<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):
            is_prime = [True] * (n + 1)
            is_prime[0] = is_prime[1] = False

            for p in range(2, int(n**0.5) + 1):
                if is_prime[p]:
                    for i in range(p * p, n + 1, p):
                        is_prime[i] = False

            primes = [i for i in range(2, n + 1) if is_prime[i]]
            return primes
        prime=set(sieve_of_eratosthenes(n))
        tree={}
        for i,j in edges:
            tree[i]=tree.get(i,set())
            tree[i].add(j)
            tree[j]=tree.get(j,set())
            tree[j].add(i)
        dp={}
        def do(cur):
            cur_vis.add(cur)
            ans=1
            for i in tree.get(cur,[]):
                if(i not in prime and i not in cur_vis):
                    ans+=do(i)
            return ans
        cur_vis=set()
        for i in range(1,n+1):
            if(i not in prime and i not in dp):
                cur_vis=set()
                t=do(i)
                for i in cur_vis:
                    dp[i]=t
        o=0
        for i in prime:
            s=[]
            crs=0
            for j in tree[i]:
                if(j in prime):
                    continue
                s.append(dp[j])
                crs+=dp[j]
                o+=dp[j]
            for i in s:
                crs-=i
                o+=crs*i
        return o
1579