<p>You are given an integer array <code>nums</code>, and you can perform the following operation <strong>any</strong> number of times on <code>nums</code>:</p>

<ul>
	<li>Swap the positions of two elements <code>nums[i]</code> and <code>nums[j]</code> if <code>gcd(nums[i], nums[j]) &gt; 1</code> where <code>gcd(nums[i], nums[j])</code> is the <strong>greatest common divisor</strong> of <code>nums[i]</code> and <code>nums[j]</code>.</li>
</ul>

<p>Return <code>true</code> <em>if it is possible to sort </em><code>nums</code><em> in <strong>non-decreasing</strong> order using the above swap method, 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 = [7,21,3]
<strong>Output:</strong> true
<strong>Explanation:</strong> We can sort [7,21,3] by performing the following operations:
- Swap 7 and 21 because gcd(7,21) = 7. nums = [<u><strong>21</strong></u>,<u><strong>7</strong></u>,3]
- Swap 21 and 3 because gcd(21,3) = 3. nums = [<u><strong>3</strong></u>,7,<u><strong>21</strong></u>]
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,2,6,2]
<strong>Output:</strong> false
<strong>Explanation:</strong> It is impossible to sort the array because 5 cannot be swapped with any other element.
</pre>

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

<pre>
<strong>Input:</strong> nums = [10,5,9,3,15]
<strong>Output:</strong> true
We can sort [10,5,9,3,15] by performing the following operations:
- Swap 10 and 15 because gcd(10,15) = 5. nums = [<u><strong>15</strong></u>,5,9,3,<u><strong>10</strong></u>]
- Swap 15 and 3 because gcd(15,3) = 3. nums = [<u><strong>3</strong></u>,5,9,<u><strong>15</strong></u>,10]
- Swap 10 and 15 because gcd(10,15) = 5. nums = [3,5,9,<u><strong>10</strong></u>,<u><strong>15</strong></u>]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>2 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class UnionFind:

    def __init__(self, n: int):
        self.root = list(range(n))
        self.rank = [1 for _ in range(n)]

    def find(self, x: int) -> int:
        if self.root[x] != x: self.root[x] = self.find(self.root[x])
        return self.root[x]

    def union(self, x: int, y: int) -> None:
        rootx, rooty = self.find(x), self.find(y)
        if rootx == rooty: return
        if self.rank[rootx] > self.rank[rooty]: self.root[rooty] = rootx
        elif self.rank[rootx] < self.rank[rooty]: self.root[rootx] = rooty
        else:
            self.root[rooty] = rootx
            self.rank[rootx] += 1

    def is_connected(self, x: int, y: int) -> bool:
        return self.find(x) == self.find(y)


class Solution:
    def gcdSort(self, nums: List[int]) -> bool:
        def get_prime_factors(n: int) -> List[int]:
            s = set()
            while n % 2 == 0:
                s.add(2)
                n //= 2
            
            for i in range(3, int(math.sqrt(n)) + 1, 2):
                while n % i == 0:
                    s.add(i)
                    n //= i
            if n > 2: s.add(n)
            return list(s)

        primes = collections.defaultdict(int)
        uf = UnionFind(len(nums))
        for i, num in enumerate(nums):
            prime = get_prime_factors(num)
            for p in prime:
                if p in primes: uf.union(primes[p], i)
                primes[p] = i
        
        for i, (_, (_, j)) in enumerate(zip(nums, sorted([(num, i) for i, num in enumerate(nums)]))):
            if not uf.is_connected(i, j): return False
        
        return True
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