<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>

import math
from collections import defaultdict
class union_find:
    def __init__(self, nn):
        self.root = [i for i in range(0,nn)]
        self.size = [1 for i in range(0,nn)]
    def find_root(self, a):
        while a!=self.root[a]:
            self.root[a] = self.root[self.root[a]]
            a = self.root[a]
        return a
    def union(self, a, b):
        root_a = self.find_root(a)
        root_b = self.find_root(b)
        if root_a==root_b:
            return False
        else:
            if self.size[root_a]<self.size[root_b]:
                self.root[root_a] = root_b
                self.size[root_a]+=self.size[root_b]
            else:
                self.root[root_b] = root_a
                self.size[root_b]+=self.size[root_a]
            return True
class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
       
        if len(nums)==1:
            return True
        nums = list(set(nums))
        n = len(nums)
        mx = max(nums)+1
        uf = union_find(mx)
        #find primary until max(nums)+1
        primary = [0]*mx
        for i in range(2,mx):
            if primary[i]==0:
                for v in range(i, mx, i):
                    primary[v] = i

        if nums == [1]:
            return False
        for i in range(0, n):
            val = nums[i]
            while val>1:
                pp = primary[val]
                uf.union(nums[i], pp)
                val = val//pp
        
        for i in range(1,n):
            if uf.find_root(nums[i])!=uf.find_root(nums[0]):
                return False
        
        return True



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