<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:
        if n == 1:
            return 0
        
        prime = [False] * 2 + [True] * (n - 1)
        p = 2
        while p * p <= n:
            if prime[p]:
                for i in range(p * p, n+1, p):
                    prime[i] = False
                    
            p += 1
        
        graph = defaultdict(set)
        for u, v in edges:
            graph[u].add(v)
            graph[v].add(u)
            
        p_cnts = {}
        n_cnts = {}
        n_sum = defaultdict(int)
        parents = list(range(n+1))
        
        def dfs(node, parent):
            if parent != -1:
                parents[node] = parent
                
            if prime[node]:
                p_cnts[node] = []
            else:
                n_cnts[node] = defaultdict(int)
            
            total_cnt = 0
            for child in graph[node]:
                if child != parent:
                    cnt = dfs(child, node)
                    total_cnt += cnt
                    
                    if cnt > 0:
                        if prime[node]:
                            p_cnts[node].append(cnt)
                        else:
                            n_cnts[node][child] = cnt
                            
            if not prime[node]:
                n_sum[node] = total_cnt
                        
            return 0 if prime[node] else total_cnt + 1
        
        dfs(2, -1)
        
        for p in p_cnts:
            l = deque([p])
            new_p = parents[p]
            while not prime[new_p]:
                l.append(new_p)
                new_p = parents[new_p]
            
            cnt = 0
            while len(l) > 1:
                node = l.pop()
                cnt += n_sum[node] - n_cnts[node][l[-1]]
                cnt += 1
                
            if cnt > 0:
                p_cnts[p].append(cnt)
                
        res = 0
        for p, l in p_cnts.items():
            total = sum(l)
            cur = 0
            for cnt in l:
                cur += (total - cnt) * cnt
            res += cur // 2 + total
            
        return res
1722