<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: List[int]) -> bool:
        m=max(nums)
        k=set(nums)
        if 1 in k and len(nums)>1:return False
        nums=[*k]
        n=len(nums)
        def unioncolor(x,y):
            xroot,yroot=findcolor(x),findcolor(y)
            if xroot==0:dad[yroot]=xroot;return
            if yroot==0:dad[xroot]=yroot;return
            if rank[xroot]>rank[yroot]:
                dad[yroot]=xroot
            else:
                if rank[xroot]==rank[yroot]:
                    rank[xroot]+=1
                    rank[yroot]+=1
                dad[xroot]=yroot
        def findcolor(x):
            if dad[x]!=x:
                dad[x]=findcolor(dad[x])
            return dad[x]
        dad=[i for i in range(m+1)]
        rank=[0]*(m+1)
        notprimes=set() #eratosthenes sieve
        for i in range(2,int(m**.3)+1):
            if i not in notprimes:
                j=i
                while j<int(m**.6)+1:
                    j+=i
                    notprimes.add(j)
        primes={*range(2,int(m**.6)+1)}-notprimes
        primes=sorted(primes)
        for e in nums:
            ecop=e
            for p in primes:
                if e%p==0:
                    unioncolor(e,p)
                    while ecop%p==0:ecop//=p
                if p>e:break #that's why I sorted them
            if ecop!=1:unioncolor(e,ecop)
        return len({*map(findcolor,nums)})==1 #check they are all equivalent
                

        
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