<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 Solution:

    def canTraverseAllPairs(self, nums):
        MAX = max(nums)
        N = len(nums)
        has = [False] * (MAX + 1)
        for x in nums:
            has[x] = True

        # edge cases
        if N == 1:
            return True
        if has[1]:
            return False

        # the general solution
        sieve = [0] * (MAX + 1)
        for d in range(2, MAX + 1):
            if sieve[d] == 0:
                for v in range(d, MAX + 1, d):
                    sieve[v] = d

        union = DSU(2 * MAX + 1)
        for x in nums:
            val = x
            while val > 1:
                prime = sieve[val]
                root = prime + MAX
                if union.find(root) != union.find(x):
                    union.merge(root, x)
                while val % prime == 0:
                    val //= prime

        cnt = 0
        for i in range(2, MAX + 1):
            if has[i] and union.find(i) == i:
                cnt += 1
        return cnt == 1


class DSU:

    def __init__(self, N):
        self.dsu = list(range(N + 1))
        self.size = [1] * (N + 1)

    def find(self, x):
        return x if self.dsu[x] == x else self.find(self.dsu[x])

    def merge(self, x, y):
        fx = self.find(x)
        fy = self.find(y)
        if fx == fy:
            return
        if self.size[fx] > self.size[fy]:
            fx, fy = fy, fx
        self.dsu[fx] = fy
        self.size[fy] += self.size[fx]
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