<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:

        # build graph
        graph = collections.defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)

        # loop each prime number
        prime = self.sieve(n)
        memo = [-1] * (n + 1)
        ans = 0
        for p in range(n + 1):
            if not prime[p]:
                continue

            num = []
            for node in graph[p]:
                if prime[node]:
                    continue
                if memo[node] > 0:
                    num.append(memo[node])
                else:
                    cur = self.dfs(node, p, graph, prime)
                    self.node_num(node, p, graph, prime, memo, cur)
                    num.append(cur)
            totalnum = sum(num)

            ans += totalnum + totalnum * (totalnum - 1) // 2 - sum([k*(k - 1)// 2 for k in num]) 
        return ans

    def sieve(self, n):
        prime = [True] * (n + 1)
        prime[0] = prime[1] = False
        for p in range(2, n + 1):
            if not prime[p]:
                continue
            for nxt in range(p * p, n + 1, p):
                prime[nxt] = False
        return prime

    def dfs(self, node, parent, graph, prime):
        sumval = 1
        for nxt in graph[node]:
            if nxt == parent or prime[nxt]:
                continue
            sumval += self.dfs(nxt, node, graph, prime)
        return sumval
    def node_num(self, node, parent, graph, prime, memo, number):
        memo[node] = number
        for nxt in graph[node]:
            if nxt == parent or prime[nxt]:
                continue
            self.node_num(nxt, node, graph, prime, memo, number)
1512