<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>

def prime_sieve(num):
    is_prime = [True for i in range(num + 1)]
    is_prime[1] = False
    p = 2
    while p * p <= num:
        if is_prime[p]:
            for i in range(p * p, num + 1, p):
                is_prime[i] = False
        p += 1
    return is_prime

        
is_prime = prime_sieve(100005)


class UnionFind:
    # 1 based
    def __init__(self, n):
        self.fa = [i for i in range(n + 1)]
        self.size = [1] * (n + 1)
    
    def find(self, x):
        if x == self.fa[x]:
            return x
        else:
            self.fa[x] = self.find(self.fa[x])
            return self.fa[x]
    
    def union(self, u, v):
        fu = self.find(u)
        fv = self.find(v)
        self.fa[fu] = fv
        self.size[fv] += self.size[fu]

        
class Solution:
    def countPaths(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        for u, v in edges:
            if not (is_prime[u] or is_prime[v]):
                uf.union(u, v)
                
        g = defaultdict(list)
        for u, v in edges:
            if is_prime[u] and not is_prime[v]:
                g[u].append(uf.find(v))
            if not is_prime[u] and is_prime[v]:
                g[v].append(uf.find(u))
        
        ans = 0
        for i in range(1, n + 1):
            if not is_prime[i]:
                continue
            tot = 0
            for node in g[i]:
                tot += uf.size[node]
            ans += tot
            tmp = 0
            for node in g[i]:
                tmp += uf.size[node] * (tot - uf.size[node])
            tmp //= 2
            ans += tmp
            
        return ans
58600