<p>There is an undirected tree with <code>n</code> nodes labeled from <code>1</code> to <code>n</code>. You are given the integer <code>n</code> and a 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> in the tree.</p>

<p>Return <em>the <strong>number of valid paths</strong> in the tree</em>.</p>

<p>A path <code>(a, b)</code> is <strong>valid</strong> if there exists <strong>exactly one</strong> prime number among the node labels in the path from <code>a</code> to <code>b</code>.</p>

<p><strong>Note</strong> that:</p>

<ul>
	<li>The path <code>(a, b)</code> is a sequence of <strong>distinct</strong> nodes starting with node <code>a</code> and ending with node <code>b</code> such that every two adjacent nodes in the sequence share an edge in the tree.</li>
	<li>Path <code>(a, b)</code> and path <code>(b, a)</code> are considered the <strong>same</strong> and counted only <strong>once</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/27/example1.png" style="width: 440px; height: 357px;" />
<pre>
<strong>Input:</strong> n = 5, edges = [[1,2],[1,3],[2,4],[2,5]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The pairs with exactly one prime number on the path between them are: 
- (1, 2) since the path from 1 to 2 contains prime number 2. 
- (1, 3) since the path from 1 to 3 contains prime number 3.
- (1, 4) since the path from 1 to 4 contains prime number 2.
- (2, 4) since the path from 2 to 4 contains prime number 2.
It can be shown that there are only 4 valid paths.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/27/example2.png" style="width: 488px; height: 384px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[1,2],[1,3],[2,4],[3,5],[3,6]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The pairs with exactly one prime number on the path between them are: 
- (1, 2) since the path from 1 to 2 contains prime number 2.
- (1, 3) since the path from 1 to 3 contains prime number 3.
- (1, 4) since the path from 1 to 4 contains prime number 2.
- (1, 6) since the path from 1 to 6 contains prime number 3.
- (2, 4) since the path from 2 to 4 contains prime number 2.
- (3, 6) since the path from 3 to 6 contains prime number 3.
It can be shown that there are only 6 valid paths.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li>The input is generated such that <code>edges</code> represent a valid tree.</li>
</ul>

class Solution:
    def countPaths(self, n: int, edges: List[List[int]]) -> int:
        def sieve_of_eratosthenes(n):
            if n < 2: return []
            isPrime = [True]*(n+1)
            p = 2
            while p * p <= n:
                if isPrime[p]:
                    for i in range(p * p, n + 1, p):
                        isPrime[i] = False
                p += 1
            return [i for i in range(2, n+1) if isPrime[i]]
        def find(i):
            while i != par[i]:
                par[i] = par[par[i]] # path compression
                i = par[i]
            return i
        def union(i, j):
            i, j = find(i), find(j)
            if i == j: 
                return 0
            else:
                if rank[i] < rank[j]:
                    par[i] = j
                    rank[j] += rank[i]
                else:
                    par[j] = i
                    rank[i] += rank[j]
            return 1
        
        primes = set(sieve_of_eratosthenes(n))
        ans = 0
        g = defaultdict(list)
        for u, v in edges:
            g[u].append(v)
            g[v].append(u)
        par = [i for i in range(n+1)]
        rank = [1]*(n+1)
        for u, v in edges:
            if u not in primes and v not in primes:
                union(u, v)
        ps = [i for i in range(1, n+1) if i in primes]
        for p in ps:
            pars = set()
            for nei in g[p]:
                if nei in primes: continue
                pars.add(par[nei])
            pars = [rank[i] for i in pars]
            tot = sum(pars)
            ans += tot
            y = 0
            for x in pars:
                y += x*(tot-x)
            ans += y//2
        return ans
1495