<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> seive(int n){
    vector<int> p(n+1,1),lp(n+1,-1);
    p[0]=0;p[1]=0;
    for(int i=2;i*i<=n;i++){
        if(!p[i]) continue;
        lp[i]=i;
        for(int j=i*i;j<=n;j+=i){
            p[j]=0;
            if(lp[j]==-1) lp[j]=i;
        }
    }
    return lp;
}

const static int N=1e5+5;
int par[N],siz[N],there[N];

void make(){
    for(int i=0;i<N;i++){
        par[i]=i;
        siz[i]=1;
        there[i]=0;
    }
}

int find_p(int v){
    if(v==par[v]) return v;
    return par[v]=find_p(par[v]);
}

void merge(int u,int v){
    u=find_p(u);v=find_p(v);
    if(u==v) return;
    if(siz[u]<siz[v]) swap(u,v);
    siz[u]+=siz[v];
    par[v]=u;
}
    bool canTraverseAllPairs(vector<int>& nums) {
        int n=nums.size();
        if(n==1) return 1;
        vector<int> lp=seive(1e5+5);
        make();
        for(auto x : nums){
            vector<int> p;
            if(x==1) return 0;
            while(x>1){
                int k=lp[x];
                if(k==-1) break;
                there[k]=1;
                p.push_back(k);
                while(x%k==0) x/=k;
            }
            if(x>1){
                there[x]=1;p.push_back(x);
            }
            for(int i=1;i<p.size();i++){
                int a=p[i-1],b=p[i];
                merge(a,b);
            }
        }
        set<int> s;
        for(int i=0;i<N;i++){
            if(!there[i]) continue;
            int v=find_p(i);
            s.insert(v);
        }
        return (s.size()==1);
    }
};
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