent = {}
        self.size = 0
        self.rank = {}
    
    def add(self, x):
        if x not in self.parent:
            self.parent[x] = None
            self.size += 1
            self.rank[x] = 1
    
    def find(self, x):
        root = x
        while self.parent[root] != None:
            root = self.parent[root]
        
        while root != x:
            self.parent[x], x = root, self.parent[x]
        return root
    
    def union(self, x, y):
        rx, ry = self.find(x), self.find(y)
        if rx == ry:
            return
        
        if self.rank[rx] > self.rank[ry]:
            self.parent[ry] = rx
        elif self.rank[rx] < self.rank[ry]:
            self.parent[rx] = ry
        else:
            self.parent[rx] = ry
            self.rank[ry] += 1

        self.size -= 1

class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        n = len(nums)
        if n == 1:
            return True

        uf = UnionFind()
        m = max(nums)
        isPrime = [0] + [1] * m
        isPrime[1] = 0
        for i in range(2, isqrt(m) + 1):
            if isPrime[i]:
                for j in range(i * i, m + 1, i):
                    isPrime[j] = 0
        
        for num in nums:
            if num == 1:
                return False

            uf.add(num)
            for a in range(2, isqrt(num) + 1):
                if num % a == 0:
                    if isPrime[a]:
                        uf.add(a)
                        uf.union(a, num)

                    b = num // a
                    if isPrime[b]:
                        uf.add(b)
                        uf.union(b, num)

        return uf.size == 1

        
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