<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>

// #pragma GCC optimize("O3", "unroll-loops")
class UnionFind {    
public:
    vector<int> root, size;
    UnionFind(int n) : root(n), size(n) {
        size.assign(n, 1);
        iota(root.begin(), root.end(), 0);
    }

    int Find(int x) {
        if (x == root[x]) return x;
        else return root[x] = Find(root[x]);
    }

    void Union(int x, int y) {
        int rX = Find(x), rY = Find(y);
        if (rX == rY)  return;
        if (size[rX] > size[rY]) swap(rX, rY);   
        root[rX] = rY;
        size[rY] += size[rX];
    }

    bool connected(int x, int y) {
        return Find(x) == Find(y);
    }

    void reset(int x){//Useful for removal
        root[x]=x;
        size[x]=1;
    }
};

class Solution {
public:
    vector<int> prime;
    vector<int> num_state;

    bitset<100001> sieved=0;
    
    void Sieve(int n){//TC: O(n log log n)
        sieved[0]=sieved[1]=1;
        int&& n_sqrt=int(sqrt(n));
        for(int i=2; i<=n_sqrt; i++){
            if (!sieved[i]){
                for(int j=i*i ; j<=n; j+=i)
                    sieved[j]=1;
            }
        }
        prime.reserve((int)log(n));
        for(int i=2; i<=n; i++)
            if (!sieved[i]) prime.push_back(i);
    }
    // Different from the one for DFS
    void build_commonFactor_path(auto& nums, int M, UnionFind& graph) {
        num_state.assign(M+1, 0);
       
        for (int x : nums)
            num_state[x] = 1;

        for (int p : prime) {
            for (int i = p*2; i <= M; i += p) {
                if (num_state[i]==1) {
                    graph.Union(i, p);
                }
            }
        }
    }

    bool canTraverseAllPairs(vector<int>& nums) {
        if (nums.size()==1) return 1;
        int M=*max_element(nums.begin(), nums.end());
        int sqrtM=sqrt(M);
        Sieve(M);
        bitset<100001> S=0;
        for (int x : nums){
            if (x==1 ) return 0;
            S[x]=1;
        }
        nums.clear();
        for (int i=2; i<=M; i++)
            if(S[i]) nums.push_back(i);
        
        int n=nums.size(), sz=0;
        UnionFind graph(M+1);
        build_commonFactor_path(nums, M, graph);
        int x0=nums[0];
        for (int i=1; i<n; i++)
            if (!graph.connected(x0, nums[i])) return 0;
        return 1;
    }
};
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