**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
import java.util.*;
public class Main
{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-- >0){
		    int n=sc.nextInt(),i,j;
		    int[] a = new int[n];
		    int[] b = new int[n];
		    for(i=0;i<n;i++) a[i]=sc.nextInt();
		    for(i=0;i<n;i++) b[i]=sc.nextInt();
		    int c=0,c1=0;
		    for(i=0;i<n;i++){
		        if(a[i]!=b[i]) c=1;
		        if(a[i]!=b[n-i-1]) c1=1;
		        if(c==1&&c1==1) break;
		    }
		    if(c==1&&c1==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:
```plaintext
3
4
1 2 3 4
4 3 2 1
5
1 2 3 4 5
5 4 3 2 1
6
1 2 3 4 5 6
6 5 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: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
#include <iostream>
#include <vector>
#include <string>
#include <set>

using namespace std;

string solve() {
  int n;
  cin >> n;

  vector<int> a(n), b(n);

  for (int i = 0; i < n; ++i)
    cin >> a[i];

  for (int i = 0; i < n; ++i)
    cin >> b[i];

  if (n == 1)
    return a[0] != b[0] ? "Alice" : "Bob";

  set<pair<int, int>> adjacent_a, adjacent_b;

  for (int i = 0; i < n - 1; ++i) {
    adjacent_a.emplace(min(a[i], a[i + 1]), max(a[i], a[i + 1]));
    adjacent_b.emplace(min(b[i], b[i + 1]), max(b[i], b[i + 1]));
  }

  for (auto& adj : adjacent_a) {
    if (!adjacent_b.count(adj))
      return "Alice";
  }

  return "Bob";
}

int main() {
  int t;
  cin >> t;

  while (t--)
    cout << solve() << 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:
To expose the bug in the provided code, we need to focus on scenarios where the code's logic for determining the winner is flawed. The buggy code uses a set to store adjacent pairs of elements in the arrays and then checks if there are any pairs in Alice's set that are not in Bob's set. This approach is incorrect because it does not consider the optimal play strategy of both players.

### Key Observations:
1. **Optimal Play Strategy**: Both players will always choose the best move to maximize their chances of winning. The code should simulate this optimal play rather than just comparing adjacent pairs.
2. **Edge Cases**: The code might fail for cases where the arrays have a specific structure that makes the optimal play different from the adjacent pair comparison.

### Challenging Test Input:
Let's create a test case where the arrays have a structure that the buggy code will fail to handle correctly. Consider the following example:

- **n = 4**
- **a = [1, 2, 3, 4]**
- **b = [4, 3, 2, 1]**

### Explanation:
- In this case, both arrays are permutations of each other but in reverse order.
- If both players play optimally, they will always remove the same elements in each turn, leading to the same remaining element in both arrays.
- The correct output should be "Bob" because the remaining elements will be the same.

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

### Expected Output:
```plaintext
Bob
```

This test input should expose the bug in the provided code because the code will incorrectly determine the winner based on adjacent pairs rather than the optimal play strategy.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

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

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ll t;
    cin >> t;
    while (t--)
    {
        ll n;
        cin >> n;
        ll m1 = 0, m2 = 0;
        while (n--)
        {
            ll a, b;
            cin >> a >> b;
            if (a == 1 && b == 1)
            {
                if (m1 < m2)
                {
                    m1++;
                }
                else
                {
                    m2++;
                }
            }
            else if (a == 1)
            {
                m1++;
            }
            else if (b == 1)
            {
                m2++;
            }
            else if (a == -1 && b == -1)
            {
                if (m1 > m2)
                {
                    m1--;
                }
                else
                {
                    m2--;
                }
            }
        }
        cout << min(m1, m2) << '\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:
To expose the bug in the provided code, we need to focus on scenarios where the logic for handling the neutral and mixed opinions might fail. The code seems to handle the cases where both movies are liked or both are disliked, but it might not handle the cases where one movie is liked and the other is disliked, or where there are neutral opinions.

Let's consider the following edge cases:
1. **All neutral opinions**: This should not change the ratings.
2. **Mixed opinions (one liked, one disliked)**: The code might not handle these correctly, especially when the counts of positive and negative reviews are close.
3. **Uneven distribution of opinions**: This can highlight issues in how the code balances the ratings.

Here is a test input that combines these edge cases:

```plaintext
3
3
0 0 0
0 0 0
4
1 -1 1 -1
-1 1 -1 1
5
1 1 0 -1 -1
-1 -1 0 1 1
```

### Explanation:
1. **Test Case 1**:
   - All opinions are neutral.
   - Expected output: 0 (since no reviews change the rating).

2. **Test Case 2**:
   - Mixed opinions where one movie is liked and the other is disliked.
   - Expected output: 0 (since the positive and negative reviews should cancel each other out).

3. **Test Case 3**:
   - Uneven distribution of opinions.
   - Expected output: 1 (since the maximum possible rating can be achieved by balancing the positive and negative reviews).

This test input should help expose any issues in the code's logic for handling neutral and mixed opinions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<int> a;
        vector<int> b;
        for(int i=0;i<n;i++){
            int l;
            cin>>l;
            a.push_back(l);
        }
        for(int i=0;i<n;i++){
            int l;
            cin>>l;
            b.push_back(l);
        }
        int m1=0;
        int m2=0;
        for(int i=0;i<n;i++){
            if(a[i]==b[i]){
                if(a[i]==-1){
                    if(m1<m2){
                        m2--;
                    }
                    else{
                        m1--;
                    }
                }
                else if(a[i]==1){
                    if(m2<m1){
                        m2++;
                    }
                    else{
                        m1++;
                    }
                }
            }
            else{
                if(m2<m1){
                    m2+=max(a[i],b[i]);
                }
                else{
                    m1+=max(a[i],b[i]);
                }
            }
        }
        cout<<min(m1,m2)<<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
3
1 0 -1
-1 0 1
1
0
0
4
1 1 -1 -1
-1 -1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int MOD = 1e9+7;
const int N = 2e5+5;

int n, k;
int a[N], b[N];

void solve(void) {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }
    for (int i = 1; i <= n; ++i) {
        cin >> b[i];
    }
    int s1 = 0, s2 = 0;
    int numNeg = 0, numPos = 0;
    for (int i = 1; i <= n; ++i) {
        if (a[i] != b[i]) {
            if (a[i] > b[i]) s1 += a[i];
            else s2 += b[i];
        }
        else if (a[i] != 0) {
            if (a[i] < 0) ++numNeg;
            else ++numPos;
        }
    }
    if (s1 > s2) swap(s1, s2);
    if (s2 - s1 < numNeg) {
        numNeg -= (s2 - s1);
        s2 = s1;
    }
    else {
        s2 -= numNeg;
        numNeg = 0;
    }
    int x = numPos - numNeg;
    if (x >= (s2 - s1)) {
        x -= (s2 - s1);
        cout << s2 + (x / 2) << "\n";
    }
    else {
        cout << s1 + x << "\n";
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int T;
    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 generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases that might cause the code to fail. The code seems to handle most cases correctly, but it might have issues with specific scenarios involving the distribution of positive, negative, and neutral reviews.

### Key Points to Consider:
1. **Distribution of Reviews**: The code might have issues when there are a significant number of negative reviews for both movies.
2. **Equal Ratings**: The code might not handle cases where the ratings of both movies are equal correctly.
3. **Neutral Reviews**: The code might not handle neutral reviews correctly, especially when they are mixed with positive and negative reviews.

### Challenging Test Input:
Let's create a test case where the number of negative reviews is high, and the distribution of positive and negative reviews is such that the code might fail to maximize the company's rating.

```plaintext
1
6
-1 -1 -1 1 1 1
1 1 1 -1 -1 -1
```

### Explanation:
- **Test Case**: There is one test case.
- **Number of People**: 6 people.
- **Attitudes towards the First Movie**: The first three people disliked the first movie, and the last three people liked the first movie.
- **Attitudes towards the Second Movie**: The first three people liked the second movie, and the last three people disliked the second movie.

### Expected Output:
- **Maximum Possible Rating**: 0

### Reasoning:
- If we assign the first three people (who disliked the first movie) to review the second movie, and the last three people (who liked the first movie) to review the first movie, both movies will have a rating of 0.
- This is because the positive and negative reviews will cancel each other out.

### Why This Test Case is Challenging:
- The code might not handle the cancellation of positive and negative reviews correctly, especially when the number of negative reviews is high.
- The code might incorrectly assume that the ratings can be maximized by simply balancing the positive and negative reviews, without considering the neutral reviews and the specific distribution of positive and negative reviews.

This test case should expose any issues in the code's logic for handling the distribution of positive, negative, and neutral reviews.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
for _ in range(int(input())):
    n=int(input())
    L=list(map(int, input().split()))
    LL=list(map(int, input().split()))
    cp=cn=0
    for i in range(n):
        if L[i] ==-1 and LL[i]==-1:
            cn+=1
            L[i]=0
            LL[i]=0
        elif L[i]==1 and LL[i]==1:
            cp+=1
            L[i]=0
            LL[i]=0
        else :
            if L[i]==-1 : L[i]=0
            elif LL[i]==-1 : LL[i]=0
    l1=sum(L)
    l2=sum(LL)
    mini= abs(l1-l2)
    if cn+cp >= mini : print((l1+l2+cp-cn)//2)
    else: print(cn+cp)

**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 highlight the issues in the logic. The main issues in the buggy code are:

1. **Improper handling of neutral attitudes**: The code sets both `L[i]` and `LL[i]` to 0 when a person has a neutral attitude towards both movies, which is not necessary and can lead to incorrect sums.
2. **Incorrect logic for balancing the ratings**: The code attempts to balance the ratings by adjusting the counts of positive and negative reviews, but it does not always do this optimally.

Given these issues, we can create a test case that includes a mix of positive, negative, and neutral attitudes, and specifically one where the neutral attitudes are not properly handled.

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

```plaintext
1
5
0 -1 1 0 1
-1 1 0 0 1
```

Explanation:
- **Test Case 1**:
  - **n = 5**: There are 5 people.
  - **a = [0, -1, 1, 0, 1]**: Attitudes towards the first movie.
  - **b = [-1, 1, 0, 0, 1]**: Attitudes towards the second movie.

This test case includes:
- Neutral attitudes (0) for both movies.
- Positive and negative attitudes for both movies.

The expected output for this test case should be **1**, as the maximum possible rating of the company can be achieved by assigning the positive reviews and neutral reviews in a way that maximizes the minimum rating of the two movies. However, the buggy code might not handle the neutral attitudes correctly, leading to an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())

    a, b = 0, 0
    pos, neg = 0, 0
    arr = list(map(int, input().split()))
    brr = list(map(int, input().split()))

    for i in range(n):
        x, y = arr[i], brr[i]
        if x < y:
            b += y
        elif y < x:
            a += x
        # now all x == y
        elif x == -1:
            neg += 1
        elif x == 1:
            pos += 1
        elif x== 0:
            continue
        else:
            print(x, y)
            assert False

    a -= neg
    b -= neg

    pos += neg
    if a > b: a, b = b, a

    # a <= b
    diff = b - a
    if diff <= pos:
        print(a + diff)
    else:
        print(a + pos + (diff - pos) // 2)

**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 given buggy code, we need to focus on edge cases where the logic in the code might fail. The key points to consider are:

1. **Balanced Neutral and Negative Reviews**: The code handles negative reviews by subtracting them from both `a` and `b`, and then adding the neutral reviews to the positive count. We need to test scenarios where the distribution of neutral and negative reviews is such that the code's logic might not handle them correctly.

2. **Equal Ratings**: The code assumes that if `x == y`, it will handle the neutral and negative reviews in a specific way. We need to test scenarios where the ratings are equal but the distribution of positive and negative reviews is such that the code's logic might fail.

3. **Large Number of Neutral Reviews**: The code might fail if there are a large number of neutral reviews, and the distribution of positive and negative reviews is such that the final ratings are not maximized correctly.

Let's construct a test input that covers these scenarios:

```plaintext
1
5
1 0 0 -1 1
0 -1 -1 1 0
```

Explanation:
- **n = 5**: We have 5 people.
- **arr = [1, 0, 0, -1, 1]**: The first movie is liked by the 1st and 5th person, neutral for the 2nd and 3rd person, and disliked by the 4th person.
- **brr = [0, -1, -1, 1, 0]**: The second movie is neutral for the 1st and 5th person, disliked by the 2nd and 3rd person, and liked by the 4th person.

This input should expose the bug in the code because:
- The 1st and 5th person can be used to maximize the ratings.
- The 2nd and 3rd person have negative reviews for both movies, which should be handled carefully.
- The 4th person has a negative review for the first movie and a positive review for the second movie, which should be handled to maximize the ratings.

Let's see how the buggy code handles this input:

1. **Initial Values**:
   - `a = 0`, `b = 0`
   - `pos = 2`, `neg = 2`

2. **Processing the Reviews**:
   - For the 1st person: `a += 1`, `b += 0`
   - For the 2nd person: `neg += 1`
   - For the 3rd person: `neg += 1`
   - For the 4th person: `b += 1`, `a -= 1`
   - For the 5th person: `a += 1`, `b += 0`

3. **Final Values**:
   - `a = 1 - 2 + 1 = 0`
   - `b = 1`

4. **Handling Neutral and Negative Reviews**:
   - `a -= neg`: `a = 0 - 2 = -2`
   - `b -= neg`: `b = 1 - 2 = -1`
   - `pos += neg`: `pos = 2 + 2 = 4`

5. **Final Calculation**:
   - `a <= b`: `a = -2`, `b = -1`
   - `diff = b - a = -1 - (-2) = 1`
   - `diff <= pos`: `1 <= 4`
   - `a + diff = -2 + 1 = -1`

The correct answer should be `1`, but the buggy code will output `-1` due to incorrect handling of neutral and negative reviews. This test input should expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
t = int(input())

for _ in range(t):
    # substr=(input())
    # subsq=input()

    # # x,y = map(int, input().split())
    n=int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    r1=0
    r2=0
    for i in range(n):
        if a[i]<b[i]:
            r2+=b[i]
        elif a[i]>b[i]:
            r1+=a[i]
        elif r1>r2 and b[i]==-1:
            r1+a[i]
        elif r1>r2 and b[i]==0:
            continue
        elif r1>r2 and b[i]==+1:
            r2+=b[i]
        elif r1==r2:
            r1+=a[i]
        elif r1<r2 and a[i]==-1:
            r2+a[i]
        elif r1<r2 and a[i]==0:
            continue
        elif r1<r2 and a[i]==+1:
            r1+=a[i]

    a1=(min(r1,r2))

    a,b=b,a

    r1=0
    r2=0
    for i in range(n):
        if a[i]<b[i]:
            r2+=b[i]
        elif a[i]>b[i]:
            r1+=a[i]
        elif r1>r2 and b[i]==-1:
            r1+a[i]
        elif r1>r2 and b[i]==0:
            continue
        elif r1>r2 and b[i]==+1:
            r2+=b[i]
        elif r1==r2:
            r1+=a[i]
        elif r1<r2 and a[i]==-1:
            r2+a[i]
        elif r1<r2 and a[i]==0:
            continue
        elif r1<r2 and a[i]==+1:
            r1+=a[i]

    a2=(min(r1,r2))

    print(max(a1,a2))

**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 0 -1
-1 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class twoMovies {
    public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            int t = sc.nextInt();
            while(t!=0){
                t--;
                int n = sc.nextInt();
                int sum1 = 0;
                int sum2 = 0;
                int[] first = new int[n];
                int[] second = new int[n];
                ArrayList<Integer> cars = new ArrayList<Integer>();
                for (int i = 0; i < n; i++) {
                    first[i] = sc.nextInt();
                }

                for (int i = 0; i < n; i++) {
                    second[i] = sc.nextInt();
                }

                for (int i = 0; i < n; i++) {
                    if (sum1 > sum2){
                        if(second[i]>0)
                            sum2+=second[i];
                        else
                            sum1+=first[i];

                    }
                    else if(sum2>sum1){
                        if(first[i]>0)
                            sum1+=first[i];
                        else
                            sum2+=second[i];
                    }else{
                        if(first[i] == second[i]){
                            cars.add(i);
                        } else if (first[i]>second[i]) {
                            sum1+=first[i];
                        }else
                            sum2+=second[i];
                    }
                }
                for (int i = 0; i < cars.size(); i++) {
                    if(sum1>sum2){
                        if(second[i] == -1){
                            sum1+=first[i];
                        }else
                            sum2+=second[i];
                    }else {
                        if(first[i] == -1)
                            sum2+=second[i];
                        else
                            sum1+=first[i];
                    }
                }
                System.out.println(Math.min(sum1,sum2));
            }
        }
}

**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 0 -1
-1 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

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

public class Vikash_45{
	static FastScanner sc = new FastScanner();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		int t = sc.nextInt();
	    while(t-->0) {
	    	vk45();
	    	out.flush();
	    }
	    out.close();
	}

/*-----------------------------------------------------------------------------------------------------------------------------*/

	static void vk45() {
		int n = sc.nextInt();
		int a[] = sc.readArray(n);
		int b[] = sc.readArray(n);

		int ra = 0;
		int rb = 0;
		int neg = 0;
		int pos = 0;
		for(int i=0; i<n; i++) {
			if(a[i]+b[i]==2) {
				pos++;
			}else if(a[i]+b[i]==-2) {
				neg++;
			}else {
				if((a[i]==-1&&b[i]==1) || (b[i]==-1&&a[i]==1)) {
					if(a[i]==1) {
						ra++;
					}else {
						rb++;
					}
				}
			}
		}

		if(neg<=Math.abs(ra-rb)) {
			if(ra>=rb) {
				ra-=neg;
			}else {
				rb-=neg;
			}
		}else {
			neg-=Math.abs(ra-rb);
			ra = Math.min(ra, rb);
			rb = ra;
			int v = neg/2;
			int k = neg/2;
			if(neg%2!=0) {
				k++;
			}
			ra-=v;
			rb-=k;
		}

		if(pos<=Math.abs(ra-rb)) {
			if(ra<=rb) {
				ra+=pos;
			}else {
				rb+=pos;
			}
		}else {
			pos-=Math.abs(ra-rb);
			ra = Math.max(ra, rb);
			rb = ra;
			int v = pos/2;
			int k = pos/2;
			if(pos%2!=0) {
				k++;
			}
			ra+=v;
			rb+=k;
		}

		out.println(Math.min(ra, rb));

	}

	static class pair implements Comparable<pair>{
		int a;
		int b;
		public pair(int a,int b) {
			this.a = a;
			this.b = b;
		}
		@Override
		public int compareTo(pair p) {
			if(this.a==p.a) {
				return this.b-p.b;
			}
			return this.a-p.a;
		}
	}

/*-----------------------------------------------------------------------------------------------------------------------------*/

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
		long[] readArrayLong(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}
		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 expose the bug in the provided code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Let's break down the problem and think about the potential issues:

1. **Handling of Neutral Opinions:**
   - The code does not handle neutral opinions (`0`) correctly. It only considers the sum of attitudes for each person, which might lead to incorrect counts.

2. **Handling of Mixed Opinions:**
   - The code has logic to handle mixed opinions (one movie liked and the other disliked), but it might not handle all such cases correctly.

3. **Balancing Positive and Negative Reviews:**
   - The code attempts to balance positive and negative reviews, but it might not do so optimally in all cases.

Let's construct a small-scale test input that can expose these issues:

### Test Input
```plaintext
1
5
1 0 -1 1 -1
-1 0 1 -1 1
```

### Explanation
- **Test Case:**
  - `n = 5`
  - `a = [1, 0, -1, 1, -1]`
  - `b = [-1, 0, 1, -1, 1]`

### Analysis
- **Person 1:**
  - Liked the first movie (`1`)
  - Disliked the second movie (`-1`)
  - This person can contribute a positive review to the first movie and a negative review to the second movie.

- **Person 2:**
  - Neutral towards both movies (`0`)
  - This person does not contribute to the rating of either movie.

- **Person 3:**
  - Disliked the first movie (`-1`)
  - Liked the second movie (`1`)
  - This person can contribute a negative review to the first movie and a positive review to the second movie.

- **Person 4:**
  - Liked the first movie (`1`)
  - Disliked the second movie (`-1`)
  - This person can contribute a positive review to the first movie and a negative review to the second movie.

- **Person 5:**
  - Disliked the first movie (`-1`)
  - Liked the second movie (`1`)
  - This person can contribute a negative review to the first movie and a positive review to the second movie.

### Expected Output
- **Maximum Possible Rating:**
  - By assigning the reviews optimally, we can achieve a company rating of `0`.

### Why This Input is Challenging
- **Neutral Opinions:**
  - The neutral opinion (Person 2) does not affect the ratings, which tests the code's handling of neutral cases.

- **Mixed Opinions:**
  - There are multiple people with mixed opinions (Person 1, Person 3, Person 4, Person 5), which tests the code's ability to balance the ratings of the two movies.

- **Balancing Positive and Negative Reviews:**
  - The input requires the code to balance the positive and negative reviews to maximize the company's rating, which is a key part of the problem.

This test input should help expose any issues in the code's handling of neutral opinions and balancing of mixed opinions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
#include <bits/stdc++.h>

using namespace std;

int main() {
    int t;
    cin >> t; // Number of test cases

    while (t--) {
        int n;
        cin >> n; // Number of viewers

        vector<int> a(n), b(n);

        // Read preferences for the first movie
        int maxa=INT_MIN;
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
            maxa=(maxa,a[i]);

        }

        // Read preferences for the second movie
        int maxb=INT_MIN;
        for (int i = 0; i < n; ++i) {
            cin >> b[i];
            maxb=max(maxb,b[i]);
        }

        // Calculate the maximum possible rating
        int ans=0;
        if((maxa==1&&maxb==1)||(maxa==1&&maxb==0)||(maxa==0&&maxb==1))ans=1;
        if(maxa==-1&&maxb==-1)ans=-1;

        // Output the result for this test case
        cout << ans << 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
3
1 0 -1
0 1 -1
2
-1 1
1 -1
1
0
0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int MOD = 998244353, N = 3e5 + 5;

int main(){

    ll t = 1;
    cin >> t;
    while (t--){
    //  ll n,m,k;
    //  cin>>n>>m>>k;
    //  ll a[n][m];
    //  char ch[n][m];
    //  for(ll i=0;i<n;i++) for(ll j=0;j<m;j++) cin>>a[i][j];
    //  for(ll i=0;i<n;i++) for(ll j=0;j<m;j++) cin>>ch[i][j];
    ll n; cin>>n;
    vector<ll> a(n),b(n);
    for(ll i=0;i<n;i++) cin>>a[i];
    for(ll i=0;i<n;i++) cin>>b[i];
    vector<ll> v1(n+1),v2(n+1);
    for(ll i=n-1;i>=0;i--){
        if(a[i] == 1) v1[i] = 1 + v1[i+1];
        else v1[i] = v1[i+1];
        if(b[i] == 1) v2[i] = 1 + v2[i+1];
        else v2[i] = v2[i+1];
    }
    ll x1 = 0,x2 = 0;
    for(ll i=0;i<n;i++){
        if(a[i] == 0 && b[i] == 0) continue;
        if(a[i] == 0){ if(b[i] == -1) continue; else x2++; continue;}
        if(b[i] == 0){ if(a[i] == -1) continue; else x1++; continue;}
        if(a[i] == 1 && b[i] == -1) {x1++; continue;}
        if(a[i] == -1 && b[i] == 1) {x2++; continue;}
        ll dif = abs(x1 - x2);
        ll one1 = v1[i+1];
        ll one2 = v2[i+1];
        if(a[i] == 1 && b[i] == 1){
            if(x1 < x2){
                if(dif + 1 <= one1) x2++;
                else x1++;
            }
            else{
                if(dif + 1 < one2) x1++;
                else x2++;
            }
        }
        else{
            if(x1 < x2){
                if(dif + 1 <= one1) x1--;
                else x2--;
            }
            else{
                if(dif + 1 < one2) x2--;
                else x1--;
            }
        }
    }
    cout<<min(x1,x2)<<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
1
4
1 0 0 -1
-1 1 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
def solve():
    n = int(input())
    a = list(int(x) for x in input().split())
    b = list(int(x) for x in input().split())

    r = 0
    for i in range(n):
        if a[i] > b[i]:
            r += a[i]
        elif a[i] < b[i]:
            r += b[i]
        else:
            r += a[i]

    print(r//2)

for _ in range(int(input())):
    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 0 -1
-1 0 1
4
1 1 1 1
-1 -1 -1 -1
5
-1 -1 0 1 1
1 1 0 -1 -1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
for _ in range(int(input())):
    input()
    movone = list(map(int , input().split()))
    movtwo = list(map(int , input().split()))

    reviewone , reviewtwo = 0 , 0

    for i in range(len(movone)):
        review = max(movone[i] , movtwo[i])
        if reviewone >= reviewtwo:
            reviewtwo += review
        else:
            reviewone += review
    print(min(reviewone,reviewtwo))

**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 maximize the failure probability. The bug in the code is that it always adds the maximum of the two ratings to the movie with the lower current rating. This approach can lead to suboptimal solutions in certain cases.

Let's consider a scenario where the bug is most likely to be exposed:

1. **Uneven Distribution of Likes and Dislikes**: If one movie has a lot of likes and the other has a lot of dislikes, the algorithm might not distribute the reviews optimally.
2. **Neutral Attitudes**: Neutrals do not affect the rating, so they can be used to balance the reviews between the two movies.

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

```plaintext
1
3
1 0 -1
-1 0 1
```

### Explanation:
- **Test Case 1**:
  - **n = 3**: There are 3 people.
  - **a = [1, 0, -1]**: The first movie is liked by the first person, neutral for the second, and disliked by the third.
  - **b = [-1, 0, 1]**: The second movie is disliked by the first person, neutral for the second, and liked by the third.

### Expected Output:
- The maximum possible rating of the company is **0**.

### Reasoning:
- **Person 1**: Likes the first movie and dislikes the second movie. Reviewing the first movie will increase the first movie's rating by 1.
- **Person 2**: Neutral towards both movies. Reviewing either movie will not change the ratings.
- **Person 3**: Dislikes the first movie and likes the second movie. Reviewing the second movie will increase the second movie's rating by 1.

If the algorithm correctly distributes the reviews:
- **First movie**: 1 (from Person 1)
- **Second movie**: 1 (from Person 3)

The company's rating is the minimum of the two ratings, which is **0**.

The buggy code might not handle this case correctly because it always adds the maximum review to the movie with the lower current rating, which can lead to suboptimal results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

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

 public class one{
	 static int recursive_upper_bound(int arr[], int low,
             int high, int key)
{
if (low > high || low == arr.length)
return low;
int mid = low + (high - low) / 2;
if (key >= arr[mid]) {
return recursive_upper_bound(arr, mid + 1, high,
                 key);
}
return recursive_upper_bound(arr, low, mid - 1,
             key);
}
	  static int lower(int array[], int key)
	    {
	        int lowerBound = 0;
	        while (lowerBound < array.length) {
	            if (key > array[lowerBound])
	                lowerBound++;
	            else
	                return lowerBound;
	        }

	        return lowerBound;
	    }
	  public static void sort(int[] a)
	    {

	        ArrayList<Integer> ls = new ArrayList<>();
	        for(int x: a)
	            ls.add(x);
	        Collections.sort(ls);
	        for(int i=0; i < a.length; i++)
	            a[i] = ls.get(i);
	    }
	 public static int[] readarr(int n) {
		  MyScanner key = new MyScanner();
	      out = new PrintWriter(new BufferedOutputStream(System.out));
		 int []a=new int[n];
		 for(int j=0;j<n;j++) {
			 a[j]=key.nextInt();
		 }
		 return a;

	 }
	  public static void sort(long[] a)
	    {

	        ArrayList<Long> ls = new ArrayList<>();
	        for(long x: a)
	            ls.add(x);
	        Collections.sort(ls);
	        for(int i=0; i < a.length; i++)
	            a[i] = ls.get(i);
	    }
	    public static long gcd(long a, long b)
	    {
	        if(a > b)
	            a = (a+b)-(b=a);
	        if(a == 0L)
	            return b;
	        return gcd(b%a, a);
	    }
	    public static HashMap<Integer, Integer>
	    sortByValue(HashMap<Integer, Integer> hm)
	    {

	        List<Map.Entry<Integer, Integer> > list
	            = new LinkedList<Map.Entry<Integer, Integer> >(
	                hm.entrySet());

	        Collections.sort(
	            list,
	            (i1,
	             i2) -> i1.getValue().compareTo(i2.getValue()));

	        HashMap<Integer, Integer> temp
	            = new LinkedHashMap<Integer, Integer>();
	        for (Map.Entry<Integer, Integer> aa : list) {
	            temp.put(aa.getKey(), aa.getValue());
	        }
	        return temp;
	    }
	    public static boolean isPrime(long n)
	    {
	        if(n < 2) return false;
	        if(n == 2 || n == 3) return true;
	        if(n%2 == 0 || n%3 == 0) return false;
	        long sqrtN = (long)Math.sqrt(n)+1;
	        for(long i = 6L; i <= sqrtN; i += 6) {
	            if(n%(i-1) == 0 || n%(i+1) == 0) return false;
	        }
	        return true;
	    }
	    public static int solve(int i,int []a,int []b) {
	    	int n=a.length;
	    	if(i>=n)
	    		return 0;
	    	if(i==n-1)
	    		return Math.min(a[i]+solve(i+1,a,b),b[i]+solve(i+1,a,b));
	    	return Math.max(a[i]+solve(i+1,a,b),b[i]+solve(i+1,a,b));
	    }

   public static void main(String[] args) {
      MyScanner key = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      int t=key.nextInt();
        while(t-->0) {
        	int n=key.nextInt();
        	int []a=new int[n];
        	int []b=new int[n];
        	for(int j=0;j<n;j++) {
        		a[j]=key.nextInt();
        	}
        	for(int g=0;g<n;g++) {
        	   b[g]=key.nextInt();
        	}
        	int sum1=0,sum2=0,etra=0;
        	for(int g=0;g<n;g++) {
        		if(b[g]<a[g]) {
        			etra++;
        		}
        		if(a[g]<b[g]) {
        			etra--;
        		}
        	}
for(int g=0;g<n;g++) {
	if(a[g]==1 && b[g]==0) {
		sum1++;
	}else if(a[g]==0 && b[g]==1) {
		sum2++;
	}else if(a[g]==0 && b[g]==0) {
		continue;
	}else if(a[g]==1 && b[g]==1) {
		if(sum1>sum2)
			sum2++;
		else if(sum2>sum1)
			sum1++;
		else if(etra>0) {
			sum2++;
			etra--;
		}else if(etra<0) {
			sum1++;
			etra++;
		}else {
			sum2++;
		}
	}else if(a[g]==-1 && b[g]==0) {
			continue;
	}else if(a[g]==0 && b[g]==-1) {
			continue;
	}else if(a[g]==-1 && b[g]==-1) {
		if(sum1>sum2)
			sum1--;
		else if(sum2>sum1)
			sum2--;
		else if(etra>0) {
			sum1--;
			etra--;
		}else if(etra<0) {
			sum2--;
			etra++;
		}else {
			sum1--;
		}
	}else if(a[g]==-1 && b[g]==1){
			sum2++;
			etra--;
	}else {
			sum1++;
			etra++;
	}
}
out.println(Math.min(sum1, sum2));
        }

    	out.close();
   }

   public static PrintWriter out;

   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;

      public MyScanner() {
         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());
      }

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

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

   }

}
 class Pair {
int x,y;

	    public Pair(int x,int y) {
	        this.x=x;
	        this.y=y;
	    }
	}
 class DSU{
		ArrayList<Integer>rank=new ArrayList<Integer>();
		ArrayList<Integer>size=new ArrayList<Integer>();
		ArrayList<Integer> parent=new ArrayList<Integer>();
		DSU(int n){
			for(int g=0;g<n+1;g++) {
				parent.add(g);
				rank.add(0);
				size.add(1);
			}
		}
		public  int uparent(int node) {
			if(parent.get(node)==node)
				return node;
		 parent.set(node,uparent(parent.get(node)));
		 return parent.get(node);
		}
		public void unionbyrank(int u,int v) {
			int ulu=uparent(u);
			int ulv=uparent(v);
			if(ulu==ulv)
				return;
			if(rank.get(ulu)<rank.get(ulv)) {
				parent.set(ulu,ulv);
			}else if(rank.get(ulv)<rank.get(ulu)) {
				parent.set(ulv, ulu);
			}else {
				parent.set(ulv,ulu);
			rank.set(ulu,rank.get(ulu)+1);
			}
		}
		public void unionbysize(int u,int v) {
			int ulu=uparent(u);
			int ulv=uparent(v);
			if(ulu==ulv)return;

			if(size.get(ulu)<size.get(ulv)) {
				parent.set(ulu,ulv);
				size.set(ulv,size.get(ulv)+size.get(ulu));
			}else {
				parent.set(ulv, ulu);
				size.set(ulu,size.get(ulv)+size.get(ulu));
			}
		}
	}

**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 that can highlight issues in the logic. The code seems to have a complex approach to managing the ratings of the two movies, and it might have issues with how it handles neutral and mixed opinions.

Here are the steps to think through the test input:

1. **Mixed Opinions**: The code needs to handle cases where people have mixed opinions (e.g., liking one movie and disliking the other) correctly.
2. **Neutral Opinions**: The code should handle neutral opinions without affecting the ratings.
3. **Balanced Ratings**: The code should be able to balance the ratings of the two movies to maximize the company's rating.

Let's create a small-scale test input that includes these elements:

- **Mixed Opinions**: People who like one movie and dislike the other.
- **Neutral Opinions**: People who are neutral about both movies.
- **Balanced Ratings**: A scenario where the ratings need to be balanced to maximize the company's rating.

Here is a test input that should expose potential bugs:

```plaintext
1
4
1 0 0 -1
-1 0 0 1
```

Explanation:
- **n = 4**: There are 4 people.
- **a = [1, 0, 0, -1]**: The first person likes the first movie, the next two are neutral, and the last one dislikes the first movie.
- **b = [-1, 0, 0, 1]**: The first person dislikes the second movie, the next two are neutral, and the last one likes the second movie.

This input should challenge the code to handle:
- The first person (likes the first movie, dislikes the second movie) should contribute to the first movie's rating.
- The last person (dislikes the first movie, likes the second movie) should contribute to the second movie's rating.
- The neutral people should not affect the ratings.

By running this input, we can observe if the code correctly balances the ratings to maximize the company's rating.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @Author congge
 * @Date 2024/6/27 23:24
 * @description https://codeforces.com/contest/1989/problem/C
 */
public class Main {
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static int N = 200010;
    static int[] a = new int[N], b = new int[N];
    static int n;

    public static void main(String[] args) {
        int T = sc.nextInt();
        while (T-- > 0) {
            n = sc.nextInt();
            for (int i = 1; i <= n; i++) a[i] = sc.nextInt();
            for (int i = 1; i <= n; i++) b[i] = sc.nextInt();
            int A = 0, B = 0;
            int allneg = 0;
            for (int i = 1; i <= n; i++) {
                if (a[i] > 0 && b[i] > 0) {
                    if (A > B) B += 1;
                    else A += 1;
                } else if (a[i] > 0 && b[i] <= 0) A += 1;
                else if (b[i] > 0 && a[i] <= 0) B += 1;
                else if (a[i] < 0 && b[i] < 0) {
                    allneg++;
                }
            }
            int res = 0;
            int sub = Math.abs(A - B);
            if (sub >= allneg) res = Math.min(A, B);
            else {
                allneg -= sub;
                res = Math.min(A, B) - (allneg + 1) / 2;
            }
            out.println(res);
        }
        out.flush();
    }
}

class FastReader {
    StringTokenizer st;
    BufferedReader br;

    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());
    }

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

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

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

    // 是否由下一个
    boolean hasNext() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = br.readLine();
                if (line == null)
                    return false;
                st = new StringTokenizer(line);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return true;
    }
}

**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 code, we need to focus on scenarios that involve the handling of negative attitudes towards both movies. The buggy code has an issue in how it handles cases where both movies are disliked by the same person. Specifically, the code does not correctly account for the number of negative reviews that can be offset by positive reviews.

Let's consider the following test case:

1. **All people dislike both movies**: This scenario should highlight the issue with the code's handling of negative reviews.
2. **A mix of positive, neutral, and negative attitudes**: This will test the code's ability to balance the ratings of the two movies.

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

```plaintext
2
5
-1 -1 -1 -1 -1
-1 -1 -1 -1 -1
5
1 0 -1 1 0
-1 0 1 -1 1
```

### Explanation:
1. **First Test Case**:
   - All 5 people dislike both movies.
   - The expected output should be `-5` because there are no positive reviews to offset the negative ones.
   - The code might incorrectly handle this case, leading to a higher (and incorrect) rating.

2. **Second Test Case**:
   - The first movie has 1 positive, 2 neutral, and 2 negative reviews.
   - The second movie has 2 negative, 2 neutral, and 1 positive review.
   - The expected output should be `0` because the positive reviews can offset the negative reviews to achieve a balanced rating.
   - This case tests the code's ability to balance the ratings and handle mixed attitudes.

This test input should help expose the bug in the code's handling of negative reviews and the overall balancing of the ratings.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
#include <iostream>

using namespace std;

const int N = 2e5 + 10;
int n,FM[N],SM[N],m;

int main(){
	cin >> n;
	while (n --){
		cin >> m;
		for (int i = 0;i < m;i ++){
			cin >> FM[i];
		}
		for (int i = 0;i < m;i ++){
			cin >> SM[i];
		}
		int tmp,M = 0xc0c0c0c0;
		for (int i = 0;i < m;i ++){
			tmp = FM[i] + SM[i];
			M = max(M,tmp);
		}
		if (M < 0){
			cout << -1 << endl;
		}else if (M == 0){
			cout << 0 << endl;
		}else{
			cout << 1 << 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:
To generate a challenging test input that exposes the bug in the provided code, we need to consider the logic of the code and identify scenarios where it might fail. The buggy code calculates the sum of attitudes for each person and uses the maximum of these sums to determine the output. This approach is flawed because it does not consider the possibility of distributing the reviews optimally between the two movies to maximize the minimum rating.

### Steps to Identify the Bug:
1. **Sum Calculation**: The code calculates the sum of attitudes for each person and takes the maximum of these sums.
2. **Output Determination**: The code then outputs -1, 0, or 1 based on the maximum sum.

### Scenario to Expose the Bug:
- **Case with Mixed Attitudes**: Consider a case where the sum of attitudes for each person is zero or negative, but by distributing the reviews optimally, we can achieve a higher rating.
- **Case with Neutral Attitudes**: Consider a case where there are many neutral attitudes, which can be used to balance the ratings of the two movies.

### Test Input:
Let's create a test case where the sum of attitudes for each person is zero, but by distributing the reviews optimally, we can achieve a higher rating.

```plaintext
1
6
1 -1 0 1 -1 0
-1 1 0 -1 1 0
```

### Explanation:
- **n = 6**: There are 6 people.
- **a = [1, -1, 0, 1, -1, 0]**: Attitudes towards the first movie.
- **b = [-1, 1, 0, -1, 1, 0]**: Attitudes towards the second movie.

### Expected Output:
- The correct output should be **0** because by distributing the reviews optimally, we can ensure that both movies have a rating of 0. However, the buggy code will output **-1** because it only considers the maximum sum of attitudes, which is 0 in this case, and incorrectly outputs -1.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Two Movies
time_limit_ms: 2000
memory_limit_mb: 256
Description: A movie company has released $2$ movies. These $2$ movies were watched by $n$ people. For each person, we know their attitude towards the first movie (liked it, neutral, or disliked it) and towards the second movie.

If a person is asked to leave a review for the movie, then:

  * if that person liked the movie, they will leave a positive review, and the movie's rating will increase by $1$;    * if that person disliked the movie, they will leave a negative review, and the movie's rating will decrease by $1$;    * otherwise, they will leave a neutral review, and the movie's rating will not change.

Every person will review exactly one movie — and for every person, you can choose which movie they will review.

The company's rating is the minimum of the ratings of the two movies. Your task is to calculate the maximum possible rating of the company.

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

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

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-1 \le a_i \le 1$), where $a_i$ is equal to $-1$ if the first movie was disliked by the $i$-th viewer; equal to $1$ if the first movie was liked; and $0$ if the attitude is neutral.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-1 \le b_i \le 1$), where $b_i$ is equal to $-1$ if the second movie was disliked by the $i$-th viewer; equal to $1$ if the second movie was liked; and $0$ if the attitude is neutral.

Additional constraint on the input: the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the maximum possible rating of the company, if for each person, choose which movie to leave a review on.

Examples:
input:
4
2
-1 1
-1 -1
1
-1
-1
5
0 -1 1 0 1
-1 1 0 0 1
4
-1 -1 -1 1
-1 1 1 1
output:
0
-1
1
1

Note:

Buggy Code:
#pragma GCC optimize("O3,unroll-loops")

#include<bits/stdc++.h>

using namespace std;
using namespace chrono;
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define MOD 1000000007
#define MOD1 998244353
#define INF 1e18
#define nl '\n'
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.141592653589793238462
#define set_bits __builtin_popcountll
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define clr(x) memset(x, 0, sizeof(x))
#define pqt priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>

#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x<<" "; _print(x); cerr << endl;
#else
#define debug(x);
#endif

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

void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
#define int long long

/*..............................................................................................................................................................................................................................................................*/
ll gcd(ll a, ll b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}
ll expo(ll a, ll b, ll mod) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1;} return res;}
void extendgcd(ll a, ll b, ll*v) {if (b == 0) {v[0] = 1; v[1] = 0; v[2] = a; return ;} extendgcd(b, a % b, v); ll x = v[1]; v[1] = v[0] - v[1] * (a / b); v[0] = x; return;} //pass an arry of size1 3
ll mminv(ll a, ll b) {ll arr[3]; extendgcd(a, b, arr); return arr[0];} //for non prime b
ll mminvprime(ll a, ll b) {return expo(a, b - 2, b);}
bool revsort(ll a, ll b) {return a > b;}
ll combination(ll n, ll r, ll m, ll *fact, ll *ifact) {ll val1 = fact[n]; ll val2 = ifact[n - r]; ll val3 = ifact[r]; return (((val1 * val2) % m) * val3) % m;}
void google(int t) {cout << "Case #" << t << ": ";}
vector<ll> sieve(int n) {int*arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}
ll mod_add(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
ll mod_mul(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
ll mod_sub(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
ll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, mminvprime(b, m), m) + m) % m;}  //only for prime m
ll phin(ll n) {ll number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (ll i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll getRandomNumber(ll l, ll r) {return uniform_int_distribution<ll>(l, r)(rng);}
/*..............................................................................................................................................................................................................................................................*/

void solve()
{
    int n;
    cin>>n;
    int arr[n];
    int brr[n];
    for(int i=0;i<n;i++){
        cin>>arr[i];
    }
    for(int i=0;i<n;i++){
        cin>>brr[i];
    }
    int a=0;
    int b=0;
    int c=0;
    int d=0;
    for(int i=0;i<n;i++){
        if(arr[i]==brr[i] && arr[i]==-1){
            c++;
        }
        else if(arr[i]==brr[i] && arr[i]==1){
            d++;
        }
        else{
            if(arr[i]>brr[i]){
                a+=arr[i];
            }
            else{
                b+=brr[i];
            }
        }
    }
    debug(a);
    debug(b);
    debug(c);
    debug(d);
    int res=a+b+d-c;
    if(res<0){
        cout<<(res-1)/2<<endl;
    }
    else{
        cout<<res/2<<endl;
    }
}

signed main()
{
    #ifndef ONLINE_JUDGE
        freopen("error.txt", "w", stderr);
    #endif
        fastio();
        auto start1 = high_resolution_clock::now();
        int tt = 1;
        cin >> tt;
        while (tt--)
        {
            solve();
        }
        auto stop1 = high_resolution_clock::now();
        auto duration = duration_cast<microseconds>(stop1 - start1);
    #ifndef ONLINE_JUDGE
        cerr << "Time: " << duration . count() / 1000 << endl;
    #endif
}

**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 the following aspects:

1. **Edge Cases**: Test cases with small values of \( n \) to ensure that the code handles small inputs correctly.
2. **Special Cases**: Test cases where the attitudes towards both movies are the same for some viewers, or where all viewers have neutral attitudes.
3. **Balanced Cases**: Test cases where the number of positive, negative, and neutral attitudes are balanced to see how the code handles such scenarios.
4. **Maximizing the Failure Probability**: Test cases where the distribution of attitudes can lead to the minimum possible company rating, as the code might have issues in handling such scenarios.

Let's construct a test case that maximizes the failure probability by ensuring that the code's logic for calculating the company's rating is exposed.

### Chosen Test Case
- **Test Case 1**: \( n = 4 \)
  - Attitudes towards the first movie: \( [1, -1, 0, 1] \)
  - Attitudes towards the second movie: \( [1, -1, 0, -1] \)

### Explanation
- **Person 1**: Likes both movies (1, 1)
- **Person 2**: Dislikes both movies (-1, -1)
- **Person 3**: Neutral towards both movies (0, 0)
- **Person 4**: Likes the first movie but dislikes the second movie (1, -1)

### Expected Output
- The maximum possible rating of the company should be 0.
  - If Person 1 and Person 4 review the first movie, and Person 2 and Person 3 review the second movie, the ratings will be:
    - First movie: 1 (from Person 1) + 1 (from Person 4) = 2
    - Second movie: -1 (from Person 2) + 0 (from Person 3) = -1
  - The company's rating is the minimum of the two ratings: min(2, -1) = -1

However, if we assign the reviews differently, we can achieve a better rating:
- Person 1 and Person 4 review the second movie, and Person 2 and Person 3 review the first movie:
  - First movie: 0 (from Person 3) - 1 (from Person 2) = -1
  - Second movie: 1 (from Person 1) - 1 (from Person 4) = 0
  - The company's rating is min(-1, 0) = -1

To maximize the company's rating, we should aim for:
- First movie: 1 (from Person 1) + 1 (from Person 4) = 2
- Second movie: 1 (from Person 1) - 1 (from Person 4) = 0
- The company's rating is min(2, 0) = 0

### Test Input
```plaintext
1
4
1 -1 0 1
1 -1 0 -1
```

This test case should expose the bug in the buggy code, as the code might not handle the distribution of reviews correctly to achieve the maximum possible rating of the company.