<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 {
public:
    bool canTraverseAllPairs(vector<int>& nums) {
        if(nums.size() == 1) return true;
        generate_primes(330);
        for(auto x:nums){
            if(x == 1) return false;
            auto factors = get_factors(x);
            connect(factors);
        }
    
        auto p = g.begin()->first;
        unordered_set<int> visited;
        dfs(p, visited);
        return visited.size() == g.size();
    }

    void dfs(int i, unordered_set<int>& visited){
        if(visited.contains(i)) return;
        visited.insert(i);
        for(auto j : g[i]){
            dfs(j, visited);
        }
    }

    void connect(const vector<int>& factors){
        for(int i=0;i<factors.size();i++){
            auto f1 = factors[i];
            g[f1];
            for(int j=0;j<i;j++){
                auto f2 = factors[j];
                g[f1].insert(f2);
                g[f2].insert(f1);
            }
        }
    }

    void generate_primes(int n){
        for(int i=2;i<=n;i++){
            bool is_prime = true;
            for(auto p:primes){
                if(i % p == 0) {
                    is_prime = false;
                    break;
                }
                if(i < p*p) break;
            }
            if(is_prime) primes.push_back(i);
        }
    }

    vector<int> get_factors(int x) const{
        vector<int> res;
        for(auto p:primes){
            if(p > x) break;
            if(x % p == 0){
                res.push_back(p);
                while(x % p ==0) x/=p;
            }
        }
        if(x > 1) res.push_back(x);
        return res;
    }

    unordered_map<int,unordered_set<int>> g;
    vector<int> primes;

};
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