<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>par,sz;
    int find(int x){
        if(x==par[x])return x;
        else return par[x]=find(par[x]);
    }
    void merge(int x,int y){
        x=par[x];
        y=par[y];
        if(sz[x]>sz[y]){
            sz[x]+=sz[y];
            par[y]=x;
        }
        else{
            sz[y]+=sz[x];
            par[x]=y;
        }
    }
    bool canTraverseAllPairs(vector<int>& v) {
        if(v.size()==1)return true;
        par.clear();
        sz.clear();
        int n=v.size();
        set<int>st;
        for(int i=0;i<n;i++)st.insert(v[i]);
        if(st.count(1)>0)return false;
        int mx=*st.rbegin();
        par.resize(mx+1);
        sz.assign(mx+1,1);
        for(int i=1;i<=mx;i++)par[i]=i;

        for(int i=2;i<=mx;i++){
            if(par[i]!=i)continue;
            for(long long j=2*i;j<=mx;j+=i){
                if(st.find(j)==st.end())continue;
                int x=find(i),y=find(j);
                if(x!=y){
                    merge(i,j);
                }
            }
        }
        set<int>t;
        for(auto i:st){
            t.insert(find(i));
            cout<<find(i)<<' ';
        }
        cout<<endl;
        if(t.size()>1)return false;
        else return true;
    }
};
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