<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 findparent(int u,vector<int>&par)
    {
        if(u==par[u])
            return u;
        return par[u]=findparent(par[u],par);
    }
    void Union(int u,int v,vector<int>&par)
    {
        int pu=findparent(u,par);
        int pv=findparent(v,par);
         if(pu!=pv)
         {
             par[pu]=pv;
         }
    }
    bool canTraverseAllPairs(vector<int>& nums) 
    {
        int n=nums.size();
        vector<int>par(100001+n);
        for(int i=1;i<100001+n;i++)
        {
            par[i]=i;
        }
        int cnt=1;
        if(n==1)
        {
            return true;
        }
        for(int i=0;i<n;i++)
        {
            if(nums[i]==1)
                return false;
            Union(100001+i,nums[i],par);
            for(int j=2;j*j<=nums[i];j++)
            {
                if(nums[i]%j==0)
                {
                    Union(100001+i,j,par);
                    if(nums[i]/j!=j)
                    {
                         Union(100001+i,nums[i]/j,par);
                    }
                }
            }
        }
        int p=findparent(100001,par);
        for(int i=1;i<nums.size();i++)
        {
            int pi=findparent(100001+i,par);
            if(p!=pi)
            {
                return false;
            }
        }
        return true;
    }
};
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