**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 os
import sys
from io import BytesIO, IOBase

def solve():
    n = int(input())
    arr = list(map(int, input().split()))

    ans = 0

    for num in arr:
        ans = abs(ans+num)

    second = 0

    for num in arr:
        second += num

    print(max(ans, abs(second)))

def main():
    tc = 1
    tc = int(input())
    for _ in range(tc):
        solve()

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin = IOWrapper(sys.stdin)
sys.stdout = IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

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 buggy code, we need to consider the following points:

1. **Understanding the Bug**: The buggy code always takes the absolute value of the sum of the array elements in one scenario and the sum of the array elements in another scenario. It then prints the maximum of these two values. The bug is that it does not consider the optimal way to use the absolute value operation to maximize the final value of `c`.

2. **Edge Cases**:
   - **All Positive Numbers**: The code will work correctly if all numbers are positive.
   - **All Negative Numbers**: The code will work correctly if all numbers are negative.
   - **Mixed Positive and Negative Numbers**: The code may fail if the array contains both positive and negative numbers, especially when the sum of the array is negative and the absolute value operation can significantly increase the final value.

3. **Challenging Scenario**:
   - A scenario where the sum of the array is negative, but by strategically using the absolute value operation, the final value of `c` can be significantly higher.

Given these points, a challenging test input would be an array where the sum of the elements is negative, but by taking the absolute value at the right time, the final value of `c` can be maximized.

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

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

**Explanation**:
- The array is `[-1, -2, 3, 4, -10]`.
- The sum of the array is `-1 - 2 + 3 + 4 - 10 = -6`.
- If we always take the absolute value, the final value of `c` will be `6`.
- However, if we strategically use the absolute value operation, we can achieve a higher value. For example:
  - Start with `c = 0`.
  - Add `-1` to `c`, `c = -1`.
  - Add `-2` to `c`, `c = -3`.
  - Add `3` to `c`, `c = 0`.
  - Add `4` to `c`, `c = 4`.
  - Add `-10` to `c`, `c = -6`.
  - Take the absolute value of `c`, `c = 6`.

The correct output for this test case should be `6`, but the buggy code will output `6` as well, which is correct in this case. To make it more challenging, consider the following input:

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

**Explanation**:
- The array is `[1, -2, -3, 4]`.
- The sum of the array is `1 - 2 - 3 + 4 = 0`.
- If we always take the absolute value, the final value of `c` will be `0`.
- However, if we strategically use the absolute value operation, we can achieve a higher value. For example:
  - Start with `c = 0`.
  - Add `1` to `c`, `c = 1`.
  - Take the absolute value of `c`, `c = 1`.
  - Add `-2` to `c`, `c = -1`.
  - Take the absolute value of `c`, `c = 1`.
  - Add `-3` to `c`, `c = -2`.
  - Take the absolute value of `c`, `c = 2`.
  - Add `4` to `c`, `c = 6`.

The correct output for this test case should be `6`, but the buggy code will output `4` because it only considers the sum and the absolute value of the sum.

Therefore, the final test input is:

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


**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 sys
# sys.setrecursionlimit(5*10**5)
input = sys.stdin.readline
from collections import defaultdict, deque, Counter
from heapq import heappop, heappush
from bisect import bisect_left, bisect_right
from math import gcd

def sol():
    n = int(input())
    a = list(map(int,input().split()))

    lr = []
    for i in range(n):
        if i == 0:
            lr.append(a[i])
        else:
            if a[i] <= 0 and a[i-1] <= 0:
                lr[-1] += a[i]
            elif a[i] >= 0 and a[i-1] >= 0:
                lr[-1] += a[i]
            else:
                lr.append(a[i])

    now = 0
    for i in lr:
        now += i
        now = abs(now)

    print(now)
    return

T = int(input())
# T = 1
for i in range(T):
    sol()

**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 for handling consecutive positive or negative values might fail. The key issue in the buggy code is that it incorrectly merges consecutive values with the same sign, which can lead to suboptimal results.

Let's consider the following steps:

1. **Identify the Bug**: The bug in the code is that it merges consecutive values with the same sign, which can lead to incorrect results. For example, if the array has consecutive negative values, the code will sum them up, which might not be the optimal way to maximize the final value of \( c \).

2. **Create a Test Case**: We need a test case where the optimal solution involves not merging consecutive values with the same sign. A good candidate is an array with alternating positive and negative values, or an array where the optimal solution involves taking the absolute value at specific points to maximize \( c \).

3. **Test Case Example**: Consider an array where the values alternate between positive and negative, and the optimal solution involves taking the absolute value at specific points to maximize \( c \).

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

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

### Explanation:
- **Array**: \([-1, 2, -3, 4, -5]\)
- **Optimal Solution**:
  - Start with \( c = 0 \).
  - Add \(-1\): \( c = -1 \).
  - Add \( 2 \): \( c = 1 \).
  - Take the absolute value: \( c = 1 \).
  - Add \(-3\): \( c = -2 \).
  - Take the absolute value: \( c = 2 \).
  - Add \( 4 \): \( c = 6 \).
  - Take the absolute value: \( c = 6 \).
  - Add \(-5\): \( c = 1 \).
  - Take the absolute value: \( c = 1 \).

The maximum value of \( c \) is \( 6 \).

### Expected Output:
```plaintext
6
```

This test case should expose the bug in the provided code because the code will incorrectly merge consecutive values with the same sign, leading to a suboptimal result.


**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 int
#define ld long double
#define ek ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

void init()
{

}

void solve()
{
    ll n;
    cin>>n;
    vector<ll> v(n);
    for(ll i=0;i<n;i++)
    {
        cin>>v[i];
    }
    ll c=0;
    ll sum=0;
    for(ll i=0;i<n;i++)
    {
        if(c<=0&&v[i]<0)
        {
            sum+=v[i];
        }
        else
        {
            c=abs(c+v[i]);
        }
    }
    c+=abs(sum);
    cout<<abs(c)<<endl;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    ek;
    ll t = 1;
    cin >> t;
    init();
    while(t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the logic of the code and identify scenarios where it might fail. The bug in the code seems to be in how it handles negative numbers and the decision-making process for when to take the absolute value.

### Analysis of the Buggy Code:
- The code initializes `c` to 0 and `sum` to 0.
- It iterates through the array and updates `c` based on the condition `if(c <= 0 && v[i] < 0)`.
- If the condition is true, it adds `v[i]` to `sum`.
- Otherwise, it sets `c` to `abs(c + v[i])`.
- Finally, it adds `abs(sum)` to `c` and prints `abs(c)`.

### Identifying the Bug:
- The code might fail when there are multiple negative numbers in the array, especially when the sequence of negative and positive numbers is not straightforward.
- The code does not consider the optimal way to maximize the value of `c` by strategically taking the absolute value at the right moments.

### Challenging Test Case:
- We need a test case where the sequence of numbers is such that the buggy code does not correctly maximize the value of `c`.
- A sequence with a mix of positive and negative numbers, where the optimal strategy is not to take the absolute value immediately, can expose the bug.

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

### Explanation:
- For the input `[-1, 2, -3, 4, -5]`:
  - The optimal strategy is to:
    1. Set `c` to `c + (-1) = -1`.
    2. Set `c` to `c + 2 = 1`.
    3. Set `c` to `|c + (-3)| = 2`.
    4. Set `c` to `c + 4 = 6`.
    5. Set `c` to `|c + (-5)| = 1`.
  - The maximum value of `c` is `6`.
- The buggy code, however, might not follow this optimal strategy, leading to a suboptimal result.

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:
//package org.example.div2;

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Magnitude {

    static Map<String,Long> map;

    public static void main(String[] args) {

        Scanner sc =  new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0) solveT(sc);

    }

    public static void solveT(Scanner sc){
        int n = sc.nextInt();
        map = new HashMap<>();

        long[] nums = new long[n];

        for(int i = 0 ; i < n ; i++){
            nums[i] = sc.nextInt();
        }
        long ans = Math.max(helper(1,nums[0],nums) , helper(1,Math.abs(nums[0]),nums));
        System.out.println(ans);

    }

    public static long helper(int i, long c, long[] nums){

        if(i >= nums.length) return c;

        if(map.containsKey(i+""+c)) return map.get(i+""+c);
        long ans  = 0;

        long option1 = helper(i+1,c+nums[i], nums ) ;
        long option2 = helper(i+1,Math.abs(c+nums[i]), nums ) ;

        ans = Math.max(option1, option2);
        map.put(i+""+c,ans);
        return 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:
To expose the bug in the provided code, we need to focus on scenarios where the dynamic programming approach might fail to correctly compute the maximum value of \( c \). The key issue in the code is the use of a map to store intermediate results, which might not handle negative values of \( c \) correctly due to the string key format.

### Key Points to Consider:
1. **Negative Values**: The map keys are strings formed by concatenating the index and the value of \( c \). This can lead to issues when \( c \) is negative, as the string representation of negative numbers might not be handled correctly.
2. **Large Values**: The code should handle large values of \( a_i \) and \( c \) correctly, which can be in the range of \([-10^9, 10^9]\).
3. **Edge Cases**: Consider edge cases where the array \( a \) has a mix of positive and negative values, or where the array is very small.

### Test Input:
To expose the bug, we can create a test case where the array \( a \) has a mix of positive and negative values, and the optimal solution involves taking the absolute value at a specific point to maximize \( c \).

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

### Explanation:
- **Test Case 1**: The array \( a \) is `[1, -2, 3, -4]`.
- **Optimal Solution**:
  - Start with \( c = 0 \).
  - Add \( a_1 = 1 \): \( c = 1 \).
  - Take the absolute value of \( c + a_2 = 1 + (-2) = -1 \): \( c = 1 \).
  - Add \( a_3 = 3 \): \( c = 4 \).
  - Take the absolute value of \( c + a_4 = 4 + (-4) = 0 \): \( c = 4 \).

The maximum value of \( c \) after the procedure is \( 4 \).

This test case should expose the bug in the dynamic programming approach if the map key handling is not robust enough to handle negative values correctly.


**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 cp {
  public static void main(String[] args) throws IOException {
    if (System.getProperty("ONLINE_JUDGE") == null) {
      PrintStream ps = new PrintStream(new File("output.txt"));
      InputStream is = new FileInputStream("input.txt");
      System.setIn(is);
      System.setOut(ps);
    }
    Scanner sc = new Scanner(System.in);
    int testcases = sc.nextInt();
    while (testcases-- > 0) {
      solve(sc);
    }
    sc.close();
  }

  public static long helper (int[] arr,int idx,long sum,HashMap<String,Long> dp) {
    if (idx == arr.length) {
      return sum;
    }
    String key = idx + "" + sum;
    if (dp.containsKey(key)) {
      return dp.get(key);
    }
    long withoutmod = helper (arr,idx+1,sum+arr[idx],dp);
    long with = helper (arr,idx+1,Math.abs(sum+arr[idx]),dp);

    long ans = Math.max(withoutmod,with);
    dp.put(key, ans);
    return ans;
  }

  public static void solve(Scanner sc) {
    int n = sc.nextInt();
    int[] arr = new int[n];
    for (int i=0;i<n;i++) {
      arr[i] = sc.nextInt();
    }
    HashMap <String,Long> dp = new HashMap<>();
    long ans = helper(arr,0,0,dp);
    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:
To generate a challenging test input that exposes the bug in the buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. The key points to focus on are:

1. **Large Values**: The input can contain very large positive and negative values to test the limits of the algorithm.
2. **Negative Values**: The input can contain a sequence of negative values to test how the algorithm handles such cases.
3. **Mixed Values**: The input can contain a mix of positive and negative values to test the decision-making process of the algorithm.
4. **Edge Cases**: The input can be very small or very large to test the algorithm's performance and correctness.

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

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

### Explanation:
1. **Test Case 1:**
   - Input: `4 -1 -2 -3 -4`
   - Expected Output: `10`
   - Reason: The maximum value of `c` can be achieved by always taking the absolute value, resulting in `10` (i.e., `|1| + |2| + |3| + |4|`).

2. **Test Case 2:**
   - Input: `4 1 -1 1 -1`
   - Expected Output: `4`
   - Reason: The maximum value of `c` can be achieved by alternating between adding and taking the absolute value, resulting in `4` (i.e., `1 + |-1 + 1| + |-1|`).

3. **Test Case 3:**
   - Input: `4 1000000000 -1000000000 1000000000 -1000000000`
   - Expected Output: `4000000000`
   - Reason: The maximum value of `c` can be achieved by always taking the absolute value, resulting in `4000000000` (i.e., `|1000000000| + |1000000000| + |1000000000| + |1000000000|`).

This test input covers a range of scenarios, including negative values, mixed values, and large values, which should help expose any bugs in the faulty solution.


**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;
using ll = long long;
using pi = pair<int, int>;
ll MOD = 998244353;

void solve() {
    int n; cin >> n;
    vector<ll> a(n+1);
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }

    vector<ll> dp_pos(n+1, -1e10), dp_neg(n+1, 1e10);
    vector<ll> ways_pos(n+1), ways_neg(n+1);
    // dp[0] = ways[0] = {0, 0};
    ways_pos[0] = ways_neg[0] = 1;
    dp_pos[0] = 0;
    dp_neg[0] = 0;
    for (int i = 1; i <= n; ++i) {
        // positive
        // can come from pos(i-1) or neg(i-1)
        // list all four options
        vector<pair<ll, ll>> ops;

        dp_pos[i] = max(dp_pos[i], dp_pos[i-1] + a[i]);
        dp_pos[i] = max(dp_pos[i], abs(dp_pos[i-1] + a[i]));
        dp_pos[i] = max(dp_pos[i], dp_neg[i-1] + a[i]);
        dp_pos[i] = max(dp_pos[i], abs(dp_neg[i-1] + a[i]));

        dp_neg[i] = min(dp_neg[i], dp_pos[i-1] + a[i]);
        dp_neg[i] = min(dp_neg[i], abs(dp_pos[i-1] + a[i]));
        dp_neg[i] = min(dp_neg[i], dp_neg[i-1] + a[i]);
        dp_neg[i] = min(dp_neg[i], abs(dp_neg[i-1] + a[i]));
    }

    // cout << ways_pos[n] % MOD << endl;
    cout << dp_pos[n] << endl;
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);

    int tc; cin >> tc;
    while (tc--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4
10 -20 10 -5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

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

int N = 998244353;
int dp[1001][55001];
signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    vector<int> f = {1,1};
    for(int i=0; i<22; i++){
        f.push_back(f[f.size()-1] + f[f.size() - 2]);
    }
    int n,x,m;
    cin >> n >> x >> m;
    unordered_set<int> p;dp[0][0] = 1;p = {0};
    for(int i=0; i<x; i++){
        //cout << i << "\n";
        for(int j=0; j<n; j++){
            for (auto d: p){
                if(dp[j][d] == 0)continue;
                dp[j+1][d + f[i]] += dp[j][d];
                p.insert(d + f[i]);
                dp[j+1][d + f[i]] %= N;
            }
        }
    }
    int ans=0;
    for(auto t : p){
        int r=0;
        int o = t;
        for(int i=f.size()-1; i>0; i--){
            if(f[i] <= o){
                o -=f[i];
                r++;
            }
        }
        if(r==m){
            ans+=dp[n][t];
            ans%=N;
        }
    }
    cout << ans << "\n";
}

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

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

Think step by step.

Response:
```plaintext
1 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

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

using namespace std;

using ll = long long;
const int INF = 1e9 + 9;
const int N = 2e5 + 7;

void solve() {
    int mod = 998244353;
    int n , m , x , X , k;
    cin >> n >> X >> m;
    x = 30;
    function<int(int , int , int)> ad_ = [&](int x , int y , int mod){
        x += y;
        if(x > mod) x -= mod;
        if(x < 0) x += mod;
        return x;
    };

    function<int(int , int , int)> mu_ = [&](int x , int y , int mod){
        return (1ll * x * y) % mod;
    };

    vector<int>fib(x + 1);
    fib[1] = 1;
    for(int i = 2;i <= x;i ++){
        fib[i] = fib[i - 1] + fib[i - 2];
    }
    int M = fib[x] * n;
    vector<int>cost(M + 1);
    for(int i = 1;i <= M;i ++){
        cost[i] = i;
        for(int j = 1;j <= x && fib[j] <= i;j ++){
            cost[i] = min(cost[i] , cost[i - fib[j]] + 1);
        }
    }
    vector<vector<int>>dp(n + 1 , vector<int>(M + 1));
    dp[0][0] = 1;
    for(int i = 1;i <= X;i ++ ){
        for(int c = 1; c <= n;c ++){
            for(int j = fib[i];j <= M;j ++){
                dp[c][j] = ad_(dp[c][j] , dp[c - 1][j - fib[i]] , mod);
            }
        }
    }
    int ans = 0;
    for(int i = 0;i <= M ; i++){
        if(cost[i] == m){
            ans = ad_(ans , dp[n][i] , mod);
        }
    }
    cout << ans << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;

    //cin >> T;

    while (T--) {
        solve();
    }

}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to consider the following aspects:

1. **Edge Cases**: Test with the smallest and largest possible values for `n`, `x`, and `m`.
2. **Boundary Conditions**: Test with `n` equal to `m` and `m` equal to 1.
3. **Large Values**: Test with large values of `n` and `x` to see if the algorithm handles them correctly.
4. **Specific Combinations**: Test with specific combinations of `n`, `x`, and `m` that might reveal issues in the dynamic programming or cost calculation.

Given the constraints:
- \(1 \le m \le n \le 1000\)
- \(2 \le x \le 10\)

Let's consider a small-scale test input that is complete and valid:

### Test Input
```plaintext
10 2 2
```

### Explanation
- **n = 10**: A moderate number of chips.
- **x = 2**: The smallest possible value for `x` other than 1, which can simplify the problem but still test the algorithm.
- **m = 2**: A specific cost value that is neither the minimum nor the maximum possible cost.

This input should be challenging enough to expose any issues in the dynamic programming and cost calculation parts of the code. It is also small enough to be manageable and verifiable.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

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

int N = 998244353;
int dp[1001][55001];
signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    vector<int> f = {1,1};
    for(int i=0; i<22; i++){
        f.push_back(f[f.size()-1] + f[f.size() - 2]);
    }
    int n,x,m;
    cin >> n >> x >> m;
    unordered_set<int> p;dp[0][0] = 1;p = {0};
    for(int i=0; i<x; i++){
        //cout << i << "\n";
        for(int j=0; j<n; j++){
            for (auto d: p){
                if(d>55000){
                    cout << d << "\n";
                }
                if(dp[j][d] == 0)continue;
                dp[j+1][d + f[i]] += dp[j][d];
                p.insert(d + f[i]);
                dp[j+1][d + f[i]] %= N;
            }
        }
    }
    int ans=0;
    for(auto t : p){
        int r=0;
        int o = t;
        for(int i=f.size()-1; i>0; i--){
            if(f[i] <= o){
                o -=f[i];
                r++;
            }
        }
        if(r==m){
            ans+=dp[n][t];
            ans%=N;
        }
    }
    cout << ans << "\n";
}

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

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

Think step by step.

Response:
```plaintext
10 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
MOD = 998244353

def solve(n, x, m):
    # Memoization dictionary
    dp = {}

    def dfs(remaining_chips, current_point, min_cost):
        # Check if the state has been computed before
        if (remaining_chips, current_point, min_cost) in dp:
            return dp[(remaining_chips, current_point, min_cost)]

        # Base case: no more chips to place
        if remaining_chips == 0:
            return 1 if min_cost == 0 else 0

        # If we run out of points or exceed the minimum cost requirement
        if current_point > x or min_cost < 0:
            return 0

        # Explore placing chips at current point
        total_ways = 0
        for i in range(min(remaining_chips + 1, current_point + 1)):
            # Place 'i' chips at current point
            next_remaining = remaining_chips - i
            # Calculate new minimum cost after placing 'i' chips
            next_min_cost = max(min_cost - i, 0)
            total_ways += dfs(next_remaining, current_point + 1, next_min_cost)

        # Store the result in the memoization dictionary
        dp[(remaining_chips, current_point, min_cost)] = total_ways % MOD
        return total_ways

    return dfs(n, 1, m)

# Example usage
if __name__ == "__main__":
    import sys
    input = lambda: sys.stdin.readline().strip()
    n, x, m = map(int, input().split())
    print(solve(n, x, m))

**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
10 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
mo = 998244353
S = 55000
fib = [0,1,1,2,3,5,8,13,21,34,55]
while True:
  x = fib[-1] + fib[-2]
  if x > 55000:
    break
  fib.append(x)

C = [0 for s in range(S + 1)]
for s in range(1,S + 1):
  for v in fib[::-1]:
    if s < v:
      continue
    C[s] = C[s - v] + 1
    break

N,x,m = map(int,input().split())
SS = S + 1
dp = [0 for _ in range(SS * (N + 1))]
dp[0] = 1
def f(n,s):
  return n * SS + s

for c in range(1,x + 1):
  v = fib[c]
  for n in range(N):
    for s in range(S + 1):
      if s + v > S:
        continue
      dp[f(n + 1,s + v)] = (dp[f(n + 1,s + v)] + dp[f(n,s)]) % mo

ans = 0
for s in range(S + 1):
  if C[s] == m:
    ans = (ans + dp[f(N,s)]) % mo
print(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 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
MOD = 998244353

def add(a, b):
    """Add two numbers modulo MOD."""
    return (a + b) % MOD

def mul(a, b):
    """Multiply two numbers modulo MOD."""
    return (a * b) % MOD

def powmod(base, exponent):
    """Calculate base^exponent modulo MOD."""
    result = 1
    while exponent > 0:
        if exponent % 2 == 1:
            result = mul(result, base)
        base = mul(base, base)
        exponent //= 2
    return result

def inverse(x):
    """Calculate the modular multiplicative inverse of x."""
    return powmod(x, MOD - 2)

def comb(n, k):
    """Calculate the binomial coefficient nCk."""
    if k < 0 or k > n:
        return 0
    numerator = 1
    denominator = 1
    for i in range(k):
        numerator = mul(numerator, n - i)
        denominator = mul(denominator, i + 1)
    return mul(numerator, inverse(denominator))

def solve():
    n, x, m = map(int, input().split())

    # Initialize DP table
    dp = [[[0] * (m + 1) for _ in range(n + 1)] for _ in range(x + 1)]

    # Base case: One way to place zero chips
    for i in range(1, x + 1):
        dp[i][0][0] = 1

    # Fill up DP table
    for i in range(1, x + 1):
        for j in range(1, n + 1):
            for k in range(m + 1):
                # Don't place any chips at point i
                dp[i][j][k] = add(dp[i][j][k], dp[i - 1][j][k])

                # Place j chips at point i
                if i <= j:
                    dp[i][j][k] = add(dp[i][j][k], comb(j - 1, i - 1))

                # Move a chip from point i to points i-1 and i-2
                if i >= 3 and j >= 2:
                    dp[i][j][k] = add(dp[i][j][k], dp[i - 2][j - 2][max(0, k - 1)])

                # Merge two chips from points i-1 and i
                if i >= 2 and j >= 2:
                    dp[i][j][k] = add(dp[i][j][k], dp[i - 1][j - 2][max(0, k - 1)])

    # Return the result
    return dp[x][n][m]

print(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 edge cases and scenarios that are likely to cause the algorithm to fail. Here are some considerations:

1. **Small Values for \( n \) and \( x \)**: Small values can help us understand the behavior of the algorithm more clearly.
2. **Special Cases for \( m \)**: Values of \( m \) that are close to \( n \) or very small can reveal issues in the algorithm's logic.
3. **Boundary Conditions**: Test the boundary conditions for \( n \), \( x \), and \( m \) to ensure the algorithm handles them correctly.

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

- **\( n = 5 \)**: A small number of chips to keep the problem manageable.
- **\( x = 3 \)**: A small number of points to simplify the placement.
- **\( m = 2 \)**: A value for \( m \) that is neither too small nor too close to \( n \), but still challenging.

This input will help us test the algorithm's ability to handle small-scale scenarios and edge cases.

```plaintext
5 3 2
```

This test input should expose any issues in the algorithm's logic for small values of \( n \), \( x \), and \( m \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

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

const int N = 1e3 + 5, INF = 1e9 + 7, MOD = 998244353;
int M = 55 * N;

int n, x, m, fib[41];

int dp1[55 * N + 1], dp2[N][55 * N + 1];

void add(int &x, int y){
    x += y;
    while(x > MOD) x -= MOD;
}
signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    #ifdef LOCAL
    freopen("local.inp", "r", stdin);
    #endif // LOCAL
    fib[1] = 1, fib[2] = 1;
    for(int i = 3; i <= 40; i++) fib[i] = fib[i - 1] + fib[i - 2];
    cin >> n >> x >> m;

    M = n * fib[x];
    fill(dp1 + 1, dp1 + M, INF);
    dp1[0] = 0;
    for(int i = 1; i <= M; i++){
        for(int j = 1; j <= 40 && fib[j] <= i; j++){
            dp1[i] = min(dp1[i], dp1[i - fib[j]] + 1);
        }
    }

    dp2[0][0] = 1;
    for(int i = 1; i <= x; i++){
        for(int j = 1; j <= n; j++){
            for(int k = fib[i]; k <= M; k++){
                add(dp2[j][k], dp2[j - 1][k - fib[i]]);
            }
        }
    }

    int ans = 0;
    for(int weight = 1; weight <= M; weight++){
        if(dp1[weight] == m){
            add(ans, dp2[n][weight]);
        }
    }
    cout << ans << '\n';
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
10 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
#include <bits/stdc++.h>
#define pb push_back
#define fst first
#define snd second
#define fore(i,a,b) for(ll i=a,jet=b;i<jet;++i)
#define forr(i,a,b) for(ll i=b-1,jet=a;i>=jet;--i)
#define SZ(x) ((int)x.size())
#define ALL(x) x.begin(),x.end()
#define mset(a,v) memset((a),(v),sizeof(a))
#define FIN ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define imp(v) {for(auto gdljh:v)cout<<gdljh<<" "; cout<<"\n";}
using namespace std;
typedef long long ll;
typedef pair<ll,ll> ii;
const ll MOD=998244353,MAXN=1005,MAXS=55005,MAXX=12,FB=24;
int add(int a, int b){a+=b;if(a>=MOD)a-=MOD;return a;}
int sub(int a, int b){a-=b;if(a<0)a+=MOD;return a;}
int mul(ll a, ll b){return a*b%MOD;}
int fpow(int a, ll b){
	int r=1;
	while(b){if(b&1)r=mul(r,a); b>>=1; a=mul(a,a);}
	return r;
}

ll N,M,x;
ll dp[2][MAXS][MAXX];
ll dp2[2][MAXS][FB];
bool can[MAXN][MAXS];

int main(){FIN;
	vector<ll>fib={1,1};
	while(fib.back()<=55000)fib.pb(fib.back()+fib.end()[-2]);
	fib.pop_back();
	// imp(fib);
	// cout<<SZ(fib)<<"\n";;
	// cout<<fib[9]<<"\n";
	cin>>N>>x>>M;
	ll S=N*fib[x-1]+3;
	fore(i,0,FB)dp2[0][0][i]=1;
	can[0][0]=1;
	fore(m,1,M+1)fore(s,0,S)forr(i,0,FB){
		ll w=m&1;
		ll &res=dp2[w][s][i];
		if(s&&fib[i]>s){res=0;continue;}
		res=0;
		if(i+1<FB)res=dp2[w][s][i+1];
		if(fib[i]<=s)res=add(res,dp2[w^1][s-fib[i]][i]);

		can[m][s]|=res>0;

	}
	vector<ll>val(S);
	fore(s,0,S){
		val[s]=dp2[M&1][s][0];
		fore(m,0,M)if(can[m][s])val[s]=0;
		// if(val[s])cout<<s<<": "<<val[s]<<"\n";
	}
	fore(i,0,x)fore(s,0,S)dp[0][s][i]=val[s];
	fore(n,1,N+1)forr(s,0,S)forr(i,0,x){
		ll w=n&1;
		ll &res=dp[w][s][i];
		res=0;
		if(i+1<x)res=dp[w][s][i+1];
		if(s+fib[i]<S)res=add(res,dp[w^1][s+fib[i]][i]);
	}

	ll res=dp[N&1][0][0];
	cout<<res<<"\n";
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int dp[56000][1005], fib[30];

signed main(){
    cin.tie(0)->sync_with_stdio(0);

    fib[0] = fib[1] = 1;
    for(int i=2;i<30;i++){
        fib[i] = fib[i-1] + fib[i-2];
    }
    int n,x,m;
    cin >> n >> x >> m;
    dp[0][0] = 1;
    for(int i=0;i<x;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k+fib[i]<=fib[x-1]*n;k++){
                dp[k+fib[i]][j+1] += dp[k][j];
                dp[k+fib[i]][j+1] %= mod;
            }
        }
        // cout << i << " " << fib[i] << "\n";
        // for(int j=0;j<=n;j++){
        //     for(int k=0;k<=fib[x-1]*n;k++){
        //         cout << dp[k][j] << " ";
        //     }
        //     cout << "\n";
        // }
    }
    int ans = 0;
    for(int i=0;i<=fib[x-1]*n;i++){
        int j = i, k = 29, l = 0;
        while(j){
            if(j >= fib[k]){
                l++;
                j -= fib[k];
            }
            k--;
        }
        // cout << i << " " << l << "\n";
        if(l == m){
            ans = (ans + dp[i][n]) % mod;
        }
    }
    cout << ans << "\n";
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <cassert>
#include <functional>
#include <stack>
#include <random>
#define all(arr) arr.begin(), arr.end()
#define ll long long
#define MS(arr, val) memset(arr, val, sizeof(arr))
#define vi vector<int>
#define pb emplace_back
#define PI pair<int, int>
#define PL pair<ll, ll>
#define lll __int128
#define ull unsigned long long
#define Kases int T; rd(T); rep(kase, 1, T)
#define rep(i, a, b) for(int i = (a), __ ## i = (b); i <= __ ## i; ++i)
#define per(i, b, a) for(int i = (b), __ ## i = (a); i >= __ ## i; --i)

using namespace std;

// static char buf[1000000],*p1=buf,*p2=buf,obuf[1000000],*p3=obuf;
// #define getchar() p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++
// #define putchar(x) (p3-obuf<1000000)?(*p3++=x):(fwrite(obuf,p3-obuf,1,stdout),p3=obuf,*p3++=x)
inline void rd(char* x){scanf("%s", x);}
inline void rd(double &x){scanf("%lf", &x);}
// inline void rd(int &x){scanf("%d", &x);}
template<typename T> inline void rd(T &x){
    x = 0; bool is = 0; char ch = getchar();
    while(!isdigit(ch)) is |= (ch == '-'), ch = getchar();
    while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    is && (x = -x);
}
// template<typename T>void write(T x){
//     x < 0 && (x = -x, putchar('-')); if(x >= 10) write(x / 10); putchar(x % 10 + '0');
// }
template<typename T, typename ...U> inline void rd(T &x, U &...y){rd(x), rd(y...);}

// std::random_device rnd;
// std::mt19937 rng(rnd());
// int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}
const ll mod = 998244353;
const int maxn = 2222;
ll cost[maxn * 60];
int a[31] = {0, 1, 1};
int n, m, x;
#define tp tuple<int, int, int>

map<PI, int> mp[11];
int mrk[maxn * 60];
int dp(int pos, int val, int cnt)
{
    if(mrk[val] > cnt) return 0;
    if(pos == 0) return (val == 0 && cnt == 0) ? 1 : 0;
    if(cnt == 0) return (val == 0 ? 1 : 0);
    if(val <= 0) return 0;
    if(a[pos] * cnt < val) return 0;
    if(mp[pos].count({val, cnt})) return mp[pos][{val, cnt}];
    return mp[pos][{val, cnt}] = (dp(pos, val - a[pos], cnt - 1) + dp(pos - 1, val, cnt)) % mod;
}
int main()
{
    rep(i, 3, 30) a[i] = a[i - 1] + a[i - 2];
    rd(n, x, m);
    MS(cost, 0x3f);
    cost[0] = 0;
    rep(i, 1, 60 * n) rep(j, 1, 30) if(a[j] <= i) cost[i] = min(cost[i], cost[i - a[j]] + 1);
    MS(mrk, 0x3f); mrk[0]= 0;
    rep(i, 1, 60 * n) rep(j, 1, x) if(a[j] <= i && mrk[i - a[j]] < n) mrk[i] = min(mrk[i], mrk[i - a[j]] + 1);

    int ans = 0;
    rep(i, 1, a[x] * n) if(cost[i] == m) ans = (ans + dp(x, i, n)) % mod;
    cout << 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
10 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
'''
'   author: gche
'   created: 2024-08-02 20:10:19
'   link:
'   tag:
'''

import sys

input = lambda: sys.stdin.readline().strip()
mod = 9998244353

n, x, m = list(map(int, input().split()))

if m > 12:
  print(0)
  exit()

w = [1, 1]
while len(w) < x:
  w.append(w[-1] + w[-2])

mx = w[-1] * n
fib = [1, 1]
while fib[-1] <= mx:
  fib.append(fib[-1] + fib[-2])
  fib[-1] %= mod

c = [0] * (mx + 1)

for i in range(1, mx + 1):
  for j in range(len(fib) - 1, -1, -1):
    if i >= fib[j]:
      c[i] = c[i - fib[j]] + 1
      c[i] %= mod
      break

dp = [[0] * (mx + 1) for _ in range(n + 1)]
dp[0][0] = 1

cnt = 0
for i in range(len(w) - 1, -1, -1):
  for j in range(1, n + 1):
    for k in range(w[i], mx + 1):
      dp[j][k] += dp[j - 1][k - w[i]]
      # dp[j][k] %= mod
      if dp[j][k] > mod:
        dp[j][k] -= mod

ans = 0
for i in range(mx + 1):
  if c[i] == m:
    ans += dp[n][i]
    # ans %= mod
    if ans > mod:
      ans -= mod

print(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
10 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
import  bisect
f=[0,1,1]
while f[-1]<56000:
    f.append(f[-1]+f[-2])
cou=0
X=[0]*(56001)
for i in range(1,56000):
    cc=bisect.bisect_right(f,i)-1
    X[i]=X[i-f[cc]]+1
mod=998244353
# from functools import cache
# @cache
def t():
    n,x,m=map(int,input().split())
    ma=n*f[x]
    dp=[[0]*(n+1) for _ in range(ma+1)]
    dp[0][0]=1
    for k in range(1,x+1):
        for j in range(n):
            for i in range(ma+1-f[k]):
                dp[i+f[k]][j+1]+=dp[i][j]
                dp[i+f[k]][j+1]%=mod
    ans=0
    for i in range(ma+1):
        if dp[i][n] and X[i]==m:
            ans+=dp[i][n]
            ans%=mod
    print(ans)

    return

if __name__=="__main__":

    # for _ in range(II()):
    #     t()
    t()

**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
10 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

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

public class E {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);
	static int mod = 998244353;
	static int [][][] memo;
	static boolean [][][] visited;

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

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = 1;

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void advice() {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			// Source of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
			// Not using the right vairable eg idx in method for i in loop
			// Accounting for special case in general logic. eg special in dp
		}

		private void solve(int t) {
			int n = sc.nextInt();

			int x = sc.nextInt();

			int m = sc.nextInt();

			int [] fib = new int [30];

			fib[1] = 1;
			fib[2] = 1;

			for (int i = 3; i < fib.length; ++i) {
				fib[i] = fib[i - 1] + fib[i - 2];
			}

			// System.out.println(Arrays.toString(fib));

			int max = fib[x] * n + 1;

			int [] dp = new int [max];

			for (int i = 1; i < max; ++i) {
				dp[i] = getNeeded(i, fib);
			}

			// System.out.println(Arrays.toString(dp));

			int res = 0;

			int sum = 0;

			memo = new int [x + 1][][];
			visited = new boolean [x + 1][][];

			for (int i = 2; i <= x; ++i) {
				int mx = fib[i] * n + 1;

				memo[i] = new int [n + 1][mx];
				visited[i] = new boolean [n + 1][mx];
			}

			for (int i = 1; i < max; ++i) {
				if (dp[i] == m) {

					res += solve(x, n, fib, i);
					res %= mod;
				}
			}

			out.println(res);
		}

		private int solve(int pos, int moves, int [] fib, int left) {
			// System.out.println(pos + " " + left + " " + moves);
			if (left == 0)
				return moves == 0 ? 1 : 0;
			else if (moves * fib[pos] < left)
				return 0;
			else if (moves == 0)
				return 0;
			else if (pos < 3)
				return moves == left ? (pos == 2 ? moves : 0) + 1 : 0;
			else if (visited[pos][moves][left] == true)
				return memo[pos][moves][left];

			int res = solve(pos - 1, moves, fib, left);

			if (left >= fib[pos])
				res += solve(pos, moves - 1, fib, left - fib[pos]);

			res %= mod;

			visited[pos][moves][left] = true;
			memo[pos][moves][left] = res;
			return res;
		}

		private int getNeeded(int left, int [] fib) {

			int total = 0;

			for (int i = fib.length - 1; i > 0; --i) {
				if (left >= fib[i]) {
					++total;
					left -= fib[i];
				}
			}

			return total;
		}

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

      public IOHandler() {
         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());
      }

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

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

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

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

   }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are the steps to think through this:

1. **Edge Cases for `n`, `x`, and `m`:**
   - Small values: `n = 1`, `x = 2`, `m = 1`
   - Large values: `n = 1000`, `x = 10`, `m = 1000`
   - Boundary values: `n = m`, `x = 2`, `m = 1`
   - Special cases: `n = 2`, `x = 2`, `m = 2`

2. **Understanding the Algorithm:**
   - The algorithm uses dynamic programming to calculate the number of valid placements.
   - The `fib` array is used to represent the Fibonacci sequence, which is crucial for the operations.
   - The `solve` method recursively calculates the number of ways to achieve a certain cost.

3. **Challenging Scenarios:**
   - **Small `n` and `x` with high `m`:** This can test the edge case where the cost is higher than the number of chips.
   - **Large `n` and `x` with small `m`:** This can test the algorithm's efficiency and correctness in handling large inputs.
   - **Equal `n` and `m`:** This can test the edge case where the number of chips equals the cost.

4. **Selecting the Test Case:**
   - Let's choose a test case that is small but challenging: `n = 5`, `x = 5`, `m = 5`.
   - This test case will force the algorithm to handle a situation where the cost is equal to the number of chips, which can be tricky.

Given these considerations, the test input that might expose the bug in the code is:

```plaintext
5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

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

public class Main {
boolean MULTI_CASE = false;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt(); //1-1000
    int axisEnd = nextInt(); //2-10
    int m = nextInt(); //1-n
    int cntN = 55010;
    long[][][] f = new long[2][n + 1][cntN];
    int[] c = new int[axisEnd];
    c[0] = c[1] = 1;
    for (int i = 2; i < c.length; i++)
        c[i] = c[i - 1] + c[i - 2];
    f[0][0][0] = 1;
    long M = 998244353;
    for (int i = 1; i <= n; i++) f[0][i][i] = 1;
    int p = 0;
    for (int tt = 1; tt < axisEnd; tt++) {
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j < cntN; j++) {
                // g[4][6] = f[4][6]+f[3][4]+f[2][2]+f[1][0]
                // g[3][4] = f[3][4]+f[2][2]+f[1][0]
                f[1 - p][i][j] = f[p][i][j];
                if (i - 1 >= 0 && j - c[tt] >= 0)
                    f[1 - p][i][j] = (f[1 - p][i][j] + f[1 - p][i - 1][j - c[tt]]) % M;
            }
        }
        p = 1 - p;
    }
    long ret = 0;
    for (int i = 1; i < cntN; i++) {
        if (b[i] == m)
            ret = (ret + f[p][n][i]) % M;
    }
    sl(ret);
}

static int cntN = 55010;
static int[] b = new int[cntN];

static {
    int[] c = new int[30];
    c[0] = c[1] = 1;
    for (int i = 2; i < c.length; i++) c[i] = c[i - 1] + c[i - 2];
    b[1] = 1;
    b[2] = 1;
    for (int i = 3; i < cntN; i++) {
        b[i] = i;
        for (int j = 0; j < c.length; j++) {
            if (i >= c[j])
                b[i] = Math.min(b[i], b[i - c[j]] + 1);
        }
    }
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

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

**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 cause the faulty logic to fail. Here are the steps to consider:

1. **Understand the Bug**: The bug in the code seems to be related to the dynamic programming (DP) state transitions and the handling of the cost array `b`. Specifically, the DP transitions might not be correctly accounting for the operations and the cost calculations.

2. **Edge Cases**:
   - **Small `n` and `x`**: Test with small values of `n` and `x` to ensure the basic logic is correct.
   - **Large `n` and small `x`**: Test with a large number of chips and a small number of points to ensure the DP transitions handle large values correctly.
   - **Special Values of `m`**: Test with `m` values that are close to `n` or 1, as these are often edge cases.

3. **Test Input**:
   - **Small `n` and `x`**: `n = 3`, `x = 2`, `m = 1`
   - **Large `n` and small `x`**: `n = 100`, `x = 2`, `m = 2`
   - **Special Values of `m`**: `n = 10`, `x = 3`, `m = 10`

Given the constraints and the need for a small-scale test input, let's choose a test case that is likely to expose the bug:

```plaintext
100 2 2
```

This input is chosen because:
- It has a large number of chips (`n = 100`), which can stress the DP transitions.
- It has a small number of points (`x = 2`), which simplifies the problem but still requires correct handling of the operations.
- The cost `m = 2` is a non-trivial value that is not the minimum or maximum possible cost, making it a good test for the cost calculation logic.

This input should help in identifying any issues with the DP state transitions and cost calculations in the buggy code.