<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 {
    static constexpr int primes[]{
        2,   3,   5,   7,   11,  13,  17,  19,  23,  29,  31,  37,  41,
        43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89,  97,  101,
        103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167,
        173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239,
        241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313};

    array<int, 100001> dsu, rank;
    int root(int x) {
        if (dsu[x] == 0) { return dsu[x] = x; }
        return dsu[x] == x ? x : dsu[x] = root(dsu[x]);
    }
    void unite(int lhs, int rhs) {
        const auto l = root(lhs), r = root(rhs);
        if (l == r) { return; }
        if (rank[l] > rank[r]) {
            dsu[r] = l;
        } else if (rank[l] < rank[r]) {
            dsu[l] = r;
        } else {
            dsu[r] = l;
            ++rank[l];
        }
    }

public:
    bool canTraverseAllPairs(vector<int> & nums) {
        if (nums.size() == 1) { return true; }
        for (auto x : nums) {
            if (x == 1) { return false; }
            root(x);
            for (auto p : primes) {
                if (p >= x) { break; }
                if (x % p == 0) {
                    unite(root(x), root(p));
                    unite(root(x), root(x / p));
                    x /= p;
                }
            }
        }
        const auto rt = root(nums[0]);
        for (auto i : nums) {
            if (rt != root(i)) { return false; }
        }
        return true;
    }
};
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