<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:
        
        parents= [i for i in range(len(nums))]
        rank = [0]* len(nums)
        count = len(nums)

        def findParent(i):
            p = parents[i]
            if p!=i:
                parents[i] = findParent(p)
            return parents[i]
        
        def union(i,j):
            nonlocal count
            p1 = findParent(i)
            p2 = findParent(j)

            if p1!=p2:
                if rank[p1] >= rank[p2]:
                    parents[p2]=p1
                    rank[p1]+=1
                elif rank[p1] < rank[p2]:
                    parents[p1]= p2
                    rank[p2]+=1

                count-=1


        tags = {}
        def MergeTags(n, index):
            f = 2
            while f * f <= n:
                if n % f == 0:
                    if f not in tags:
                        tags[f]=index
                    else:
                        union(tags[f], index)

                    while n % f==0:
                        n = n//f
                f+=1
            if n > 1:
                if n in tags:
                    union(tags[n], index)
                else:
                    tags[n]=index
        

        

        for index, value in enumerate(nums):
            MergeTags(value, index)
        
      
          
        return True if count==1 else False
        
       

     
        
'''
0,  1, 2, 3
[14,30,30,49]

[0][1] = true
[0][2] = true
[0][3] =true
[1][2]=true
[1][3]=false
[2][3] = false

0->1 true
0->2 true
0->3 true
1->2 true
1->3 1-2-3 true
2->3 2-0->3 true
'''




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