**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

vector<pair<int, int>> movements = {
    {0, 1}, {0, -1}, {1, 0}, { -1, 0},
    {1, 1}, {1, -1}, { -1, 1}, { -1, -1}
};

bool isvalid(int i, int j) {
    return i >= 0 && j >= 0;
}

int hd(const string& s1, const string& s2) {
    int dist = 0;
    for (int i = 0; i < s1.length(); ++i) {
        if (s1[i] != s2[i]) {
            ++dist;
        }
    }
    return dist;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        vector<int> arr(3);
        for (int i = 0; i < 3; i++) {
            cin >> arr[i];
        }

        sort(arr.begin(), arr.end());
        int n = 5;
        int ans = 0;

        while (n > 0) {
            sort(arr.begin(), arr.end()); // keep the array sorted
            if (arr[2] - arr[0] > n) {
                arr[0] += n;
                n = 0;
            } else {
                arr[0] += 1;
                n -= 1;
            }
        }
        ans = arr[0] * arr[1] * arr[2];

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

1. **Edge Cases**: The code should handle the smallest and largest possible values for `a`, `b`, and `c` within the given constraints.
2. **Optimal Distribution**: The code should correctly distribute the increments to maximize the product `a * b * c`.
3. **Sorting and Incrementing**: The code sorts the array and increments the smallest element, which might not always be the optimal strategy.

Given the constraints \(1 \le a, b, c \le 10\) and the operations are performed at most 5 times, a good test case would be one where the initial values are close to each other, but not identical, and the optimal strategy is not immediately obvious.

Let's consider the following test case:

```plaintext
1
3 4 5
```

### Explanation:
- **Initial Values**: \(a = 3\), \(b = 4\), \(c = 5\)
- **Optimal Strategy**: The optimal strategy is to increment the smallest value (3) as much as possible, but we need to check if incrementing the middle value (4) might yield a higher product.

### Steps to Verify:
1. **Initial Product**: \(3 \times 4 \times 5 = 60\)
2. **Incrementing the Smallest Value (3) 5 Times**:
   - After 1 increment: \(4 \times 4 \times 5 = 80\)
   - After 2 increments: \(5 \times 4 \times 5 = 100\)
   - After 3 increments: \(6 \times 4 \times 5 = 120\)
   - After 4 increments: \(7 \times 4 \times 5 = 140\)
   - After 5 increments: \(8 \times 4 \times 5 = 160\)

3. **Incrementing the Middle Value (4) 5 Times**:
   - After 1 increment: \(3 \times 5 \times 5 = 75\)
   - After 2 increments: \(3 \times 6 \times 5 = 90\)
   - After 3 increments: \(3 \times 7 \times 5 = 105\)
   - After 4 increments: \(3 \times 8 \times 5 = 120\)
   - After 5 increments: \(3 \times 9 \times 5 = 135\)

4. **Mixed Strategy**:
   - Increment the smallest value (3) 3 times: \(6 \times 4 \times 5 = 120\)
   - Increment the middle value (4) 2 times: \(6 \times 6 \times 5 = 180\)

### Correct Output:
The correct output for the test case `3 4 5` should be `180`.

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

This test case should expose the bug in the provided code, as it does not consider the mixed strategy of incrementing both the smallest and middle values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

typedef long long ll;
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
#define fi first
#define se second
#define en '\n'
#define file "test"
#define log 17
#define memset(a,x) memset(a, x, sizeof(a))
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
template<class T1,class T2>
bool maximize(T1 &x,const T2 &y) {
    if(x < y) {
        x = y;
        return 1;
    }
    return 0;
}
template<class T1,class T2>
bool minimize(T1 &x, const T2 &y) {
    if(x > y) {
        x = y;
        return 1;
    }
    return 0;
}
const double pi = 3.141592653589793;
const int dx[] = {1, -1, 0,0};
const int dy[] = {0, 0, 1, -1};
const int modd = 1e9 + 7;

int a[5];

signed main () {
    #ifndef ONLINE_JUDGE
        freopen(file".inp", "r", stdin);
        freopen(file".out", "w", stdout);
    #endif
    fast_cin();
    int te;
    cin >> te;
    while (te--) {
        cin >> a[1] >> a[2] >> a[3];
        sort(a + 1, a + 4);
        if (a[3] - a[1] >= 5) {
            cout<<(a[1] + 5) * a[2] * a[3];
        }
        else {
            if (a[3] - a[2] >= 5 - a[3] + a[1]) {
                cout<<a[3] * a[3] * (a[2] + 5 - a[3] + a[1]);
            }
            else {
                int r = 5 - 2 * a[3] + a[1] + a[2];
                int cc = r / 3;
                r = r % 3;
                if (r == 0) {
                    cout<<pow((a[3] + cc), 3);
                }
                else if (r == 1) {
                    cout<<pow((a[3] + cc),  2) * (a[3] + cc + r);
                }
                else {
                    cout<<(a[3] + cc) * pow((a[3] + cc + 1), 2);
                }
            }
        }
        cout<<en;
    }

    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The code seems to have issues with how it handles the distribution of the 5 increments across the three variables. Let's consider the following scenarios:

1. **Equal values**: When all three values are equal, the code might not distribute the increments optimally.
2. **Very close values**: When the values are very close but not equal, the code might not handle the distribution correctly.
3. **One value much smaller**: When one value is significantly smaller than the others, the code might not handle the distribution correctly.
4. **One value much larger**: When one value is significantly larger than the others, the code might not handle the distribution correctly.

Let's generate a test input that includes these scenarios:

```plaintext
4
1 1 1
2 2 3
1 2 10
1 9 9
```

### Explanation:
1. **1 1 1**: All values are equal. The optimal solution is to increase all three values by 1, resulting in \(2 \times 2 \times 2 = 8\).
2. **2 2 3**: Values are close but not equal. The optimal solution is to increase the smallest value (2) twice and the second smallest value (2) once, resulting in \(4 \times 3 \times 3 = 36\).
3. **1 2 10**: One value is much smaller. The optimal solution is to increase the smallest value (1) five times, resulting in \(6 \times 2 \times 10 = 120\).
4. **1 9 9**: One value is much smaller. The optimal solution is to increase the smallest value (1) five times, resulting in \(6 \times 9 \times 9 = 486\).

This test input should expose any issues in the buggy code's handling of these scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
for  i in range(int(input())):
  m=list(map(int,input().split()))
  m.sort()
  a=m[0];b=m[1];c=m[2]
  l=[(a+1)*(b+4)*c,(a+2)*(b+3)*c,(a+3)*(b+2)*c,(a+4)*(b+1)*c,(a+0)*(b+5)*c,(a+5)*(b+0)*c]
  print(max(l))

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
9 9 9
5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
x = int(input())
lst=[]
ans=[]
for i in range(x):
    lst.clear()
    m= input()
    q = m.split()
    lst.append(int(q[0]))
    lst.append(int(q[1]))
    lst.append(int(q[2]))

    lst.sort()
    count=0
    while(not(lst[2]+1==lst[0]) and count<5):
        lst[0] = lst[0]+1
        count=count+1

    while(not(lst[2]+1==lst[1]) and count<5):
        lst[1] = lst[1]+1
        count=count+1

    ans.append((lst[0]*lst[1]*lst[2]))

for i in ans:
    print(i)

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
9 9 9
10 1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
import java.util.*;

public class Solution {
	public static void main(String[] args) {
		Scanner s =  new Scanner(System.in);

		int T = s.nextInt();
		while(T -- > 0) {
		  //  int a = s.nextInt(), b = s.nextInt(), c = s.nextInt();
		    int ar[] = new int[3];
		    for(int i=0;i<3;i++)
		        ar[i] = s.nextInt();

		    Arrays.sort(ar);

		    if(ar[1] == ar[2]) {
		        ar[0] += 5;
		    }
		    else {
    		    ar[0] += 3;
    		    ar[1] += 2;
		    }
		    System.out.println(ar[0]*ar[1]*ar[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:
```plaintext
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: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

public class MyClass {

    public static void main(String[] args) {

        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();

        for(int i = 0; i<n; i++)
        {
            int a = scan.nextInt();
            int b = scan.nextInt();
            int c = scan.nextInt();

            int min = Math.min(a,b);
            min = Math.min(min,c);

            int max = Math.max(a,b);
            max = Math.max(min,c);

            int mid = a+b+c-min-max;

            for(int j = 0; j<5; j++)
            {
                min++;
                if(mid<min)
                {
                    int t = min;
                    min = mid;
                    mid = t;

                    if(max<mid)
                    {
                        t=max;
                        max=mid;
                        mid = t;
                    }
                }
            }

            System.out.println(min*mid*max);

        }
    }
}

**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
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

int main(){

    int t;
    cin >>t;
    while(t--){
        int a,b,c;
        vector<int> v(3);

        cin>> a>>b>>c;
        v[0] = a;
        v[1] = b;
        v[2] = c;
        int ans1 = 1;
        sort(v.begin(), v.end());
        int an = v[0]*v[1]*v[2];
        an = max(an,(v[0]+5)*v[1]*v[2]);
        an = max(an,(v[0]+4)*(v[1]+1)*v[2]);
        an = max(an,(v[0]+3)*(v[1]+2)*v[2]);
        an = max(an,(v[0]+2)*(v[1]+3)*v[2]);
        cout<<an<<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 generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following points:

1. **Edge Cases**: The input values should be at the boundaries of the allowed range.
2. **Optimal Strategy**: The input should test the logic of the code, especially the part where it decides how to distribute the 5 increments.

The buggy code sorts the array and only considers a few specific ways to distribute the 5 increments. However, it does not consider all possible combinations, which can lead to suboptimal results.

Let's consider a test case where the numbers are close to each other but not identical, and where the optimal strategy is not one of the specific cases considered by the code.

### Chosen Test Case
- \( a = 3 \)
- \( b = 3 \)
- \( c = 3 \)

### Reasoning
- The initial product is \( 3 \times 3 \times 3 = 27 \).
- The optimal strategy is to distribute the 5 increments in a way that maximizes the product. One optimal way is to add 2 to one number and 1 to another, resulting in \( (3+2) \times (3+1) \times 3 = 5 \times 4 \times 3 = 60 \).

However, the buggy code only considers a few specific ways to distribute the increments, and it might miss this optimal distribution.

### Test Input
```plaintext
1
3 3 3
```

This test case will help expose the bug in the code by showing that the optimal distribution of increments is not being considered correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

using namespace std;

int a[10], b[10], maxx;

int main(){
  int t;
  cin >> t;
  while(t--){
    for(int i = 1; i <= 3; i++){
      cin >> a[i];
    }
    for(int i = 0; i <= 5; i++){
      for(int j = 0; j <= 5; j++){
        for(int k = 0; k <= 5; k++){
          if(i + j + k <= 5){
            int ans = (a[1] + i) * (a[2] + j) * (a[3] + k);
            maxx = max(maxx, ans);
          }
        }
      }
    }
    cout << maxx << '\n';
  }
  return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the code might fail to correctly compute the maximum value of \(a \times b \times c\) after the allowed operations. The main issue in the code is the initialization of the `maxx` variable, which is not reset for each test case. This can lead to incorrect results if the maximum value from a previous test case is carried over to the next one.

To create a test input that exposes this bug, we can use a test case where the initial values of \(a\), \(b\), and \(c\) are small, and the expected maximum value after the operations is also small. This will make it easier to see if the code is carrying over values from a previous test case.

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

```plaintext
2
1 1 1
10 10 10
```

### Explanation:
1. **First Test Case (1 1 1):**
   - Initial values: \(a = 1\), \(b = 1\), \(c = 1\)
   - After 5 operations, the best we can do is to increase each of \(a\), \(b\), and \(c\) by 1 (since \(1 + 1 + 1 + 1 + 1 = 5\)).
   - The maximum value is \((1+1) \times (1+1) \times (1+1) = 2 \times 2 \times 2 = 8\).

2. **Second Test Case (10 10 10):**
   - Initial values: \(a = 10\), \(b = 10\), \(c = 10\)
   - After 5 operations, the best we can do is to increase one of the values by 5.
   - The maximum value is \((10+5) \times 10 \times 10 = 15 \times 10 \times 10 = 1500\).

### Expected Output:
```plaintext
8
1500
```

### Bug Exposure:
- The `maxx` variable is not reset after each test case.
- If the first test case results in a maximum value of 8, and the second test case should result in 1500, the code might output 1500 for both test cases if `maxx` is not reset.

This test input should clearly show the issue with the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
def max_bananas(a, b, c):
    # Sort the numbers so that we always work with the smallest first
    nums = sorted([a, b, c])

    # Distribute the increments as evenly as possible while prioritizing small values
    for i in range(min(5, sum(nums) - 3)):
        nums[0] += 1
        nums.sort()

    return nums[0] * nums[1] * nums[2]

# Read the number of test cases
t = int(input())
for _ in range(t):
    # Read the values for a, b, and c
    a, b, c = map(int, input().split())
    # Output the maximum number of bananas
    print(max_bananas(a, b, c))

**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
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
from math import prod
for i in range(int(input())):
    banana = list(map(int, input().split()))
    maxabanana = []
    odinbanana = 0
    odenbanana2 = 0
    for a in range(3):
        banana[odenbanana2] += 5
        h = prod(banana)
        maxabanana.append(h)
        banana[odenbanana2] -= 5
        odenbanana2 += 1
    odenbanana2 = 0
    for r in range(3):
        banana[odenbanana2] += 3
        banana[odenbanana2-1] += 2
        h = prod(banana)
        maxabanana.append(h)
        banana[odenbanana2] -= 3
        banana[odenbanana2 - 1] -= 2
        odenbanana2 -= 1
    odenbanana2 = 0
    banana[odenbanana2] += 2
    banana[odenbanana2 + 1] += 2
    banana[odenbanana2 + 2] += 1
    h = prod(banana)
    maxabanana.append(h)
    banana[odenbanana2] -= 2
    banana[odenbanana2 + 1] -= 2
    banana[odenbanana2 + 2] -= 1
    for f in range(3):
        banana[odinbanana] += 1
        odinbanana += 1
    odinbanana = 0
    for p in range(3):
        banana[odinbanana] += 2
        h = prod(banana)
        maxabanana.append(h)
        banana[odinbanana] -= 2
        odinbanana += 1
    print(max(maxabanana))

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
9 9 9
5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;
public class OnlyPulses {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int[] arr = new int[3];
            for (int j = 0; j < 3; j++) {
                int a = sc.nextInt();
                arr[j]=a;
            }
            int ans = helper(arr);
            System.out.println(ans);
        }
    }
    public static int helper(int[]arr){
        Arrays.sort(arr);
        int max1 =(arr[0]+3)*(arr[1]+2)*(arr[2]);
        int max2 = (arr[0]+5)*(arr[1])*(arr[2]);
        int max3 = (arr[0])*(arr[1]+5)*(arr[2]);
        int max4 = (arr[0])*(arr[1])*(arr[2]+5);
        int maxi1 = Math.max(max1,max2);
        int maxi2 = Math.max(max3,max4);
        return Math.max(maxi1,maxi2);
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
9 9 9
5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

/**
 * @author Tushar Rastogi
 **/

public class Main {
    /**
     * Global Variables
     */
    FastReader sc = new FastReader();
    PrintWriter out = new PrintWriter(System.out);

    /**
     * Global Variables
     */

    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    // Program kickstart
    void run() throws IOException {
        int t;
        t = 1;
        t = sc.nextInt();
        for (int i = 1; i <= t; i++) {
            // level sabke niklenge
            solve();
        }
        out.flush();
    }

    public int gcd(int a, int b) {

        if (b == 0) return a;

        return gcd(b, a % b);

    }

    public List<Integer> primeFactors(int n) {

        List<Integer> list = new ArrayList<>();

        int i;

        while (n % 2 == 0) {
            list.add(2);
            n /= 2;
        }

        for (i = 3; i * i <= n; i += 2) {

            while (n % i == 0) {
                list.add(i);
                n /= i;
            }
        }

        if (n > 2)
            list.add(n);

        return list;
    }

    public void solve() throws IOException {

           int a=sc.nextInt();
           int b=sc.nextInt();
           int c=sc.nextInt();

             int arr[]=new int[3];
              arr[0]=a;
              arr[1]=b;
              arr[2]=c;

              Arrays.sort(arr);

               a=arr[0];
               b=arr[1];
               c=arr[2];

                if(a==b&&b==c)
                {

                    System.out.println((a+5)*b*c);
                    return ;
                }
                int x=5;

                 while(x>0) {
                     if (b - a <= x) {

                          if(b-a==0)
                          {
                              b++;
                              x--;
                          }

                          else {
                              x -= (b - a);
                              a = b;
                          }
                     }

                     else{
                         a+=5;
                         break;
                     }

                      if(x<=0)
                          break;

                     if(c-b<=x)
                     {
                          if(c-b==0)
                          {
                              c++;
                              x--;
                          }
                          else {
                              x -= (c - b);
                              b = c;
                          }
                     }

                     else{
                         b+=5;
                         break;
                     }
                 }

                 System.out.println(a*b*c);

    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}

/*

If the remainder of  number is -ve ,add the number from which the number is divided to its remainder to get the ans
e.g remainder of -36 % 5  is -1
 ans = -1 +5 = 4
 -a % b =  r + b
 */

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
5 5 5
2 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

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

#define nline     "\n"
#define all(x)       (x).begin(), (x).end()
#define input(v)  for (auto &i : v) cin >> i;
#define print(v)  for (auto &j : v) cout << j<<" ";
#define FAST()       ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

typedef long long ll;

void solve(int);
void precal();

int32_t main()
{
   FAST();
   cout << fixed;
   cout << setprecision(15);
   precal();
   int t = 1;
   cin>>t;
   for (int i = 1; i <= t; i++)
   {
      solve(i);
      cout << nline;
   }
}

void precal()
{

}

ll nxt()
{
    ll x;
    cin >> x;
    return x;
}

void solve(int kase)
{
  ll n=nxt();
  vector<ll> v(n),av,pre(n);
  input(v);
  ll c=0;
  ll temp=v[0];
  for(int i=1;i<n;i++)
  {
    if(((temp>=0)&&(v[i]>=0))||((temp<=0)&&(v[i]<=0)))
    {
        temp+=v[i];
    }
    else
    {
        av.push_back(temp);
        temp=v[i];
    }
  }
  av.push_back(temp);
  pre[av.size()-1]=av[av.size()-1];
  for(int i=av.size()-2;i>=0;i--)
  {
    pre[i]=pre[i+1]+av[i];
  }
  for(int i=0;i<av.size();i++)
  {
    if(i==av.size()-1)
    {
        c=abs(c+av[i]);
    }
    else
    {
        if(pre[i+1]<0)
        {
            c+=v[i];
        }
        else
        {
            c=abs(c+av[i]);
        }
    }
  }
  cout<<c;
}

**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
5
4
1 2 -3 4
3
-1 2 -3
2
-1 -1
4
1000000000 -1000000000 1000000000 -1000000000
3
-1000000000 1000000000 -1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
#include<iostream>
#include<vector>
#include<algorithm>
#include <climits>
#include <string>
#include <cmath>
#include <set>
#include <numeric>
#include <map>

using namespace std;

#define INF 10000000000000000
#define ll long long
#define vl vector<long long>
#define vs vector<string>
#define vc vector<char>

int main() {

    long long T;
    cin >> T;

    while (T--){
        ll n;
        cin >> n;
        vl a(n);
        for (ll i = 0; i < n; i++){
            cin >> a[i];
        }

        vector<vl> b(n);
        ll c=0;
        for(ll i=0;i<n;i++){
            if(i==0){
                b[i].push_back(max(c+a[i],abs(c+a[i])));
                b[i].push_back(min(c+a[i],abs(c+a[i])));

            }
            else{
                for(ll j=0;j<b[i-1].size();j++){
                    b[i].push_back(max(b[i-1][j]+a[i],abs(b[i-1][j]+a[i])));
                    b[i].push_back(min(b[i-1][j]+a[i],abs(b[i-1][j]+a[i])));

                }
            }
        }

        sort(b[n-1].begin(),b[n-1].end());
        cout<<b[n-1][b[n-1].size()-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:
```plaintext
3
2
-1 1
4
-1 -2 -3 -4
5
1 -2 3 -4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

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

#define ll long long
ll solve(ll idx, ll c, ll n, vector<ll> &a)
{
    if(idx>=n)
    {
        return c;
    }

    ll op1 = solve(idx+1,c+a[idx],n,a);
    ll op2 = solve(idx+1,abs(c+a[idx]),n,a);

    return max(op1,op2);
}
int main()
{
    ll t;
    cin>>t;

    while(t--)
    {
        ll n;
        cin>>n;

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

        int neg = 0;
        for(auto it : a)
        {
            if(it<0)
            {
                neg++;
            }
        }

        if(neg==0)
        {
            ll sum = accumulate(a.begin(),a.end(),0);
            cout<<sum<<endl;
            continue;
        }

        cout<<solve(0,0,n,a)<<endl;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
def solve(idx, value, array):
    if idx == len(array):
        return value

    a = solve(idx + 1, value + array[idx], array)
    b = solve(idx + 1, abs(value + array[idx]), array)
    return max(a, b)

def main():
    t = int(input())
    for _ in range(t):
        n = int(input())
        array = list(map(int, input().split()))

        result = solve(0, 0, array)

        print(result)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on scenarios where the decision between taking the absolute value or not can significantly impact the final result. The buggy code does not handle these scenarios correctly, leading to incorrect results.

### Steps to Create the Test Input:
1. **Edge Cases:**
   - Arrays with a mix of positive and negative values.
   - Arrays with large values to test the limits of the algorithm.
   - Arrays with zeros to test how the algorithm handles neutral values.

2. **Optimal Path:**
   - Create a scenario where the optimal path involves a mix of taking absolute values and not taking them.
   - Ensure that the final result is significantly different from the result obtained by always taking absolute values or never taking them.

### Example Test Input:
```plaintext
4
5
10 -10 10 -10 10
5
-10 10 -10 10 -10
3
0 0 0
4
1 -2 3 -4
```

### Explanation:
1. **Test Case 1:**
   - Array: `[10, -10, 10, -10, 10]`
   - Optimal path: `10 + (-10) = 0`, `|0 + 10| = 10`, `10 + (-10) = 0`, `|0 + 10| = 10`
   - Expected result: `10`

2. **Test Case 2:**
   - Array: `[-10, 10, -10, 10, -10]`
   - Optimal path: `|-10 + 10| = 0`, `0 + (-10) = -10`, `|-10 + 10| = 0`, `0 + (-10) = -10`
   - Expected result: `10`

3. **Test Case 3:**
   - Array: `[0, 0, 0]`
   - Optimal path: `0 + 0 = 0`, `0 + 0 = 0`, `0 + 0 = 0`
   - Expected result: `0`

4. **Test Case 4:**
   - Array: `[1, -2, 3, -4]`
   - Optimal path: `1 + (-2) = -1`, `|-1 + 3| = 2`, `2 + (-4) = -2`
   - Expected result: `2`

This test input should effectively expose the bug in the provided code by forcing it to make suboptimal decisions in the recursive function.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
ii = lambda: list(map(int, input().split()))
i = lambda: int(input())

for _ in range(i()):
    n = i()
    a = ii()
    mx, mn = 0, 0
    for x in a:
        mx = max(mx + x, abs(mn + x))
        mn = min(mn + x, abs(mx + x))
    # print(mx, mn)
    print(max(mx, abs(mn)))

**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 focus on edge cases and scenarios that can maximize the failure probability. The buggy code is supposed to calculate the maximum final value of \( c \) after applying the given operations. Let's analyze the code and identify potential issues:

1. **Initialization of `mx` and `mn`**: The code initializes `mx` and `mn` to 0, which is correct.
2. **Update logic for `mx` and `mn`**:
   - `mx` is updated to the maximum of `mx + x` and `abs(mn + x)`.
   - `mn` is updated to the minimum of `mn + x` and `abs(mx + x)`.

The bug in the code lies in the update logic for `mn`. Specifically, `abs(mx + x)` is always non-negative, which means `mn` will never be a negative value if `mx + x` is positive. This can lead to incorrect results in cases where the optimal solution involves maintaining a negative value in `mn`.

### Challenging Test Input
To expose this bug, we can create a test case where the optimal solution involves maintaining a negative value in `mn` and using it to compute the maximum value of `c`.

Consider the following test input:
```plaintext
1
5
1 -2 3 -4 5
```

### Explanation
- **Initial values**: `mx = 0`, `mn = 0`.
- **First element (1)**:
  - `mx = max(0 + 1, abs(0 + 1)) = 1`
  - `mn = min(0 + 1, abs(1 + 1)) = 1`
- **Second element (-2)**:
  - `mx = max(1 - 2, abs(1 - 2)) = max(-1, 1) = 1`
  - `mn = min(1 - 2, abs(1 - 2)) = min(-1, 1) = -1`
- **Third element (3)**:
  - `mx = max(1 + 3, abs(-1 + 3)) = max(4, 2) = 4`
  - `mn = min(-1 + 3, abs(4 + 3)) = min(2, 7) = 2`
- **Fourth element (-4)**:
  - `mx = max(4 - 4, abs(2 - 4)) = max(0, 2) = 2`
  - `mn = min(2 - 4, abs(4 - 4)) = min(-2, 0) = -2`
- **Fifth element (5)**:
  - `mx = max(2 + 5, abs(-2 + 5)) = max(7, 3) = 7`
  - `mn = min(-2 + 5, abs(7 + 5)) = min(3, 12) = 3`

The final value of `mx` is 7, which is the correct maximum final value of `c`.

However, the buggy code will produce a different result because it does not correctly handle the negative values in `mn`. This test case will help 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: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
t=int(input())
for i in range(t):
    n=int(input())
    l=list(map(lambda x:int(x),input().split()))
    c=0
    k=0
    j=0
    p=0
    k1=0
    for i in range(0,n):
        k=c+l[i]
        j=abs(c+l[i])
        if(k>=j):
            c=k
        else:
            c=j
        p+=l[i]

        k1+=abs(l[i])
    p=abs(p)
    if(p==0):
        print(k1)
    elif(c>p):
        print(c)
    else:
        print(p)

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

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

Think step by step.

Response:
```plaintext
1
5
-1 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: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

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

public class C1 {
    public static int[] arr;public static long max=0;
    public static void main(String [] args)throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(br.readLine());
        ArrayList<Long> arrlist=new ArrayList<>();
        for(int i=0;i<t;i++){
            max=0;
            int n=Integer.parseInt(br.readLine());
            StringTokenizer st=new StringTokenizer(br.readLine());

             arr=new int[n];
            for(int x=0;x<n;x++){
                arr[x]=Integer.parseInt(st.nextToken());
            }
            recurse(0, 0);
            arrlist.add(max);

        }
        for(long i: arrlist){
            System.out.println(i);
        }

    }

    public static void recurse(int i, long c){
        if(i==arr.length){
            max=Math.max(c, max);
        }
        else{
            long temp1=c+arr[i];
            recurse(i+1, temp1);
            long temp2=(long)Math.abs(c+arr[i]);
            recurse(i+1, temp2);
        }
    }

}
/*
1
4
-1000000000 1000000000 1000000000 1000000000
 */

**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 can highlight the issues in the recursive approach. The main issue in the buggy code is that it uses a simple recursive method without memoization, which can lead to exponential time complexity and potential stack overflow for large inputs.

Here are the steps to create a challenging test input:

1. **Large Number of Test Cases**: While the input constraints allow up to \(10^4\) test cases, we can start with a smaller number to keep the input manageable.
2. **Edge Cases for Array Length**: Test with the minimum and maximum allowed lengths for the array.
3. **Edge Cases for Array Values**: Include arrays with very large positive and negative values, and arrays with both positive and negative values.
4. **Specific Patterns**: Create arrays that force the algorithm to explore many paths, such as alternating positive and negative values.

Given these considerations, a small-scale but challenging test input can be:

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

### Explanation:
1. **Test Case 1**:
   - Array: \([-1000000000, 1000000000]\)
   - This tests the handling of very large positive and negative values.
   - The maximum value of \(c\) is \(1000000000\).

2. **Test Case 2**:
   - Array: \([-1, -2, -3]\)
   - This tests the handling of all negative values.
   - The maximum value of \(c\) is \(6\).

3. **Test Case 3**:
   - Array: \([1, 2, -3, 4]\)
   - This tests a mix of positive and negative values.
   - The maximum value of \(c\) is \(4\).

This input is small enough to be manageable but complex enough to potentially expose issues in the recursive approach of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
import java.util.*;
public final class Main
{
    static long max(long[]arr,int i,long c)
    {
        if(i==arr.length-1)

        {
            return Math.max(c+arr[i],Math.abs(c+arr[i]));
        }
       long noAbs=max(arr,i+1,c+arr[i]);
        long abs=max(arr,i+1,Math.abs(c+arr[i]));
        return Math.max(abs,noAbs);

    }
  public static void main (String[]args)
  {
	Scanner scr = new Scanner (System.in);
	int t = scr.nextInt ();
	for (int i = 0; i < t; i++)
	  {
	      int n=scr.nextInt();
	      long[]arr=new long[n];
	      for(int j=0;j<n;j++)
	      arr[j]=scr.nextLong();
	      long ans=max(arr,0,0);
        System.out.println(ans);

	  }
  }
}

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

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

Think step by step.

Response:
```plaintext
1
4
-10 10 -10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
for _ in range(int(input())):
  n = int(input())
  a = list(map(int, input().split()))
  c = [[0, 0] for _ in range(n)]
  c[0][0] = a[0]
  c[0][1] = abs(a[0])
  for i in range(1, n):
    c[i][0] = (c[i - 1][0] + a[i])
    c[i][1] = abs(c[i - 1][1] + a[i])
  print(max(c[-1][1], abs(c[-1][0])))

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

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

Think step by step.

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

1. **Edge Cases**: The code needs to handle cases where the array has negative and positive values, and where the absolute value operation might change the result significantly.
2. **Maximize Failure Probability**: We need to create a scenario where the buggy code's logic of always taking the absolute value at each step might lead to a suboptimal solution.

### Analysis:
- The buggy code calculates two values at each step: one without taking the absolute value and one with taking the absolute value.
- The final result is the maximum of the last values of these two arrays.
- The bug is that it might not always be optimal to take the absolute value at each step.

### Challenging Test Case:
- Consider an array where the initial values are negative, and the subsequent values are positive. This can force the buggy code to take the absolute value too early, leading to a suboptimal result.
- For example, an array like `[-1, -2, -3, 10, 10]` can expose the bug. If the code takes the absolute value after the first few negative values, it will not maximize the final value of `c`.

### Test Input:
```plaintext
1
5
-1 -2 -3 10 10
```

### Explanation:
- The correct solution should consider the option to delay taking the absolute value until the positive values are added, which can significantly increase the final value of `c`.
- The expected output for this test case is `20`, which is the maximum value of `c` after applying the optimal operations. The buggy code, however, might output a smaller value because it takes the absolute value too early.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
import heapq
t = int(input())
for i in range(t):
    n = int(input())
    a = list(map(int , input().split()))
    possible_c = [0]
    k = 1000

    for i in range(n):
        new_possible_c = []
        for c in possible_c:
            new_possible_c.append(c + a[i])
            new_possible_c.append(abs(c + a[i]))

        if len(new_possible_c) > k:
            new_possible_c = heapq.nlargest(k, new_possible_c)

        possible_c = new_possible_c

    print(max(possible_c))

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

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

Think step by step.

Response:
```plaintext
1
5
-1000000000 1000000000 -1000000000 1000000000 -1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

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

public class Main {
  static class FastReader {
      private BufferedReader br;
      private StringTokenizer st;

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

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

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

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

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

      public String nextLine() {
          String str = "";
          try {
              if(st.hasMoreTokens()) {
                  str = st.nextToken("\n");
              }
              else {
                  str = br.readLine();
              }
          }
          catch (IOException e) {
              e.printStackTrace();
          }
          return str;
      }
  }

    static class Pair {
        private int a, b;

        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }

        public int hashCode() {
            return Objects.hash(a, b);
        }

        public boolean equals(Object o) {
            if (o == null || getClass() != o.getClass()) {
                return false;
            }
            Pair pair = (Pair) o;
            return pair.a == a && pair.b == b;
        }
    }

    static class Triple {
        private int a, b, c;

        public Triple(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

        public int hashCode() {
            return Objects.hash(a, b, c);
        }

        public boolean equals(Object o) {
            if (o == null || getClass() != o.getClass()) {
                return false;
            }
            Triple triple = (Triple) o;
            return triple.a == a && triple.b == b && triple.c == c;
        }
    }

    private static int solve(int a[], int index) {
        if(index == -1)
            return 0;
        int next = solve(a, index - 1);
        return Math.max(next + a[index], Math.abs(next + a[index]));
    }
  public static void main(String[] args) {
    FastReader reader = new FastReader();
    StringBuilder answer = new StringBuilder();
    int t1 = reader.nextInt(), i, j, k, l;
    for(i = 0; i < t1; i += 1) {
        int n = reader.nextInt();
        int[] a = new int[n];
        int lastNegative = -1;
        for(j = 0; j < n; j += 1) {
            a[j] = reader.nextInt();
            if(a[j] < 0)
                lastNegative = j;
        }
        long current = 0;
        for(j = 0; j < n; j += 1) {
            current += a[j];
            if(j == lastNegative)
                current = Math.abs(current);
        }
        answer.append(current);
        answer.append("\n");
    }
    System.out.println(answer);
  }
}

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

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

Think step by step.

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

**Explanation:**

1. **Test Case 1:**
   - Array: `[1, -2, 1]`
   - The optimal strategy is to take the absolute value after the first and third additions:
     - Step 1: `c = 1`
     - Step 2: `c = |1 - 2| = 1`
     - Step 3: `c = 1 + 1 = 2`
   - Final value of `c` is `2`.

2. **Test Case 2:**
   - Array: `[-1, 1, -1, 1]`
   - The optimal strategy is to take the absolute value after each addition:
     - Step 1: `c = -1`
     - Step 2: `c = |-1 + 1| = 0`
     - Step 3: `c = |0 - 1| = 1`
     - Step 4: `c = |1 + 1| = 2`
   - Final value of `c` is `2`.

3. **Test Case 3:**
   - Array: `[1, -2, 2, -1, 1]`
   - The optimal strategy is to take the absolute value after the second and fourth additions:
     - Step 1: `c = 1`
     - Step 2: `c = |1 - 2| = 1`
     - Step 3: `c = 1 + 2 = 3`
     - Step 4: `c = |3 - 1| = 2`
     - Step 5: `c = 2 + 1 = 3`
   - Final value of `c` is `3`.

These test cases are designed to expose potential issues in the buggy code, especially in handling negative numbers and the timing of taking the absolute value.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
import java.util.*;
import static java.lang.System.clearProperty;
import static java.lang.System.out;
//ik its fucked up(the code) but noe more than my life, so
public class new1 {
     static int[] dp ;
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();

        while (t-- > 0) {
            solve(scanner);
        }
    }

    private static void solve(Scanner sc) {
        int n = sc.nextInt();
        int[] a = new int[n];
        long positive = 0; long neagtives = 0;
        for(int i = 0; i < n; i++ ) {
            a[i] = sc.nextInt();
            if( a[i] < 0) neagtives += a[i];
            else positive += a[i];
        }
//        a[n-1] = sc.nextInt();
        long c = 0;
        if( (-neagtives ) > positive ) { //negaties dominate
//            int c = 0;
            for( int i = 0; i < n; i++ ) {
                c = c + a[i];
            }
            out.println(Math.abs(c));
        }
        else {
            for( int i = 0; i < n ; i++ ) {
                c = Math.abs(c + a[i]);
//                out.print (c + " ");
            }
            out.println(Math.abs(c));
        }
        //we cant flip the value of c if its negative

//        if(possible || (x >= 5 && x <= 18)) out.println("YES");
//        else out.println("NO");
//        int[] a = new int[n];
//        for( int i = 0; i < n ; i++ ) a[i] = sc.nextInt();
//        boolean reduce = false;
//        int firstDigit = 0;
//        int lastDigit = (int)(x%10);
//        while( x >= 10 ) {
//            int digit = (int)(x%10);
//            if( reduce ) digit--;
//            if( x/10 < 10 && x /10 >= 1) firstDigit = (int)x;
//            if( digit != 0 ) reduce = true;
//            else reduce = false;
//            x = x/10;
//        }
////        out.println(firstDigit);
//        if( reduce) {
//            firstDigit--;
//        }
//        if( firstDigit >= 10 && firstDigit <= 18 && lastDigit != 9 ) out.println("YES");
//        else out.println("NO");
//            x = x/10;

        }
//        rec(a, 0, n-1);
//        for( int i = 0; i < n ; i++ ) {
//            out.print(dp[i] + " ");
//        }

    private static int rec(int[] a, int c, int i) {
        if( i == -1) return 0;
        int c1 = rec(a, c, i-1);
        if(dp[i] != (int) (-(1e9)-1) ) return dp[i];
        if( c1 + a[i] >= Math.abs(c1 + a[i])) {
            dp[i] = (c1 + a[i]);
        }
        else {
            dp[i] = Math.abs(c1 + a[i]);
        }
        return dp[i];
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code seems to have issues with handling negative and positive values correctly, especially when the negative values dominate the positive values.

Here are the steps to think through the test input:

1. **Negative Values Dominating**: The code incorrectly handles cases where the sum of negative values is greater than the sum of positive values. It sums all values and then takes the absolute value, which is not always the optimal approach.

2. **Positive Values Dominating**: The code correctly handles cases where positive values dominate, but it still takes the absolute value at each step, which can lead to suboptimal results.

3. **Mixed Values**: Cases where both positive and negative values are present and the optimal solution involves a combination of both options.

4. **Single Negative Value**: A single negative value in the array can also expose issues.

5. **Large Values**: Large positive and negative values can help test the limits of the code.

Given these considerations, let's create a small-scale test input that can expose the bug:

```plaintext
3
5
-10 10 10 10 -10
4
-1 -2 -3 -4
3
10 -20 10
```

### Explanation:
1. **Test Case 1**: `5 -10 10 10 10 -10`
   - This case has a mix of positive and negative values. The optimal solution is to take the absolute value after adding the first negative value to maximize the final result.
   - Expected output: `30`

2. **Test Case 2**: `4 -1 -2 -3 -4`
   - This case has only negative values. The optimal solution is to take the absolute value after the last addition.
   - Expected output: `10`

3. **Test Case 3**: `3 10 -20 10`
   - This case has a mix of positive and negative values with a single large negative value. The optimal solution is to take the absolute value after the first negative value.
   - Expected output: `20`

This test input should help expose the bug in the buggy code by testing various scenarios where the current implementation might fail.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;

public class exp {
    // use sc.next() instead of sc.nextLine()

    static long M = (long) (1e9 + 7);

    public static Scanner sc = new Scanner(System.in);

    public static class Pair implements Comparable<Pair> {
        long first;
        long second;

        Pair(long a, long b) {
            this.first = a;
            this.second = b;
        }

        @Override
        public int compareTo(Pair p2) {
            // if same fisrt occur sort on basis of the second
            if (this.first != p2.first) {
                return Long.compare(this.first, p2.first); // first based sorting (desc order)
            } else {
                return Long.compare(this.second, p2.second); // second based sorting (asc order)
            }
        }
    }

    public static void solve() {
        int n = sc.nextInt();
        long a[] = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
        }
        long c = 0;
        for (int i = 0; i < n - 1; i++) {
            c += a[i];
            if (c < 0 && a[i + 1] > 0) {
                c *= -1;
            }
        }
        c = Math.max(c + a[n - 1], Math.abs(c + a[n - 1]));
        System.out.println(c);
    }

    public static void main(String[] args) {
        long t = sc.nextLong();
        while (t-- > 0) {
            solve();
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic might fail. The main issue in the code is that it only checks if `c` is negative and the next element is positive to decide whether to take the absolute value. This logic can be flawed in certain edge cases.

Here are the steps to create a challenging test input:

1. **Identify the Bug**: The bug is in the condition `if (c < 0 && a[i + 1] > 0) { c *= -1; }`. This only considers the current value of `c` and the next element, but it doesn't consider the overall impact on the final value of `c`.

2. **Edge Cases**:
   - **Negative Sequence Followed by a Large Positive**: This can make the code take the absolute value prematurely, leading to a suboptimal result.
   - **Alternating Positive and Negative Values**: This can cause the code to take the absolute value multiple times, which might not be the best strategy.
   - **Large Initial Negative Values**: This can cause the code to take the absolute value early, which might not be optimal.

3. **Construct the Test Case**:
   - **Example 1**: A sequence with a large initial negative value followed by a large positive value.
   - **Example 2**: A sequence with alternating positive and negative values.

Let's construct a test input that includes these edge cases:

```plaintext
3
4
-10 10 -10 10
5
-100 50 -50 100 200
3
-1 1 -1
```

### Explanation:
1. **Test Case 1**: `4 -10 10 -10 10`
   - The code might take the absolute value after the first element, leading to a suboptimal result. The optimal result is `10` by taking the absolute value at the end.

2. **Test Case 2**: `5 -100 50 -50 100 200`
   - The code might take the absolute value after the first element, leading to a suboptimal result. The optimal result is `200` by taking the absolute value at the end.

3. **Test Case 3**: `3 -1 1 -1`
   - The code might take the absolute value multiple times, leading to a suboptimal result. The optimal result is `1` by taking the absolute value at the end.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Magnitude (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions of the problem are different. You may want to read both versions. You can make hacks only if both versions are solved.

You are given an array $a$ of length $n$. Start with $c = 0$. Then, for each $i$ from $1$ to $n$ (in increasing order) do exactly one of the following:

  * Option $1$: set $c$ to $c + a_i$.    * Option $2$: set $c$ to $|c + a_i|$, where $|x|$ is the absolute value of $x$.

Let the maximum final value of $c$ after the procedure described above be equal to $k$. Find $k$.

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

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

The second line of each case contains $n$ integers $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ ($-10^9 \leq a_i \leq 10^9$).

The sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the value of $k$.

Examples:
input:
5
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
4
-1000000000 1000000000 1000000000 1000000000
4
1 9 8 4
output:
6
24
6
4000000000
22

Note: In the first test case, if we set $c$ to its absolute value every time we add to it, we end up with $6$. It can be shown that this is the maximum result.

In the second test case, taking the absolute value will never change anything, so we can just sum the array without doing anything to get $24$.

In the third test case, it is optimal to wait until the end to set $c$ to its absolute value, resulting in an answer of $6$.

Buggy Code:
import java.util.Scanner;

public class magnitude {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int nt=sc.nextInt();
        for(int i=0;i<nt;i++){
         int len=sc.nextInt();

         int arr[]=new int[len];

         long tneg=0;
         long tpos=0;

         for(int j=0;j<len;j++){
           arr[j]=sc.nextInt();
           if(arr[j]>0){
            tpos+=arr[j];
           }
           else{
            tneg+=arr[j];
           }
        }

         long c=0;

         for(int j=0;j<len;j++){
         long temp=arr[j];
         if(temp==0){
          continue;
         }
         if(temp>0){
          tpos=tpos-temp;
         }
         else{
          tneg=tneg-temp;
         }

         c+=temp;

         if(Math.abs(tneg)<tpos){
          c=Math.abs(c);
         }

         }

         c=Math.abs(c);

         System.out.println(c);

        }
    }
}

**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 code, we need to focus on scenarios where the logic for deciding when to take the absolute value of `c` is flawed. The code seems to have issues with handling negative and positive values correctly, especially when the sum of negative values (`tneg`) and positive values (`tpos`) are compared.

Here's a step-by-step approach to generating a challenging test input:

1. **Mixed Positive and Negative Values**: Ensure the array contains both positive and negative values to test the logic of when to take the absolute value.
2. **Large Magnitudes**: Use large values to test the limits of the code and ensure it handles large sums correctly.
3. **Zero Values**: Include zero values to test how the code handles them.
4. **Edge Cases**: Test with small arrays to ensure the code works for edge cases.

Given these considerations, here is a small-scale test input that should expose the bug:

```plaintext
3
4
10 -9 -3 4
8
1 4 3 4 1 4 3 4
3
-1 -2 -3
```

This input includes:
- A test case with mixed positive and negative values.
- A test case with all positive values.
- A test case with all negative values.

These cases should help identify any issues in the logic of the buggy code.