<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):
        self.parent = {}
        self.count = 0
    def find(self,x): 
        if self.parent[x] == x:
            return x
        self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
    def union(self,x,y): 
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx != rooty:
            self.parent[rootx] = self.parent[rooty]
            self.count -= 1


class Solution:
    def gcdSort(self, nums: List[int]) -> bool:
        cnt = set()
        def primes2(n):
            primfac = set()
            d = 2
            while d * d <= n:
                while (n % d) == 0:
                    primfac.add(d)
                    n //= d
                d += 1
            if n > 1:
                primfac.add(n)
            return primfac
        n = max(nums)
        uf = UnionFind(n+1)
        for i in range(n+1):
            uf.parent[i] = i
            uf.count += 1

        for num in nums:
            primfac = primes2(num)
            for f in primfac:
                uf.union(num,f)
        for x, y in zip(nums, sorted(nums)):
            if uf.find(x) != uf.find(y):
                return False

        return True
    
    


        
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