**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
from collections import defaultdict, deque
import sys
sys.setrecursionlimit(10000)

def build_tree(n, edges):
    graph = defaultdict(list)
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)
    return graph

def dfs_tree(graph, node, parent, depth, par):
    par[node] = parent
    depth[node] = depth[parent] + 1 if parent != 0 else 0
    for neighbor in graph[node]:
        if neighbor != parent:
            dfs_tree(graph, neighbor, node, depth, par)

def get_path(a, b, par, depth):
    """
    利用父节点和深度信息，求出从 a 到 b 的唯一路径。
    算法：将 a 和 b 分别往上爬，直到相遇（LCA），记录路径。
    返回的路径顺序为从 a 到 b。
    """
    path_a = []
    path_b = []
    while a != b:
        if depth[a] >= depth[b]:
            path_a.append(a)
            a = par[a]
        else:
            path_b.append(b)
            b = par[b]
    path_a.append(a)  # LCA 节点
    return path_a + path_b[::-1]  # 将 a 的路径和 b 的反向路径拼接

def count_components(n, graph, removed):
    """
    计算在移除节点集合 removed 后，剩余图的连通分量数。
    使用 BFS 遍历未被移除的节点。
    """
    visited = set()
    components = 0
    for node in range(1, n+1):
        if node not in visited and node not in removed:
            components += 1
            dq = deque([node])
            while dq:
                cur = dq.popleft()
                if cur in visited:
                    continue
                visited.add(cur)
                for nei in graph[cur]:
                    if nei not in visited and nei not in removed:
                        dq.append(nei)
    return components

def solve():
    t = int(input())
    results = []

    for _ in range(t):
        n = int(input())
        edges = []
        for _ in range(n-1):
            u, v = map(int, input().split())
            edges.append((u, v))

        graph = build_tree(n, edges)
        # 以 1 为根节点，预处理父节点和深度信息
        par = [0] * (n+1)
        depth = [0] * (n+1)
        dfs_tree(graph, 1, 0, depth, par)

        max_components = 1
        # 枚举所有 a < b 对（路径 a->b 与 b->a 相同）
        for a in range(1, n+1):
            for b in range(a+1, n+1):
                removed_nodes = set(get_path(a, b, par, depth))
                comp = count_components(n, graph, removed_nodes)
                max_components = max(max_components, comp)

        results.append(max_components)

    for res in results:
        print(res)

if __name__ == "__main__":
    solve()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
4
1 2
2 3
3 4
5
1 2
2 3
3 4
4 5
6
1 2
2 3
3 4
4 5
5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import sys
import random
from math import *
from string import ascii_lowercase,ascii_uppercase
from collections import Counter, defaultdict, deque
from itertools import accumulate, combinations, permutations
from heapq import heappushpop, heapify, heappop, heappush
from bisect import bisect_left,bisect_right
from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc
inp = lambda: sys.stdin.buffer.readline().decode().strip()
out=sys.stdout.write
def S():
    return inp()
def I():
    return int(inp())
def MI():
    return map(int, inp().split())
def MF():
    return map(float, inp().split())
def MS():
    return inp().split()
def LS():
    return list(inp().split())
def LI():
    return list(map(int,inp().split()))
def print1(x):
    return out(str(x)+"\n")
def print2(x,y):
    return out(str(x)+" "+str(y)+"\n")
def print3(x,y,z):
    return out(str(x)+" "+str(y)+" "+str(z)+"\n")
def Query(i,j):
    print3('?', i, j)
    sys.stdout.flush()
    qi=I()
    return qi
def print_arr(arr):
    for num in arr:
        out(str(num)+" ")
    out(str("\n"))
@bootstrap
def dfs(node,par,val):
    global ans
    mmax,curr=0,max(0,len(graph[node])-2)
    for nei in graph[node]:
        if nei==par: continue
        x=yield dfs(nei,node,val+curr)
        mmax=max(x,mmax)
    mmax+=curr
    ans=max(ans,mmax+val+2)
    yield mmax
for _ in range(I()):
    n=I()
    graph=[[] for _ in range(n+1)]
    for _ in range(n-1):
        u,v=MI()
        graph[u].append(v)
        graph[v].append(u)
    if n==2: print1(1); continue;
    ans=max([len(graph[i]) for i in range(n+1)])
    dfs(1,-1,0)
    print1(ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following steps:

1. **Understand the Bug**: The buggy code seems to have issues with how it calculates the maximum number of connected components. Specifically, it might not correctly handle the removal of the path between two nodes.

2. **Edge Cases**:
   - **Linear Tree**: A linear tree where all nodes are connected in a straight line.
   - **Star Tree**: A tree where one central node is connected to all other nodes.
   - **Balanced Tree**: A balanced binary tree where each node has two children.
   - **Path with a Branch**: A tree where most nodes form a path, but a few nodes have additional branches.

3. **Challenging Scenarios**:
   - **Long Path with a Branch**: This can test the code's ability to handle long paths and ensure it correctly accounts for branches.
   - **Multiple Paths with Branches**: This can test the code's ability to handle multiple branches and ensure it correctly maximizes the number of connected components.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
3
4
1 2
2 3
3 4
5
1 2
2 3
3 4
4 5
6
1 2
2 3
3 4
4 5
5 6
```

This test input includes:
1. A linear tree with 4 nodes.
2. A linear tree with 5 nodes.
3. A linear tree with 6 nodes.

These scenarios will help expose any issues in the code's handling of linear paths and ensure it correctly calculates the maximum number of connected components.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
    {
	static FastScanner fs = null;
	static ArrayList<ArrayList<Integer>> ll;
	static int N;
	static int res;
	public static void main(String[] args) {
		fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = fs.nextInt();
        while(t-->0){
            ll = new ArrayList<>();
            N = fs.nextInt();
            res = 0;
            for(int i=0;i<=N;i++){
                ll.add(new ArrayList<Integer>());
            }
            for(int i = 1;i<N;i++){
                int u = fs.nextInt();
                int v = fs.nextInt();
                ll.get(u).add(v);
                ll.get(v).add(u);
            }
            dfs(1,-1,0);
            res-=1;
            out.println(res);
        }
		out.close();
	}
	static void dfs(int node,int par,int mx){
	    int sz = ll.get(node).size();
	    //System.out.println(node+" "+par+" "+mx+" "+sz);
	    res = Math.max(res,sz+1);
	    res = Math.max(res,mx+sz);
	    for(Integer p:ll.get(node)){
	        if(p==par)
	        continue;
	        dfs(p,node,Math.max(mx,sz-1));
	    }
	}
	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Understanding the Bug**: The buggy code calculates the maximum number of connected components by using a depth-first search (DFS) to traverse the tree. However, it seems to have an issue with correctly calculating the maximum number of components that can be formed by removing a path.

2. **Edge Cases**:
   - **Linear Tree**: A tree where all nodes are in a single path. Removing any edge should split the tree into exactly two components.
   - **Star Tree**: A tree where one node is connected to all other nodes. Removing the central node should split the tree into `n-1` components.
   - **Balanced Tree**: A tree with a balanced structure. Removing an edge should split the tree into two roughly equal components.

3. **Test Input**:
   - We will create a star tree with a central node connected to multiple leaf nodes. This should expose the issue in the DFS logic where the maximum number of components is not correctly calculated.

Here is a small-scale test input that should expose the bug:

```plaintext
1
6
1 2
1 3
1 4
1 5
1 6
```

### Explanation:
- **Test Case 1**:
  - **Tree Structure**: A star tree with 6 nodes, where node 1 is the central node connected to nodes 2, 3, 4, 5, and 6.
  - **Expected Output**: If we remove the central node 1, the tree will be split into 5 connected components (each of the leaf nodes 2, 3, 4, 5, and 6 will become a separate component).
  - **Correct Output**: 5
  - **Buggy Code Output**: The buggy code might not correctly handle the star tree structure, leading to an incorrect output.

This test input should help in identifying the issue in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import java.util.*;
import java.io.*;
public class Main
{
    // static int M=(int)(1e9+7);
    static int M=998244353;
    static FastReader in  = new FastReader();
    static FastWriter out = new FastWriter();

    static double PI=3.1415926535897932384626433832795;
    @SuppressWarnings("unchecked")

    // हर हर महादेव
    // हरे कृष्णा

    //10^6 space of int is allowed not of long
    // Yesterday was history
    // Tomorrow will be a mystery
    // but today is a gift
    // That is why it is called the present

    /* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
    */

    //Credit Striver
    // Remember you were also a novice when you started,
    // hence never be rude to anyone who wants to learn something
    // Never open a ranklist untill and unless you are done with solving problems, wastes 3/4 minuts
    // Donot treat CP as a placement thing, love it and ejoy it, you will suceed for sure.

    public static void main(String[] args)throws Exception
    {
        // in = new FastReader("input11.txt");
            // out = new FastWriter("mootube.out");
            int testcases=1;

            testcases=in.nextInt();

            // if(testcases==10000)
            // {

            //     for(int i1=1;i1<=testcases;i1++)
            //     {
            //         int n=in.nextInt();
            //         graph=new ArrayList<>();
            //         for(int i=0;i<n;++i)graph.add(new ArrayList<>());
            //         for(int i=0;i<n-1;i++)
            //         {
            //             int a=in.nextInt();int b=in.nextInt();
            //             a--;b--;
            //             graph.get(a).add(b);
            //             graph.get(b).add(a);
            //         }
            //         if(i1==50)
            //         {
            //             out.println(n);
            //             out.println(graph);
            //             out.close();
            //             return;
            //         }
            //     }
            // }
            // testcases=sc.nextInt();//For interactive problems;
            outer:
            for(int tt=1;tt<=testcases;tt++)
            {
                solve();
                // YES(solve());
                // Yes(solve());
                // out.println("Case #"+tt+": "+solve());
            }
            out.close();

    }

    static ArrayList<ArrayList<Integer>> graph;
    static ArrayList<ArrayList<Integer>> graph1;
    static ArrayList<ArrayList<Edge>> g1;
    static ArrayList<ArrayList<Edge>> g2;

    static boolean visited[];
    static int color[];
    static ArrayList<ArrayList<Pair>> graphh=new ArrayList<>();
    static Scanner sc=new Scanner(System.in);
    static int parent[];

    static long dp[];
    static long ans=0;
    @SuppressWarnings("unchecked")
    public static void solve()throws Exception
    {
        int n=in.nextInt();
        graph=new ArrayList<>();
        for(int i=0;i<n;++i)graph.add(new ArrayList<>());
        for(int i=0;i<n-1;i++)
        {
            int a=in.nextInt();int b=in.nextInt();
            a--;b--;
            graph.get(a).add(b);
            graph.get(b).add(a);
        }
        dp=new long[n];
        ans=0;
        dfx(0,-1);
        // printArray(dp);

        for(int i=0;i<n;i++)

        {
            ans=max(ans,dp[i]);
            ans=max(ans,graph.get(i).size());

        }
        out.println(ans);
    }

    public static void dfx(int u,int p) throws Exception
    {
        TreeMap<Long,Integer> t=new TreeMap<>();
        for(int child:graph.get(u))
        {
            if(child!=p)
            {
                dfx(child,u);
                dp[u]=max(dp[u],dp[child]+graph.get(u).size()-1);
                addMap(t,dp[child]+graph.get(u).size()-1);
                if(p==-1)
                {
                    // System.err.println(graph.get(u).size());
                    // dp[u]=max(dp[u],dp[child]+graph.get(u).size()-1);
                }
            }
        }
        // if(u==1)out.println(t);

        for(int child:graph.get(u))
        {
            if(child!=p)
            {
                removeMap(t,dp[child]+graph.get(u).size()-1);
                if(t.size()==0)
                {
                    addMap(t,dp[child]+graph.get(u).size()-1);
                    continue;
                }

                Long x=t.lastKey();
                if(x!=null)
                {
                    x--;
                    ans=max(ans,graph.get(child).size()-1+x);
                    if(ans==4)
                    {
                    //    out.println(child+1+" "+(u+1)+" "+x);
                    }
                }
                addMap(t,dp[child]+graph.get(u).size()-1);
            }
        }

        dp[u]=max(graph.get(u).size()-1,dp[u]);
        return;
    }

    public static long abs(long a)
    {
        return Math.abs(a);
    }
    public static int abs(int a)
    {
        return Math.abs(a);
    }
    public static double abs(double a)
    {
        return Math.abs(a);
    }
    public static int first(Integer ar[],int k)
    {
        int n=ar.length;
        if(k>ar[n-1])return n;
        int ans=0;
        int min=0;
        int max=n-1;
        while(max>=min)
        {
            int mid=min+(max-min)/2;
            if(ar[mid]>=k)
            {
                ans=mid;
                max=mid-1;
            }
            else min=mid+1;
        }
        return ans;
    }

    public static int last(Integer ar[],int k)
    {
        if(k<ar[0])
        {
            return 0;
        }
        int n=ar.length;
        if(k>=ar[n-1])return n;
        int ans=n-1;
        int min=0;
        int max=n-1;
        while(max>=min)
        {
            int mid=min+(max-min)/2;

            if(ar[mid]>k)
            {
                ans=mid;
                max=mid-1;
            }
            else min=mid+1;
        }

        return ans;
    }

    public static long gcd(long a,long b)//Note a>b
    {
        if(b==0)return a;
        return gcd(b,a%b);
    }

    public static long lcm(long a,long b)
    {
        return a*b/gcd(a,b);
    }

    public static long pow(long a, long n, long M)
    {
        if (n == 0) return 1;
        long ans = 1;
        a %= M;
        while (n != 0) {
            if ((n & 1) == 1) {
                ans = (ans % M * a % M) % M;
            }
            a = (a % M * a % M) % M;
            n /= 2;
        }
        return ans;
    }

    public static int max(int ... a)
    {
        int max = a[0];
        for (int i : a)
        {
            if (i > max)
            {
                max = i;
            }
        }
        return max;
    }

    public static long max(long ... a)
    {
        long max = a[0];
        for (long i : a) {
            if (i > max) {
                max = i;
            }
        }
        return max;
    }

    public static double max(double ... a)
    {
        double max = a[0];
        for (double i : a) {
            if (i > max) {
                max = i;
            }
        }
        return max;
    }

    // Minimum functions
    public static int min(int ... a)
    {
        int min = a[0];
        for (int i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static long min(long ... a)
    {
        long min = a[0];
        for (long i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static double min(double ... a)
    {
        double min = a[0];
        for (double i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static boolean isSorted(long ar[])
    {
        int n=ar.length;
        for(int i=0;i<n-1;i++)
        {
            if(ar[i]>ar[i+1])return false;
        }
        return true;
    }

    public static int maxIntArray(ArrayList<Integer> ar)
    {
        int max=ar.get(0);
        for(int i=0;i<ar.size();i++)max=Math.max(max,ar.get(i));
        return max;
    }

    private static int countDigits(long l)
    {
        if (l >= 1000000000000000000L) return 19;
        if (l >= 100000000000000000L) return 18;
        if (l >= 10000000000000000L) return 17;
        if (l >= 1000000000000000L) return 16;
        if (l >= 100000000000000L) return 15;
        if (l >= 10000000000000L) return 14;
        if (l >= 1000000000000L) return 13;
        if (l >= 100000000000L) return 12;
        if (l >= 10000000000L) return 11;
        if (l >= 1000000000L) return 10;
        if (l >= 100000000L) return 9;
        if (l >= 10000000L) return 8;
        if (l >= 1000000L) return 7;
        if (l >= 100000L) return 6;
        if (l >= 10000L) return 5;
        if (l >= 1000L) return 4;
        if (l >= 100L) return 3;
        if (l >= 10L) return 2;
        return 1;
    }
    public static int minIntArray(ArrayList<Integer> ar)
    {

        int n=ar.size();
        int max=ar.get(0);
        for(int i=0;i<n;i++)max=Math.min(max,ar.get(i));
        return max;
    }
    public static int maxIntArray(int ar[])
    {
        int max=ar[0];
        for(int i=0;i<ar.length;i++)max=Math.max(max,ar[i]);
        return max;
    }

    public static int minIntArray(int ar[])
    {

        int max=ar[0];
        for(int i=0;i<ar.length;i++)max=Math.min(max,ar[i]);
        return max;
    }

    public static void endl()throws Exception
    {
        out.println();
    }
    public static boolean isSorted(int ar[],int n)
    {
        if(n==1)return true;
        int diff=ar[1]-ar[0];
        for(int i=1;i<n;i++)
        {
            diff=Math.min(diff,ar[i]-ar[i-1]);
        }
        if(diff<0)return false;
        return true;
    }
    public static boolean isSorted(Integer ar[],int n)

    {
        if(n==1)return true;
        int diff=ar[1]-ar[0];
        for(int i=1;i<n;i++)
        {
            diff=Math.min(diff,ar[i]-ar[i-1]);
        }
        if(diff<0)return false;
        return true;
    }

    // public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }

    public static <K> void removeMap(Map<K, Integer> map, K element)
    {
        if (!map.containsKey(element)) return;
        int currentCount = map.get(element);
        if (currentCount == 1) {
            map.remove(element);
        } else {
            map.put(element, currentCount - 1);
        }
    }

    public static <K> void addMap(Map<K, Integer> map, K element)
    {
        if (map.containsKey(element)) {
            map.put(element, map.get(element) + 1);
        } else {
            map.put(element, 1);
        }
    }

   // public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }
    public static void removeMap(TreeMap<Integer,Integer> map,int element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(TreeMap<Long,Integer> map,long element)
    {
        if(!map.containsKey(element))return;
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(HashMap<Long,Integer> map,long element)
    {
        if(!map.containsKey(element))return;
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(HashMap<Integer,Integer> map,int element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }

    public static void removeMap(HashMap<String,Integer> map,String element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    // public static void addMap(TreeMap<Object,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }

    public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static void addMap(TreeMap<String,Integer> map,String element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static void addMap(TreeMap<Long,Integer> map,long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Integer,Integer> map,int element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Long,Long> map,long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1L);
    }
    public static void addMap(HashMap<Object,Integer> map,Integer element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Long,Integer> map,Long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<String,Integer> map,String element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static int[] nextIntArray(int n)throws Exception
    {
        int ar[]=new int[n];
        for(int i=0;i<n;i++)ar[i]=in.nextInt();
        return ar;
    }
    public static long[] nextlongArray(int n)throws Exception//For long
    {
        long ar[]=new long[n];
        for(int i=0;i<n;i++)ar[i]=in.nextLong();
        return ar;
    }
    public static Integer[] nextIntegerArray(int n)throws Exception
    {
        Integer ar[]=new Integer[n];
        for(int i=0;i<n;i++)ar[i]=in.nextInt();
        return ar;
    }
    public static Long[] nextLongArray(int n)//For Long array
    {
        Long ar[]=new Long[n];
        for(int i=0;i<n;i++)ar[i]=in.nextLong();
        return ar;
    }
    public static String[] nextStringArray(int n)
    {
        String ar[]=new String[n];
        for(int i=0;i<n;i++)ar[i]=in.next();
        return ar;
    }
    public static void Yes(boolean flag)throws Exception
    {
        if(flag)out.println("Yes");
        else  out.println("No");
    }
    public static void YES(boolean flag)throws Exception
    {
        if(flag)out.println("YES");
        else  out.println("NO");
    }
    public static void YES()throws Exception
    {
        out.println("YES");
    }
    public static void NO()throws Exception
    {
        out.println("NO");
    }
    public static void Yes()throws Exception
    {
        out.println("Yes");
    }
    public static void No()throws Exception
    {
        out.println("No");
    }

    public static void printArrayListPair(ArrayList<Pair> al)throws Exception
    {
        for(int i=0;i<al.size();i++)
        {
            out.println(al.get(i).a+" "+al.get(i).b);
        }
    }

    public static void printArray(boolean ar[])throws Exception{
        for(boolean x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(char ar[])throws Exception{
        for(char x:ar)out.print(x+" ");
        out.println();
    }

    public static void printArray(int ar[])throws Exception
    {
        for(int x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(Object ar[])throws Exception
    {
        for(Object x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(long ar[])throws Exception
    {
        for(long x:ar)out.print(x+" ");
        out.println();
    }
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(String filename) throws FileNotFoundException
        {
            br = new BufferedReader(new FileReader(filename));
        }
        public FastReader()
        {
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next()
        {
            while(st==null || !st.hasMoreTokens())
            {
                try
                {
                    st=new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt()
        {
            return Integer.parseInt(next());
        }
        long nextLong()
        {
            return Long.parseLong(next());
        }
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
        int[] nextIntArray(int n)
        {
            int ar[]=new int[n];
            for(int i=0;i<n;i++)ar[i]=Integer.parseInt(next());
            return ar;
        }
        long[] nextlongArray(int n)
        {
            long ar[]=new long[n];
            for(int i=0;i<n;i++)ar[i]=Long.parseLong(next());
            return ar;
        }
        String nextLine()
        {
            String str="";
            try
            {
                str=br.readLine().trim();
            } catch (Exception e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    static class FastWriter
    {
        private final BufferedWriter bw;
        public FastWriter(String filename) throws IOException
        {
            this.bw = new BufferedWriter(new FileWriter(filename));
        }
    public FastWriter()
        {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

    public void print(Object object) throws IOException
        {
            bw.append("" + object);
        }
    public void println(Object object) throws IOException
        {
            print(object);
            bw.append("\n");
        }
    public void println() throws IOException
        {
            bw.append("\n");
        }
    public void close() throws IOException
        {
            bw.close();
        }
    }
}

class Edge
{
    int a,b;long weight;
    Edge(int a,int b,long weight)
    {
        this.a=a;
        this.b=b;
        this.weight=weight;
    }
    Edge(int a,long weight)
    {
        this.a=a;
        this.weight=weight;
    }
}

class compareEdge implements Comparator<Edge>
{
    public int compare(Edge p1,Edge p2)
    {
        return Long.compare(p1.weight, p2.weight);
    }
}

class Pair
{
   int a,b,c;
// long a;
// long b;
    int d;
    boolean flag;
    // Pair(int a,int b,boolean flag)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.flag=flag;
    // }
    // Pair(long a,long b)
    // {
    //     this.a=a;
    //     this.b=b;
    // }
    // Pair(long a,long b,long c)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    // }
    // Pair(int a,int b)
    // {
    //     this.a=a;
    //     this.b=b;
    // }
    Pair(int a,int b,int c)
    {
        this.a=a;
        this.b=b;
        this.c=c;
    }
    // Pair (int a,int b,int c)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    // }

    // Pair (int a,int b,int c,int d)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    //     this.d=d;
    // }
    @Override
    public String toString() {
        // return a + " " + b + " " + c;
        return a+" "+b;
    }

    public boolean equals(Object obj) {
        if (this == obj) return true;
        if (obj == null || getClass() != obj.getClass()) return false;
        Pair pair = (Pair) obj;
        // return a == pair.a && b == pair.b && c == pair.c && d == pair.d && flag == pair.flag;

        return a==pair.a && b==pair.b;
    }

    // @Override
    // public int hashCode() {
    //     int result = a;
    //     result = 31 * result + b;
    //     result = 31 * result + c;
    //     result = 31 * result + d;
    //     result = 31 * result + (flag ? 1 : 0);
    //     return result;
    // }
}

class compare1 implements Comparator<Pair>
{
    public int compare(Pair p1,Pair p2)
    {
        // if(p1.a==p2.a && p2.b==p1.b)return 0;
        // return p2.a-p1.a;//decreasing
        // return p1.c-p2.c;
        // if(p1.a!=p2.a)
        // return p1.a-p2.a;//increasing
        // if(p1.a==p2.b)return p1.b-p2.b;

        return p1.a-p2.a;
        // return p1.b-p2.b;
        // return Long.compare(p1.a,p2.a);

        // if(p1.a!=p2.a)
        // {
        //     return p1.a-p2.a;
        // }
        // else
        // return p1.b-p2.b;

        // return Long.compare((p2.a+1)/2*p2.b ,(p1.a+1)/2*p1.b);

        // return p2.a-p1.b;
        // return p1.a-p2.a;
        // return p2.c-p1.c;
        // else
        // return p1.b-p2.b;
        // if(p1.a==p2.a)
        // {
        //     return p2.b-p1.b;
        // }
        // else return +p1.a-p2.a;
        // if(p1.a!=p2.a)
        //     return p1.a-p2.a;
        // else
        //     return p2.b-p1.b;

        // if(p1.a!=p2.a)
        // {
        //     return p2.a-p1.a;
        // }
        // else
        // {
            // return p1.b-p2.b;
        // }
    }

    public boolean equals(Pair p1,Pair p2)
    {
        return p1.a==p2.a && p2.b==p1.b;
    }
}

class MO

{
    int a,b;
    int c;
    static int size;

    MO(int a,int b,int c)
    {
        this.a=a;
        this.b=b;
        this.c=c;
    }
}
class compareMO implements Comparator<MO>
{
    public int compare(MO a,MO b)
    {
        int size=MO.size;
        if(a.a/size!=b.a/size)
        {
            return a.a/size-b.a/size;
        }
        return a.b-b.b;
    }
}

class SegmentTree {
    int[] tree;     // Segment tree array
    int[] arr;      // Original array
    int n;          // Size of the original array

    // Constructor to build the segment tree
    public SegmentTree(int[] input) {
        n = input.length;
        arr = input;
        // Allocate memory for segment tree
        // The maximum size of segment tree array is 2*n - 1 for n leaf nodes
        tree = new int[4 * n];
        build(0, 0, n - 1);
    }

    // Build the segment tree
    private void build(int node, int start, int end) {
        if (start == end) {
            // Leaf node: store the value of the array element
            tree[node] = arr[start];
        } else {
            int mid = (start + end) / 2;
            // Recursively build the left and right subtrees
            build(2 * node + 1, start, mid);
            build(2 * node + 2, mid + 1, end);
            // Internal node: store the minimum of the two children
            tree[node] = Math.min(tree[2 * node + 1], tree[2 * node + 2]);
        }
    }

    // Query for the minimum in range [L, R]
    public int query(int L, int R) {
        return query(0, 0, n - 1, L, R);
    }

    private int query(int node, int start, int end, int L, int R) {
        if (R < start || L > end) {
            // Range [L, R] is completely outside the range [start, end]
            return Integer.MAX_VALUE;
        }
        if (L <= start && end <= R) {
            // Range [L, R] completely covers the range [start, end]
            return tree[node];
        }
        // Partial overlap
        int mid = (start + end) / 2;
        int leftMin = query(2 * node + 1, start, mid, L, R);
        int rightMin = query(2 * node + 2, mid + 1, end, L, R);
        return Math.min(leftMin, rightMin);
    }

    // Update an element in the array
    public void update(int index, int value) {
        update(0, 0, n - 1, index, value);
    }

    private void update(int node, int start, int end, int index, int value) {
        if (start == end) {
            // Leaf node: update the value of the array element
            arr[index] = value;
            tree[node] = value;
        } else {
            int mid = (start + end) / 2;
            if (index <= mid) {
                // Update the left child
                update(2 * node + 1, start, mid, index, value);
            } else {
                // Update the right child
                update(2 * node + 2, mid + 1, end, index, value);
            }
            // Update the current node after updating the children
            tree[node] = Math.min(tree[2 * node + 1], tree[2 * node + 2]);
        }
    }
}

class GCDSegmentTree {
    private int[] tree;
    private int[] arr;

    public GCDSegmentTree(int[] inputArr) {
        int n = inputArr.length;
        arr = inputArr;
        int treeSize = 4 * n;
        tree = new int[treeSize];
        buildTree(0, 0, n - 1);
    }

    // Build segment tree
    private void buildTree(int node, int start, int end) {
        if (start == end) {
            tree[node] = arr[start];  // Leaf node
        } else {
            int mid = (start + end) / 2;
            buildTree(2 * node + 1, start, mid);   // Left child
            buildTree(2 * node + 2, mid + 1, end); // Right child
            tree[node] = gcd(tree[2 * node + 1], tree[2 * node + 2]); // Combine GCD
        }
    }

    // GCD Query
    private int query(int node, int start, int end, int L, int R) {
        if (R < start || end < L) {
            return 0;  // Out of range
        }
        if (L <= start && end <= R) {
            return tree[node];  // Current segment fully in range
        }
        int mid = (start + end) / 2;
        int leftGCD = query(2 * node + 1, start, mid, L, R);
        int rightGCD = query(2 * node + 2, mid + 1, end, L, R);
        return gcd(leftGCD, rightGCD);  // Combine GCD from left and right children
    }

    public int getGCD(int L, int R) {
        return query(0, 0, arr.length - 1, L, R);
    }

    // Utility GCD function
    private int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

}

class DSU {
    private int[] parent;
    private int[] rank;
    private int[] size; // Array to track the size of each set
    private int groups; // Counter for the number of groups

    // Constructor to initialize the DSU
    public DSU(int n) {
        parent = new int[n];
        rank = new int[n];
        size = new int[n];
        groups = n; // Initially, every element is its own group

        // Initialize each node as its own parent (self-loop)
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            rank[i] = 0; // Initial rank is 0
            size[i] = 1; // Each set initially contains only one element
        }
    }

    // Find operation with path compression
    public int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]); // Path compression
        }
        return parent[x];
    }

    // Union operation with union by rank
    public void union(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);

        if (rootX != rootY) {
            if (rank[rootX] > rank[rootY]) {
                parent[rootY] = rootX;
                size[rootX] += size[rootY]; // Update size of rootX
            } else if (rank[rootX] < rank[rootY]) {
                parent[rootX] = rootY;
                size[rootY] += size[rootX]; // Update size of rootY
            } else {
                parent[rootY] = rootX;
                size[rootX] += size[rootY]; // Update size of rootX
                rank[rootX]++;
            }
            groups--; // Decrease the number of groups as two groups are merged
        }
    }

    // Check if two elements are in the same set
    public boolean isConnected(int x, int y) {
        return find(x) == find(y);
    }

    // Get the size of the set containing the element x
    public int size(int x) {
        return size[find(x)];
    }

    // Get the number of groups
    public int numberOfGroups() {
        return groups;
    }

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes bugs in the provided code, we need to focus on edge cases and scenarios that can lead to incorrect results. Here are the steps to generate such a test input:

1. **Small-scale Test Case**: Since the output length is limited, we should create a small-scale test case with a manageable number of vertices and edges. This will help in easily verifying the correctness of the output.

2. **Edge Cases**:
   - **Linear Tree**: A tree where all vertices are connected in a straight line. This can test the algorithm's ability to handle simple, linear structures.
   - **Star Tree**: A tree where one central vertex is connected to all other vertices. This can test the algorithm's ability to handle a central node with many children.
   - **Balanced Tree**: A tree where each node has approximately the same number of children. This can test the algorithm's ability to handle more complex, balanced structures.

3. **Maximizing Connected Components**:
   - Choose vertices for removal that can split the tree into the maximum number of connected components. For example, removing the central vertex in a star tree can split it into many components.

Given these considerations, let's create a test case that includes a star tree and a linear tree, which are both challenging scenarios.

### Test Input
```plaintext
2
5
1 2
1 3
1 4
1 5
4
1 2
2 3
3 4
```

### Explanation
1. **Test Case 1 (Star Tree)**:
   - **Vertices**: 5
   - **Edges**: 1-2, 1-3, 1-4, 1-5
   - **Expected Output**: 4
   - **Reason**: Removing the central vertex (1) will split the tree into 4 connected components: {2}, {3}, {4}, {5}.

2. **Test Case 2 (Linear Tree)**:
   - **Vertices**: 4
   - **Edges**: 1-2, 2-3, 3-4
   - **Expected Output**: 2
   - **Reason**: Removing any two adjacent vertices (e.g., 2 and 3) will split the tree into 2 connected components: {1}, {4}.

This test input is small-scale and includes two challenging scenarios that can help expose bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <random>

using namespace __gnu_pbds;
using namespace std;

#define sp << ' '
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using vi = vector<int>;
#define endl '\n'
#define sign(x) (x < 0 ? -1 : x > 0 ? 1 : 0)
typedef vector<ll> vll;
typedef array<int, 2> ii;
typedef vector<array<int, 2>> vii;
typedef array<ll, 2> pll;
#define nl <<'\n'
#define out(txt) {cout << txt << '\n'; return;}
#define MOD 1000000007ll
#define EPS 1e-9

template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> using multi_ordered_set = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using min_queue = priority_queue<T, vector<T>, greater<>>;

template<typename Container>
void printVec(Container &v) {
    for (const auto &i: v) { cout << i << ' '; }
    cout nl;
}

//template<typename Container>
//void printVec(vector<Container> &v) {
//    for (const auto &item: v) printVec(item);
//}

void solve();

void precompute();

bool myPC = false;

int main(int argc, char *argv[]) {
    if (argc > 1 && argv[1][0] - '0') {
        freopen("in.txt", "r", stdin), freopen("out.txt", "w", stdout), myPC = true;
    }
    ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
//    precompute();
    int t = 1;
    cin >> t;
    for (int i = 0; i < t; ++i) {
//        cout << "Case " << i + 1 << ": ";
        solve();
    }
}

void solve() {
    int n;
    cin >> n;
    vector<vi> adj(n + 1);
    vi deg(n + 1);
    for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
        deg[u]++;deg[v]++;
    }
    // 0 if in path, 1 if start, 2 if end, 3 not child
    vector<vi> dp(n + 1, vi(4));
    int ans = 1;
    function<void(int, int, int)> dfs = [&](int s, int p, int ds){
        ans = max(ans, deg[s]);
        dp[s][0] = deg[s] - 2;
        priority_queue<int> pq;
        pq.push(0);
        pq.push(0);
        for (auto u: adj[s]){
            if (u != p) {
                dfs(u, s, ds + deg[s]);
                pq.push(max(dp[u][0], dp[u][2]));
            }
        }
        dp[s][1] = dp[s][2] = deg[s] - 1;

        dp[s][0] += pq.top();
        dp[s][3] += pq.top();
        dp[s][1] += pq.top();
        pq.pop();
        dp[s][3] += pq.top();
        pq.pop();

    };

    dfs(1, 0, 0);

    for (int i = 1; i <= n; ++i) {
        ans = max({ans, dp[i][0], dp[i][1], dp[i][2], dp[i][3], deg[i]});
    }

    cout << ans nl;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that can maximize the failure probability. The code is supposed to find the maximum number of connected components after removing a path from a tree. Here are the steps to create a challenging test input:

1. **Small Tree with a Linear Structure**:
   - A linear tree (a chain) can be a good test case because it has a clear path that, when removed, should result in a specific number of components.

2. **Star Tree**:
   - A star tree, where one central vertex is connected to all other vertices, can be challenging because removing the central vertex should result in the maximum number of components.

3. **Balanced Tree**:
   - A balanced tree (like a complete binary tree) can be used to test the algorithm's ability to handle more complex structures.

4. **Tree with a Long Path and a Few Branches**:
   - A tree that has a long path with a few branches can test the algorithm's ability to handle mixed structures.

Let's create a small-scale test input that includes a linear tree, a star tree, and a balanced tree:

```plaintext
3
4
1 2
2 3
3 4
5
1 2
1 3
1 4
1 5
7
1 2
2 3
3 4
4 5
5 6
6 7
```

Explanation:
- **Test Case 1**: A linear tree with 4 vertices. Removing the path from 1 to 4 should result in 1 component.
- **Test Case 2**: A star tree with 5 vertices. Removing the central vertex (1) should result in 4 components.
- **Test Case 3**: A balanced tree with 7 vertices. Removing the path from 1 to 7 should result in 3 components.

This test input should be challenging enough to expose potential issues in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(v) v.begin(),v.end()
#define sz(s)    (int)(s.size())
#define e "\n"
const ll mod = 1e9 + 7, N = 2e5 + 5, OO = 0x3f3f3f3f, OOL = 0x3f3f3f3f3f3f3f3f;
void H() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin),
            freopen("output.txt", "w", stdout);
#endif
}

struct DSU {
    vector<int> parent, Group_sz;
    DSU(int n) {
        Group_sz.resize(n + 5), parent.resize(n + 5);
        for (int i = 1; i <= n; i++) {
            parent[i] = i, Group_sz[i] = 1;
        }
    }
    int find_leader(int x) {
        if (parent[x] == x)return x;
        return parent[x] = find_leader(parent[x]);
    }
    void union_sets(int x, int y) {
        x = find_leader(x);
        y = find_leader(y);
        if (x == y) return;
        if (Group_sz[x] < Group_sz[y]) swap(x, y);
        Group_sz[x] += Group_sz[y];
        parent[y] = x;
    }
    bool is_same_set(int x, int y){return find_leader(x) == find_leader(y);}
    int get_size(int x){return Group_sz[find_leader(x)];}
};
vector<set<int>> adj(N);
vector<int>par(N);
int x=1,mx=0;
void dfs(int u,int p,int d) {
    if (d > mx) {
        mx = d, x = u;
    }
    for (auto v: adj[u]) {
        if (v == p)continue;
        dfs(v, u, d + sz(adj[u])-1);
    }
}
void dfspar(int u,int p) {
    par[u] = p;
    for (auto v: adj[u]) {
        if (v == p)continue;
        dfspar(v, u);
    }
}
void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        adj[u].insert(v);
        adj[v].insert(u);
    }
    int z;
    for(int i=1;i<=n;i++) {
        if (sz(adj[i]) == 1)z=i;
    }
    mx = -1;
    dfs(z, -1, 0);
    mx = -1;
    int y = x;
//    cout<<x<<e;
    dfs(x, -1, 0);
//    cout << x << " " << y << e;
    y = *adj[y].begin();
    x = *adj[x].begin();
//    cout << x << " " << y << e;
    dfspar(x, -1);
    set<int> blocked;
    while (true) {
        blocked.insert(y);
        if (y == x)break;
        y = par[y];
    }
    DSU dsu(n);
    for (int i = 1; i <= n; i++) {
        if (blocked.count(i))continue;
        for (auto v: adj[i]) {
            if (blocked.count(v))continue;
            dsu.union_sets(i, v);
        }
    }
    set<int> ans;
    for (int i = 1; i <= n; i++) {
        if (blocked.count(i))continue;
        ans.insert(dsu.find_leader(i));
    }
    if(n==2)cout<<1;
    else cout << sz(ans);
    for (int i = 1; i <= n; i++) {
        adj[i].clear();
        par[i] = -1;
    }
}
int main() {
    H();
    int t = 1; cin >> t;
    while (t--) {
        solve();
        if(t)cout<<e;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Understand the Bug**: The code is supposed to find the maximum number of connected components after removing a path from a tree. The bug might be in how it handles specific tree structures or edge cases.

2. **Identify Potential Issues**:
   - **Linear Tree**: A tree where all nodes are connected in a straight line.
   - **Star Tree**: A tree where one central node is connected to all other nodes.
   - **Balanced Tree**: A tree where the nodes are evenly distributed.
   - **Small Trees**: Trees with very few nodes (e.g., 2 or 3 nodes).
   - **Large Trees**: Trees with a large number of nodes (up to the limit).

3. **Generate Test Inputs**:
   - **Linear Tree**: This can test the code's ability to handle a simple, straightforward path.
   - **Star Tree**: This can test the code's ability to handle a tree with a central node.
   - **Balanced Tree**: This can test the code's ability to handle more complex structures.
   - **Small Trees**: These can test the code's edge cases.
   - **Large Trees**: These can test the code's performance and correctness with large inputs.

4. **Choose a Small-Scale Test Input**:
   - For this task, we will choose a small-scale test input that is complete and valid.

Let's consider a **Star Tree** with 5 nodes, as it is a simple yet challenging structure that can expose issues in the code.

```plaintext
1
5
1 2
1 3
1 4
1 5
```

This input represents a star tree with node 1 as the center and nodes 2, 3, 4, and 5 as the leaves. The expected output for this input should be 4, as removing the path from any leaf to the center will result in 4 connected components.

Here is the test input:

```plaintext
1
5
1 2
1 3
1 4
1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import sys

def main():
    input = sys.stdin.read
    data = input().split()

    T = int(data[0])
    idx =1
    res = []
    for _ in range(T):
        n = int(data[idx]); idx +=1
        adj = [[] for _ in range(n +1)]
        for __ in range(n -1):
            u = int(data[idx]); idx +=1
            v = int(data[idx]); idx +=1
            adj[u].append(v)
            adj[v].append(u)
        L = 0
        for i in range(1, n +1):
            if len(adj[i]) ==1:
                L +=1
        max_components = min(L, n -1)
        res.append(str(max_components))
    print('\n'.join(res))

if __name__ == '__main__':
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The bug in the code is that it incorrectly calculates the maximum number of connected components by only considering the number of leaf nodes (L) and not the structure of the tree.

### Step-by-Step Reasoning:

1. **Understanding the Bug**:
   - The code calculates the maximum number of connected components as `min(L, n - 1)`, where `L` is the number of leaf nodes.
   - This approach is incorrect because it does not consider the structure of the tree and the effect of removing a path on the number of connected components.

2. **Edge Cases**:
   - **Star Tree**: A tree where one central node is connected to all other nodes. Removing the central node will result in `n - 1` connected components.
   - **Path Tree**: A tree where all nodes are connected in a linear sequence. Removing any node (except the endpoints) will split the tree into two parts.
   - **Balanced Tree**: A tree where the nodes are evenly distributed. Removing a path that goes through the center can significantly increase the number of connected components.

3. **Challenging Test Case**:
   - We will use a star tree with a central node connected to all other nodes. This will expose the bug because the code will incorrectly output `1` instead of `n - 1`.

### Test Input:

```plaintext
1
5
1 2
1 3
1 4
1 5
```

### Explanation:

- **Test Case**:
  - **Number of Test Cases**: 1
  - **Number of Nodes**: 5
  - **Edges**:
    - Node 1 is connected to Node 2
    - Node 1 is connected to Node 3
    - Node 1 is connected to Node 4
    - Node 1 is connected to Node 5

- **Expected Output**:
  - The maximum number of connected components should be `4` because removing the central node (Node 1) will result in 4 separate components (Nodes 2, 3, 4, and 5).

- **Actual Output from Buggy Code**:
  - The buggy code will output `1` because it only considers the number of leaf nodes (which is 4) and takes the minimum with `n - 1` (which is 4), but it does not correctly handle the structure of the tree.

This test input will effectively expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import sys
input = sys.stdin.readline

t = int(input())

for _ in range(t):
    n = int(input())
    degree = [0] * (n + 1)
    for _ in range(n - 1):
        u,v = map(int,input().split())
        degree[u] += 1
        degree[v] += 1
    ans = 0
    for x in range(1, n + 1):
        if degree[x] == 1:
            ans += 1
    if ans == 2:
        if n == 2:
            print(1)
        else:
            print(2)
    else:
        print(ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
2
1 2
5
1 2
2 3
3 4
4 5
4
1 2
2 3
3 4
5
1 2
2 3
3 4
4 5
6
1 2
2 3
3 4
4 5
5 6
6
1 2
2 3
3 4
4 5
5 6
```

### Explanation:
1. **Test Case 1**: A simple tree with 2 nodes. The expected output is 1 because removing the only edge will leave one connected component.
2. **Test Case 2**: A linear tree with 5 nodes. The expected output is 3 because removing the middle edge (2-3) will split the tree into 3 components.
3. **Test Case 3**: A linear tree with 4 nodes. The expected output is 2 because removing the middle edge (2-3) will split the tree into 2 components.
4. **Test Case 4**: A linear tree with 5 nodes, similar to Test Case 2. The expected output is 3.
5. **Test Case 5**: A linear tree with 6 nodes. The expected output is 4 because removing the middle edge (3-4) will split the tree into 4 components.
6. **Test Case 6**: A linear tree with 6 nodes, similar to Test Case 5. The expected output is 4.

These test cases are designed to cover different scenarios, including small and medium-sized trees, to ensure the code handles various edge cases correctly. The linear structure of the trees in Test Cases 2, 4, and 5 is particularly challenging for the buggy code, as it will need to correctly identify the number of components after removing the middle edge.