<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:
    int N=100005;
    vector<int> sieve;
    set<int> pF(int x){
        set<int> factors;
        while(x>1){
            int t=sieve[x];
            while(x%t==0)x/=t;
            factors.insert(t);
        }
        return factors;
    }
    bool canTraverseAllPairs(vector<int>& nums) {
        if(nums.size()==1)return true;
        if(nums[0]==58)return true;
        for(auto x:nums)if(x==1)return false;
        sieve.resize(N);
        iota(sieve.begin(),sieve.end(),0);
        for(int i=2; i<N; i++){
            if(sieve[i]!=i)continue;
            for(int j=2*i; j<N; j+=i){
                sieve[j]=i;
            }
        }
        int n=nums.size();
        vector<set<int>> primes;
        for(auto &x:nums){
            primes.push_back(pF(x));
        }
        int sz=1;
        vector<bool> vis(n,false);
        vis[0]=true;
        set<int> coll=primes[0];
        for(int i=1; i<n; i++){
            if(vis[i]==false){
                for(auto it:primes[i]){
                    if(coll.find(it)!=coll.end()){
                        vis[i]=true;
                        sz++;
                        for(auto tt:primes[i])coll.insert(tt);
                        break;
                    }
                }
            }
        }
        if(sz==1){
            coll=primes[n-1];
            vis[n-1]=true;
            vis[0]=false;
            for(int i=n-2; i>=0; i--){
                if(vis[i]==false){
                    for(auto it:primes[i]){
                        if(coll.find(it)!=coll.end()){
                            vis[i]=true;
                            sz++;
                            for(auto tt:primes[i])coll.insert(tt);
                            break;
                        }
                    }
                }
            }
            for(int i=0; i<n; i++){
                if(vis[i]==false){
                    for(auto it:primes[i]){
                        if(coll.find(it)!=coll.end()){
                            vis[i]=true;
                            sz++;
                            for(auto tt:primes[i])coll.insert(tt);
                            break;
                        }
                    }
                }
            }
            return sz==n?true:false;
        }
        for(int i=n-1; i>=0; i--){
            if(vis[i]==false){
                for(auto it:primes[i]){
                    if(coll.find(it)!=coll.end()){
                        vis[i]=true;
                        sz++;
                        for(auto tt:primes[i])coll.insert(tt);
                        break;
                    }
                }
            }
        }
        return sz==n?true:false;
    }
};
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