<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 get_prime(n): # 埃拉特斯特尼筛,时间复杂度为 O(nloglogn)
    prime = [True] * n
    prime[0] = prime[1] = True # not necessarily needed if we are only interested in the prime numbers which are kept in ans[]
    ans = []

    for i in range(2, n):
        if prime[i]:
            ans.append(i)
            for j in range(i*i, n, i):
                prime[j] = False
    return ans

primes = get_prime(10**5*2)
Primes = set(primes)

class Solution:
    def countPaths(self, n: int, edges: List[List[int]]) -> int:
        graph = collections.defaultdict(set)
        for a, b in edges:
            graph[a].add(b)
            graph[b].add(a)
            
        
        branches = collections.defaultdict(lambda: collections.Counter())
        total = {}
        def count_node(node, parent):
            summ = 0
            for nxt in graph[node]:
                if nxt != parent:
                    v = count_node(nxt, node)
                    branches[node][nxt] = v # total number of non-Prime numbers in the branch
                    summ += v
                    
            total[node] = summ # total number of non-Prime numbers in the children
            
            if node in Primes:
                return 0
            
            else:
                return summ + 1
        
        count_node(1, -1)
        
        #print(branches)
        #print(total)
            
        
        res = 0
        def dfs(node, parent, summ):
            nonlocal res
            
            if node in Primes: # update res
                children = list(branches[node].values())
                
                temp = 0
                children.append(summ)
                tot = total[node] + summ
                for child in children:
                    temp += child * (tot - child)
                
                res += tot + temp // 2
                
                #print(node, summ, children, res, tot, temp)
                
                for nxt in graph[node]:
                    if nxt != parent:
                        dfs(nxt, node, 0)
                
                
            else:
                for nxt in graph[node]:
                    if nxt != parent:
                        dfs(nxt, node, total[node] + summ + 1 - branches[node][nxt])
            
            
        dfs(1, -1, 0)
        return res
1755