<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>

#define ll long long

class DSU{
public:
    vector<ll> parents;
    vector<ll> rank;
    DSU(ll n){
        parents.resize(n, 0);
        rank.resize(n, 0);
        for(int i=1; i<n; i++){
            parents[i] = i;
        }
    }
    ll findparent(ll a){
        if(a==parents[a]) return a;
        return parents[a] = findparent(parents[a]);
    }
    void Unite(ll a, ll b){
        ll ap = findparent(a);
        ll bp = findparent(b);
        if(rank[ap]<rank[bp]) parents[ap] = bp;
        else if (rank[bp]<rank[ap]) parents[bp] = ap;
        else{
            parents[ap] = bp;
            rank[bp]++;
        }
    }
};

class Solution {
public:
    ll getrt(ll num){
        ll s = 1;
        ll e = num;
        while(s!=e){
            ll mid = (s+e)/2;
            if(mid*mid<num) s = mid+1;
            else e = mid;
        }
        return s;
    }
    
    bool canTraverseAllPairs(vector<int>& nums) {
        // sort(nums.begin(), nums.end());
        ll mx = nums[0];
        for(auto x: nums) mx = max(1ll*x, mx);
        // vector<ll> primes = Seive(mx);
        ll poss = getrt(mx);
        DSU D(nums.size());
        map<ll, vector<ll>> mp;
        for(ll num=1; num<=poss; num++){
            for(int i=0; i<nums.size(); i++){
                if(nums[i]%num==0){
                    mp[num].push_back(i);
                    mp[nums[i]/num].push_back(i);
                }
            }
        }
        for(auto x: mp){
            if(x.first==1) continue;
            bool found = false;
            int ind = -1;
            for(auto y: x.second){
                if(!found){
                    found = true;
                    ind = y;
                }
                else{
                    D.Unite(y, ind);
                }
            }
        }

        // for(ll num=2; num<=poss; num++){
        //     ll found = false;
        //     ll ind = -1;
        //     for(int i=0; i<nums.size(); i++){
        //         if(nums[i]%num==0){
        //             if(found){
        //                 D.Unite(i, ind);
        //             }
        //             else{
        //                 found = true;
        //                 ind = i;
        //             }
        //         }
        //     }
        // }
        for(int i=0; i<D.parents.size(); i++){
            cout<<D.parents[i]<<" ";
        }
        for(int i=1; i<nums.size(); i++){
            if(D.findparent(i) != D.findparent(i-1)) return false;
        }
        return true;
    }
};
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