<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>

from collections import defaultdict

class Solution:
    def countPaths(self, n: int, edges: List[List[int]]) -> int:            
        primes = self.find_primes(n)
        prime_neighbors = defaultdict(list)
        for a, b in edges:
            if a in primes and b not in primes:
                prime_neighbors[a].append(b)
            elif b in primes and a not in primes:
                prime_neighbors[b].append(a)

        uf = UnionFind(n)
        for a, b in edges:
            if a not in primes and b not in primes:
                uf.union(a, b)
        
        return sum(self.valid_through(p, prime_neighbors, uf) for p in primes)

    def find_primes(self, n):
        is_prime = [True]*(n+1)
        is_prime[0] = False
        is_prime[1] = False
        i = 2
        while i*i <= n:
            if not is_prime[i]:
                i += 1
                continue
            j = 2
            while j*i <= n:
                is_prime[j*i] = False
                j += 1
                
            i += 1
                
        return {i for i in range(n+1) if is_prime[i]}
    
    def valid_through(self, p, prime_neighbors, uf):
        vt = 0
        st_sum = sum(uf.size(a) for a in prime_neighbors[p])
        for a in prime_neighbors[p]:
            vt += uf.size(a)*(st_sum-uf.size(a))
            
        vt //= 2
        vt += st_sum
        
        return vt
    

class UnionFind:
    def __init__(self, n):
        self._table = {i: i for i in range(1, n+1)}
        self._size = {i: 1 for i in range(1, n+1)}

    def union(self, a, b):
        a_size = self.size(a)
        b_size = self.size(b)
        self._table[self.find(b)] = self.find(a)
        self._size[self.find(a)] = a_size + b_size

    def find(self, a):
        if a != self._table[a]:
            self._table[a] = self.find(self._table[a])

        return self._table[a]

    def size(self, i):
        return self._size[self.find(i)]
                
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