<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:
        mx=max(nums)
        def sieve(n):
            is_prime = [True] * (n + 1)
            is_prime[0] = is_prime[1] = False
            
            p = 2
            while (p * p <= n):
                if is_prime[p]:
                    for i in range(p * p, n + 1, p):
                        is_prime[i] = False
                p += 1
            primes = [p for p in range(n + 1) if is_prime[p]]
            return primes
        pr=sieve(mx)
        s=set(nums)
        if len(nums)==1:
            return True
        if 1 in s:
            return False
        par=[0]
        rank=[1]
        for i in range(mx+1):
            par.append(1+i)
            rank.append(1)
        def unio(x,y):
            x=get(x)
            y=get(y)
            if rank[x]==rank[y]:
                rank[x]+=1
            if rank[x]>rank[y]:
                par[y]=par[x]
            else:
                par[x]=par[y]
        def get(x):
            if x==par[x]:
                return x
            par[x]=get(par[x])
            return par[x]
        for i in pr:
            fl=False
            par1=-1
            for j in range(i,mx+1,i):
                if j in s and not fl:
                    par1=j
                    fl=True
                elif j in s:
                    unio(j,par1)
        ans=0
        par1=-1
        for i in range(mx+1):
            if i in s:
                if par1==-1:
                    par1=get(i)
                else:
                    if get(i)!=par1:
                        return False
        return True

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