<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:
    bool canTraverseAllPairs(vector<int>& nums) {
        int n = nums.size();
        std::vector<int> p(n);
        std::vector<int> sz(n, 1);
        std::iota(p.begin(), p.end(), 0);
        const auto find = [&](int x){
            while (p[x] != x) {
                x = p[x] = p[p[x]];
            }
            return x;
        };
        const auto join = [&](int a, int b){
            a = find(a);
            b = find(b);
            if (a == b) return;
            if (sz[a] < sz[b]) std::swap(a, b);
            sz[a] += sz[b];
            p[b] = a;
        };
        std::map<int, int> nset;
        auto nsadd = [&](int v, int x){
            if (v <= 1) return;
            auto it = nset.find(v);
            if (it == nset.end()) {
                nset[v] = x;
            } else {
                join(x, it->second);
                it->second = x;
            }
        };
        for (int i = 0; i < n; ++i) {
            nsadd(nums[i], i);
        }
        const auto sieve_max = [&]{return nset.begin() == nset.end() ? 0 : std::prev(nset.end())->first;};
        std::vector<bool> iscomp(sieve_max() + 1);
        for (int i = 2; i * i <= sieve_max() && i <= sieve_max() / i; ++i) {
            if (iscomp[i]) continue;
            int px = -1;
            for (int j = i; j <= sieve_max(); j += i) {
                if (iscomp[j]) continue;
                iscomp[j] = true;
                auto it = nset.find(j);
                if (it == nset.end()) continue;
                int v = j;
                int x = it->second;
                nset.erase(it);
                if (px >= 0) {
                    join(px, x);
                }
                px = x;
                while (v % i == 0) {
                    v /= i;
                }
                nsadd(v, x);
            }
        }
        return sz[find(0)] == n;
    }
};
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