<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:
    class DSU {
        vector<int> parent;
        vector<int> size;

    public:
        DSU(int n) {
            for (int i = 0; i < n; i++) {
                parent.push_back(i);
                size.push_back(1);
            }
        }
        int find(int n) {
            if (n == parent[n])
                return n;
            return parent[n]=find(parent[n]);
        }
        void merge(int a, int b) {
            int p1 = find(a);
            int p2 = find(b);
            if (p1 != p2) {
                if(size[p1]>size[p2])
                    swap(p1,p2);
                parent[p1] = p2;
                size[p1]+=size[p2];
            }
        }
    };
    bool canTraverseAllPairs(vector<int>& nums) {
        if(nums.size()==1)return true;
        int MAX = 1e5 + 1;
        int sieve[100001] = {0};
        for (int i = 2; i < MAX; i++) {
            if (sieve[i] == 0) {
                for (int d = i; d < MAX; d += i) {
                    sieve[d] = i;
                }
            }
        }
        DSU unionfind(2*MAX+1);
        for (auto i : nums) {
            if(i==1) return false;
            int val = i;
            while (val > 1) {
                int prime = sieve[val];
                int root = prime + MAX;
                if(unionfind.find(root)!=unionfind.find(i)){
                    unionfind.merge(root,i);
                }
                while(val%prime==0 && val>1){
                    val/=prime;
                }
            }
        }
        int cnt = 0;
        unordered_map<int,int>mp;
        for(auto i:nums){
            if(mp[i]) continue;
            mp[i]=9;
            if(unionfind.find(i)==i){
                cnt++;
                if(cnt>1) return false;
            }
        }
        return true;
    }
};
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