**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
def sol(L):

    L.sort()
    M=[L[0]]
    for i in L:
        if i!=M[-1]:
            M+=[i]

    index=-1
    for i in range(len(M)-1):
        if M[i+1]-M[i]==1:
            index=i
            break

    if index==-1:

        if M[0]==1:
            print('Bob')
        else:
            print('Alice')

    else:

        if M[0]!=1:
            print('Alice')

        else:
            ind=len(M)-1
            for i in range(index+1,len(M)-1):
                if M[i]-M[i-1]>1:
                    ind=i
                    break
            if ind%2==0:
                print('Alice')
            else:
                print('Bob')

t=int(input())
for _ in range(t):
    n=int(input())
    L=list(map(int,input().split()))
    if t==10000 and _==52:
        print(L)
    sol(L)

**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 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    a = list(map(int,input().split()))
    a = set(a)
    a = list(a)
    a.sort()
    n = len(a)
    tem = 0
    if a[0] == 1:
        tem+=1
        for i in range(1,n-1):
            if a[i] == a[0]+i:
                tem+=1
            else:
                break
        if tem%2 ==1 :
            print("Bob")
        else:
            print("Alice")
    else:
        now = 0
        for i in range(n-1):
            if a[i] == a[i+1]-1:
                for j in range(i,n-1):
                    if a[j] == a[i]+j-i:
                        now+=1
                    else:
                        break
                break
        else:
            print("Alice")
            continue
        if now%2 ==1:
            print("Bob")
        else:
            print("Alice")

**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
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
import java.util.*;
import java.util.stream.Collectors;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        Integer testCase = scanner.nextInt();
        while (testCase-->0) {
            Integer n = scanner.nextInt();
            ArrayList<Integer> arrayList = new ArrayList<>(n);
            for(int i = 0; i<n; i++){
                Integer number = scanner.nextInt();
                arrayList.add(number);
            }
            arrayList = (ArrayList<Integer>) arrayList.stream().distinct().sorted().collect(Collectors.toList());

            Integer count = 0;
            for(int i = 0; i<arrayList.size()-1; i++){
                if(i == 0){
                    if(arrayList.get(i) == 1){
                        count++;
                    }
                } else{
                    if(arrayList.get(i) - arrayList.get(i-1) == 1){
                        count ++;
                    }
                }

            }
            if(count%2 == 0 || arrayList.get(0) != 1){
                System.out.println("Alice");
            } else {
                System.out.println("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 edge cases that can highlight the issues in the logic. The code seems to have a flawed approach to determining the winner by counting the number of consecutive differences of 1 in the sorted and distinct list of pile sizes. This approach does not correctly account for the optimal play strategy in the game of Nim.

### Key Points to Consider:
1. **Piles with the same size**: The code should handle cases where multiple piles have the same size.
2. **Piles with small differences**: The code should correctly handle cases where the differences between pile sizes are small but not necessarily 1.
3. **Piles with large differences**: The code should handle cases where the differences between pile sizes are large.

### Test Input:
We will create a test input that includes:
- Piles with the same size.
- Piles with small differences.
- Piles with large differences.

```plaintext
3
3
1 1 1
4
1 2 3 4
2
1 1000000000
```

### Explanation:
1. **First Test Case:**
   - Piles: `1 1 1`
   - The code should recognize that all piles have the same size and that Alice can win by removing 1 stone from each pile on her first turn.
   - Expected Output: "Alice"

2. **Second Test Case:**
   - Piles: `1 2 3 4`
   - The code should recognize that the differences between consecutive piles are 1, but this does not necessarily mean Bob will win.
   - Expected Output: "Alice"

3. **Third Test Case:**
   - Piles: `1 1000000000`
   - The code should recognize that the difference between the two piles is very large, and Alice can win by removing 1 stone from each pile on her first turn.
   - Expected Output: "Alice"

This test input should help expose the bug in the provided code by challenging its logic in handling various pile sizes and differences.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
import static java.lang.Math.*;
import java.util.*;
import java.util.jar.JarEntry;
import java.io.*;

public class Main
{
    // static int dfs(ArrayList<Integer>[] tree, int node, boolean[] visited,int[] dp) {
    //     visited[node] = true;
    //     if(node!=0&&tree[node].size()==1){
    //         return dp[node] = 1;
    //     }
    //     for (int neighbor : tree[node]) {
    //         if (!visited[neighbor]) {
    //             dp[node] += dfs(tree, neighbor, visited,dp);
    //         }
    //     }
    //     return dp[node];
    // }
    // static int dfs_height(ArrayList<Integer>[] tree, int node, boolean[] visited) {
    //     visited[node] = true;
    //     if(node!=0&&tree[node].size()==1){
    //         return 1;
    //     }
    //     int a = 1;
    //     for (int neighbor : tree[node]) {
    //         if (!visited[neighbor]) {
    //              a = Math.max(a,dfs_height(tree, neighbor, visited));
    //         }
    //     }
    //     return a+1;
    // }
    // public static int maxLengthSubarray(ArrayList<Integer> nums, int k) {
    //     int maxLength = 0;
    //     int currentSum = 0;
    //     int left = 0;

    //     for (int right = 0; right < nums.size(); right++) {
    //         currentSum += nums.get(right);

    //         while (currentSum > k && left <= right) {
    //             currentSum -= nums.get(left);
    //             left++;
    //         }

    //         if (currentSum <= k) {
    //             maxLength = Math.max(maxLength, right - left + 1);
    //         }
    //     }

    //     return maxLength;
    // }
    // public static boolean sameCharCount(String str1, String str2) {
    //     // Convert strings to maps of character counts
    //     Map<Character, Integer> count1 = new HashMap<>();
    //     Map<Character, Integer> count2 = new HashMap<>();

    //     for (char c : str1.toCharArray()) {
    //         count1.put(c, count1.getOrDefault(c, 0) + 1);
    //     }
    //     for (char c : str2.toCharArray()) {
    //         count2.put(c, count2.getOrDefault(c, 0) + 1);
    //     }

    //     // Check if the maps are equal
    //     return count1.equals(count2);
    // }

    // public static int highestonebit(int num,int size){
    //     int ans =0 ;
    //     for(int i=1;i<=size;i++){

    //         if(((num) & 1)==1){
    //             ans=i;
    //         }
    //         num = num>>1;
    //     }
    //     return ans;
    // }
    // public static int highestzerobit(int num,int size){
    //     int ans =0 ;
    //     for(int i=1;i<=size;i++){

    //         if(((num) & 1)==0){
    //             ans=i;
    //         }
    //         num = num>>1;
    //     }
    //     return ans;
    // }

    // public static int findIndexJustAboveOrEqual(ArrayList<ArrayList<Integer>> listOfLists, int targetValue) {
    //     int low = 0;
    //     int high = listOfLists.size() - 1;
    //     int result = -1;

    //     while (low <= high) {
    //         int mid = low + (high - low) / 2;
    //         int valueAtIndexZero = listOfLists.get(mid).get(0);

    //         if (valueAtIndexZero == targetValue) {
    //             return mid;
    //         } else if (valueAtIndexZero < targetValue) {
    //             low = mid + 1;
    //         } else {
    //             result = mid;
    //             high = mid - 1;
    //         }
    //     }
    //     return result;
    // }

    public static void main(String[] args) throws Exception
    {

        Scanner scanner = new Scanner(System.in);
        int t =scanner.nextInt();
        while(t-->0){
            int n=scanner.nextInt();
            int a[] =new int[n];
            int t1= 0;
            for(int i=0;i<n;i++){
                a[i]=scanner.nextInt();
                if(a[i]==1){
                    t1++;
                }
            }
            if(t1%2==0){
                if(n%4==0){
                    System.out.println("Alice");
                }else if(n%4==1){
                    System.out.println("Alice");
                }
                else if(n%4==2){
                    System.out.println("Alice");
                }
                else if(n%4==3){
                    System.out.println("Bob");
                }
            }else{
                if(n%4==0){
                    System.out.println("Bob");
                }else if(n%4==1){
                    System.out.println("Bob");
                }
                else if(n%4==2){
                    System.out.println("Bob");
                }
                else if(n%4==3){
                    System.out.println("Alice");
                }
            }

        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can highlight the incorrect logic. The provided code seems to have issues with how it handles the number of piles and the presence of piles with a single stone. Here are the steps to create a test input:

1. **Identify the Bug**: The code incorrectly handles the number of piles and the presence of piles with a single stone. It uses a modulo operation to determine the winner, which can be misleading in certain edge cases.

2. **Edge Cases**:
   - **Single Pile with a Large Number of Stones**: This should be a straightforward case where Alice wins.
   - **Multiple Piles with a Single Stone**: This should be a case where the number of piles with a single stone is odd or even, which can influence the outcome.
   - **Multiple Piles with Different Numbers of Stones**: This should be a case where the number of piles and the distribution of stones can affect the outcome.

3. **Generate the Test Input**:
   - A single pile with a large number of stones.
   - Multiple piles with a single stone, where the number of such piles is odd.
   - Multiple piles with different numbers of stones, where the number of piles is a multiple of 4.

Here is a small-scale test input that covers these edge cases:

```plaintext
3
1
1000000000
3
1 1 1
4
1 2 3 4
```

Explanation:
1. **Test Case 1**: A single pile with 1,000,000,000 stones. Alice wins because she can remove all stones in one move.
2. **Test Case 2**: Three piles, each with 1 stone. Bob wins because Alice must remove 1 stone from each pile, and Bob can then remove the remaining stones.
3. **Test Case 3**: Four piles with different numbers of stones. Alice wins because she can remove 1 stone from each pile, and the remaining configuration will still allow her to win if both play optimally.

This test input should expose the bug in the provided code by challenging its incorrect handling of the number of piles and the presence of piles with a single stone.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
def play(stones, i, debt, dp):
    if i >= len(stones):
        return False, True
    diff = stones[i] - debt
    state = 0 if diff == 1 else 1
    if dp[i][state] != None:
        return dp[i][state]
    if diff == 1:
        notTurnWins, turnWins = play(stones, i+1, debt+1, dp)
        dp[i][state] = [turnWins, notTurnWins]
    else:
        notTurnWinsA, turnWinsA = play(stones, i, debt+diff-1, dp)
        notTurnWinsB, turnWinsB = play(stones, i+1, debt+diff, dp)
        dp[i][state] = [turnWinsA, notTurnWinsA] if turnWinsA else [turnWinsB, notTurnWinsB]
    return dp[i][state]

def stonesGame(stones):
    stones = [int(n) for n in set(stones)]
    stones.sort()
    dp = [[None, None] for i in range(len(stones))]
    return "Alice" if play(stones, 0, 0, dp)[0] else "Bob"

cases = int(input())
results = []
for i in range(cases):
    input()
    res = stonesGame(input().split())
    results.append(res)
print("\n".join(results))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can highlight the incorrect logic. The buggy code has issues with how it handles the state transitions and the memoization. Here are the steps to create a suitable test input:

1. **Identify the Bug**: The bug in the code seems to be related to how the state transitions are handled, particularly in the recursive calls and the memoization.

2. **Edge Cases**: Consider scenarios where the piles have very specific configurations that could lead to incorrect state transitions. For example, piles with very similar or identical values, or piles with a small number of stones.

3. **Small-Scale Input**: Since the output is limited, we should create a small-scale input that is still challenging enough to expose the bug.

Given these considerations, let's create a test input that includes:
- A test case with a small number of piles where the values are close to each other.
- A test case with a single pile.
- A test case with piles that have a mix of small and large values.

Here is the test input:

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

### Explanation:
1. **Test Case 1**: `3 2 2 2`
   - Alice can win by choosing `k=2` on her first turn, which will leave all piles with 0 stones, and Bob will have no move.
   - This tests the scenario where all piles have the same value.

2. **Test Case 2**: `1 1`
   - Alice must choose `k=1` on her first turn, which will leave the pile with 0 stones, and Bob will have no move.
   - This tests the scenario with a single pile.

3. **Test Case 3**: `4 1 2 3 4`
   - Alice can win by choosing `k=1` on her first turn, which will leave the piles as `0 1 2 3`. Bob will then have to choose `k=1` again, leaving `0 0 1 2`. Alice can then choose `k=1` again, leaving `0 0 0 1`. Bob will have no move left.
   - This tests the scenario with a mix of small and large values.

This test input should expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
# -*- coding: utf-8 -*-
"""
Created on Sun May  5 15:39:58 2024

@author: dell
"""

t=int(input())
for i in range(t):
    n=int(input())
    l=list(map(int,input().split()))
    l1=[]
    for i in l:
        if i not in l1:
            l1.append(int(i))
    l1.sort()
    if len(l1)==1 or l1[0]!=1:
        print("Alice")
    else:
        x=1
        for i in range(len(l1)-1):
            if l1[i]+1==l1[i+1]:
                x+=1
            else:
                break
        if x==len(l1):
            if x%2==0:
                print("Bob")
            else:
                print("Alice")
        else:
            if x%2==0:
                print("Alice")
            else:
                print("Bob")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = true;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static List<Integer> d;
    static void solve(PrintWriter o) {
        try {
            int n = nextInt();
            Integer[] a = new Integer[n+1];
            a[0] = 0;
            for(int i=1;i<=n;i++) a[i] = nextInt();
            Arrays.sort(a);
            d = new ArrayList<>();
            for(int i=1;i<=n;i++) {
                int diff = a[i]-a[i-1];
                if(diff == 0) continue;
                else if(diff == 1) d.add(1);
                else d.add(2);
            }
            boolean ok = win(0, 2);
            if(ok) o.println("Alice");
            else o.println("Bob");
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    static boolean win(int i, int s) {
        boolean ok = false;
        if(i >= d.size()) return ok;
        if(s == 1 || d.get(i) == 1) ok = !win(i+1, 2);
        else {
            ok |= !win(i+1, 2);
            ok |= !win(i, 1);
        }
        return ok;
    }
    public static int upper_bound(List<Integer> a, int val){
        int l = 0, r = a.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a.get(mid) <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(List<Integer> a, int val){
        int l = 0, r = a.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a.get(mid) < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

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

import java.lang.Math.*;

import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.abs;

//Cses-N

public class Quickie {
    static FastReader sc;
    static PrintWriter out;
    static final String yes = "YES";
    static final String no = "NO";
    static final long mod = (long) (1e9+7);
    static final long modi = 998244353;

//    private record Pair(long first, long second){
//
//    }

//    private record Pair(int first, int second, int third, int forth){
//
//    }

    static class Pair{
        int first;
        int second;
        //int third;

        Pair(int first, int second){
            this.first=first;
            this.second=second;
            //this.third=third;

        }

    }

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

        return gcd(b,a%b);

    }

    static void primeFact(int N, HashMap<Integer,Integer> hm){

        for (int i = 2 ; i*i <= N ; i++) {
            if (N%i==0){
                int cnt = 0 ;
                while (N%i==0){
                    cnt++;
                    N = N/i;
                    hm.put(i,hm.getOrDefault(i,0)+1);
                }
                //print(i+"^"+cnt);
            }
        }
        if (N>1) {
            hm.put(N,hm.getOrDefault(N,0)+1);
            //println(N+"^"+1);
        }
        //return  hm;
    }

//    static  int dfs(int[][] mat , int k){
//
//        if (mat[k][0]==0 && mat[k][1]==0) return 0;
//        if (k==0) return 10000000;
//
//        if (mat[k][2]==0){
//            return Math.min(1 + dfs(mat,mat[k][0]),1+dfs(mat,mat[k][1]));
//        }
//        else if (mat[k][2]==-1){
//            return Math.min(dfs(mat,mat[k][0]),1+dfs(mat,mat[k][1]));
//        }
//
//        else if (mat[k][2]==1){
//            return Math.min(1 + dfs(mat,mat[k][0]),dfs(mat,mat[k][1]));
//        }
//
//
//
//
//        return 0;
//    }

    static int mex(int[] arri, int N)
    {

        // sort the array
        int[] arr = Arrays.copyOf(arri,N) ;
        Arrays.sort(arr);

        int mex = 0;
        for (int idx = 0; idx < N; idx++) {
            if (arr[idx] == mex) {
                // Increment mex
                mex += 1;
            }
        }

        // Return mex as answer
        return mex;
    }

    static int[] prexor(int[]arr){
        int[] ans = new int[arr.length+1];
        ans[0] = 0;
        ans[1] = arr[0];
        for (int i = 2; i < arr.length+1; i++) {
            ans[i] = arr[i-1]^ans[i-1];
        }

        return ans;
    }

    static long dfs(ArrayList<ArrayList<Integer>> al , int i , long[]arr , long[] ans){
        //if (vis[i]) return ans[i];
        if (al.get(i).size()==0) return ans[i] = arr[i];
        long sum = 0;

        for (int it : al.get(i)) {
            sum +=  dfs(al, it , arr , ans);
        }
        return ans[i]= min(arr[i],sum);
    }

    static void solve(int itt) throws IOException {  //extra

        int n = sc.nextInt();
//        int[] arr = sc.readIntArray(n);
//        ruffleSort(arr);

        //TreeSet<Integer> ts = new TreeSet<>();
        ArrayList<Integer> al = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            //ts.add(arr[i]);
            int x = sc.nextInt();
            if(!al.contains(x)) al.add(x);
        }
        Collections.sort(al);

        int chk = 0 ;
        //int ind = 0 ;
        //int last = arr[n];

        for (int it : al) {
            if (it==chk+1){
                chk++;
            }
            else break;
        }

        if (chk==al.size()){
            if (chk%2==0) println("Bob");
            else println("Alice");
        }
        else {
            if (chk%2==0) println("Alice");
            else println("Bob");
        }

    }

    static long pow(long a, long b, long mod) {
        if (b == 0)
            return 1L;

        long val = pow(a, b / 2, mod);

        if (b % 2 == 0)
            return val * val % mod;
        else
            return val * val % mod * a % mod;
    }

    static int dp(int n , int m ){
        if (n==m) return 0;

        int max = max(n,m);
        int min = min(n,m);

        return 1+dp(max-min,min);
    }

    /*   short notes
            // PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
            // Q.DIFFERENT DIVISORS
             / TWO CASES:
             / 1,P,Q,A WHERE P AND Q PRIMES AND P >= 1+D AND Q >=P+D AND A=P*Q; (THOUGHT MYSELF)(BUILT LOGIC MYSELF)
             /     OR
             / 1,P,P^2,A WHERE P >=1+D AND A=P^3; (REALIZED AFTER READING EDI)(Although not required in tc's)
             / SO FIND MIN(P*Q,P^3);
            // Q.TO Become Max: start = mid+1 not start++ : end = mid-1 , not end--
            // gcd() just like min/max like maxi = arr[0] , forloop(maxi = Math.max(maxi,arr[i));
            //'0' is 48
            // use queues instead of two pointers in C. Pair Programming rating:1100
            // In 1574B - Combinatorics Homework (rated:1100), instead of trying to find possible combinations(which is difficult), just check if the given m falls
             / in the range of the possible m's that can be accepted with the given a's, b's and c's. lb: c-(a+b+1)<=m ;; ub: m<=(a+b+c-3).
            // In B. Kalindrome Array (rated:1100), start pointers from left and from right simultaneously, check the first time there is an inequality,
             / then check one by one if deleting all the no.==arr[i] or no.==arr[n-1-i] results in palindrome (then break,else tle), since deleting all the same no.s in palindrome still gives palindrome.
            // 1843D - Apple Tree :
             / 1st error : algo done without dfs (it only counted number of child of a node rather than specifying which child), mp caused wrong ans in tc-3
             / 2nd error : algo done using dfs, but it assumed the shorted node to be the parent node, mp caused wrong ans in tc-4 but not sure
             / 3rd error : answers overflowed int datatype else it was correct;
            // 1520D - Same Differences : Here res is long (all others are int) since length of array can be 10^5 and if all numbers are having same difference as their indices
             / (such as 1,2,3,4,5...), then no of pairs will be 10^5C2 which is greater than int type.
            // 1832C - Contrast Value : The "NOT EQUAL SWITCH (uses 0,1,-1)", use not equal to turn on equal to turn off, used to count/segregate ascending and descending subarrays in an array,
             / very simple and saved a lot from complex algorithms that were anyway giving error/tle.
            // {au−av ≥ bu−bv --> au-bu >= av-bv}. In D. Strong Vertices. While I got the trick, couldn't get the maths/logic behind it. Forgot that I did a similar question back then, shouldn't forget next time.
            // cnt += hp.getOrDefault(res-arr[i],0); and hp.put(arr[i],hp.getOrDefault(arr[i],0)+1); to count no. of pairs having sum = res
            // D - Sorting By Multiplication : if (i+1<n){if (arr[i]>=arr[i+1]) cnt1--;//[distinguisher line]} and if (i>0){if (arr[i-1]<=arr[i]) cnt2++;} : used to divide an array into two subarrays
            // If in an array question , you can't think of anything to do with elements, think if you can do something involving their indices
            // CF906,Div-2,C : Use StringBuilder.insert(index,string); . Don't fear going the brute/long code way if nothing hits. It may be the correct and more optimal than it seems.
             / when the string which is being traversed changes everytime, the iterator j corrects itself each time since it doesn't depend on original string but iteration numbers.
             / For eg in str = "11" str[0] = (first)1 but int (new) str = "0111", str[1] = (still first)1.
            // Codeton round 6 C: array 'h' stores the latest position/index of the elements in the array 'a'.pointers-like address/indexes storing array 'nxt'.
            // Traversing the hashmap hm: for (Map.Entry<Integer, Integer> entry : hm.entrySet()) {}. Use entry.getKey() or entry.getValue().
            // Given any integer x, we can see that its lowest significant bit is a divisor of x
            // CF915,Div 2,C. Largest Subsequence : Recheck-pop-last method to beat TLE. Brings nested for-loop to single for-loop since popping is O(N).
             / Use it when you want to have next best things in an array.
            // Pinely Round 3 (Div. 1 + Div. 2) Q.C Heavy Intervals: Use mountain-plateau comparison minimizing (mountain)
     */

    static final Random random = new Random();

    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);
    }

    public static StringBuilder insertString(String string1, String string2, int index) {
        if (index == -1)
            return new StringBuilder(string2 + string1);

        StringBuilder sb = new StringBuilder(string1);
        sb.insert(index + 1, string2);
        return sb;
    }

    public static void main(String[] args) throws IOException {
        sc = new FastReader();
        out = new PrintWriter(System.out);
        int t = sc.nextInt();
        //int t = 1;
        int itt = 1; //extra
        while (t-- > 0) {

            solve(itt); //extra
            itt++; //extra
        }
        out.close();
    }

    static <E> void print(E[] a) {
        out.println(Arrays.toString(a));
    }

    static <E> void print(E res) {
        out.print(res);
    }

    static <E> void println(E res) {
        out.println(res);
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] readIntArray(int n) {
            int[] res = new int[n];
            for (int i = 0; i < n; i++)
                res[i] = nextInt();
            return res;
        }

        long[] readLongArray(int n) {
            long[] res = new long[n];
            for (int i = 0; i < n; i++)
                res[i] = nextLong();
            return res;
        }
    }

}

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

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

Think step by step.

Response:
```plaintext
3
1
1
2
1 1
3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

#define int long long
#define BANKAI ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

signed main() {
    BANKAI;
    int t; cin >> t;
    for(int f=0; f<t; f++){
        int n; cin >> n; // piles
        vector<int> a(n+1);
        int maxi = 0; // largest pile size
        for(int i=0; i<n; i++) {
            cin >> a[i];
            maxi = max(maxi, a[i]);
        }
        vector<bool> vec(maxi+2, false);
        for(int i=0; i<n; i++){
            vec[a[i]] = true;
        }
        // sort(a.begin(), a.end());
        int b; // b minda smallest positive integer isnt pile size isnt 0;
        for(int i=1; i<maxi+2; i++){ // yvela zomis unda mosinjos ra tu 1 2 3 gvaq maxi = 3 da b = 4
            bool tmp = vec[i];
            // for(int j=0; j<a.size(); j++){
            //     if(i == a[j]){
            //         tmp = true;
            //         break;
            //     }
            // }
            if(!tmp) {b = i; break;}
            // auto it = find(a.begin(), a.end(), i);
            // if(it == a.end()) { b = i; break; }
        }
        // if(b == -1) b = maxi + 1;

        bool first = true;

        if(b > maxi){
            if(maxi % 2 == 0) first = false;
        }else{
            if(b % 2 == 0) first = false;
        }

        if(first) cout << "Alice" << endl;
        else cout << "Bob" << endl;
    }
    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
1
1
2
1 1
3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Everything Nim
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game on $$$n$$$ piles of stones. On each player's turn, they select a positive integer $$$k$$$ that is at most the size of the smallest nonempty pile and remove $$$k$$$ stones from each nonempty pile at once. The first player who is unable to make a move (because all piles are empty) loses.

Given that Alice goes first, who will win the game if both players play optimally?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of piles in the game.

The next line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the initial number of stones in the $$$i$$$-th pile.

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

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print "Alice", otherwise print "Bob" (without quotes).

Examples:
input:
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
output:
Alice
Bob
Alice
Alice
Bob
Alice
Alice

Note: In the first test case, Alice can win by choosing $$$k=3$$$ on her first turn, which will empty all of the piles at once.

In the second test case, Alice must choose $$$k=1$$$ on her first turn since there is a pile of size $$$1$$$, so Bob can win on the next turn by choosing $$$k=6$$$.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define fori(n) for(int i=0; i<n; i++)
#define testcases int t; cin>>t; while (t--)
#define ld long double
#define vl vector<long long>
#define fast ios::sync_with_stdio(0); cin.tie(0);
#define endl '\n'

using namespace std;

int main()
{
    fast
    testcases {
        int n;
        cin>>n;
        int a[n];
        fori(n) {
            cin>>a[i];
        }
        sort(a, a+n);
        if (a[0]!=1) {
            cout<<"Alice"<<endl;
            continue;
        }
        bool check=true;
        for (int i=1; i<n; i++) {
            if (a[i]==a[i-1]) {
                continue;
            }
            if (a[i]!=a[i-1]+1) {
                check=false;
                if (a[i-1]%2==1) {
                    cout<<"Bob"<<endl;
                }
                else {
                    cout<<"Alice"<<endl;
                }
                break;
            }
        }
        if (check) {
            if (n%2==1) {
                cout<<"Alice"<<endl;
            }
            else {
                cout<<"Bob"<<endl;
            }
        }
    }
}

**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
7
5
3 3 3 3 3
2
1 7
7
1 3 9 7 4 2 100
3
1 2 3
6
2 1 3 4 2 4
8
5 7 2 9 6 3 3 2
1
1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int t;
int n,ans;
int a[5003];
int b[5003];
int vis[5003];
void dfs(int x,int re){
    if(x==n+1){
        ans=max(ans,re);
        return;
    }
    for(int i=1;i<=n;i++){
        if(!vis[i]){
            vis[i]=1;
            dfs(x+1,min(re,abs(a[x]-b[i])));
            vis[i]=0;
        }
    }
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin>>t;
    while(t--){
        cin>>n;
        ans=0;
        for(int i=1;i<=n;i++){
            vis[i]=0;
            cin>>a[i];
        }
        for(int i=1;i<=n;i++){
            cin>>b[i];
        }
        dfs(1,0x7fffffff);
        cout<<ans<<"\n";
    }
    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
4
1 2 3 4
4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
/**
 *   Author : Mahbub_Ahmed
 *   Created: 2024-03-25 18:00:34
 **/
#include <bits/stdc++.h>
using namespace std;

#define FastIO                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);
#define en '\n'
#define YES cout << "YES" << en
#define NO cout << "NO" << en
#define yes cout << "Yes" << en
#define no cout << "No" << en
#define fill_number(x, n) setfill('0') << setw(n) << (x)
#define precision_number(x, n) fixed << setprecision(n) << (x)
template <typename T>
T lcm(T a, T b)
{
    return (a * (b / __gcd(a, b)));
}

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;

ll calc(const vector<ll> &a, const vector<ll> &b)
{
    ll mn = INT_MAX;
    for (ll i = 0; i < a.size(); i++)
    {
        mn = min(mn, abs(a[i] - b[i]));
    }
    return mn;
}

void TEST_CASES(ll testCase)
{
    ll n;
    cin >> n;
    vector<ll> a(n), b(n);
    for (ll i = 0; i < n; i++)
    {
        cin >> a[i] >> b[i];
    }
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    ll ans = 0;
    for (ll i = 0; i < n; i++)
    {
        rotate(b.begin(), b.begin() + 1, b.end());
        ans = max(ans, calc(a, b));
    }
    cout << ans << en;
}
int32_t main()
{
    FastIO;

    ll t = 1;
    cin >> t;
    for (ll test = 1; test <= t; test++)
    {
        TEST_CASES(test);
    }

    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 3
3 2 1
4
1 2 3 4
4 3 2 1
5
1 3 5 7 9
9 7 5 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
#include <iostream>
#include <bits/stdc++.h>
#define int long long int
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define all(x) x.begin(), x.end()
#define read(arr, n) for (int i = 0; i < n; i++) cin >> arr[i]
#define w(arr, n) for (int i = 0; i < n; i++) cout << arr[i] << " "
#define rep(i, a, n) for (int i = a; i < n; i++)
#define pii vector<pair<int, int>>
#define pb push_back
#define vi vector<int>
#define f first
#define sec second
using namespace std;
///////////////////////////////////////////////////////WRITE MAIN LOGIC/////////////////////////////////////////////////////////////////////////////////////////////////////////////

bool check(int mid,multiset<int>m1,multiset<int>m2,int n){
  int cnt=0;
  for(auto it=m1.rbegin();it!=m1.rend();it++){
    int val=*it;
    auto it2=m2.upper_bound(val-mid);
    if(it2!=m2.begin()){
      it2--;
      int v=*it2;
      if(abs(v-val)>=mid){
        m2.erase(m2.find(v));cnt++;
      }
    }
  }
  for(auto it=m2.rbegin();it!=m2.rend();it++){
    int val=*it;
    auto it2=m1.upper_bound(val-mid);
    if(it2!=m1.begin()){
      it2--;
      int v=*it2;
      if(abs(v-val)>=mid){
        m1.erase(m1.find(v));cnt++;
      }
    }
  }
  return cnt==n;
}
void solve(){
  int n;cin>>n;
  multiset<int>m1,m2;
  rep(i,0,n){
    int a;cin>>a;m1.insert(a);
  }
  rep(i,0,n){
    int a;cin>>a;m2.insert(a);
  }
  int ans=0;
  int l=0,r=1e9;
  while(l<=r){
    int mid=l+(r-l)/2;
    if(check(mid,m1,m2,n)){
      ans=mid;l=mid+1;
    }else r=mid-1;
  }
  cout<<ans<<endl;
}

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    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:
```plaintext
1
4
1 2 3 4
4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
s = int(input())
for i in range(s):
    n = int(input())
    a = [int(x) for x in input().split()]
    b = [int(y) for y in input().split()]
    a.sort()
    b.sort()
    c= [0]* n

    minim =0
    biggest =0
    for t in range(1, n+1):
        for k in range(1 , t+1):
            c[k-1]=abs(a[k-1] - b[k+(n-t)-1])
        for k in range(t+1 ,n+1):
            c[k-1]=abs(a[k-1]-b[k-t-1])
        sum =0
        for j in range (n):
            sum += c[j]
        if sum >= biggest:
            biggest = sum
            minim = min(c)
    print(minim)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
def read_int():
    return int(input())

def read_ints():
    return list(map(int, input().split()))

# Increase recursion limit to handle deep recursions in DFS
import sys
sys.setrecursionlimit(1 << 25)

t = read_int()
for _ in range(t):
    n = read_int()
    a = read_ints()
    b = read_ints()

    # Generate all unique charm differences
    charm_set = set()
    for ai in a:
        for bi in b:
            charm_set.add(abs(ai - bi))
    charm_values = sorted(charm_set)

    low = 0
    high = len(charm_values) - 1
    answer = 0

    while low <= high:
        mid = (low + high) // 2
        c = charm_values[mid]

        # Build the bipartite graph
        adj = [[] for _ in range(n)]
        for i in range(n):
            for j in range(n):
                if abs(a[i] - b[j]) >= c:
                    adj[i].append(j)

        # Matching arrays
        match_to = [-1] * n

        # DFS function for matching
        def bpm(u, vis):
            for v in adj[u]:
                if not vis[v]:
                    vis[v] = True
                    if match_to[v] == -1 or bpm(match_to[v], vis):
                        match_to[v] = u
                        return True
            return False

        # Try to find a perfect matching
        result = 0
        for u in range(n):
            vis = [False] * n
            if bpm(u, vis):
                result += 1
            else:
                break  # Early exit if matching is not possible

        if result == n:
            # Perfect matching found
            answer = c
            low = mid + 1
        else:
            high = mid - 1

    print(answer)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
from collections import deque;
for _ in range (int(input())):
    n = int(input()); ma = 0;
    lit = sorted(list(map(int, input().split())));
    lit2 = deque(sorted(list(map(int, input().split()))));
    for _ in range (n):
        u = lit2[0];
        lit2.popleft();
        lit2.append(u);
        z = float('inf');
        for i in range (n):
            z = min(z, abs(lit[i]-lit2[i]));
        ma = max(ma, z);
    print (ma);

**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
1
1000000000
0
2
0 1000000000
1000000000 0
4
1 2 3 4
5 6 7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

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

public class D_Main {
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	//贪心，枚举折点，证明不会^_^
	static int[] arr1=new int[5001];
	static int[] arr2=new int[5001];
	public static void main(String[] args) throws IOException {
		int t=nextInt();
		while(t-->0) {
			int n=nextInt();
			for(int i=0;i<n;i++)arr1[i]=nextInt();
			for(int i=0;i<n;i++)arr2[i]=nextInt();
			int ans=Integer.MIN_VALUE;
			for(int i=0;i<n;i++) {
				int temp=Integer.MAX_VALUE;
				for(int j=0;j<i;j++) {
					temp=Math.min(temp, Math.abs(arr1[j]-arr2[n-i+j-1]));
				}
				for(int j=i;j<n;j++) {
					temp=Math.min(temp, Math.abs(arr1[j]-arr2[j-i]));
				}
				ans=Math.max(ans, temp);
			}
			out.println(ans);
		}
		out.close();
	}

	static int nextInt() throws IOException {
		in.nextToken();
		return (int) in.nval;
	}
}

**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 and scenarios that can maximize the failure probability. Here are the steps to think through the problem:

1. **Identify the Bug**: The buggy code seems to be using a greedy approach that does not necessarily find the optimal pairing. It calculates the minimum charm for each possible pairing and then takes the maximum of these minimums. This approach might not consider all possible pairings, especially in cases where the spiciness values are not uniformly distributed.

2. **Edge Cases**:
   - **All appetizers and main dishes have the same spiciness**: This should be easy to handle, but let's check it.
   - **Appetizers and main dishes have very different spiciness values**: This can expose issues in the greedy approach.
   - **Mixed spiciness values**: This can be more challenging and might require a more sophisticated approach to find the optimal pairing.

3. **Constructing the Test Case**:
   - **Test Case 1**: All appetizers and main dishes have the same spiciness.
   - **Test Case 2**: Appetizers and main dishes have very different spiciness values.
   - **Test Case 3**: Mixed spiciness values that require careful pairing to maximize the minimum charm.

Let's construct a small-scale test input that includes these scenarios:

```plaintext
3
3
1 1 1
1 1 1
5
0 100 200 300 400
400 300 200 100 0
6
0 10 20 30 40 50
50 40 30 20 10 0
```

Explanation:
- **Test Case 1**: All appetizers and main dishes have the same spiciness. The charm for each meal will be 0, which is the minimum possible charm.
- **Test Case 2**: Appetizers and main dishes have very different spiciness values. The optimal pairing will be to pair the smallest spiciness appetizer with the largest spiciness main dish, and so on. The minimum charm will be 100.
- **Test Case 3**: Mixed spiciness values. The optimal pairing will require careful consideration to maximize the minimum charm. The minimum charm will be 10.

This test input should help expose the bug in the given buggy code.