<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:
    vector<int>P;
    int sets;
    bool canTraverseAllPairs(vector<int>& nums) {
        int n=nums.size();
        if(P.size()==0)build_Prime();
        build_set(n);
        vector<vector<int>>groups(100005);
        for(int i=0;i<nums.size();i++)if(nums[i]!=1){
            for(int p:P){
                if(nums[i]%p==0){
                    divide(nums[i],p);
                    groups[p].push_back(i);
                }
                if(nums[i]==1||p*p>nums[i])break;
            }
            if(nums[i]>1)groups[nums[i]].push_back(i);
        }
        for(int i=0;i<100005;i++)if(groups[i].size()>1){
            for(int j=1;j<groups[i].size();j++){
                uni(groups[i][0],groups[i][j]);
                if(sets==1)return 1;
            }
        }
        return (sets==1);
    }

private:
    using pii=pair<int,int>;
    vector<pii>dsu;
    
    void build_Prime(){
        bitset<100005>notP;
        notP[1]=1;
        for(int i=2;i<100005;i++)if(!notP[i]){
            P.push_back(i);
            for(long long j=(long long)i*i;j<100005;j+=i)notP[j]=1;
        }
        // cout<<P.size()<<"\n";
    }
    void divide(int &a,long long p){
        if(p<=1)return;
        if((long long)a%(long long)(p*p)==0)divide(a,p*p);
        if(a%p==0)a/=p;
    }
    void build_set(int n){
        dsu.resize(n);
        sets=n;
        for(int i=0;i<n;i++)dsu[i]=pii(i,1);//parent,size
    }
    int find_set(int i){
        if(dsu[i].first!=i)
            dsu[i].first=find_set(dsu[i].first);
        return dsu[i].first;
    }
    void uni(int i,int j){
        i=find_set(i),j=find_set(j);
        if(i==j)return;
        sets--;
        if(dsu[i].second<dsu[j].second)swap(i,j);
        dsu[j].first=i;
        dsu[i].second+=dsu[j].second;
        dsu[j].second=0;
    }
};
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