<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 UnionFind:
    def __init__(self, n):
        self.ar = [-1] * n
    
    def find(self, x):
        if self.ar[x] < 0:return x
        self.ar[x] = self.find(self.ar[x])
        return self.ar[x]
    
    def union(self, x , y):
        p1, p2 = self.find(x), self.find(y)
        if p1 == p2:return
        if p1 > p2:
            self.ar[p2] += self.ar[p1]
            self.ar[p1] = p2
        else:
            self.ar[p1] += self.ar[p2]
            self.ar[p2] = p1


class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        if len(nums) == 1:return True
        l = len(nums)
        uf = UnionFind(l)
        m = max(nums)
        self.memo = {}

        def sieve_of_eratosthenes(n):
            # Step 1: Create a boolean array "prime[0..n]" and initialize all entries as true.
            prime = [True for _ in range(n+1)]
            p = 2  # Start from the first prime number

            # Step 2: Mark non-primes as False
            while (p * p <= n):
                if (prime[p] == True):
                    for i in range(p * p, n+1, p):
                        prime[i] = False
                p += 1

            # Step 3: Collect and return all prime numbers
            primes = [p for p in range(2, n+1) if prime[p]]
            return primes
        
        primes = sieve_of_eratosthenes(m)

        def pf(n):
            if n in self.memo:
                return self.memo[n]
            original_n = n
            factors = []
            for prime in primes:
                if prime * prime > n:
                    break
                if n % prime == 0:
                    factors.append(prime)
                while n % prime == 0:
                    n //= prime
                    if n in self.memo:
                        factors += self.memo[n]
                        self.memo[original_n] = factors
                        return factors
            if n > 1:
                factors.append(n)
            self.memo[original_n] = factors
            return factors
        
        d = {}
        for i in range(l - 1, -1, -1):
            if nums[i] == 1:return False
            temp = pf(nums[i])
            for j in temp:
                if j not in d:
                    d[j] = i
                else:
                    uf.union(i, d[j])
                    d[j] = i
        count = 0
        for i in uf.ar:
            if i < 0:
                count += 1
        if count > 1:return False
        return True
                
        

        
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