**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
import sys
sys.setrecursionlimit(1 << 25)
input = sys.stdin.read
data = input().split()

ptr = 0
T = int(data[ptr])
ptr += 1

def read_int():
    global ptr
    res = int(data[ptr])
    ptr += 1
    return res

def read_array(n):
    global ptr
    arr = list(map(int, data[ptr:ptr+n]))
    ptr += n
    return arr

for _ in range(T):
    n = read_int()
    adj = [[] for _ in range(n)]
    for _ in range(n - 1):
        a = read_int() - 1
        b = read_int() - 1
        adj[a].append(b)
        adj[b].append(a)
    u = read_int() - 1
    v = read_int() - 1

    # Find path from u to v
    parent = [-1] * n
    queue = [u]
    found_v = False
    while queue:
        curr = queue.pop(0)
        for next_node in adj[curr]:
            if next_node != parent[curr]:
                parent[next_node] = curr
                queue.append(next_node)
                if next_node == v:
                    found_v = True
                    break
        if found_v:
            break
    if not found_v:
        raise ValueError("Path not found")

    # Reconstruct path
    path = []
    curr = v
    while curr != u:
        path.append(curr)
        curr = parent[curr]
    path.append(u)
    path.reverse()

    # Now, perform DFS from root 0 to compute sizes and depths
    size = [0] * n
    depth = [0] * n

    def dfs(node, par):
        size[node] = 1
        depth[node] = 0 if par == -1 else depth[par] + 1
        for child in adj[node]:
            if child != par:
                size[node] += dfs(child, node)
        return size[node]

    dfs(0, -1)

    # For each p_j in path, compute the result
    for p_j in path:
        # Compute the LCA of 0 and p_j
        # Since 0 is the root, LCA is the node where paths from 0 and p_j diverge
        # Which is p_j itself, since 0 is the root

        # The size of the subtree rooted at p_j is size[p_j]
        # The part of the tree accessible to Alice is n - size[p_j]
        # The part accessible to Bob is size[p_j]

        # However, since Bob starts at p_j, we need to account for overlapping

        # Total moves = (number of nodes Alice can reach) + (number of nodes Bob can reach) - 2
        # But needs precise calculation

        # Instead, observe the parity of the distance between 0 and p_j
        # From observations, if depth[p_j] is even, Alice wins; else, Bob wins
        if depth[p_j] % 2 == 0:
            print("Alice")
        else:
            print("Bob")

**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
1
5
1 2
1 3
2 4
2 5
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
from collections import defaultdict, deque

def dfs(graph, start, target, path):
    stack = [(start, [start])]
    visited = set()

    while stack:
        node, current_path = stack.pop()
        if node == target:
            return current_path
        if node not in visited:
            visited.add(node)
            for neighbor in graph[node]:
                if neighbor not in visited:
                    stack.append((neighbor, current_path + [neighbor]))
    return []

def solve_case(n, edges, u, v):
    # Build the graph
    graph = defaultdict(list)
    for a, b in edges:
        graph[a].append(b)
        graph[b].append(a)

    # Find the path from u to v
    path = dfs(graph, u, v, [])

    # Determine the winners
    results = []
    for p_i in path:
        # Calculate the distance from 1 to p_i
        queue = deque([(1, 0)])
        visited = set([1])

        found = False
        while queue:
            node, dist = queue.popleft()
            if node == p_i:
                found = True
                break
            for neighbor in graph[node]:
                if neighbor not in visited:
                    visited.add(neighbor)
                    queue.append((neighbor, dist + 1))

        if found:
            if dist % 2 == 0:
                results.append("Alice")
            else:
                results.append("Bob")
        else:
            raise ValueError(f"Node {p_i} not reachable from 1")

    return results

# Read input
import sys
input = sys.stdin.read
data = input().split()

index = 0
t = int(data[index])
index += 1
results = []

for _ in range(t):
    n = int(data[index])
    index += 1
    edges = []
    for _ in range(n - 1):
        a = int(data[index])
        b = int(data[index + 1])
        index += 2
        edges.append((a, b))
    u = int(data[index])
    v = int(data[index + 1])
    index += 2

    result = solve_case(n, edges, u, v)
    results.extend(result)

# Output results
for res in results:
    print(res)

**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
3
1 2
2 3
2 3
6
1 2
1 3
2 4
2 5
1 6
4 5
4
1 2
1 3
2 4
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
from collections import defaultdict, deque

def dfs(graph, start, target, visited, path):
    stack = [(start, [start])]
    while stack:
        node, current_path = stack.pop()
        if node == target:
            return current_path
        visited.add(node)
        for neighbor in graph[node]:
            if neighbor not in visited:
                stack.append((neighbor, current_path + [neighbor]))
    return []

def solve(n, edges, u, v):
    # Build the graph
    graph = defaultdict(list)
    for a, b in edges:
        graph[a].append(b)
        graph[b].append(a)

    # Find the path from u to v
    visited = set()
    path = dfs(graph, u, v, visited, [])

    # Determine the winners
    results = []
    for i in range(len(path)):
        # Alice starts at 1, Bob starts at path[i]
        if (len(path) - i) % 2 == 0:
            results.append("Bob")
        else:
            results.append("Alice")

    return results

# Read input
import sys
input = sys.stdin.read
data = input().split()

index = 0
t = int(data[index])
index += 1

results = []
for _ in range(t):
    n = int(data[index])
    index += 1

    edges = []
    for _ in range(n - 1):
        a = int(data[index])
        b = int(data[index + 1])
        index += 2
        edges.append((a, b))

    u = int(data[index])
    v = int(data[index + 1])
    index += 2

    result = solve(n, edges, u, v)
    results.extend(result)

# Output results
print("\n".join(results))

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
import java.io.*;
import java.util.*;

public class Main {
    static ArrayList<Integer>[] graph;
    static int[] depth;
    static int[] parent;
    static int[] subtreeSize;
    static int[] distanceFromOne;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int t = Integer.parseInt(br.readLine());

        while (t-- > 0) {
            int n = Integer.parseInt(br.readLine());
            graph = new ArrayList[n + 1];
            for (int i = 1; i <= n; i++) {
                graph[i] = new ArrayList<>();
            }

            for (int i = 0; i < n - 1; i++) {
                String[] edge = br.readLine().split(" ");
                int a = Integer.parseInt(edge[0]);
                int b = Integer.parseInt(edge[1]);
                graph[a].add(b);
                graph[b].add(a);
            }

            String[] uv = br.readLine().split(" ");
            int u = Integer.parseInt(uv[0]);
            int v = Integer.parseInt(uv[1]);

            depth = new int[n + 1];
            parent = new int[n + 1];
            subtreeSize = new int[n + 1];
            distanceFromOne = new int[n + 1];

            dfs(1, 0);

            List<Integer> path = getPath(u, v);

            for (int node : path) {
                int aliceMoves = distanceFromOne[node] - 1;
                int bobMoves = n - distanceFromOne[node] - subtreeSize[node] + 1;

                if ((aliceMoves <= bobMoves && aliceMoves % 2 == 0) ||
                    (aliceMoves > bobMoves && bobMoves % 2 == 1)) {
                    pw.println("Bob");
                } else {
                    pw.println("Alice");
                }
            }
        }

        pw.close();
    }

    static void dfs(int v, int p) {
        parent[v] = p;
        subtreeSize[v] = 1;
        if (p != 0) {
            distanceFromOne[v] = distanceFromOne[p] + 1;
        }
        for (int u : graph[v]) {
            if (u != p) {
                depth[u] = depth[v] + 1;
                dfs(u, v);
                subtreeSize[v] += subtreeSize[u];
            }
        }
    }

    static List<Integer> getPath(int u, int v) {
        List<Integer> pathU = new ArrayList<>();
        List<Integer> pathV = new ArrayList<>();
        int nodeU = u;
        int nodeV = v;

        while (nodeU != 0) {
            pathU.add(nodeU);
            nodeU = parent[nodeU];
        }
        while (nodeV != 0) {
            pathV.add(nodeV);
            nodeV = parent[nodeV];
        }

        Collections.reverse(pathU);
        Collections.reverse(pathV);

        int lca = 1;
        for (int i = 0; i < Math.min(pathU.size(), pathV.size()); i++) {
            if (pathU.get(i).equals(pathV.get(i))) {
                lca = pathU.get(i);
            } else {
                break;
            }
        }

        List<Integer> path = new ArrayList<>();
        for (int i = pathU.size() - 1; i >= 0 && pathU.get(i) != lca; i--) {
            path.add(pathU.get(i));
        }
        path.add(lca);
        for (int i = pathV.indexOf(lca) + 1; i < pathV.size(); i++) {
            path.add(pathV.get(i));
        }

        return path;
    }
}

**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
1
5
1 2
1 3
2 4
2 5
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        StringBuilder results = new StringBuilder();
        while (t-- > 0) {
            int n = scanner.nextInt();
            List<List<Integer>> tree = new ArrayList<>();
            for (int i = 0; i <= n; i++) {
                tree.add(new ArrayList<>());
            }
            for (int i = 0; i < n - 1; i++) {
                int a = scanner.nextInt();
                int b = scanner.nextInt();
                tree.get(a).add(b);
                tree.get(b).add(a);
            }
            int u = scanner.nextInt();
            int v = scanner.nextInt();
            List<Integer> path = new ArrayList<>();
            boolean[] visited = new boolean[n + 1];
            findPath(tree, u, v, visited, path);
            for (int start : path) {
                int distanceFrom1 = getDistanceFrom1(tree, start);
                if (distanceFrom1 % 2 == 0) {
                    results.append("Alice\n");
                } else {
                    results.append("Bob\n");
                }
            }
        }
        System.out.print(results);
        scanner.close();
    }
    private static boolean findPath(List<List<Integer>> tree, int u, int v, boolean[] visited, List<Integer> path) {
        visited[u] = true;
        path.add(u);
        if (u == v) {
            return true;
        }
        for (int neighbor : tree.get(u)) {
            if (!visited[neighbor]) {
                if (findPath(tree, neighbor, v, visited, path)) {
                    return true;
                }
            }
        }
        path.remove(path.size() - 1);
        return false;
    }
    private static int getDistanceFrom1(List<List<Integer>> tree, int start) {
        Queue<int[]> queue = new LinkedList<>();
        boolean[] visited = new boolean[tree.size()];
        queue.add(new int[]{1, 0});
        visited[1] = true;
        while (!queue.isEmpty()) {
            int[] current = queue.poll();
            int vertex = current[0];
            int distance = current[1];
            if (vertex == start) {
                return distance;
            }
            for (int neighbor : tree.get(vertex)) {
                if (!visited[neighbor]) {
                    visited[neighbor] = true;
                    queue.add(new int[]{neighbor, distance + 1});
                }
            }
        }
        return -1;
    }
}

**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
1
6
1 2
1 3
2 4
2 5
1 6
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
// Problem: F2. Game in Tree (Hard Version)
// Contest: Codeforces - Codeforces Round 973 (Div. 2)
// URL: https://codeforces.com/contest/2013/problem/F2
// Memory Limit: 256 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define uint unsigned
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 200100;
const int logn = 20;

int n, U, V, a[maxn], m, b[maxn], g[logn][maxn], h[logn][maxn], ans[maxn], tot;
vector<int> G[maxn];

int fa[maxn], f[maxn], dep[maxn], p[maxn], q[maxn];
bool vis[maxn];

inline int qmax1(int l, int r) {
	int k = __lg(r - l + 1);
	return max(g[k][l], g[k][r - (1 << k) + 1]);
}

inline int qmax2(int l, int r) {
	int k = __lg(r - l + 1);
	return max(h[k][l], h[k][r - (1 << k) + 1]);
}

void dfs(int u, int t) {
	fa[u] = t;
	f[u] = 0;
	for (int v : G[u]) {
		if (v == t) {
			continue;
		}
		dep[v] = dep[u] + 1;
		dfs(v, u);
		f[u] = max(f[u], f[v] + 1);
	}
}

inline int qlca(int x, int y) {
	while (dep[x] > dep[y]) {
		x = fa[x];
	}
	while (dep[y] > dep[x]) {
		y = fa[y];
	}
	while (x != y) {
		x = fa[x];
		y = fa[y];
	}
	return x;
}

struct DSU {
	int fa[maxn];

	inline void init() {
		for (int i = 1; i <= m + 1; ++i) {
			fa[i] = i;
		}
	}

	int find(int x) {
		return fa[x] == x ? x : fa[x] = find(fa[x]);
	}
} D;

struct node {
	int l, r, x;
	node(int a = 0, int b = 0, int c = 0) : l(a), r(b), x(c) {}
} c[maxn], d[maxn];

inline void work(int U, int V) {
	for (int i = 1; i <= n; ++i) {
		vis[i] = 0;
	}
	m = 0;
	for (int i = U; i; i = fa[i]) {
		vis[i] = 1;
		a[++m] = i;
	}
	reverse(a + 1, a + m + 1);
	int t = -1;
	for (int i = 1; i <= m; ++i) {
		if (a[i] == V) {
			t = i;
			break;
		}
	}
	for (int i = 1; i <= m; ++i) {
		b[i] = 0;
		p[i] = q[i] = -1;
		int u = a[i];
		for (int v : G[u]) {
			if (v == fa[u] || vis[v]) {
				continue;
			}
			b[i] = max(b[i], f[v] + 1);
		}
		g[0][i] = b[i] - i;
		h[0][i] = b[i] + i - 1;
	}
	for (int j = 1; (1 << j) <= m; ++j) {
		for (int i = 1; i + (1 << j) - 1 <= m; ++i) {
			g[j][i] = max(g[j - 1][i], g[j - 1][i + (1 << (j - 1))]);
			h[j][i] = max(h[j - 1][i], h[j - 1][i + (1 << (j - 1))]);
		}
	}
	int t1 = 0, t2 = 0;
	for (int i = 1; i <= m; ++i) {
		int l = i + 1, r = m, pos = -1;
		while (l <= r) {
			int mid = (l + r) >> 1;
			int k = mid + i - 1;
			if (b[i] + i - 1 > qmax1(i + 1, mid) + k) {
				pos = mid;
				l = mid + 1;
			} else {
				r = mid - 1;
			}
		}
		if (pos == -1) {
			continue;
		}
		l = i * 2;
		r = min(m, pos + i - 1);
		if (l <= r) {
			c[++t1] = node(l, r, i - 1);
		}
	}
	for (int i = 2; i <= m; ++i) {
		int l = 2, r = i - 1, pos = -1;
		while (l <= r) {
			int mid = (l + r) >> 1;
			int k = mid + i - 2;
			if (b[i] - i >= qmax2(mid, i - 1) - k) {
				pos = mid;
				r = mid - 1;
			} else {
				l = mid + 1;
			}
		}
		if (pos == -1) {
			continue;
		}
		l = pos + i - 2;
		r = min(m, i * 2 - 3);
		if (l <= r) {
			d[++t2] = node(l, r, -i);
		}
	}
	sort(c + 1, c + t1 + 1, [&](const node &a, const node &b) {
		return a.x < b.x;
	});
	D.init();
	for (int i = 1; i <= t1; ++i) {
		for (int j = D.find(c[i].l); j <= c[i].r; j = D.find(j)) {
			p[j] = c[i].x;
			D.fa[j] = j + 1;
		}
	}
	sort(d + 1, d + t2 + 1, [&](const node &a, const node &b) {
		return a.x < b.x;
	});
	D.init();
	for (int i = 1; i <= t2; ++i) {
		for (int j = D.find(d[i].l); j <= d[i].r; j = D.find(j)) {
			q[j] = j + d[i].x;
			D.fa[j] = j + 1;
		}
	}
	for (int i = m; i >= t; --i) {
		if (p[i] == -1) {
			if (q[i] == -1) {
				ans[++tot] = (i & 1);
			} else {
				ans[++tot] = 0;
			}
		} else if (q[i] == -1) {
			ans[++tot] = 1;
		} else {
			ans[++tot] = (p[i] <= q[i]);
		}
	}
}

void solve() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		vector<int>().swap(G[i]);
	}
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v);
		G[u].pb(v);
		G[v].pb(u);
	}
	scanf("%d%d", &U, &V);
	dfs(1, 0);
	int lca = qlca(U, V);
	tot = 0;
	work(U, lca);
	for (int i = 1; i <= tot; ++i) {
		puts(ans[i] ? "Alice" : "Bob");
	}
	tot = 0;
	work(V, lca);
	reverse(ans + 1, ans + tot + 1);
	for (int i = 2; i <= tot; ++i) {
		puts(ans[i] ? "Alice" : "Bob");
	}
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}

**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
3
1 2
2 3
2 3
6
1 2
1 3
2 4
2 5
1 6
4 5
4
1 2
1 3
2 4
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int MAXN=2e5+5;
inline int bit(int x) { return 1<<x; }
struct ST {
	int v[MAXN],f[MAXN][18];
	int cmp(int x,int y) { return v[x]>v[y]?x:y; }
	void upd(int x,int z) {
		v[x]=z,f[x][0]=x;
		for(int k=1;bit(k)<=x;++k) f[x][k]=cmp(f[x][k-1],f[x-bit(k-1)][k-1]);
	}
	int qry(int l,int r) {
		int k=__lg(r-l+1);
		return cmp(f[l+bit(k)-1][k],f[r][k]);
	}
}	A,B;
void upd(int d,int w) { A.upd(d,w+d-1),B.upd(d,w-d); }
bool sol(int n) {
	int mid=(n+1)>>1;
	for(int l=A.qry(1,mid),r=B.qry(mid+1,n);;) {
		if(l-1<=n-r) {
			if(A.v[l]>B.qry(l+1,n-l+1)+n) return 1;
			if(l>=mid) return 0;
			l=A.qry(l+1,mid);
		} else {
			if(B.v[r]+n>=A.qry(n-r+2,r-1)) return 0;
			if(r<=mid+1) return 1;
			r=B.qry(mid+1,r-1);
		}
	}
}
vector <int> G[MAXN];
int n,d[MAXN],f[MAXN],fa[MAXN];
bool ans[MAXN];
void dfs1(int u,int fz) {
	d[u]=d[fz]+1,f[u]=0,fa[u]=fz;
	for(int v:G[u]) if(v^fz) dfs1(v,u),f[u]=max(f[u],f[v]+1);
}
void dfs2(int u) {
	int mx=0,sx=0;
	for(int v:G[u]) if(v^fa[u]) {
		if(f[v]+1>mx) sx=mx,mx=f[v]+1;
		else sx=max(sx,f[v]+1);
	}
	if(u>1) upd(d[u],mx),ans[u]=sol(d[u]);
	for(int v:G[u]) if(v^fa[u]) upd(d[u],f[v]+1==mx?sx:mx),dfs2(v);
}
void prt(int u) { cout<<(ans[u]?"Alice":"Bob")<<"\n"; }
void out(int u,int v) {
	if(u==v) return prt(u);
	if(d[u]>d[v]) prt(u),out(fa[u],v);
	else out(u,fa[v]),prt(v);
}
void solve() {
	cin>>n;
	for(int i=1;i<=n;++i) G[i].clear();
	for(int i=1,u,v;i<n;++i) cin>>u>>v,G[u].push_back(v),G[v].push_back(u);
	dfs1(1,0),dfs2(1);
	int s,t;
	cin>>s>>t,out(s,t);
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int _; cin>>_;
	while(_--) solve();
	return 0;
}

**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 focus on edge cases and scenarios that are likely to cause issues. The buggy code involves dynamic programming and tree traversal, so we should consider cases where the tree structure or the path from \(u\) to \(v\) can lead to unexpected behavior.

### Step-by-Step Reasoning:

1. **Tree Structure**: We need a tree that is not trivial. A linear chain or a balanced binary tree can be good candidates.
2. **Path from \(u\) to \(v\)**: The path should not pass through vertex 1, and it should be long enough to cause potential issues in the code.
3. **Test Cases**: We should include multiple test cases to ensure the code handles different scenarios.

### Chosen Test Input:
- **Tree**: A linear chain of vertices.
- **Path from \(u\) to \(v\)**: A path that avoids vertex 1 and is long enough to cause issues in the code.

### Test Input:
```plaintext
3
5
1 2
2 3
3 4
4 5
2 4
6
1 2
2 3
3 4
4 5
5 6
2 5
7
1 2
2 3
3 4
4 5
5 6
6 7
2 6
```

### Explanation:
1. **First Test Case**:
   - **Tree**: A linear chain of 5 vertices: 1-2-3-4-5.
   - **Path from \(u\) to \(v\)**: 2-3-4.
   - **Expected Output**:
     - Bob (if Bob starts at vertex 2)
     - Alice (if Bob starts at vertex 3)
     - Alice (if Bob starts at vertex 4)

2. **Second Test Case**:
   - **Tree**: A linear chain of 6 vertices: 1-2-3-4-5-6.
   - **Path from \(u\) to \(v\)**: 2-3-4-5.
   - **Expected Output**:
     - Bob (if Bob starts at vertex 2)
     - Alice (if Bob starts at vertex 3)
     - Alice (if Bob starts at vertex 4)
     - Bob (if Bob starts at vertex 5)

3. **Third Test Case**:
   - **Tree**: A linear chain of 7 vertices: 1-2-3-4-5-6-7.
   - **Path from \(u\) to \(v\)**: 2-3-4-5-6.
   - **Expected Output**:
     - Bob (if Bob starts at vertex 2)
     - Alice (if Bob starts at vertex 3)
     - Alice (if Bob starts at vertex 4)
     - Bob (if Bob starts at vertex 5)
     - Alice (if Bob starts at vertex 6)

This test input is small-scale and complete, focusing on edge cases that can expose the bug 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: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
from collections import deque, defaultdict
import sys
input = sys.stdin.read
data = input().split()

index = 0
t = int(data[index])
index += 1
results = []

def dfs(node, parent, dist):
    distances[node] = dist
    for neighbor in tree[node]:
        if neighbor != parent:
            dfs(neighbor, node, dist + 1)

for _ in range(t):
    n = int(data[index])
    index += 1
    tree = defaultdict(list)
    for __ in range(n - 1):
        a = int(data[index])
        b = int(data[index + 1])
        index += 2
        tree[a].append(b)
        tree[b].append(a)
    u = int(data[index])
    v = int(data[index + 1])
    index += 2

    # Calculate distances from vertex 1
    distances = [0] * (n + 1)
    dfs(1, -1, 0)

    # Find the path from u to v
    queue = deque([u])
    parent = [-1] * (n + 1)
    while queue:
        node = queue.popleft()
        if node == v:
            break
        for neighbor in tree[node]:
            if parent[node] != neighbor:
                parent[neighbor] = node
                queue.append(neighbor)

    path = []
    current = v
    while current != -1:
        path.append(current)
        current = parent[current]
    path.reverse()

    winners = []
    for vertex in path:
        if distances[vertex] % 2 == 0:
            winners.append("Alice")
        else:
            winners.append("Bob")

    results.extend(winners)

print("\n".join(results))

**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
3
1 2
2 3
2 3
6
1 2
1 3
2 4
2 5
1 6
4 5
4
1 2
1 3
2 4
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
import sys
from collections import deque

input = sys.stdin.read
data = input().split()
idx = 0

def bfs(start, n, adj):
    dist = [-1] * (n + 1)
    queue = deque([start])
    dist[start] = 0

    while queue:
        node = queue.popleft()
        for neighbor in adj[node]:
            if dist[neighbor] == -1:
                dist[neighbor] = dist[node] + 1
                queue.append(neighbor)

    return dist

def find_path(u, v, parent):
    path = []
    cur = v
    while cur != u:
        path.append(cur)
        cur = parent[cur]
    path.append(u)
    path.reverse()
    return path

# Read number of test cases
t = int(data[idx])
idx += 1
results = []

for _ in range(t):
    # Read number of vertices
    n = int(data[idx])
    idx += 1

    adj = [[] for _ in range(n + 1)]

    # Read edges
    for __ in range(n - 1):
        a = int(data[idx])
        b = int(data[idx + 1])
        idx += 2
        adj[a].append(b)
        adj[b].append(a)

    # Read u and v
    u = int(data[idx])
    v = int(data[idx + 1])
    idx += 2

    # Perform BFS from node 1 to get distances and parents
    dist_from_1 = bfs(1, n, adj)
    parent = [0] * (n + 1)
    dist_from_u = bfs(u, n, adj)

    # Reconstruct the path from u to v
    cur = v
    while cur != u:
        for neighbor in adj[cur]:
            if dist_from_u[neighbor] == dist_from_u[cur] - 1:
                parent[cur] = neighbor
                cur = neighbor
                break

    path = find_path(u, v, parent)

    # Calculate results for each position on the path
    result = []
    for pos in path:
        if dist_from_1[pos] % 2 == 0:
            result.append("Alice")
        else:
            result.append("Bob")

    results.extend(result)

print("\n".join(results))

**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 code, we need to focus on edge cases or scenarios that can highlight issues in the algorithm. Here are the steps to think through this:

1. **Identify the Bug**: The bug in the provided code is in the way it determines the winner. It uses the distance from vertex 1 to the current position to decide the winner, which is not always correct. The correct approach should consider the alternating moves and the structure of the path.

2. **Edge Cases**: We need to consider edge cases such as:
   - Paths that are very short.
   - Paths that are very long.
   - Paths that involve vertices with high degrees.
   - Paths that involve vertices with low degrees.
   - Paths that are linear (i.e., a straight line).

3. **Small-Scale Test Input**: We need to create a small-scale test input that is complete and valid. This will help us ensure that the test input is manageable and can be easily verified.

Given these considerations, let's create a test input:

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

### Explanation:
1. **Test Case 1**:
   - Tree: `1 - 2 - 3 - 4`
   - Path: `2 - 4`
   - Expected Output: `Bob` (Alice starts at 1, Bob starts at 2), `Alice` (Alice starts at 1, Bob starts at 3), `Alice` (Alice starts at 1, Bob starts at 4)

2. **Test Case 2**:
   - Tree: `1 - 2 - 3 - 4 - 5`
   - Path: `2 - 5`
   - Expected Output: `Bob` (Alice starts at 1, Bob starts at 2), `Alice` (Alice starts at 1, Bob starts at 3), `Alice` (Alice starts at 1, Bob starts at 4), `Bob` (Alice starts at 1, Bob starts at 5)

3. **Test Case 3**:
   - Tree: `1 - 2 - 3 - 4 - 5 - 6`
   - Path: `2 - 6`
   - Expected Output: `Bob` (Alice starts at 1, Bob starts at 2), `Alice` (Alice starts at 1, Bob starts at 3), `Alice` (Alice starts at 1, Bob starts at 4), `Alice` (Alice starts at 1, Bob starts at 5), `Bob` (Alice starts at 1, Bob starts at 6)

This test input should help expose the bug in the provided code by testing different lengths of paths and different positions of vertices.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N  = 2e5+10;
struct sparse
{
	int mx[20][N],lg[N];
	int n;
	void build(vector<int> &a)
	{
		n = a.size();
		for(int i =0;i<n;i++)
			mx[0][i] = a[i];
		lg[0] = lg[1] = 0;
		for(int i=3;i<N;i++)
		lg[i] = lg[i/2] + 1;
		for(int k=1;k<20;k++)
		{
			for(int i=0;i+(1<<k) - 1 < n ;i++)
			{
				mx[k][i] = max(mx[k-1][i],mx[k-1][i+(1<<(k-1))]);
			}
		}
	}
	int get(int l,int r)
	{
		if(l>r) return -1e9;
		int k = lg[r-l+1];
		return max(mx[k][l],mx[k][r-(1<<k) + 1]);
	}
}st_a,st_b;
vector<int> e[N];
vector<int> ord;
int dp[N];
int calc(int u,int p,int en)
{
	int tag = 0 ;
	if(u == en) tag = 1 ;
	dp[u] = 1;
	for(auto v:e[u])
	{
		if(v == p)
			continue ;
		int now = calc(v,u,en);
		tag |=now ;
		if(now == 0 )
		{
			dp[u] = max(dp[u],dp[v] + 1);
		}

	}
	if(tag)
		ord.push_back(u);
	return tag;
}
vector<int> del[N],add[N];
int ans[N];

void solve(int x)
{
	ord.clear();
	calc(1,0,x);
	reverse(ord.begin(),ord.end());
	int m = ord.size();
	/*
	处理 fa  && fb

	*/
	vector<int> a(m+1),b(m+1);
	vector<int> fa(m+1,1e9),fb(m+1,-1e9);
	for(int i=0;i<m;i++)
	{
		a[i] = dp[ord[i]] + i;
		b[i] = dp[ord[i]] - i;
		del[i].clear();
		add[i].clear();
	}
	st_a.build(a);
	st_b.build(b);
	multiset<int> s;
	for(int i=1;i<m;i++)
	{
		int pos = i;
		for(int l = i+1,r = m-1;l<=r;)
		{
			int mid = (l+r)>>1;
			/*
			选择第 mid 点开始的p_mid 是否成立
			p_mid + mid 若当前在初始点，就+ mid
			初始点+1时，由于在处理 b 数组的时候，比mid 少减了一个 1 所以 加mid 即可。
			然后和 f1 一样判一下
			*/

			if(st_b.get(i+1,mid) + mid + i< a[i])
			{
				pos = mid ;
				l = mid + 1;
			}
			else
				r = mid - 1;

		}
		if( i < pos )
		{
			add[min(m,2*i)].push_back(i);
			del[min(m,pos+i)].push_back(i);
		}
		for(auto x:add[i])
		{
			s.insert(x);
		}
		if(s.size()) fa[i] = *s.begin();
		for(int x:del[i])
		{
			s.erase(s.find(x));
		}
	}
	s.clear();
	for(int i=0;i<=m;i++)
	{
		add[i].clear();
		del[i].clear();
	}
	for(int i=1;i<m;i++)
	{
		int pos = i;
		for(int l = 1,r=i-1;l<=r;)
		{
			int mid = (l+r)>>1;
			if(st_a.get(mid,i-1) - mid + 1 <= b[i] + i)
			{
				pos = mid ;
				r = mid - 1;
			}
			else
				l = mid + 1;
		}
		pos -- ;
		if(pos >= 0)
		{
			add[min(m,pos+i)].push_back(i);
			del[min(m,2*i - 1)].push_back(i);
		}
		for(int x : add[i])
			s.insert(x);
		if(s.size()) fb[i] = *s.rbegin();
		for(int x : del[i])
			s.erase(s.find(x));

	}
	for(int i = m-1;i>0;i--)
	{
		b[i]= max(b[i+1] + 1 , dp[ord[i]]);
		// 判断如果 i 为末尾是否符合
		if(b[i] >= st_a.get(1,i-1))
			fb[i] = i;
		// 判断 i 为末尾时候，
		if(a[0] > max(st_b.get(1,i-1) + i ,b[i]) )
			fb[i] = 0;
		ans[ord[i]]=0;
		if(fa[i] <= i - fb[i])
			ans[ord[i]] = 1;
	}
}
void solve()
{

	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
		e[i].clear();

	for(int i=1;i<n;i++)
	{
		int a,b;
		cin>>a>>b;
		e[a].push_back(b);
		e[b].push_back(a);
	}
	int u,v;
	cin>>u>>v;
	solve(u),solve(v);
	ord.clear();
	calc(v,0,u);
	auto p = ord;
	for(int x : p)
	{
		if(ans[x] == 1)
			cout<<"Alice"<<'\n';
		else
			cout<<"Bob"<<'\n';
	}
}

signed main (){
	std::ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int t;
	cin>>t;
	while(t--)
		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
3
1 2
2 3
2 3
6
1 2
1 3
2 4
2 5
1 6
4 5
4
1 2
1 3
2 4
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
//#include "stdafx.h"
#include <vector>
#include <iostream>
#include <math.h>
#include <string>

using namespace std;
void ras(int count, bool** tree1, int* tree3, int chek, int z, int t, int r);
string rek(int count, int* tree2, int* tree3);
void ret(int* tree, int count, bool** tree1, int* tree2, int x, int* y, int t, bool* buff);
void rasb(int count, bool** tree1, int* tree3, int chek, int z, int r);

int main()
{
	int u = 0;
	bool *buff = new bool;
	string tyemp;
	int size;
	int count;
    int x, y;
	int chek = 0;
	cin >> size;
	for (; size > 0; size--) {
		cin >> count;
        bool** tree1 = new bool*[count];
        for (int i = 0; i < count; i++)
            tree1[i] = new bool[count];
		for (int i = 0; i < count; i++)
			for (int l = 0; l < count; l++)
				tree1[i][l] = false;
		int* tree2 = new int[count];
		int* tree3 = new int[count];
		for (int i = 0; i < count - 1; i++) {
			cin >> x;
			cin >> y;
            tree1[x - 1][y - 1] = true;
			tree1[y - 1][x - 1] = true;
		}
        cin >> x;
        cin >> y;
		for (int i = 0; i < count; i++) {
			tree3[i] = -1;
			tree2[i] = -1;
		}
		int* ctree2 = new int[count];
		int* ctree3 = new int[count];
		int* BOB = new int[count];
		BOB[0] = x - 1;
		int t = 0;
		int* y1 = new int;
		*y1 = y - 1;
		*buff = false;
		tree2[BOB[u]] = 0;
		rasb(count, tree1, tree2, chek, BOB[u], chek);
		tree3[0] = 0;
		ras(count, tree1, tree3, chek, 0, BOB[u], chek);
		ret(BOB, count, tree1, tree2, x - 1, y1,t, buff);
		*buff = false;
		for(; BOB[u] + 1 != y;){
			for (int l = 0; l < count; l++) {
				ctree2[l] = tree2[l];
				ctree3[l] = tree3[l];
			}
			tyemp += rek(count, ctree2, ctree3) + "\n";
			u++;
			for (int i = 0; i < count; i++) {
				tree2[i] = -1;
				tree3[i] = -1;
			}
			tree2[BOB[u]] = 0;
			rasb(count, tree1, tree2, chek, BOB[u], chek);
			tree3[0] = 0;
			ras(count, tree1, tree3, chek, 0, BOB[u], chek);
		}
		u = 0;
		for (int i = 0; i < count; i++) {
			ctree2[i] = tree2[i];
			ctree3[i] = tree3[i];
		}
		tyemp += rek(count, ctree2, ctree3) + "\n";
		delete[] tree2, tree3, ctree2, ctree3, y1, buff, BOB, tree1;

	}
	cout << tyemp;
	return 0;
}

void ret(int* tree, int count, bool** tree1,int* tree2, int x, int* y,int t, bool* buff) {
	int yq = tree2[*y];
	for (int i = 0; i < yq; i++) {
		for (int  l = 1; l < count; l++) {
			if ((tree1[*y][l] == true || tree1[l][*y] == true) && tree2[l] == tree2[*y] - 1) {
				tree[yq - i] = *y;
				*y = l;
				l = count;
			}
		}
	}
	tree[0] = *y;
}

string rek(int count, int* tree2, int* tree3) {
	bool t;
	for (int i = 0; i < count; i++) {
		t = true;
		for (int l = 0; l < count; l++) {
			if (tree3[l] == 0)
				tree3[l] = -1;
			if ((tree3[l] == i + 1) && (tree2[l] != 0)) {
				t = false;
				tree3[l] = 0;
				tree2[l] = - 1;
			}
		}
		if (t)
			return "Bob";
		t = true;
		for (int l = 0; l < count; l++) {
			if (tree2[l] == 0)
				tree2[l] = -1;
			if ((tree2[l] == i + 1) && (tree3[l] != 0)) {
				t = false;
				tree2[l] = 0;
				tree3[l] = -1;
			}
		}
		if (t)
			return "Alice";
	}
}

void ras(int count, bool** tree1, int* tree3,int chek, int z, int t, int r) {
	chek++;
	for (int i = 0; i < count; i++) {
		if ((tree1[z][i] == true || tree1[i][z] == true)) {
			if ( tree3[i] < chek && i != r && tree3[i] != 0)  {
				ras(count, tree1, tree3, chek, i, t, z);
				tree3[i] = chek;
			}
		}
	}
}

void rasb(int count, bool** tree1, int* tree3, int chek, int z, int r) {
	chek++;
	for (int i = 0; i < count; i++) {
		if (tree1[z][i] == true || tree1[i][z] == true) {
			if ( tree3[i] < chek && i != r && tree3[i] != 0) {
				tree3[i] = chek;
				rasb(count, tree1, tree3, chek, i, z);
			}
		}
	}
}

**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
2
3
1 2
1 3
2 3
4
1 2
1 3
2 4
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
#include<bits/stdc++.h>
#define MASK(i) (1ll << (i))
#define BIT(x, i) (((x) >> (i)) & 1)
#define FOR(i, a, b) for(int i = (a), _b = (b); i <= _b; ++i)
#define FORD(i, a, b) for(int i = (a), _b = (b); i >= _b; --i)
#define FORE(i, v) for(__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i)
#define REP(i, n) for(int i = 0, _n = (n); i < _n; ++i)
#define ALL(v) (v).begin(), (v).end()
#define SZ(v) (int)(v).size()
#define PATHWA "F"
#define fi first
#define se second
#define ll long long
using namespace std;

const int N = (int)2e5 + 5;
const int LOG = 18;

struct spare_table {
    int lg2[N];
    int st[LOG][N];
    spare_table(void) {
        lg2[0] = lg2[1] = 0;
        FOR(i, 2, N - 1) {
            lg2[i] = lg2[i / 2] + 1;
        }
    }
    void build(const vector<int>& a) {
        FOR(i, 0, SZ(a) - 1) {
            st[0][i] = a[i];
        }
        FOR(l, 0, LOG - 2) FOR(i, 0, SZ(a) - MASK(l + 1)) {
            st[l + 1][i] = max(st[l][i], st[l][i + MASK(l)]);
        }
    }
    int get(int l, int r) {
        int log_value = lg2[r - l + 1];
        return max(st[log_value][l], st[log_value][r - MASK(log_value) + 1]);
    }
};

int n, u, v, dp[N], da[N], db[N], ans[N];
vector<int> adj[N], ord, add[N], del[N];
spare_table sta, stb;
set<int> storage;

bool dfs(int u, int p, int f) {
    dp[u] = 1;
    bool ok = (u == f);
    for(int v : adj[u]) if(v != p) {
        bool tmp = dfs(v, u, f);
        if(tmp) {
            ok = true;
        }
        else {
            dp[u] = max(dp[u], dp[v] + 1);
        }
    }
    if(ok) ord.push_back(u);
    return ok;
}

void get_ord(int i, int f) {
    ord.clear();
    dfs(i, 0, f);
    reverse(ALL(ord));
}

void solve(int init, int finish) {
    get_ord(init, finish);
    vector<int> da(SZ(ord)), db(SZ(ord));
    FOR(i, 0, SZ(ord) - 1) {
        da[i] = dp[i] + i;
        db[i] = dp[i] - i;
    }

    sta.build(da);
    stb.build(db);

    vector<int> fa(SZ(ord), (int)1e9), fb(SZ(ord), -(int)1e9);

    FOR(i, 1, SZ(ord) - 1) {
        int res = -1;
        for(int lo = i + 1, hi = SZ(ord) - 1; lo <= hi; ) {
            int mid = lo + hi >> 1;
            if(da[i] - i > stb.get(i + 1, mid) + mid) {
                res = mid;
                lo = mid + 1;
            }
            else {
                hi = mid - 1;
            }
        }

        if(res != -1) {
            if(i + i < SZ(ord)) add[i + i].push_back(i);
            if(res + i - 1 < SZ(ord)) del[res + i - 1].push_back(i);
        }

        for(int j : add[i]) storage.insert(j);
        if(SZ(storage)) fa[i] = *storage.begin();
        for(int j : del[i]) storage.erase(j);

        add[i].clear();
        del[i].clear();
    }

    FORD(i, SZ(ord) - 2, 1) {
        dp[i] = max(dp[i], dp[i + 1] + 1);
        db[i] = max(db[i], dp[i] - i);
    }

    fb[1] = 1;
    FOR(i, 2, SZ(ord) - 1) {
        int res = -1;
        for(int lo = 1, hi = i - 1; lo <= hi; ) {
            int mid = lo + hi >> 1;
            if(sta.get(mid, i - 1) - mid + 1 <= db[i] + i) {
                res = mid;
                hi = mid - 1;
            }
            else {
                lo = mid + 1;
            }
        }

        if(res != -1) {
            if(i + res - 1 < SZ(ord)) add[i + res - 1].push_back(i);
            if(i + i - 1 < SZ(ord)) del[i + i - 1].push_back(i);
        }

        for(int j : add[i]) storage.insert(j);
        if(SZ(storage)) fb[i] = *storage.rbegin();
        for(int j : del[i]) storage.erase(j);
    }

    FOR(i, 1, SZ(ord) - 1) {
        if(da[0] > max(stb.get(1, i), db[i]) + i) fa[i] = 0;
        ans[ord[i]] = (fa[i] <= (i - fb[i]) ? 0 : 1);
    }
}

void sol(void) {
    cin >> n;
    FOR(i, 1, n) {
        dp[i] = 0;
        adj[i].clear();
        add[i].clear();
        del[i].clear();
    }
    FOR(i, 1, n - 1) {
        int a, b; cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    cin >> u >> v;

    solve(1, u);
    solve(1, v);
    get_ord(u, v);

    for(int x : ord) {
        cout << (ans[x] == 0 ? "Alice\n" : "Bob\n");
    }
}

int main(void) {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    if(fopen(PATHWA".INP", "r")) {
        freopen(PATHWA".INP", "r", stdin);
        freopen(PATHWA".OUT", "w", stdout);
    }
    int t = 1;
    cin >> t;
    while(t--) {
        sol();
    }
    return 0;
}

**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, let's consider a few key points:

1. **Tree Structure**: The tree should be structured in a way that maximizes the complexity of the game. This can include long paths, branches, and cycles (though cycles are not possible in a tree, we can mimic the effect with long paths).

2. **Start and End Vertices**: The vertices \( u \) and \( v \) should be chosen such that the path from \( u \) to \( v \) is not trivial and does not pass through vertex 1.

3. **Multiple Test Cases**: Since the input can contain multiple test cases, we can include a few different scenarios to cover various edge cases.

4. **Edge Cases**: Consider scenarios where the path from \( u \) to \( v \) is very long, where the tree is highly branched, and where the starting vertices for Bob are close to the end of the path.

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

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

### Explanation:

1. **First Test Case**:
   - **Tree Structure**: A linear path from 1 to 6.
   - **Vertices**: \( u = 2 \) and \( v = 6 \).
   - **Path from \( u \) to \( v \)**: \( 2, 3, 4, 5, 6 \).
   - **Starting Vertices for Bob**: 2, 3, 4, 5, 6.
   - **Expected Output**:
     ```
     Bob
     Alice
     Alice
     Alice
     Alice
     ```

2. **Second Test Case**:
   - **Tree Structure**: A star-like structure with vertex 1 as the center and vertices 2, 3, 4, 5, and 6 as leaves.
   - **Vertices**: \( u = 4 \) and \( v = 6 \).
   - **Path from \( u \) to \( v \)**: \( 4, 2, 6 \).
   - **Starting Vertices for Bob**: 4, 2, 6.
   - **Expected Output**:
     ```
     Alice
     Alice
     Alice
     ```

This test input includes both a linear path and a star-like structure, which should help expose any issues 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: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
//Billions of bilious blue blistering barnacles in a thundering typhoon!
//Yesterday is history, tomorrow is a mystery, today is a gift of God, which is why we call it the present.
#include<bits/stdc++.h>
using namespace std;
void solve() {
    int n;
    cin >> n;
    vector<vector<int>> adj(n + 1);
    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v), adj[v].push_back(u);
    }
    vector<int> mx(n + 1), b(n + 1), in(n + 1), out(n + 1), pr(n + 1), dep(n + 1);
    int t = 0;
    dep[0] = -1;
    vector<int> valA(n + 1), valB(n + 1);
    function<void(int, int)> dfs = [&](int v, int par) {
        in[v] = ++t;
        dep[v] = dep[par] + 1;
        pr[v] = par;
        valA[v] = mx[v] = dep[v];
        b[t] = v;
        for (int u: adj[v]) {
            if (u == par) continue;
            dfs(u, v);
            mx[v] = max(mx[v], mx[u]);
        }
        out[v] = t;
    };
    dfs(1, 0);
    int v1, v2;
    cin >> v1 >> v2;
    vector<int> path1, path2;
    int cur = v1;
    while (!(in[cur] <= in[v2] && out[v2] <= out[cur])) {
        path1.push_back(cur);
        cur = pr[cur];
    }
    path1.push_back(cur);
    cur = v2;
    while (!(in[cur] <= in[v1] && out[v1] <= out[cur])) {
        path2.push_back(cur);
        cur = pr[cur];
    }
    auto ok = [&](int v) {
        t = 0;
        vector<int> nodes;
        dfs(1, 0);
        valA[v] = mx[v];
        int u = v;
        while (u) {
            for (int j: adj[pr[u]]) {
                if (j == u || j == pr[pr[u]]) continue;
                valA[pr[u]] = max(valA[pr[u]], mx[j]);
            }
            valB[u] = valA[u] + dep[v] - 2 * dep[u];
            nodes.push_back(u);
            u = pr[u];
        }
        multiset<int> seA, seB;
        for (int j = 0; j < (int)nodes.size(); j++) {
            seA.insert(valA[nodes[j]]);
            seB.insert(valB[nodes[j]]);
        }
        int l = 0, r = (int)nodes.size() - 1;
        seA.insert(0), seB.insert(0);
        while (l <= r) {
            seB.erase(seB.find(valB[nodes[r]]));
            if (*seB.rbegin() < valA[nodes[r]]) {
                return 0;
            }
            seA.erase(seA.find(valA[nodes[r--]]));
            if (l == r) {
                return 1;
            }
            seA.erase(seA.find(valA[nodes[l]]));
            if (*seA.rbegin() <= valB[nodes[l]]) {
                return 1;
            }
            seB.erase(seB.find(valB[nodes[l++]]));
            if (l == r) {
                return 0;
            }
        }
    };
    int l = -1, r = (int)path1.size();
    while (r - l > 1) {
        int mid = (l + r) / 2;
        if (ok(path1[mid])) {
            r = mid;
        } else {
            l = mid;
        }
    }
    for (int j = 0; j < r; j++) cout << "Alice\n";
    for (int j = r; j < (int)path1.size(); j++) cout << "Bob\n";
    l = -1, r = (int)path2.size();
    while (r - l > 1) {
        int mid = (l + r) / 2;
        if (ok(path2[mid])) {
            r = mid;
        } else {
            l = mid;
        }
    }
    for (int j = r; j < (int)path2.size(); j++) cout << "Bob\n";
    for (int j = 0; j < r; j++) cout << "Alice\n";
}
int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int tt = 1;
    cin >> tt;
    while (tt--) {
        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
2
4
1 2
1 3
2 4
2 4
6
1 2
1 3
2 4
2 5
1 6
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
from collections import defaultdict

def dfs(graph, node, parent, path, target):
    path.append(node)
    if node == target:
        return True
    for neighbor in graph[node]:
        if neighbor != parent and dfs(graph, neighbor, node, path, target):
            return True
    path.pop()
    return False

def find_path(graph, u, v):
    path = []
    dfs(graph, u, -1, path, v)
    return path

def determine_winner(n, edges, u, v):
    # Build the graph
    graph = defaultdict(list)
    for a, b in edges:
        graph[a].append(b)
        graph[b].append(a)

    # Find the path from u to v
    path = find_path(graph, u, v)

    # Determine the winner for each starting position of Bob
    results = []
    for j in range(len(path)):
        # Calculate the distance from 1 to the current Bob's starting position
        dist_to_bob = len(find_path(graph, 1, path[j])) - 1

        # Calculate the total length of the path from 1 to v
        total_dist = len(find_path(graph, 1, v)) - 1

        # Determine the winner based on the parity of the distances
        if (total_dist - dist_to_bob) % 2 == 0:
            results.append("Alice")
        else:
            results.append("Bob")

    return results

# Read input
import sys
input = sys.stdin.read
data = input().split()

index = 0
t = int(data[index])
index += 1

results = []

for _ in range(t):
    n = int(data[index])
    index += 1

    edges = []
    for _ in range(n - 1):
        a = int(data[index])
        b = int(data[index + 1])
        index += 2
        edges.append((a, b))

    u = int(data[index])
    v = int(data[index + 1])
    index += 2

    winners = determine_winner(n, edges, u, v)
    results.extend(winners)

# Output results
for result in results:
    print(result)

**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 test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that maximize the failure probability in faulty solutions. The primary issue in the code is that it calculates the distance from vertex 1 to Bob's starting position and the total distance from vertex 1 to vertex v, and then determines the winner based on the parity of these distances. This approach can fail in certain scenarios, especially when the path from vertex 1 to Bob's starting position is not the same as the path from vertex 1 to vertex v.

Let's consider a scenario where the tree structure and the path from u to v are such that the distances calculated by the code do not correctly reflect the actual game state. Specifically, we can create a tree where the path from 1 to v is longer than the path from 1 to u, and the path from u to v does not pass through vertex 1.

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

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

Explanation:
- The tree has 7 vertices.
- The edges are arranged such that vertex 1 is the root, and vertices 2 and 3 are its children.
- Vertices 4 and 5 are children of vertex 2, and vertices 6 and 7 are children of vertex 3.
- The path from u (4) to v (6) does not pass through vertex 1.
- The path from 1 to 4 is 1 -> 2 -> 4.
- The path from 1 to 6 is 1 -> 3 -> 6.

In this setup, the distances calculated by the code will be:
- Distance from 1 to 4: 2
- Distance from 1 to 6: 2
- Total distance from 1 to 6: 2

For each position of Bob (4 and 6), the code will calculate the distance from 1 to Bob and the total distance from 1 to 6, and determine the winner based on the parity of these distances.

Let's see how the code will behave:
- If Bob starts at 4:
  - Distance from 1 to 4: 2
  - Total distance from 1 to 6: 2
  - (2 - 2) % 2 == 0, so Alice wins.
- If Bob starts at 6:
  - Distance from 1 to 6: 2
  - Total distance from 1 to 6: 2
  - (2 - 2) % 2 == 0, so Alice wins.

However, in reality, if Bob starts at 4, Alice can move to 2, and Bob will have no valid move (since 1 is already visited by Alice). If Bob starts at 6, Alice can move to 3, and Bob will have no valid move (since 1 is already visited by Alice).

Thus, the correct output should be:
```plaintext
Bob
Bob
```

But the buggy code will output:
```plaintext
Alice
Alice
```

This test input should 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: Game in Tree (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the hard version of the problem. In this version, it is not guaranteed that $u = v$. You can make hacks only if both versions of the problem are solved.

Alice and Bob are playing a fun game on a tree. This game is played on a tree with $n$ vertices, numbered from $1$ to $n$. Recall that a tree with $n$ vertices is an undirected connected graph with $n - 1$ edges.

Alice and Bob take turns, with Alice going first. Each player starts at some vertex.

On their turn, a player must move from the current vertex to a neighboring vertex that has not yet been visited by anyone. The first player who cannot make a move loses.

You are given two vertices $u$ and $v$. Represent the simple path from vertex $u$ to $v$ as an array $p_1, p_2, p_3, \ldots, p_m$, where $p_1 = u$, $p_m = v$, and there is an edge between $p_i$ and $p_{i + 1}$ for all $i$ ($1 \le i < m$).

You need to determine the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_j$ for each $j$ (where $1 \le j \le m$).

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

Each of the following $n - 1$ lines contains two integers $a$ and $b$ ($1 \le a, b \le n$), denoting an undirected edge between vertices $a$ and $b$. It is guaranteed that these edges form a tree.

The last line of each test case contains two integers $u$ and $v$ ($2 \le u, v \le n$).

It is guaranteed that the path from $u$ to $v$ does not pass through vertex $1$.

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

Output: For each test case, output $m$ lines.

In the $i$-th line, print the winner of the game if Alice starts at vertex $1$ and Bob starts at vertex $p_i$. Print "Alice" (without quotes) if Alice wins, or "Bob" (without quotes) otherwise.

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

Note: ![](CDN_BASE_URL/45f5cd537988c3a64939e74c3b13efab) Tree from the first example.

In the first test case, the path will be ($2,3$). If Bob starts at vertex $2$, Alice will not be able to move anywhere on her first turn and will lose.

However, if Bob starts at vertex $3$, Alice will move to vertex $2$, and Bob will have no remaining vertices to visit and will lose.

Buggy Code:
from collections import deque, defaultdict
import sys

def find_path(tree, u, v):
    parent = {}
    queue = deque([u])
    visited = set([u])

    while queue:
        node = queue.popleft()
        if node == v:
            break
        for neighbor in tree[node]:
            if neighbor not in visited:
                visited.add(neighbor)
                parent[neighbor] = node
                queue.append(neighbor)

    path = []
    while v != u:
        path.append(v)
        v = parent[v]
    path.append(u)
    return path[::-1]

def simulate_game(tree, path):
    results = []
    for bob_start in path:
        alice_position = 1
        visited = set([alice_position, bob_start])
        turn = 'A'

        while True:
            if turn == 'A':
                found = False
                for neighbor in tree[alice_position]:
                    if neighbor not in visited:
                        alice_position = neighbor
                        visited.add(neighbor)
                        found = True
                        break
                if not found:
                    results.append('Bob')
                    break
                turn = 'B'
            else:
                found = False
                for neighbor in tree[bob_start]:
                    if neighbor not in visited:
                        bob_start = neighbor
                        visited.add(neighbor)
                        found = True
                        break
                if not found:
                    results.append('Alice')
                    break
                turn = 'A'

    return results

def main():
    input_lines = sys.stdin.read().splitlines()
    index = 0
    t = int(input_lines[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(input_lines[index])
        index += 1
        tree = defaultdict(list)

        for _ in range(n - 1):
            a, b = map(int, input_lines[index].split())
            tree[a].append(b)
            tree[b].append(a)
            index += 1

        u, v = map(int, input_lines[index].split())
        index += 1

        path = find_path(tree, u, v)
        result = simulate_game(tree, path)
        results.extend(result)

    for res in results:
        print(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 buggy code, we need to consider scenarios that can cause the algorithm to fail. The buggy code has issues with how it handles the game simulation, particularly in how it manages the visited nodes and the turns.

### Key Points to Consider:
1. **Path from \( u \) to \( v \) not passing through vertex 1**: The path from \( u \) to \( v \) should not include vertex 1, as this is explicitly stated in the problem.
2. **Correct handling of turns**: The code should correctly alternate between Alice and Bob's turns and correctly determine when a player cannot make a move.
3. **Edge cases**: Consider scenarios where the path is very short or very long, or where the tree structure is complex.

### Chosen Test Case:
- **Tree Structure**: A simple tree with a linear structure, ensuring that the path from \( u \) to \( v \) does not pass through vertex 1.
- **Path from \( u \) to \( v \)**: A path that is not trivial, but not too long to manage in a small-scale test.
- **Multiple Test Cases**: To ensure the test is comprehensive, include multiple test cases with different paths and tree structures.

### Test Input:
```plaintext
3
5
1 2
2 3
3 4
4 5
2 5
6
1 2
1 3
2 4
2 5
1 6
4 5
7
1 2
1 3
2 4
2 5
1 6
6 7
3 7
```

### Explanation:
1. **First Test Case**:
   - Tree: \( 1-2-3-4-5 \)
   - Path: \( 2-3-4-5 \)
   - Expected Output:
     ```
     Bob
     Alice
     Alice
     Bob
     ```

2. **Second Test Case**:
   - Tree:
     ```
     1 - 2 - 4
     |     |
     3     5
     |     |
     6     5
     ```
   - Path: \( 4-2-5 \)
   - Expected Output:
     ```
     Bob
     Alice
     Bob
     ```

3. **Third Test Case**:
   - Tree:
     ```
     1 - 2 - 4
     |     |
     3 - 6 - 7
     |     |
     3     7
     ```
   - Path: \( 3-6-7 \)
   - Expected Output:
     ```
     Bob
     Alice
     Bob
     ```

This test input covers various scenarios, including linear and branching paths, and should help expose any issues in the buggy code.