<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 UF:
    def __init__(self, n: int):
        self.parents = [i for i in range(n)]
        self.sizes = [1] * n

    def find(self, i: int):
        if self.parents[i] != i:
            self.parents[i] = self.find(self.parents[i])
        return self.parents[i]

    def groupSize(self, i: int):
        return self.sizes[self.find(i)]

    def union(self, i: int, j: int):
        i = self.find(i)
        j = self.find(j)

        if i == j:
            return
        
        if self.sizes[i] < self.sizes[j]:
            i, j = j, i
        
        self.parents[j] = i
        self.sizes[i] += self.sizes[j]

class Solution:
    def calcIsPrime(self, n: int) -> List[bool]:
        is_prime = [True] * (n + 1)
        is_prime[1] = False

        for i in range (2, n):
            if not is_prime[i]:
                continue
            
            j = i + i
            while j <= n:
                is_prime[j] = False
                j += i
        
        return is_prime


    def countPaths(self, n: int, edges: List[List[int]]) -> int:
        graph = {i: list() for i in range(n + 1)}
        for a, b in edges:
            graph[a].append(b)
            graph[b].append(a)
        
        is_prime = self.calcIsPrime(n)

        uf = UF(n + 1)
        for a, b in edges:
            if not is_prime[a] and not is_prime[b]:
                uf.union(a, b)

        paths = 0
        for i in range(2, n + 1):
            if not is_prime[i]:
                continue
                        
            other = 0
            for neigh in graph[i]:
                if is_prime[neigh]:
                    continue

                x = uf.groupSize(neigh)
                paths += x + x * other
                other += x
            
        return paths
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