<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:
        #sieve method
        is_prime = [True] * (n + 1)
        is_prime[1] = False
        
        primes = []
        for i in range(2, n + 1):
            if is_prime[i]:
                primes.append(i)
            for x in primes:
                temp = i * x
                if temp > n:
                    break
                is_prime[temp] = False
                if i % x == 0:
                    break
        primes = set(primes)
        # print("primes", primes)
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)

        dp = [0 for i in range(n+1)]
        def dfs1(node, parent):
            if node not in primes:
                ans = 1
                for neigh in graph[node]:
                    if neigh != parent:
                        count = dfs1(neigh, node)
                        ans += count
                dp[node] = ans
                return ans
            else:
                for neigh in graph[node]:
                    if neigh != parent:
                        _ = dfs1(neigh, node)
                return 0
        _ = dfs1(1, -1)
        # print("dp", dp)
        self.res = 0
        def dfs2(node, parent, par_contrib):
            # print("node", node)
            # print("par_contrib", par_contrib)
            if node in primes:
                total = par_contrib + 1
                child_sum = 0
                for neigh in graph[node]:
                    if neigh != parent:
                        child_arm_len = dp[neigh]
                        child_sum += comb(child_arm_len, 2)
                        total += child_arm_len
                ans = comb(total, 2) - child_sum - comb(par_contrib, 2)
                self.res += ans
                for neigh in graph[node]:
                    if neigh != parent:
                        dfs2(neigh, node, 0)
            else:
                for neigh in graph[node]:
                    if neigh != parent:
                        dfs2(neigh, node, par_contrib + dp[node] - dp[neigh])
        _ = dfs2(1, -1, 0)
        return self.res
                
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