<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 par[100001] ;
    int siz[100001] ;
    void make(int i){
        par[i] = i ;
        siz[i] = 1 ;
    }
    int find(int i){
        if(par[i]==i)return i ;
        return par[i] = find(par[i]) ;
    }
    void Union(int i, int j){
        i = find(i) ;
        j = find(j) ;
        if(i==j)return ;
        if(siz[i]<siz[j])swap(i,j) ;
        siz[i] += siz[j] ;
        par[j] = i ;
    }
    bool canTraverseAllPairs(vector<int>& nums) {
        int nmax = 0 ;
        for(int &it : nums)nmax = max(nmax,it) ;
        vector<int>ct(nmax+1) ;
        int dis = 0  ;
        for(int &i : nums){
            ct[i]++ ;
            make(i) ;
            if(ct[i]==1)dis++ ;
        }
        
        for(int i = 2;i<=nmax;i++){
            int prev = -1 ;
            for(int j = i;j<=nmax;j+=i){
                if(ct[j]){
                    if(prev!=-1){
                        Union(j,prev) ;
                    }
                    prev = j ;
                }
            }
        }
        if(nums[0]==1 && nums.size() >=2)return false ;
        return (siz[find(nums[0])]==dis?true:false) ;
        
    }
};
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