<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>

MAXN = 100001

# stores smallest prime factor for
# every number = 1
spf = [1] * (MAXN + 1)
# Calculating SPF (Smallest Prime Factor)
# for every number till MAXN.
# Time Complexity : O(nloglogn)


def sieve():
    spf[0] = 0
    for i in range(2, MAXN + 1):
        if spf[i] == 1:  # if the number is prime, mark
                         # all its multiples who havent
                         # gotten their spf yet
            for j in range(i, MAXN + 1, i):
                if spf[j] == 1:  # if its smallest prime factor is
                                 # 1 means its spf hasnt been
                                 # found yet so change it to i
                    spf[j] = i

# A O(log n) function returning prime
# factorization by dividing by smallest
# prime factor at every step


def f(x):
    ret = []
    while (x != 1):
        ret.append(spf[x])
        x = x // spf[x]

    return ret

class Solution:
    def gcdSort(self, nums: List[int]) -> bool:
        sieve()

        parent = [i for i in range(max(nums) + 1)]
        def find(node):
            p = parent[node]
            while p != parent[p]:
                parent[p] = parent[parent[p]]
                p = parent[p]
            return p

        def uunion(a, b):
            p1, p2 = find(a), find(b)
            if p1 == p2: return

            parent[p1] = p2

        for num in nums:
            for fac in f(num):
                uunion(num, fac)
        for i, num in enumerate(sorted(nums)):
            if find(num) != find(nums[i]): return False
        return True
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