<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 DSU{
    vector<int> par, rank;
    public:
    DSU(int N){
        for(int i = 0; i <= N; i++){
            par.push_back(i);
            rank.push_back(1);
        }
    }
    int find(int i){
        return i == par[i] ? par[i] : find(par[i]);
    }
    void connect(int a, int b){
        int pa = find(a);
        int pb = find(b);
        if(pa == pb) return;
        if(rank[pa] >= rank[pb]){
            rank[pa] += rank[pb];
            par[pb] = pa;
        }
        else{
            rank[pb] += rank[pa];
            par[pa] = pb;
        }
    }
};
class Solution {
    const int maxN = 1e5 + 5;
    vector<int> isPrime;
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        if(n == 1) return true;
        sieveOfEratosthenes();  

        vector<bool> has(maxN, false); 
        for(auto x : nums){
            has[x] = true;
        }       
        if(has[1]) return false;

        DSU uf(2 * maxN + 1);

        for(auto x : nums){

            int val = x;
            while(val > 1){
                int prime = isPrime[val];
                int root = prime + maxN;
                uf.connect(root, x);
                while(val % prime == 0){
                    val /= prime;
                }
            }
        }

        int cnt = 0;
        unordered_set<int> par_cnt;
        for(auto x : nums){
            par_cnt.insert(uf.find(x));
        }
        return par_cnt.size() == 1;


    }
    void sieveOfEratosthenes(){
        isPrime = vector<int>(maxN + 1, 0);

        for(int p = 2; p <= maxN; p++){

            if(isPrime[p] == 0){
                for(int i = p; i <= maxN; i += p){
                    isPrime[i] = p;
                }
            }
        }
    }
};
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