**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scrambled Scrabble
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are playing a word game using a standard set of $$$26$$$ uppercase English letters: A — Z. In this game, you can form vowels and consonants as follows.

- The letters A, E, I, O, and U can only form a vowel.
- The letter Y can form either a vowel or a consonant.
- Each of the remaining letters other than A, E, I, O, U, and Y can only form a consonant.
- The string NG can form a single consonant when concatenated together.

Denote a syllable as a concatenation of a consonant, a vowel, and a consonant in that order. A word is a concatenation of one or more syllables.

You are given a string $$$S$$$ and you want to create a word from it. You are allowed to delete zero or more letters from $$$S$$$ and rearrange the remaining letters to form the word. Find the length of the longest word that can be created, or determine if no words can be created.

Input: A single line consisting of a string $$$S$$$ ($$$1 \leq |S| \leq 5000$$$). The string $$$S$$$ consists of only uppercase English letters.

Output: If a word cannot be created, output 0. Otherwise, output a single integer representing the length of longest word that can be created.

Examples:
input:
ICPCJAKARTA
output:
9

input:
NGENG
output:
5

input:
YYY
output:
3

input:
DANGAN
output:
6

input:
AEIOUY
output:
0

Note: Explanation for the sample input/output #1

A possible longest word is JAKCARTAP, consisting of the syllables JAK, CAR, and TAP.

Explanation for the sample input/output #2

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant NG, the vowel E, and the consonant NG.

Explanation for the sample input/output #3

The whole string $$$S$$$ is a word consisting of one syllable which is the concatenation of the consonant Y, the vowel Y, and the consonant Y.

Explanation for the sample input/output #4

The whole string $$$S$$$ is a word consisting of two syllables: DAN and GAN.

Buggy Code:
import java.io.DataInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
import java.util.function.Predicate;

public class Main {
    public static void main(String[] args) {
 		FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        int testcases = 1;
        for(int testcase = 1; testcase <= testcases; testcase++) {
            new Solution(fs, out).solve(testcase);
        }

        out.close();
    }
}

class Solution {

    public final int mod = 1000000007;

	int getCount(String s, Predicate<Character> predicate) {
		int count = 0;
		for(int i = 0; i < s.length(); ++i) {
			if(predicate.test(s.charAt(i))) {
				count++;
			}
		}
		return count;
	}

    void solve(int testCaseNumber) {
        //Let the magic happen
		String s = fs.next(5000);

		int vowels = getCount(s, c -> List.of('A', 'E', 'I', 'O', 'U').contains(c));
		int y = getCount(s, c -> c.equals('Y'));
		int n = getCount(s, c -> c.equals('N'));
		int g = getCount(s, c -> c.equals('G'));
		int otherConsonants = getCount(s, c -> !List.of('A', 'E', 'I', 'O', 'U', 'Y', 'N', 'G').contains(c));

		int maxNG = Math.min(n, g);

		int ans = 0;

		for(int i = 0; i <= maxNG; ++i) {
			for(int j = 0; j <= y; ++j) {
				int aggVowels = vowels + j;
				int aggConsonants = otherConsonants + y - j - i + n + g;
				ans = Math.max(ans, 3 * Math.min(aggVowels, aggConsonants / 2) + i);
			}
		}
		out.println(ans);
    }

    Solution(FastScanner fs, PrintWriter out) {
        this.fs = fs;
        this.out = out;
    }
    private final FastScanner fs;
    private final PrintWriter out;
}

class Pair<U,V> {
    private U first;
    private V second;

    Pair(U first, V second) {
        this.first = first;
        this.second = second;
    }
    U first() {
        return this.first;
    }
    V second() {
        return this.second;
    }
    static <S,T> Pair<S,T> of(S s, T t) {
        return new Pair<S,T>(s, t);
    }
	@Override
	public String toString() {
		return "{\n" +
				"\tfirst: " + first.toString() + ",\n" +
			   	"\tsecond: " + second.toString() + "\n" +
				"}";
	}
}

class FastScanner {
   final private int BUFFER_SIZE = 1 << 16;
   private DataInputStream din;
   private byte[] buffer;
   private int bufferPointer, bytesRead;

   public FastScanner() {
       din = new DataInputStream(System.in);
       buffer = new byte[BUFFER_SIZE];
       bufferPointer = bytesRead = 0;
   }

   public String next(int n) {
       byte[] buf = new byte[n]; // line length
       int cnt = 0, c;
       while ((c = read()) != -1) {
           if (c <= 32) {
               if (cnt != 0) {
                   break;
               }
               else {
                   continue;
               }
           }
           buf[cnt++] = (byte)c;
       }
       return new String(buf, 0, cnt);
   }

   public int nextInt() {
       int ret = 0;
       byte c = read();
       while (c <= ' ') {
           c = read();
       }
       boolean neg = (c == '-');
       if (neg)
           c = read();
       do {
           ret = ret * 10 + c - '0';
       } while ((c = read()) >= '0' && c <= '9');

       if (neg)
           return -ret;
       return ret;
   }

   public int[] nextIntArray(int n) {
   	int arr[] = new int[n];
   	for(int i = 0; i < n; ++i) {
   		arr[i] = nextInt();
   	}
   	return arr;
   }

   public long nextLong() {
       long ret = 0;
       byte c = read();
       while (c <= ' ')
           c = read();
       boolean neg = (c == '-');
       if (neg)
           c = read();
       do {
           ret = ret * 10 + c - '0';
       } while ((c = read()) >= '0' && c <= '9');
       if (neg)
           return -ret;
       return ret;
   }

   public double nextDouble() {
       double ret = 0, div = 1;
       byte c = read();
       while (c <= ' ')
           c = read();
       boolean neg = (c == '-');
       if (neg)
           c = read();

       do {
           ret = ret * 10 + c - '0';
       } while ((c = read()) >= '0' && c <= '9');

       if (c == '.') {
           while ((c = read()) >= '0' && c <= '9') {
               ret += (c - '0') / (div *= 10);
           }
       }

       if (neg)
           return -ret;
       return ret;
   }

   private byte read() {
       if (bufferPointer == bytesRead)
           fillBuffer();
       return buffer[bufferPointer++];
   }

   private void fillBuffer() {
       try {
           bytesRead = din.read(buffer, bufferPointer = 0,
                               BUFFER_SIZE);
           if (bytesRead == -1)
               buffer[0] = -1;
       } catch (IOException e) {
           System.err.println(e.getMessage());
       }
   }

   public void close() {
       if (din == null)
           return;
       try {
	       din.close();
       } catch (IOException e) {
           System.err.println(e.getMessage());
       }
   }
}

class Helper {
    static final Random random = new Random();
    static final int mod = 1_000_000_007;

    static void ruffleSort(int[] a) {
        int n = a.length;// shuffle, then sort
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n), temp = a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }

    static long add(long a, long b) {
        return (a + b) % mod;
    }

    static long sub(long a, long b) {
        return ((a - b) % mod + mod) % mod;
    }

    static long mul(long a, long b) {
        return (a * b) % mod;
    }

    static long exp(long base, long exp) {
        if (exp == 0)
            return 1;
        long half = exp(base, exp / 2);
        if (exp % 2 == 0)
            return mul(half, half);
        return mul(half, mul(half, base));
    }

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

    static long[] factorials = new long[2_000_001];
    static long[] invFactorials = new long[2_000_001];

    static void precompFacts() {
        factorials[0] = invFactorials[0] = 1;
        for (int i = 1; i < factorials.length; i++)
            factorials[i] = mul(factorials[i - 1], i);
        invFactorials[factorials.length - 1] = exp(factorials[factorials.length - 1], mod - 2);
        for (int i = invFactorials.length - 2; i >= 0; i--)
            invFactorials[i] = mul(invFactorials[i + 1], i + 1);
    }

    static long nCk(int n, int k) {
        return mul(factorials[n], mul(invFactorials[k], invFactorials[n - k]));
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that might cause the algorithm to fail. Here are the key points to consider:

1. **Handling of 'NG' as a single consonant**: The code should correctly handle the 'NG' pair as a single consonant.
2. **Correct counting of vowels and consonants**: Ensure that the code correctly counts the vowels, consonants, and the 'Y' character, which can be either a vowel or a consonant.
3. **Maximizing the length of the word**: The input should be designed to maximize the number of syllables that can be formed, but also include cases where the input cannot form any syllables.

Given these considerations, let's create a test input that includes:
- A string with multiple 'NG' pairs.
- A string with a mix of vowels and consonants, including 'Y'.
- A string that cannot form any syllables.

Here is a small-scale test input that meets these criteria:

```plaintext
NGYNGYNG
```

This input includes:
- Three 'NG' pairs, which should be counted as three consonants.
- Three 'Y' characters, which can be either vowels or consonants.

This input is designed to test the following:
- Whether the code correctly counts 'NG' as a single consonant.
- Whether the code correctly counts 'Y' as both a vowel and a consonant.
- Whether the code can form the maximum number of syllables from the given characters.

By using this input, we can verify if the code handles these edge cases correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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 int long long
#define ed '\n'
#define IOS ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
using namespace std;
const int mod_=1e18;
const int mod=1e9+7;
const int g=1e6+10;
vector<int>ve[g];
int deep1[g],deep2[g];
bool bf[g];

int v=0;
void bfs1()
{
	memset(bf,0,sizeof bf);
	memset(deep1,0,sizeof deep1);
	queue<int>qu;
	qu.push(1);
	while(qu.size())
	{
		auto k=qu.front();qu.pop();
		if(bf[k]==1)continue;
		bf[k]=1;
		for(int i=0;i<ve[k].size();i++)
		{
			int to=ve[k][i];
			if(bf[to])continue;
		    deep1[to]=deep1[k]+1;
		    qu.push(to);
		}
		//cout<<ed;
	}
}
void bfs2()
{
	memset(bf,0,sizeof bf);
    memset(deep2,0,sizeof deep2);
	queue<int>qu;
	qu.push(v);
	deep1[v]=0;
	while(qu.size())
	{
		auto k=qu.front();qu.pop();
		if(bf[k]==1)continue;
		bf[k]=1;
		for(int i=0;i<ve[k].size();i++)
		{
			int to=ve[k][i];
			if(bf[to])continue;
			if(deep2[k]+1<deep1[to])
			{
				deep2[to]=deep2[k]+1;
				deep1[to]=0;
				qu.push(to);
			}

		}
	}
}
void solve()
{
	int n=0;cin>>n;
	for(int i=1;i<n;i++)
	{
		int a=0,b=0;
		cin>>a>>b;
		ve[a].push_back(b);
		ve[b].push_back(a);
	}

    cin>>v;
	cin>>v;

	bfs1();
	bfs2();

	int ans1=0,ans2=0;
	for(int i=1;i<=n;i++)
	{
		ans1=max(ans1,deep1[i]);
		ans2=max(ans2,deep2[i]);
		ve[i].clear();
	//	cout<<deep1[i]<<" "<<deep2[i]<<ed;
	}

	if(ans1>ans2)cout<<"Alice"<<ed;
	else cout<<"Bob"<<ed;

	return;
}
signed main()
{
	IOS;

  int t=1;

  cin>>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:
To create a challenging test input that exposes the bug in the given code, we need to consider the following points:

1. **Tree Structure**: The tree should be designed in a way that the path from \( u \) to \( v \) (where \( u = v \)) is not trivial and does not pass through vertex 1.
2. **Edge Cases**: Consider edge cases where the path from \( u \) to \( v \) is a single vertex or where the tree structure is such that the BFS calculations might be incorrect.
3. **BFS Behavior**: The BFS should be tested to ensure that the depths are calculated correctly and that the logic for determining the winner is sound.

Given the constraints and the nature of the problem, let's create a test case where the tree is a linear chain and \( u = v \) is a vertex that is not directly connected to vertex 1. This will help us test the BFS and the logic for determining the winner.

Here is the test input:

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

### Explanation:
- **Test Case 1**:
  - **Number of vertices**: 4
  - **Edges**:
    - 1 is connected to 2
    - 2 is connected to 3
    - 3 is connected to 4
  - **u and v**: Both are 4

In this test case:
- The path from \( u = 4 \) to \( v = 4 \) is a single vertex, which is 4.
- Alice starts at vertex 1, and Bob starts at vertex 4.
- Since Bob starts at the end of the chain, Alice will move to vertex 2, and Bob will have no remaining vertices to visit and will lose.

This test case should help expose any issues in the BFS calculations or the logic for determining the winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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;

struct vert{
    int num;
    int depth;
    int alice;
};

int main(){
    int t, n;
    cin >> t;
    if (t == 10000) {
        for (int i = 0;i<t;i++) {
            cin >> n;
            for (int j = 0 ; j < n-1; j++){
                int temp1, temp2;
                cin >> temp1 >> temp2;
                if (i == 1325) cout << temp1 <<" " << temp2<<"\n";
            }
            int u1,u2;
            cin >> u1 >> u2;
            if (i == 1325) cout << u1 << "\n";
        }
        return 0;
    }
    while (t--){
        bool* visited = new bool[200200]{};
        long long n;
        cin >> n;
        vector<vert>* verts = new vector<vert>[n+2];
        for (int i = 0 ; i < n-1; i++){
            int temp1, temp2;
            cin >> temp1 >> temp2;
            verts[temp1].push_back({temp2,0,-1});
            verts[temp2].push_back({temp1,0,-1});
        }
        int u1,u2;
        cin >> u1 >> u2;
        queue<int> alice;
        queue<int> bob;
        alice.push(1);
        visited[1] = true;
        bob.push(u1);
        visited[u1] = true;
        int curdepth = 1;
        int bobmax = 0;
        int alicemax = 0;
        int vst = 2;
        while (true){
            //cout << vst << ' ';
            int anum = alice.size();
            for (int i = 0; i < anum;i++){
                int cur = alice.front();
                alice.pop();
                for (vert &index : verts[cur])
                    {
                        if(visited[index.num] == 1 || index.alice == 0) continue;
                        visited[index.num] = 1;
                        vst++;
                        index.depth = curdepth;
                        alicemax = curdepth;
                        index.alice = 1;
                        alice.push(index.num);
                    }
            }
            if (alice.size() == 0) {cout << "Bob" << '\n'; break;}
            int bnum = bob.size();
            for (int i = 0; i < bnum;i++){
                int cur = bob.front();
                bob.pop();
                for (vert &index : verts[cur])
                    {
                        if(visited[index.num] == 1 || index.alice == 1) continue;
                        visited[index.num] = 1;
                        vst++;
                        bobmax = curdepth;
                        index.depth = curdepth;
                        index.alice = 0;
                        bob.push(index.num);
                    }
            }
            if (bob.size() == 0) {cout << "Alice" << '\n'; break;}
            curdepth++;
        }
        //if (alicemax > bobmax)
        //cout << "Alice";
        //else
        //cout <<"Bob";
        //cout <<'\n';
        delete[] visited;
        delete[] verts;
    }
    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
1
3
1 2
2 3
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// #include <iostream>
// #include <stack>
#include <random>
using namespace std;
// using namespace __gnu_pbds;

template<class T> using prior_q = priority_queue<T>;
template<class T> using prior_qg = priority_queue<T, vector<T>, greater<T> >;

// #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>

#define int long long
#define pii pair<int,int>
#define sz size()
#define ln length()
#define ft first
#define sd second
#define pb push_back
#define pop_b(x) if(x.sz)x.pop_back();else cout<<"empty vec\n";
#define ins insert
#define stand_out(x) cout<<"\n#################\n\n";for(auto i:x)cout<<i<<" ";cout<<"\n\n#############\n";

#define Mod 1000000007
#define cyes cout<<"YES\n"
#define cno cout<<"NO\n"

#define arrin(a,n) for(int i=0;i<n;i++)cin>>a[i];
#define matin(a,n,m) for(int i=0;i<n;i++)for(int j=0;j<m;j++)cin>>a[i][j];
#define all(v) v.begin(),v.end()
#define vsort(v) sort(all(v))
#define rsort(v) sort(v.rbegin(),v.rend())

template<typename t1>
void pv(t1& a,int n=-1){if(n==-1)n=a.sz;for(int z=0;z<n;z++)cout<<a[z]<<" ";cout<<"\n";}
void print(pii a){cout<<"("<<a.ft<<","<<a.sd<<") ";}
void pm(map<int,int>& m){for(auto i:m)print(i);cout<<"\n";}
int Power(int b, int e){
    int r=1;b%=Mod;
    while(e){if(e%2==1)r=(r*b)%Mod;b=(b*b)%Mod;e/=2;}
    return r;
}
int Mod_Inv(int a){return Power(a,Mod-2);}
int gcd(int a,int b){
    if(a>b)swap(a,b);
    if(b%a==0)return a;
    else return gcd(b%a,a);
}
int hcf(int a,int b){return (a*b)/gcd(a,b);}
vector<int> prefvec(vector<int>& a){
    vector<int> pref;
    pref.pb(a[0]);
    for(int i=1;i<(int)a.sz;i++)pref.pb(a[i]+pref.back());
    return pref;
}
const int mod = 1000012339;
const int hash_n = 3;
vector<int> p_pow;
void init_hash(int m){
    p_pow.pb(1);
    for(int i=1;i<=m;i++)p_pow.pb((p_pow.back()*hash_n)%mod);
}
int string_hash(string s){
    int val = 0;
    for(int i=0;i<(int)s.ln;i++)val = (val + p_pow[i]*(s[i]-'0'+1))%mod;
    return val;
}
vector<int> string_hash_prefix(string s){
    int val = 0;
    vector<int> v;
    for(int i=0;i<(int)s.ln;i++){
        val = (val + p_pow[i]*(s[i]-'0'+1))%mod;
        v.pb(val);
    }
    return v;
}
unordered_map<int,int> comp_m;
void compress_array(vector<int>& a){
    int c = 1;
    for(auto& i:a){
        if(!comp_m.count(i))comp_m[i]=c++;
        i = comp_m[i];
    }
}
pii intersect(pii a,pii b){return pii(max(a.ft,b.ft),min(a.sd,b.sd));}

vector<vector<int> > adj;
vector<int> parent;
vector<int> max_level,level;
void dfs(int a,int b){
    level[a] = level[b]+1;
    parent[a] = b;
    if(adj[a].sz==1)max_level[a] = level[a];
    for(auto i:adj[a]){
        if(i==b)continue;
        dfs(i,a);
        max_level[a] = max(max_level[a],max_level[i]);
    }
}

void init_prob(int n){
    adj.clear();adj.resize(n+1);
    level.clear();level.resize(n+1);
    parent.clear();parent.resize(n+1);
    max_level.clear();max_level.resize(n+1);
}

int i,ii,j,l,k,x,tc = 0;
void solve(){
    int n;cin>>n;
    init_prob(n);
    for(i=0;i<n-1;i++){
        int a,b;cin>>a>>b;
        adj[a].pb(b);
        adj[b].pb(a);
    }
    int u,v;cin>>u>>v;
    dfs(1,1);
    int node1 = u;
    while(!parent[node1]==1)node1 = parent[node1];
    int k = 0;
    for(auto i: adj[1])if(i!=node1){k = max(k,max_level[i]-1);}
    int ali_away = k;
    int ali_same = 0;
    int node2 = parent[u],pnode = u,bob_same = max_level[u]-level[u];
    int bob_steps = 1;
    while(!parent[node2]==1){
        int ali_steps = level[node2];
        if(ali_steps > bob_steps){
            bob_same = max(bob_same,bob_steps+max_level[node2]-level[node2]);
        }
        else{
            int k = 0;
            for(auto i:adj[node2]){
                if(i!=pnode){
                    ali_same = max(ali_same,max_level[i]-1);
                }
            }
        }
        pnode = node2;
        node2 = parent[node2];
        bob_steps++;
    }
    int bob_away = max_level[u]-level[u];
    // cout<<ali_same<<" "<<ali_away<<" "<<bob_same<<" "<<bob_away<<"\n";
    if(bob_away >= ali_away && bob_away >= ali_same){
        cout<<"Bob\n";
    }
    else if(bob_same >= ali_away && bob_same >= ali_same){
        cout<<"Bob\n";
    }
    else if(ali_away >= bob_away && ali_away >= bob_same){
        cout<<"Alice\n";
    }
    else if(ali_same >= bob_away && ali_same >= bob_same){
        cout<<"Alice\n";
    }
}

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    // init_hash(1e5+3);
    int t = 1;cin>>t;
    while(t--){tc++;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 2
4
1 2
2 3
3 4
3 3
6
1 2
1 3
2 4
2 5
3 6
4 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 (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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

def solve():
    input = sys.stdin.read
    data = input().splitlines()

    idx = 0
    t = int(data[idx])
    idx += 1
    result = []

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

        # Tree adjacency list
        tree = [[] for _ in range(n+1)]

        # Building the tree
        for __ in range(n-1):
            a, b = map(int, data[idx].split())
            tree[a].append(b)
            tree[b].append(a)
            idx += 1

        # Read u, v
        u, v = map(int, data[idx].split())
        idx += 1

        # BFS to find distances from vertex 1
        dist_from_1 = [-1] * (n+1)
        dist_from_1[1] = 0

        q = deque([1])
        while q:
            node = q.popleft()
            for neighbor in tree[node]:
                if dist_from_1[neighbor] == -1:
                    dist_from_1[neighbor] = dist_from_1[node] + 1
                    q.append(neighbor)

        # The result depends on the distance from 1 to u
        if dist_from_1[u] % 2 == 0:
            result.append("Alice")
        else:
            result.append("Bob")

    # Output all results at once
    sys.stdout.write("\n".join(result) + "\n")

# Entry point
if __name__ == "__main__":
    solve()

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

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

Think step by step.

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

1. **Tree Structure**: The tree should be structured in a way that maximizes the complexity of the BFS and the distance calculations.
2. **Path from \(u\) to \(v\)**: Since \(u = v\), the path will be a single vertex.
3. **Distance from Vertex 1 to \(u\)**: The distance from vertex 1 to \(u\) should be carefully chosen to expose the bug in the logic that determines the winner.

Given that the code incorrectly determines the winner based on the distance from vertex 1 to \(u\) modulo 2, we can create a test case where this logic fails.

### Test Case:
- **Tree Structure**: A simple linear tree where vertex 1 is connected to vertex 2, and vertex 2 is connected to vertex 3, and so on.
- **Path from \(u\) to \(v\)**: \(u = v = 3\).

### Test Input:
```plaintext
1
3
1 2
2 3
3 3
```

### Explanation:
- **Tree Structure**: The tree consists of 3 vertices: 1, 2, and 3. The edges are (1, 2) and (2, 3).
- **Path from \(u\) to \(v\)**: \(u = v = 3\).

### Expected Output:
- **Distance Calculation**: The distance from vertex 1 to vertex 3 is 2 (1 -> 2 -> 3).
- **Winner Determination**: Since the distance from vertex 1 to vertex 3 is 2 (even), the code should output "Alice".

However, the buggy code will output "Bob" because it only checks the distance from vertex 1 to \(u\) and not the path dynamics. This test case will expose the bug in the logic.

### Output:
```plaintext
1
3
1 2
2 3
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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:
"""
For God so loved the world that he gave his one and only Son,
that whoever believes in him shall not perish but have eternal life.
JOHN 3:16
"""
import sys
#sys.setrecursionlimit(110000)
"""
#For large input
input = sys.stdin.read
data = input().split()
"""
import heapq as hp
import bisect as bs
from bisect import bisect_left as bl
from bisect import bisect_right as br
from math import *
from collections import Counter,deque,defaultdict
from functools import lru_cache
from itertools import accumulate
from fractions import Fraction
import random
# constants
MOD= 10**9 + 7
ANS = ['Bob','Alice']
#print(f"{len(res)}\n{res}")
# Shortcut for sys.stdin.readline()
def I(): return int(sys.stdin.readline())
def S(): return sys.stdin.readline().strip()
def MI(): return map(int, sys.stdin.readline().split())
def MS(): return map(str, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LS(): return list(map(str, sys.stdin.readline().split()))

#defaultdict(<class 'set'>, {1: {2, 3, 4, 5}, 2: {1, 6}, 3: {1}, 4: {1}, 5: {1}, 6: {2, 7, 8}, 7: {6}, 8: {6}})
#defaultdict(<class 'set'>, {1: {2, 3, 4}, 2: {1, 5}, 3: {1, 6}, 4: {1}, 5: {2, 7}, 6: {3}, 7: {5, 8}, 8: {7}}) 7

def bfs(n,u,g):
    q = deque([1])
    p = [-1]*(n+1)
    b = []
    p[1] = 0
    while q:
        node = q.popleft()
        b.append(node)
        for child in g[node]:
            if p[child] == -1:
                q.append(child)
                p[child] = node
    pathb = [u]
    while pathb[-1] != 1:
        pathb.append(p[pathb[-1]])
    pa = set(pathb)
    dp = [0]*(n+1)
    for i in range(len(b)-1,0,-1):
        node = b[i]
        if node not in pa:
            dp[p[node]] =max(dp[p[node]] ,(1 + dp[node]))

    if len(pa)-2 == 0:
        return dp[1] > dp[u]
    #max alice score if not used path between alice to bob - dp[1]
    #max bob score if not used path between alice to bob - dp[u]
    #bob using total path between a-b
    bob = []
    for i in range(1,len(pathb)-1):
        bob.append(i+dp[pathb[i]])
    #alice using total path between a-b
    patha = pathb[::-1]
    alice = []
    for i in range(1,len(patha)-1):
        alice.append(i+dp[patha[i]])
    #bob using total length a-b if alice used other than a-b and become > bob he is winner
    if dp[1] > max(dp[u],max(bob)):
        return 1
    #bob using total length a-b if alice used other than a-b and become > bob he is winner
    if dp[u] > max(dp[1],max(alice)):
        return 0
    #both are comping face to face in path
    r = len(alice)//2
    l = r+1 if len(alice)&1 else r
    maxa = 0
    for i in range(l):
        maxa = max(maxa,alice[i])
    maxb = 0
    for i in range(r):
        maxb = max(maxb,bob[i])
    return maxa > maxb

#1-indexed tree
t = I()
for i in range(t):
    n = I()
    g = defaultdict(set)
    for _ in range(n-1):
        a,b = MI()
        g[a].add(b)
        g[b].add(a)
    #u=v b's location[node]
    u,v = MI()
    print(ANS[bfs(n,u,g)])

**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 edge cases and scenarios that can cause the algorithm to fail. Here are the steps to create such a test input:

1. **Understanding the Bug**: The buggy code seems to have issues with the way it calculates the path and the scores for Alice and Bob. Specifically, the way it handles the path and the scores for each player might be incorrect.

2. **Edge Cases**:
   - **Single Path**: A simple path where \( u = v \) and the path does not pass through vertex 1.
   - **Multiple Paths**: A more complex tree structure where the path from \( u \) to \( v \) has multiple branches and vertices.

3. **Test Case**:
   - **Tree Structure**: A tree with a simple path from \( u \) to \( v \) that does not pass through vertex 1.
   - **Multiple Test Cases**: To ensure the code handles multiple test cases correctly.

Here is a small-scale test input that fits the requirements:

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

Explanation:
- **Test Case 1**:
  - Tree: 1-2-3-4-5
  - \( u = 4 \), \( v = 4 \)
  - Expected Output: Bob (since Bob starts at vertex 4 and Alice cannot move to vertex 4).

- **Test Case 2**:
  - Tree: 1-2-3-4-5
  - \( u = 5 \), \( v = 5 \)
  - Expected Output: Bob (since Bob starts at vertex 5 and Alice cannot move to vertex 5).

- **Test Case 3**:
  - Tree: 1-2-3-4-5-6
  - \( u = 6 \), \( v = 6 \)
  - Expected Output: Bob (since Bob starts at vertex 6 and Alice cannot move to vertex 6).

This test input should expose any issues in the buggy code by focusing on the simple path scenario where \( u = v \) and the path does not pass through vertex 1.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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:
t=int(input())
for _ in range(t):
    n=int(input())
    for a in range(n):
        a,b=map(int,input().split())
    if((a-1)-1>=(n-a)):
        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:
To expose the bug in the provided buggy code, we need to focus on scenarios where the code's logic fails. The key issue in the buggy code is that it only checks the last edge read and uses a simple comparison to determine the winner, which is incorrect for the game's rules.

### Steps to Create a Challenging Test Input:

1. **Identify the Bug**: The code only considers the last edge read and uses a simple comparison to determine the winner. This does not account for the actual game dynamics where players move along the tree and the path from \( u \) to \( v \).

2. **Create a Test Case**:
   - **Tree Structure**: Create a tree where the path from \( u \) to \( v \) is non-trivial and involves multiple vertices.
   - **Path from \( u \) to \( v \)**: Ensure the path does not pass through vertex 1, as specified in the problem.
   - **Multiple Test Cases**: Include multiple test cases to cover different scenarios.

3. **Ensure Validity**: The test case should be small enough to be manageable but complex enough to expose the bug.

### Challenging Test Input:

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

### Explanation:

1. **First Test Case**:
   - Tree with 6 vertices.
   - Edges: (1, 2), (1, 3), (2, 4), (2, 5), (1, 6).
   - Path: (4, 4).
   - Expected Output: "Alice" (since the path is trivial and Bob starts at the same vertex).

2. **Second Test Case**:
   - Tree with 3 vertices.
   - Edges: (1, 2), (2, 3).
   - Path: (3, 3).
   - Expected Output: "Alice" (Bob starts at vertex 3, Alice can move to vertex 2, and Bob will have no move).

3. **Third Test Case**:
   - Tree with 4 vertices.
   - Edges: (1, 2), (1, 3), (2, 4).
   - Path: (4, 4).
   - Expected Output: "Alice" (Bob starts at vertex 4, Alice can move to vertex 2, and Bob will have no move).

This test input should expose the bug in the buggy code by providing multiple scenarios where the path from \( u \) to \( v \) is non-trivial and the simple comparison in the code fails to determine the correct winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {
		//stack=new Stack<Integer>();
		//set=new HashSet<List<Integer>>();
		//f(0, new int[3]);System.out.println(set.size());
		//p1=false;
		//System.out.println("aaa");

		  int t=sc.nextint(); while (t-->0) {

		  sovle(); }

		  out.close();

	}
	static long[] ram,v;
	static int []size,arr;
	static TreeSet<Integer>treeSet;
	static long gcd(long a,long b) {
		return b==0?a:gcd(b, a%b);
	}
	static Map<Integer, Integer>vpo;
	static void sovle2() {

	}
	static void ask(String x) {
		System.out.println("? "+x);
		System.out.flush();
	}
	static int yu,ans;
	static boolean p;

	static int []dp;

	static int[]zt;
	static boolean p1;
	static int[][] va,l,r;
	static int n,m,k;
	static void sovle() {
		int n=sc.nextint();
		map=new HashMap<>();
		for (int i = 0; i < n; i++) {
			map.put(i+1, new ArrayList<>());
		}
		for (int i = 1; i < n; i++) {
			int u=sc.nextint();
			int v=sc.nextint();
			map.get(u).add(v);
			map.get(v).add(u);
		}
		int u=sc.nextint();
		int v=sc.nextint();
		h=new int[n+1];
		dfs(v, 0);
		int max=0;int ge=1;
		for (int i : map.get(v)) {
			if (h[i]>max) {
				max=h[i];ge=1;
			}else if (h[i]==max) {
				ge++;
			}
		}
		//System.out.println(max+" "+ge);
		//System.out.println(Arrays.toString(h));
		if (ge>=2) {
			out.println("Bob");
		}else {
			out.println("Alice");
		}
	}
	static int []h;
	static void dfs(int x,int fu){

		for (int i : map.get(x)) {
			if (i!=fu) {
				dfs(i, x);
				h[x]=Math.max(h[x], h[i]);
			}
		}
		h[x]++;
	}
	static int[][]tu;
	static void addbian(int x,int y,int v) {

		tu[x][y]=v;
		map.get(x).add(y);
	}

	static int get(int x,int []shu) {
		int ans=(int) (shu[x]+sum(Math.min(n, x*2))-sum(Math.min(n, x*2-1)));

		return ans;
	}

	static long  pow(long x,long a) {
		if (a==0) {
			return 1;
		}else {
			long c=pow(x, a/2);
			if (a%2==0) {
				return c*c%mod;
			}else {
				return c*c%mod*x%mod;
			}
		}
	}
	static long mod = (long) (1e9+7);

	static void sovle1() {

	}

	static long []c;
	static int lowbit(int x) {
		return x&-x;
	}
	static void add(int x,long v) {
		while (x<c.length) {
			//System.out.println(x);
			c[x]+=v;
			x+=lowbit(x);
		}
	}
	static long sum(int x) {
		long ans=0;
		//System.out.println(x+"dasdsafadf");
		while (x>0) {
			//System.out.println(x);
			ans+=c[x];
			x-=lowbit(x);
		}
		return ans;
	}
	static Map<Integer, List<Integer>>map;

}

class kattio extends PrintWriter {

	BufferedReader r;
	StringTokenizer st;

	public kattio() {
		this(System.in, System.out);
	}

	public kattio(InputStream i, OutputStream o) {
		super(o);
		r = new BufferedReader(new InputStreamReader(i));
	}

	public kattio(String in, String out) throws IOException {
		super(out);
		r = new BufferedReader(new FileReader(in));
	}

	public String next() {
		try {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(r.readLine());
			}
			return st.nextToken();
		} catch (Exception e) {
			// TODO: handle exception
			return null;
		}
	}

	public int nextint() {
		return Integer.parseInt(next());
	}

	public long nextlong() {
		return Long.parseLong(next());
	}

	public double nextdouble() {
		return Double.parseDouble(next());
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

1. **Path from \( u \) to \( v \)**: The problem states that \( u = v \) and the path from \( u \) to \( v \) does not pass through vertex 1. This means \( u \) and \( v \) are the same vertex, and it is not the starting vertex for Alice.

2. **DFS and Height Calculation**: The code uses a depth-first search (DFS) to calculate the height of each vertex. It then determines the maximum height and the number of vertices with that maximum height.

3. **Bug Analysis**:
   - The code assumes that if there are multiple vertices with the maximum height, Bob wins. However, this logic is flawed because it does not consider the specific path from \( u \) to \( v \) (which is just \( u \) itself in this case).

4. **Edge Case**:
   - The code does not correctly handle the case where \( u = v \) and both are not equal to 1. We need to create a tree where \( u = v \) and the path from 1 to \( u \) is such that the game logic should be tested.

Let's create a small-scale test input:

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

**Explanation**:
- **Test Case**: There is one test case.
- **Tree**:
  - Vertex 1 is connected to vertex 2.
  - Vertex 2 is connected to vertex 3.
  - Vertex 3 is connected to vertex 4.
- **Starting Vertices**:
  - \( u = 4 \) and \( v = 4 \).
- **Path**:
  - The path from 4 to 4 is just the single vertex 4, which does not pass through vertex 1.

This test case should expose the bug because the code's logic for determining the winner based on the maximum height and the number of vertices with that height is flawed when \( u = v \). The correct output should be "Alice" because Alice starts at vertex 1 and Bob starts at vertex 4, and Alice can always move to vertex 2, leaving Bob with no valid move.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios where the code might fail. The key points to consider are:

1. **Tree Structure**: The tree should be structured in a way that maximizes the number of vertices and edges, but still fits within the constraints.
2. **Path from u to v**: Since \( u = v \), the path from \( u \) to \( v \) should be trivial (a single vertex).
3. **Distance Calculation**: The code calculates the distance from vertex 1 to the starting vertex of Bob. We need to ensure that this distance calculation is correct and that the parity of the distance affects the outcome.

Given the constraints and the nature of the problem, a good test case would be a tree with a linear structure, where vertex 1 is at one end and the path from \( u \) to \( v \) is a single vertex. This will test the distance calculation and the parity logic.

Here is a small-scale test input that meets these criteria:

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

Explanation:
- **Test Case 1**:
  - **Tree Structure**: The tree is a linear chain of vertices: 1-2-3-4-5.
  - **Path from \( u \) to \( v \)**: Since \( u = v = 4 \), the path is simply [4].
  - **Distance Calculation**:
    - The distance from vertex 1 to vertex 4 is 3 (1 -> 2 -> 3 -> 4).
    - Since the distance (3) is odd, Bob should win.

This test case will help expose any issues with the distance calculation and the parity logic in the provided 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 (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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;

#ifdef LOCAL
ifstream in("in.in");
ofstream out("out.out");
#else
#define in cin
#define out cout
#endif

int n,b;
vector<vector<int>> g;
vector<int> path,subSize;
vector<int> vPath;
vector<int> aValue;
bool dfs(int node, int target, int parent) {
    if (node == target) {
        path[node] = 1;
        vPath.push_back(node);
        return true;
    }
    for(auto k : g[node]) {
        if (k == parent) {
            continue;
        }
        if (dfs(k, target, node)) {
            vPath.push_back(node);
            path[node] = 1;
            return true;
        }
    }
    return false;
}

void get_sizes(int node, int parent) {
    subSize[node] = 1;
    for (auto k : g[node]) {
        if (k == parent) {
            continue;
        }
        get_sizes(k , node);
        subSize[node] = max(subSize[node], subSize[k] + 1);
    }
}
bool is_winnable = false;
void get_values(int node, int parent) {
    for (auto k : g[node]) {
        if (k == parent || path[k] == 1) {
            continue;
        }
        aValue[node] = max(aValue[node],subSize[k]);
    }
    for (auto k : g[node]) {
        if (k == parent) {
            continue;
        }
        if (path[k] == 1) {
            get_values(k, node);
        }
    }
}
int solve() {
    in >> n;
    vPath.clear();
    path = vector<int>(n + 1);
    aValue = vector<int>(n + 1, 0);
    subSize = vector<int>(n + 1, 0);
    g = vector<vector<int>>(n + 1);
    for (int i = 1; i < n; i++) {
        int u,v;
        in >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    in >> b >> b;
    dfs(1, b, -1);
    get_sizes(1, -1);
    get_values(1, -1);

    reverse(vPath.begin(), vPath.end());

    vector<int> bOptions(vPath.size());
    bOptions[(int)bOptions.size() - 1] = aValue[vPath[vPath.size() - 1]];
    for (int i = (int)bOptions.size() - 2; i >= 0; i--) {

        bOptions[i] = max(aValue[vPath[i]], bOptions[i + 1]);
    }

    for (int i = 0; i < vPath.size() / 2 + vPath.size() % 2; i++) {
        int aOptions = i  + aValue[vPath[i]];
        int bVal = ((int)vPath.size() - i - 2) + bOptions[i + 1];
        if (aOptions > bVal) {
            return true;
        }
    }
    return false;
}

int32_t main() {
    int t;
    in >> t;
    while(t--) {
        bool verdict = solve();
        if (verdict) {
            out << "Alice\n";
        } else {
            out << "Bob\n";
        }
    }
}

**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
2 3
3 4
4 5
4 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 (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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 <iostream>
#include <fstream>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <vector>
#include <iomanip>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <numeric>
#include <cmath>

#pragma warning(disable : 4996)

using namespace std;

#define f(i, st, en) for (int i = st; i < en; i++)
#define fm(i, st, en) for (int i = st - 1; i >= en; i--)
#define uchar unsigned char
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define M int(1e9 + 7)
#define PI 3.14159265358979323846

void decide()
{
    int n;
    cin >> n;
    vector<vector<int>> g(n, vector<int>());
    vector<bool> used(n, 0);
    for (int i = 0; i < n - 1; i++)
    {
        int a, b;
        cin >> a >> b;
        a--; b--;
        g[a].push_back(b);
    }
    int u, v;
    cin >> u >> v;
    u--;
    queue<int> qa, qb;

    qa.push(0);
    used[0] = 1;

    qb.push(u);
    used[u] = 1;

    while (!qa.empty() && !qb.empty())
    {
        vector<int> tmp;

        while (!qa.empty())
        {
            tmp.push_back(qa.front());
            qa.pop();
        }

        for (int i = 0; i < tmp.size(); i++)
        {
            for (int y : g[tmp[i]])
            {
                if (!used[y])
                {
                    used[y] = 1;
                    qa.push(y);
                }
            }
        }
        if (qa.empty())
        {
            cout << "Bob" << endl;
            return;
        }
        tmp.clear();

        while (!qb.empty())
        {
            tmp.push_back(qb.front());
            qb.pop();
        }

        for (int i = 0; i < tmp.size(); i++)
        {
            for (int y : g[tmp[i]])
            {
                if (!used[y])
                {
                    used[y] = 1;
                    qb.push(y);
                }
            }
        }

    }
    if (qa.empty()) cout << "Bob" << endl;
    else cout << "Alice" << endl;
    return;
}

int main()
{

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) decide();
    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 buggy code, let's consider the following steps:

1. **Identify the Bug**: The buggy code does not correctly handle the case where Alice and Bob are on the same path and the path does not pass through vertex 1. The code seems to assume that the path from `u` to `v` is always valid and does not handle the case where the path is a single vertex (i.e., `u = v`).

2. **Edge Case**: Since the problem states that `u = v` and the path from `u` to `v` does not pass through vertex 1, we need to create a test case where `u` and `v` are the same and the tree structure is such that the path does not involve vertex 1.

3. **Small-Scale Test Case**: We will create a small tree with a few vertices to ensure that the test input is valid and complete.

Let's construct a tree with 4 vertices and the path from `u` to `v` not passing through vertex 1. Here's the test input:

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

### Explanation:
- **Test Case 1**:
  - **Number of vertices**: 4
  - **Edges**:
    - 1 is connected to 2
    - 2 is connected to 3
    - 3 is connected to 4
  - **Vertices u and v**: Both are 3 (i.e., `u = v = 3`)

### Expected Output:
- **Output**:
  - Bob

### Reasoning:
- **Path from u to v**: The path from vertex 3 to vertex 3 is just the single vertex 3.
- **Game Simulation**:
  - Alice starts at vertex 1.
  - Bob starts at vertex 3.
  - Alice moves to vertex 2.
  - Bob has no valid move (since vertex 3 is already visited and there is no other unvisited neighbor for vertex 3).
  - Therefore, Bob loses.

This test case should expose the bug in the provided code, as it will not correctly handle the situation where `u = v` and the path does not pass through vertex 1.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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 sys import stdin

class FastIO:
    def __init__(self):
        self.random_seed = 0
        self.flush = False
        self.inf = 1 << 32
        self.dire4 = [(0, -1), (0, 1), (1, 0), (-1, 0)]
        self.dire8 = [(0, -1), (0, 1), (1, 0), (-1, 0)] + [(1, -1), (1, 1), (-1, -1), (-1, 1)]
        return

    @staticmethod
    def read_int():
        return int(stdin.readline().rstrip())

    @staticmethod
    def read_float():
        return float(stdin.readline().rstrip())

    @staticmethod
    def read_list_ints():
        return list(map(int, stdin.readline().rstrip().split()))

    @staticmethod
    def read_list_ints_minus_one():
        return list(map(lambda x: int(x) - 1, stdin.readline().rstrip().split()))

    @staticmethod
    def read_str():
        return stdin.readline().rstrip()

    @staticmethod
    def read_list_strs():
        return stdin.readline().rstrip().split()

    def get_random_seed(self):
        import random
        self.random_seed = random.randint(0, 10 ** 9 + 7)
        return

    def st(self, x):
        return print(x, flush=self.flush)

    def yes(self, s=None):
        self.st("Yes" if not s else s)
        return

    def no(self, s=None):
        self.st("No" if not s else s)
        return

    def lst(self, x):
        return print(*x, flush=self.flush)

    def flatten(self, lst):
        self.st("\n".join(str(x) for x in lst))
        return

    @staticmethod
    def max(a, b):
        return a if a > b else b

    @staticmethod
    def min(a, b):
        return a if a < b else b

    @staticmethod
    def ceil(a, b):
        return a // b + int(a % b != 0)

    @staticmethod
    def accumulate(nums):
        n = len(nums)
        pre = [0] * (n + 1)
        for i in range(n):
            pre[i + 1] = pre[i] + nums[i]
        return pre

class Solution:
    def __init__(self):
        return

    @staticmethod
    def main(ac=FastIO()):
        """
        url: url of the problem
        tag: algorithm tag
        """
        for _ in range(ac.read_int()):
            n = ac.read_int()
            dct = [[] for _ in range(n)]
            for _ in range(n - 1):
                i, j = ac.read_list_ints_minus_one()
                dct[i].append(j)
                dct[j].append(i)
            u, v = ac.read_list_ints_minus_one()
            stack = [0]
            parent = [-1] * n
            depth = [-1] * n
            while stack:
                x = stack.pop()
                if x >= 0:
                    stack.append(~x)
                    for y in dct[x]:
                        if y != parent[x]:
                            parent[y] = x
                            stack.append(y)
                else:
                    x = ~x
                    for y in dct[x]:
                        if y != parent[x]:
                            depth[x] = max(depth[x], depth[y] + 1)
            lst = [u]
            while lst[-1]:
                lst.append(parent[lst[-1]])
            pre = set(lst)
            m = len(lst)
            alice = cnt = 0
            for x in lst[:m // 2 + m % 2]:
                cur = [1] + [depth[y] + 1 for y in dct[x] if y != parent[x] and y not in pre]
                alice = max(alice, max(cur) + cnt)
                cnt += 1

            bob = cnt = 0
            for x in lst[m // 2 + m % 2:][::-1]:
                cur = [1] + [depth[y] + 1 for y in dct[x] if y != parent[x]]
                bob = max(bob, max(cur) + cnt)
                cnt += 1
            if alice > bob:
                ac.st("Alice")
            else:
                ac.st("Bob")
        return

Solution().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:
```plaintext
1
5
1 2
2 3
3 4
4 5
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game in Tree (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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

def bfs_distance(n, edges):
    graph = [[] for _ in range(n+1)]
    for a, b in edges:
        graph[a].append(b)
        graph[b].append(a)

    dist = [-1] * (n + 1)
    dist[1] = 0
    q = deque([1])

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

    return dist

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

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

        dist = bfs_distance(n, edges)

        if dist[u] % 2 == 0:
            results.append("Alice")
        else:
            results.append("Bob")

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

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 2
4
1 2
2 3
3 4
4 4
5
1 2
2 3
3 4
4 5
5 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 (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, $\mathbf{u = v}$. 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 $v$ ($2 \le u, v \le n$, $\mathbf{u = v}$).

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

Note: ![](CDN_BASE_URL/6df79b98ba203b10924734a14ee457df) Tree from the first and second examples.

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

In the second test case, the path will be ($3,3$). 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.ArrayList;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Main {
    static ArrayList<Integer>[]e;
    static boolean[]vis;
    static int[]dis;
    private static void solve(Kattio sc) {
        int n = sc.nextInt();
        e=new ArrayList[n+1];
        vis=new boolean[n+1];
        dis=new int[n+1];
        for (int i = 1; i <=n; i++) {
            e[i]=new ArrayList<>();
        }
        for (int i = 1; i < n; i++) {
            int u=sc.nextInt();
            int v=sc.nextInt();
            e[u].add(v);
            e[v].add(u);
        }
        int u=sc.nextInt();
        int v=sc.nextInt();
        LinkedList<Integer>queue=new LinkedList<>();
        queue.offer(1);
        while (!queue.isEmpty()){
            int x=queue.poll();
            if (vis[x])continue;
            vis[x]=true;
            for (Integer V : e[x]) {
                dis[V]=dis[x]+1;
                queue.offer(V);
            }
        }
        if (dis[u]%2==0) System.out.println("Alice");
        else System.out.println("Bob");

    }

    public static void main(String[] args) {
        Kattio sc = new Kattio();
        int t = 1;
        t = sc.nextInt();
        while (t-- > 0) solve(sc);
        sc.close();
    }

    static class Kattio extends PrintWriter {
        private BufferedReader r;
        private StringTokenizer st;

        public Kattio() {
            this(System.in, System.out);
        }

        public Kattio(InputStream i, OutputStream o) {
            super(o);
            r = new BufferedReader(new InputStreamReader(i));
        }

        public Kattio(String intput, String output) throws IOException {
            super(output);
            r = new BufferedReader(new FileReader(intput));
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(r.readLine());
                return st.nextToken();
            } catch (Exception e) {
            }
            return null;
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios where the code might fail. The key points to consider are:

1. **Tree Structure**: The tree should be structured in a way that the path from \( u \) to \( v \) (where \( u = v \)) does not pass through vertex 1, and the game dynamics should be interesting.
2. **Path from \( u \) to \( v \)**: Since \( u = v \), the path should be a single vertex.
3. **Game Dynamics**: The game should be such that the result is not trivially determined by the distance from vertex 1 to \( u \).

Given these considerations, let's construct a small-scale test input:

1. **Test Case 1**: A simple tree with 3 vertices where the path from \( u \) to \( v \) does not pass through vertex 1.
2. **Test Case 2**: A more complex tree with 4 vertices to ensure the code handles more complex structures correctly.

Here is the test input:

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

### Explanation:
1. **Test Case 1**:
   - Tree: 1 - 2 - 3
   - Path from \( u \) to \( v \): 2 (since \( u = v = 2 \))
   - Expected Output: "Bob"
   - Reason: Alice starts at vertex 1, and Bob starts at vertex 2. Alice cannot move anywhere on her first turn, so Bob wins.

2. **Test Case 2**:
   - Tree: 1 - 2 - 3 - 4
   - Path from \( u \) to \( v \): 3 (since \( u = v = 3 \))
   - Expected Output: "Alice"
   - Reason: Alice starts at vertex 1, and Bob starts at vertex 3. Alice can move to vertex 2, and Bob cannot move anywhere, so Alice wins.

This test input should help expose the bug in the given code by testing the game dynamics and the path conditions.