**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
# Visit my repo: https://github.com/Yawn-Sean/Daily_CF_Problems
standard_input, packages, output_together = 1, 1, 0
dfs, hashing, read_from_file = 1, 0, 0
de = 1

if 1:

    if standard_input:
        import io, os, sys
        input = lambda: sys.stdin.readline().strip()

        import math
        inf = math.inf

        def I():
            return input()

        def II():
            return int(input())

        def MII():
            return map(int, input().split())

        def LI():
            return list(input().split())

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

        def LFI():
            return list(map(float, input().split()))

        def GMI():
            return map(lambda x: int(x) - 1, input().split())

        def LGMI():
            return list(map(lambda x: int(x) - 1, input().split()))

    if packages:
        from io import BytesIO, IOBase

        import random
        import os

        import bisect
        import typing
        from collections import Counter, defaultdict, deque
        from copy import deepcopy
        from functools import cmp_to_key, lru_cache, reduce
        from heapq import merge, heapify, heappop, heappush, heappushpop, nlargest, nsmallest
        from itertools import accumulate, combinations, permutations, count, product
        from operator import add, iand, ior, itemgetter, mul, xor
        from string import ascii_lowercase, ascii_uppercase, ascii_letters
        from typing import *
        BUFSIZE = 4096

    if output_together:
        class FastIO(IOBase):
            newlines = 0

            def __init__(self, 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.stdout = IOWrapper(sys.stdout)

    if dfs:
        from types import GeneratorType

        def bootstrap(f, stk=[]):
            def wrappedfunc(*args, **kwargs):
                if stk:
                    return f(*args, **kwargs)
                else:
                    to = f(*args, **kwargs)
                    while True:
                        if type(to) is GeneratorType:
                            stk.append(to)
                            to = next(to)
                        else:
                            stk.pop()
                            if not stk:
                                break
                            to = stk[-1].send(to)
                    return to
            return wrappedfunc

    if hashing:
        RANDOM = random.getrandbits(20)
        class Wrapper(int):
            def __init__(self, x):
                int.__init__(x)

            def __hash__(self):
                return super(Wrapper, self).__hash__() ^ RANDOM

    if read_from_file:
        file = open("input.txt", "r", encoding='utf-16').readline().strip()
        fin = open(file, 'r')
        input = lambda: fin.readline().strip()
        output_file = open("output.txt", "w")
        def fprint(*args, **kwargs):
            print(*args, **kwargs, file=output_file)

    if de:
        def debug(*args, **kwargs):
            print('\033[92m', end='')
            print(*args, **kwargs)
            print('\033[0m', end='')

    fmax = lambda x, y: x if x > y else y
    fmin = lambda x, y: x if x < y else y

    class lst_lst:
        def __init__(self, n) -> None:
            self.n = n
            self.pre = []
            self.cur = []
            self.lst = [-1] * n

        def append(self, i, j):
            self.pre.append(self.lst[i])
            self.lst[i] = len(self.cur)
            self.cur.append(j)

        def iterate(self, i):
            tmp = self.lst[i]
            while tmp != -1:
                yield self.cur[tmp]
                tmp = self.pre[tmp]

def fl():
    ans = [0] * m
    ans[-1] = nums[-1]
    for i in range(m - 2, -1, -1):
        ans[i] = fmax(ans[i + 1], 0) + nums[i]
    return ans

def fr():
    ans = [0] * m
    ans[0] = nums[0]
    for i in range(1, m):
        ans[i] = fmax(ans[i - 1], 0) + nums[i]
    return ans

inf = 10 ** 15

t = II()
outs = []

for _ in range(t):
    n, m = MII()
    nums = LII()

    dpl = fl()
    dpr = fr()

    if _ == 7:
        print(n, m)
        print(*nums)

    for idx in range(n - 1):
        nums = LII()
        if _ == 7:
            print(*nums)

        nl = fl()
        nr = fr()

        ndpl = [-inf] * m
        ndpr = [-inf] * m

        for i in range(m - 1):
            ndpl[i] = dpr[i] + nl[i + 1] + nums[i]

        for i in range(1, m):
            ndpr[i] = dpl[i] + nr[i - 1] + nums[i]

        for i in range(m - 3, -1, -1):
            ndpl[i] = fmax(ndpl[i], ndpl[i + 1] + nums[i])

        for i in range(2, m):
            ndpr[i] = fmax(ndpr[i], ndpr[i - 1] + nums[i])

        v = -inf
        for i in range(1, m):
            ndpr[i] = fmax(ndpr[i], v + nums[i])
            v = fmax(v + nums[i], dpr[i] + nr[i])

        v = -inf
        for i in range(m - 2, -1, -1):
            ndpl[i] = fmax(ndpl[i], v + nums[i])
            v = fmax(v + nums[i], dpl[i] + nl[i])

        dpl = ndpl
        dpr = ndpr

    outs.append(max(dpl))

print('\n'.join(map(str, outs)))

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

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

Think step by step.

Response:
To create a challenging test input that can expose potential bugs in the provided code, we need to consider edge cases and scenarios that might not be handled correctly. Here are some key considerations:

1. **Edge Cases for `n` and `m`**:
   - `n = 1` and `m = 3` (smallest possible values).
   - `n = 2` and `m = 3` (smallest non-trivial case with multiple days).

2. **Negative Profits**:
   - Cases where all profits are negative.
   - Cases with a mix of positive and negative profits, where the optimal solution might not be immediately obvious.

3. **Special Patterns**:
   - All profits are the same.
   - Profits form a pattern that might be misinterpreted by the algorithm.

4. **Boundary Conditions**:
   - `m = 3` and `n = 2` (smallest non-trivial case).
   - `m = 2` (smallest valid `m`).

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

```plaintext
2
1 3
1 -1000 1
2 3
100 -1000 100
-1000 100 -1000
```

### Explanation:
1. **First Test Case**:
   - `n = 1`, `m = 3`
   - Profits: `1 -1000 1`
   - The optimal solution is to sell drinks of type 1 and 3, generating a profit of `1 + 1 = 2`.

2. **Second Test Case**:
   - `n = 2`, `m = 3`
   - Day 1: `100 -1000 100`
   - Day 2: `-1000 100 -1000`
   - The optimal solution is:
     - Day 1: Sell drinks of type 1 and 3, generating a profit of `100 + 100 = 200`.
     - Day 2: Sell drinks of type 2, generating a profit of `100`.
   - Total profit: `200 + 100 = 300`.

This test input is designed to challenge the code by including edge cases and negative profits, which can help identify any issues in the logic or boundary condition handling.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
/*
 * Author: @Jemin.s
 */
import java.io.*;
import java.util.NoSuchElementException;

public class TRY {

    public static void main(String[] args) {
        fastScanner sc = new fastScanner();
        int t = sc.nextInt();

        while (t-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();

            int[][] profit = new int[n][m];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    profit[i][j] = sc.nextInt();
                }
            }

            System.out.println(maximize(n, m, profit));
        }
    }

    private static long maximize(int n, int m, int[][] profitGrid) {
        long totalMaxProfit = Long.MIN_VALUE;
        long[][][] dp = new long[n][m][m];

        for (int i = 0; i < m; i++) {
            for (int j = i; j < m; j++) {
                long sum = 0;
                for (int k = i; k <= j; k++) {
                    sum += profitGrid[0][k];
                }
                dp[0][i][j] = sum;
            }
        }

        for (int i = 1; i < n; i++) {
            for (int j = 0; j < m; j++) {
                for (int k = j; k < m; k++) {
                    long cp = 0;
                    for (int l = j; l <= k; l++) {
                        cp += profitGrid[i][l];
                    }

                    for (int prevJ = 0; prevJ < m; prevJ++) {
                        for (int prevK = prevJ; prevK < m; prevK++) {
                            if (isValidTransition(prevJ, prevK, j, k)) {
                                dp[i][j][k] = Math.max(dp[i][j][k], dp[i - 1][prevJ][prevK] + cp);
                            }
                        }
                    }
                }
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = i; j < m; j++) {
                totalMaxProfit = Math.max(totalMaxProfit, dp[n - 1][i][j]);
            }
        }

        return totalMaxProfit;
    }

    private static boolean isValidTransition(int prevJ, int prevK, int j, int k) {
        boolean common = false;
        boolean different = false;

        for (int l = prevJ; l <= prevK; l++) {
            if (l >= j && l <= k) {
                common = true;
                break;
            }
        }

        for (int l = j; l <= k; l++) {
            if (l < prevJ || l > prevK) {
             different = true;
                break;
            }
        }

        return common && different;
    }
}

class fastScanner {

    private final InputStream in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    public fastScanner(InputStream in) {
        this.in = in;
    }

    public fastScanner() {
        this(System.in);
    }

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            return buflen > 0;
        }
    }

    private int readByte() {
        return hasNextByte() ? buffer[ptr++] : -1;
    }

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

    public boolean hasNext() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
            ptr++;
        }
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) {
            throw new NoSuchElementException();
        }
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

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

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

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through the problem:

1. **Understand the Bug**: The provided code seems to have an issue with the `isValidTransition` function, which checks if the selection of drink types on day `i` meets the required conditions relative to day `i-1`. The function should ensure that:
   - At least one drink type sold on day `i` must also have been sold on day `i-1`.
   - At least one drink type sold on day `i` must not have been sold on day `i-1`.

2. **Edge Cases**:
   - **Single Day**: Ensure the code handles a single day correctly.
   - **Single Drink Type**: Ensure the code handles the case where there is only one drink type.
   - **All Negative Profits**: Ensure the code handles the case where all profits are negative.
   - **All Positive Profits**: Ensure the code handles the case where all profits are positive.
   - **Mixed Profits**: Ensure the code handles the case where profits are mixed (both positive and negative).

3. **Complex Scenarios**:
   - **Minimum and Maximum Values**: Use the minimum and maximum values for `n` and `m` to ensure the code handles large inputs.
   - **Invalid Transitions**: Create a scenario where the `isValidTransition` function is likely to fail.

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

```plaintext
1
3 4
10 -10 10 -10
-10 10 -10 10
10 -10 10 -10
```

### Explanation:
- **Test Case**: 1
- **Number of Days**: 3
- **Number of Drink Types**: 4
- **Profit Grid**:
  - Day 1: [10, -10, 10, -10]
  - Day 2: [-10, 10, -10, 10]
  - Day 3: [10, -10, 10, -10]

This input is designed to:
- Test the transition logic between days.
- Ensure that the `isValidTransition` function correctly handles the conditions for valid transitions.
- Use mixed profits to challenge the algorithm's ability to find the optimal solution.

This input should help expose any issues in the `isValidTransition` function and the overall logic of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

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

public class DrinkProfit {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        StringBuilder result = new StringBuilder();

        while (t-- > 0) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            long[][] profits = new long[n][m];

            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    profits[i][j] = scanner.nextLong();
                }
            }

            long[][] dp = new long[n][m];
            long maxProfit = Long.MIN_VALUE;

            long currentDayMax = Long.MIN_VALUE;
            for (int j = 0; j < m; j++) {
                if (j == 0) {
                    dp[0][j] = profits[0][j];
                } else {
                    dp[0][j] = dp[0][j - 1] + profits[0][j];
                }
                currentDayMax = Math.max(currentDayMax, dp[0][j]);
            }
            dp[0][m - 1] = currentDayMax;

            for (int i = 1; i < n; i++) {
                for (int start = 0; start < m; start++) {
                    long currentProfit = 0;
                    for (int end = start; end < m; end++) {
                        currentProfit += profits[i][end];
                        for (int prevEnd = 0; prevEnd < m; prevEnd++) {
                            if (prevEnd >= start && prevEnd <= end) {
                                dp[i][end] = Math.max(dp[i][end], dp[i - 1][prevEnd] + currentProfit);
                            }
                        }
                    }
                }

                for (int j = 0; j < m; j++) {
                    maxProfit = Math.max(maxProfit, dp[i][j]);
                }
            }

            result.append(maxProfit - 5).append("\n");
        }

        System.out.print(result);
        scanner.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
#include<random>
#include<iostream>
#include<iomanip>
#include<cmath>
#include<ctime>
#include<cctype>
#include<cstdio>
#include<cstdlib>
#include<climits>
//#define NDEBUG
#include<cassert>
#include<cstring>
#include<complex>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<bitset>
//#define LL __int128
#define LL long long
#define ULL unsigned LL
#define uint unsigned int
//#define int LL
//#define double long double
#define mkp make_pair
#define par pair<int,int>
#define psb push_back
#define epb emplace_back
#define f(x) ((x).first)
#define s(x) ((x).second)
using namespace std;
#define Lbt(x) ((x)&(-(x)))
#define Swap(x,y) (x^=y^=x^=y)
const int Mxxx=1e5;
inline char gc()
{
//	return getchar();
	static char buf[Mxxx],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,Mxxx,stdin),p1==p2)?EOF:*p1++;
}
inline char pc(char ch,bool fl=false)
{
//	return fl?0:putchar(ch),0;
	static char buf[Mxxx],*p1=buf,*p2=buf+Mxxx;
	return (fl||((*p1++=ch)&&p1==p2))&&(fwrite(buf,1,p1-buf,stdout),p1=buf),0;
}
#define output pc('!',true)
inline int read()
{
	char ch=gc();
	int gans=0,gflag=0;
	for(;ch<'0'||'9'<ch;gflag|=ch=='-',ch=gc());
	for(;'0'<=ch&&ch<='9';gans=(gans<<1)+(gans<<3)+(ch^48),ch=gc());
	return gflag?-gans:gans;
}
template<typename T>
inline char read(T&gans)
{
	char ch=gc();
	int gflag=0;gans=0;
	for(;ch<'0'||'9'<ch;gflag|=ch=='-',ch=gc());
	for(;'0'<=ch&&ch<='9';gans=(gans<<1)+(gans<<3)+(ch^48),ch=gc());
	return gans=gflag?-gans:gans,ch;
}
template<typename T>
inline void write(T x)
{
	if(x>9)write(x/10);
	pc(x%10^48);
}
template<typename T>
inline void writenum(T x,char ch)
{
	if(x<0)pc('-'),x=-x;
	write(x);pc(ch);
}
inline void writechar(char x,char ch)
{
	pc(x);pc(ch);
}
template<typename T>
inline T Max(T x,T y)
{
	return x>y?x:y;
}
template<typename T>
inline T Min(T x,T y)
{
	return x<y?x:y;
}
template<typename T>
inline T Abs(T x)
{
	return x<0?-x:x;
}
template<typename T>
inline void ckmx(T&x,T y)
{
	x=Max(x,y);
}
template<typename T>
inline void ckmn(T&x,T y)
{
	x=Min(x,y);
}
int TT;
const int Mx=2e5;
int n,m;vector<int>a[Mx+5];
LL ans;vector<LL>b[Mx+5],c[Mx+5],sm[Mx+5];
signed main()
{
	#ifndef ONLINE_JUDGE
	freopen("_.in","r",stdin);
//	freopen("_.out","w",stdout);
	#endif
	int i,j;LL mn,mx;
	for(TT=read();TT;TT--)
	{
		n=read();m=read();
		for(i=0;i<=n;i++)a[i].resize(m+1),b[i].resize(m+1),c[i].resize(m+1),sm[i].resize(m+1);
		for(i=1;i<=n;i++)for(j=1;j<=m;j++)a[i][j]=read(),sm[i][j]=sm[i][j-1]+a[i][j],c[i][j]=Min(c[i][j-1],sm[i][j]);
		for(i=1;i<=n;i++)
		{
			for(j=0;j<=m;j++)b[i][j]=LLONG_MIN;
			mx=0;if(i==1)ckmx(b[i][1],(LL)a[i][1]);
			for(j=2;j<=m;j++)
			{
				ckmx(mx,b[i-1][j-1]-c[i][j-2]);
				ckmx(b[i][j],sm[i][j]+mx);
				if(i==1)ckmx(b[i][j],(LL)a[i][j]);
			}
			mn=mx=0;
			for(j=m-1;j;j--)
			{
				ckmx(mx,b[i-1][j]-mn);
				ckmn(mn,sm[i][m]-sm[i][j]);
				ckmx(b[i][j-1],sm[i][m]-sm[i][j-1]+mx);
				if(i==1)ckmx(b[i][j],(LL)a[i][j+1]);
			}
		}
		ans=LLONG_MIN;for(j=0;j<=m;j++)ckmx(ans,b[n][j]);
		writenum(ans,10);
	}
	return output;
}
/*
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
*/

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

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

Think step by step.

Response:
```plaintext
1
3 3
100 -100 100
-100 100 -100
100 -100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+5,inf=1e18;
int read(){
	int num=0,typ=1,c=getchar();
	while('0'>c||c>'9'){
		if(c=='-') typ=-1;
		c=getchar();
	}while('0'<=c&&c<='9'){
		num=num*10+c-48;
		c=getchar();
	}return num*typ;
}
int t;
void solve(){
	int n=read(),m=read(),ans=0;
	vector<vector<int>>f(n+10,vector<int>(m+10,-inf));
	vector<vector<int>>g(n+10,vector<int>(m+10,-inf));
	vector<vector<int>>a(n+10,vector<int>(m+10));
	vector<int>s(m+10);
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			a[i][j]=read();
		}
	}
	for(int j=1;j<=m;++j){
		s[j]=s[j-1]+a[1][j];
	}
	for(int j=m,mx=-inf;j;--j){
		mx=max(mx,s[j]);
		f[1][j]=mx-s[j-1];
	}
	for(int j=1,mi=inf;j<=m;++j){
		mi=min(mi,s[j-1]);
		g[1][j]=s[j]-mi;
	}
	for(int i=2;i<=n;++i){
		for(int j=1;j<=m;++j){
			s[j]=s[j-1]+a[i][j];
		}
		for(int j=m,mx=-inf,sx=-inf;j;--j){
			if(j+1<=m) sx=max(sx,s[j+1]);
			mx=max(mx,max(f[i-1][j+1]+sx,g[i-1][j]+sx));
			f[i][j]=mx-s[j-1];
		}
		for(int j=1,mx=-inf,mi=inf;j<=m;++j){
			if(j-2>=0)mi=min(mi,s[j-2]);
			mx=max(mx,max(f[i-1][j]-mi,g[i-1][j-1]-mi));
			g[i][j]=mx+s[j];
		}
	}
	for(int i=1;i<=m;++i){
		ans=max(ans,max(f[n][i],g[n][i]));
	}printf("%lld\n",ans);
}
signed main(){
	for(t=read();t;t--){
		solve();
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
import sys
from itertools import combinations as com
from itertools import permutations as per
from collections import deque,defaultdict
from math import factorial, gcd,lcm,sqrt
from heapq import heappop, heappush,heapify
from random import randrange
mod1 = 1000000007
mod = 998244353
inf = float('inf')
def input():return sys.stdin.readline().rstrip('\r\n')
def MIS():return map(int, input().split())
def LMIS():return list(MIS())
dx,dy=(0,0,1,-1,1,1,-1,-1),(1,-1,0,0,1,-1,1,-1)
ss1="RLUD"
def union(a,b):
    a1=find(a)
    b1=find(b)
    if a1<b1:
        parent[b1]=a1
    else:
        parent[a1]=b1
def find(target):
    tl=[]
    while target!=parent[target]:
        tl.append(target)
        target=parent[target]
    for t in tl:
        parent[t]=target
    return target

def unionr(a,b):
    a1=findr(a)
    b1=findr(b)
    if a1>b1:
        parentr[b1]=a1
    else:
        parentr[a1]=b1
def findr(target):
    tl=[]
    while target!=parentr[target]:
        tl.append(target)
        target=parentr[target]
    for t in tl:
        parentr[t]=target
    return target
def exgcd(a, b):
    if a == 0:
        return b,0, 1
    else:
        gcd, x, y = exgcd(b % a, a)
        return gcd,y - (b // a) * x,x
class SegmentTree:
    def __init__(self, data, default=0, func=sum):
            """initialize the segment tree with data"""
            self._default = default
            self._func = func
            self._len = len(data)
            self._size = _size = 1 << (self._len - 1).bit_length()

            self.data = [default] * (2 * _size)
            self.data[_size:_size + self._len] = data
            for i in reversed(range(_size)):
                    self.data[i] = func([self.data[i + i], self.data[i + i + 1]])

    def __delitem__(self, idx):
            self[idx] = self._default

    def __getitem__(self, idx):
            return self.data[idx + self._size]

    def __setitem__(self, idx, value):
            idx += self._size
            self.data[idx] = value
            idx >>= 1
            while idx:
                    self.data[idx] = self._func([self.data[2 * idx], self.data[2 * idx + 1]])
                    idx >>= 1

    def __len__(self):
            return self._len

    def query(self, start, stop):
            """func of data[start, stop)"""
            start += self._size
            stop += self._size

            res_left = res_right = self._default
            while start < stop:
                    if start & 1:
                            res_left = self._func([res_left, self.data[start]])
                            start += 1
                    if stop & 1:
                            stop -= 1
                            res_right = self._func([self.data[stop], res_right])
                    start >>= 1
                    stop >>= 1

            return self._func([res_left, res_right])

    def __repr__(self):
            return "SegmentTree({0})".format(self.data)

def solve():
    n,m=MIS()
    dpl=[0]*(m)
    dpr=[0]*(m)
    sm=0
    nums=LMIS()
    for i in range(m):
        sm=max(0,sm)+nums[i]
        dpr[i]=sm
    sm=0
    for i in range(m-1,-1,-1):
        sm=max(0,sm)+nums[i]
        dpl[i]=sm

    for iii in range(n-1):
        nums=LMIS()
        dpcl=[0]*(m)
        dpcr=[0]*(m)
        sm=nums[0]
        for i in range(1,m):
            dpcr[i]=max(dpcr[i-1],sm+max(dpr[i-1],dpl[i]))+nums[i]
            sm=max(0,sm)+nums[i]
        sm=nums[-1]
        for i in range(m-2,-1,-1):
            dpcl[i]=max(dpcl[i+1],sm+max(dpr[i],dpl[i+1]))+nums[i]
            sm=max(0,sm)+nums[i]
        dpr=dpcr[:]
        dpl=dpcl[:]

    print(max(dpr+dpl))

t=int(input())
for i in range(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:
```plaintext
1
3 3
1 -2 3
-1 2 -3
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
def max_profit(t, test_cases):
    results = []

    for case in test_cases:
        n, m, profits = case
        # Initialize a list to store maximum profits for each day
        max_profit = [[0] * (m + 1) for _ in range(n + 1)]

        # Process the first day
        for j in range(1, m + 1):
            for k in range(1, j + 1):
                max_profit[1][j] = max(max_profit[1][j], sum(profits[0][k-1:j]))

        # Process subsequent days
        for i in range(2, n + 1):
            current_day_profits = profits[i - 1]
            max_so_far = [0] * (m + 1)
            # Calculate max profit for previous day
            for j in range(1, m + 1):
                max_so_far[j] = max(max_so_far[j - 1], max_profit[i - 1][j])

            for j in range(1, m + 1):
                # Check all subarrays from current day
                for k in range(1, j + 1):
                    current_profit = sum(current_day_profits[k-1:j])
                    # Ensure we pick at least one shared and one different
                    if max_so_far[j] > 0:  # At least one type sold on previous day
                        max_profit[i][j] = max(max_profit[i][j], max_so_far[j - 1] + current_profit)

        # Get the final answer for this case
        results.append(max(max_profit[n][1:]))

    return results

# Reading input
t = int(input())
test_cases = []
for _ in range(t):
    n, m = map(int, input().split())
    profits = [list(map(int, input().split())) for _ in range(n)]
    test_cases.append((n, m, profits))

# Get results and print
results = max_profit(t, test_cases)
for result in results:
    print(result)

**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
2
1 3
100 200 300
2 4
-100 100 200 -100
100 -200 -100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
import java.util.*;
public class Main {

    static long mod = 1000000007;

    static long dp[][], dp2[][];

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int a[][] = new int[n + 1][m + 1];
            dp = new long[n + 1][m + 1];
            dp2 = new long[n + 1][m + 1];
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= m; j++) {
                    a[i][j] = sc.nextInt();
//                   p[i][j]= p[i][j-1]+(long)a[i][j];
//                   pm[i][j] = Math.min(p[i][j],p[i][j-1]);
                }
//                s[i][m] = a[i][m];
//                for(int j=m-1;j>=1;j--)
//                {
//                  s[i][j] = s[i][j+1]+(long)a[i][j];
//                  sm[i][j] = Math.min(sm[i][j],sm[i][j+1]);
//                }
            }
            for (int i = 1; i <= n; i++) {
                Arrays.fill(dp[i], Long.MIN_VALUE);
            }
            long ans = n != 1 ? Long.MIN_VALUE : Math.max(a[1][1], a[1][m]);
            for (int i = 1; i <= n; i++) {
                long pre = a[i][1];
                long pm = 0;
                long cur = Long.MIN_VALUE;
                if (i == 1) dp[i][1] = a[i][1];
                for (int j = 2; j <= m; j++) {
                    long val = Math.max(dp[i - 1][j - 1], dp2[i - 1][j]) - pm;
                    cur = Math.max(cur, val);
                    pm = Math.min(pm, pre);
                    pre = pre + (long) a[i][j];
                    dp[i][j] = cur + pre;
                    if (i == n) ans = Math.max(ans, dp[i][j]);
                }

                long suf = a[i][m];
                long sm = 0;
                long cur2 = Long.MIN_VALUE;
                if (i == 1) dp2[i][m] = a[i][m];
                for (int j = m - 1; j >= 1; j--) {
                    long val = Math.max(dp[i - 1][j], dp2[i - 1][j + 1]) - sm;
                    cur2 = Math.max(cur2, val);
                    sm = Math.min(sm, suf);
                    suf = suf + (long) a[i][j];
                    dp2[i][j] = cur2 + suf;
                    if (i == n) ans = Math.max(ans, dp2[i][j]);
                }
            }
//            System.out.println(dp[1][3]);
            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 provided buggy code, let's consider the following:

1. **Small Scale**: The test case should be small enough to be manageable but still able to expose the bug.
2. **Edge Cases**: Focus on edge cases such as:
   - Single day sales (n = 1).
   - Minimum number of drink types (m = 3).
   - Negative profits.
   - Situations where the optimal solution involves selecting a single drink type or a small subarray.

Given these considerations, let's create a test case that includes a single day (n = 1) with three drink types (m = 3), where the profits are both positive and negative to test the handling of different values.

```plaintext
1
1 3
-10 20 -15
```

### Explanation:
- **Test Case**: 1
- **Number of Days (n)**: 1
- **Number of Drink Types (m)**: 3
- **Profits**:
  - Day 1: -10, 20, -15

### Expected Output:
- The maximum profit that can be achieved by selling a subarray of drinks on day 1 is 20 (by selling only the second drink type).

This test case is small and manageable, and it includes both positive and negative profits, which can help expose any issues in the buggy code's handling of such scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
/*
*Author: @Jemin.s
*/
import java.util.*;
import java.io.*;

public class TRY {
    static final int MAX_SIZE = 200005;
    static final long LONG_INF = 1000000000000000000L;
    static long[] prev = new long[MAX_SIZE];
    static long[] curr = new long[MAX_SIZE];
    static long[] val = new long[MAX_SIZE];
    static long[] presum = new long[MAX_SIZE];

    public static void main(String[] args) throws IOException {
        FastScanner scanner = new FastScanner();
        PrintWriter writer = new PrintWriter(System.out);

        int testCases = scanner.nextInt();
        while (testCases-- > 0) {
            int rows = scanner.nextInt();
            int cols = scanner.nextInt();

            Arrays.fill(prev, 0);
            long max = -LONG_INF;

            for (int i = 1; i <= rows; i++) {
                for (int j = 1; j <= cols; j++) {
                    val[j] = scanner.nextLong();
                    presum[j] = presum[j - 1] + val[j];
                    curr[j] = Math.max(curr[j], val[j]);
                }

                Arrays.fill(curr, -LONG_INF);

                // if (i == 1) {
                //     for (int j = 1; j <= cols; j++) {
                //         curr[j - 1] = Math.max(curr[j - 1], val[j]);
                //         curr[j] = Math.max(curr[j], val[j]);
                //     }
                // }

                long maxpre = -LONG_INF;
                long temp = -LONG_INF;

                for (int j = 2; j <= cols; j++) {
                    maxpre = Math.max(maxpre, -presum[j - 2]);
                    temp = Math.max(temp , maxpre + prev[j - 1]);
                    curr[j] = Math.max(curr[j], temp + presum[j]);
                }

                maxpre = temp = -LONG_INF;

                for (int j = cols - 1; j > 0; j--) {
                    maxpre = Math.max(maxpre, presum[j + 1]);
                    temp = Math.max(temp, maxpre + prev[j]);
                    curr[j - 1] = Math.max(curr[j - 1], temp - presum[j - 1]);
                }

                if (i == rows) {
                    for (int j = 0; j <= cols; j++) {
                        max = Math.max(max, curr[j]);
                    }
                }

                System.arraycopy(curr, 1, prev, 1, cols - 1);
            }

            writer.println(max);
        }

        writer.flush();
        scanner.close();
    }
}

class FastScanner {
    private final InputStream inputStream;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int bufLen = 0;

    public FastScanner(InputStream inputStream) {
        this.inputStream = inputStream;
    }

    public FastScanner() {
        this(System.in);
    }

    private boolean hasNextByte() {
        if (ptr < bufLen) {
            return true;
        } else {
            ptr = 0;
            try {
                bufLen = inputStream.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            return bufLen > 0;
        }
    }

    private int readByte() {
        return hasNextByte() ? buffer[ptr++] : -1;
    }

    public boolean hasNext() {
        while (hasNextByte() && (buffer[ptr] < 33 || buffer[ptr] > 126)) {
            ptr++;
        }
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) {
            throw new NoSuchElementException();
        }
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (b >= 33 && b <= 126) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong() {
        if (!hasNext()) {
            throw new NoSuchElementException();
        }
        long num = 0;
        boolean isNegative = false;
        int b = readByte();
        if (b == '-') {
            isNegative = true;
            b = readByte();
        }
        if (b < '0' || b > '9') {
            throw new NumberFormatException();
        }
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else if (b == -1 || (b < 33 || b > 126)) {
                return isNegative ? -num : num;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }

    public int nextInt() {
        long num = nextLong();
        if (num < Integer.MIN_VALUE || num > Integer.MAX_VALUE) {
            throw new NumberFormatException();
        }
        return (int) num;
    }

    public long[] nextLongArray(int length) {
        long[] array = new long[length];
        for (int i = 0; i < length; i++) {
            array[i] = this.nextLong();
        }
        return array;
    }

    public int[] nextIntArray(int length) {
        int[] array = new int[length];
        for (int i = 0; i < length; i++) {
            array[i] = this.nextInt();
        }
        return array;
    }

    public void close() throws IOException {
        inputStream.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the constraints and the logic of the problem. The key points to consider are:

1. The selection of drink types must form a subarray.
2. On each day, at least one drink type must be the same as the previous day, and at least one drink type must be different.
3. The code uses dynamic programming to calculate the maximum profit, but there might be an issue with how it handles the transitions between days.

A good approach is to create a test case where the constraints are tight and the transitions between days are challenging. Specifically, we can create a test case where the profits are such that the optimal solution requires careful handling of the subarray selection and the transition conditions.

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

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

Explanation:
- **Day 1**: The profit for each drink type is positive and increasing.
- **Day 2**: The profit for each drink type is negative and decreasing.
- **Day 3**: The profit for each drink type is positive and increasing again.

This test case is designed to:
- Ensure that the optimal solution for Day 1 is to sell all drink types (since all profits are positive).
- Ensure that the optimal solution for Day 2 is to sell the smallest subarray that meets the transition conditions (since all profits are negative).
- Ensure that the optimal solution for Day 3 is to sell all drink types again, while maintaining the transition conditions from Day 2.

This test case should challenge the dynamic programming logic and expose any issues in handling the transitions between days.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
template <class T>
using max_heap = priority_queue<T>;
template <class T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
#define rep(i, l, r) for (ll i = (l); i < (r); i++)
#define rrep(i, r, l) for (ll i = (r); i-- > (l);)
#define all(x) begin(x), end(x)
template <class T>
bool chmin(T& lhs, T rhs) {
    return lhs > rhs ? (lhs = rhs, true) : false;
}
template <class T>
bool chmax(T& lhs, T rhs) {
    return lhs < rhs ? (lhs = rhs, true) : false;
}
struct IOIO {
    IOIO() {
        cin.tie(0)->sync_with_stdio(0);
    }
} ioio;
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T>& p) {
    os << '(' << p.first << ", " << p.second << ')';
    return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& vs) {
    os << '{';
    rep(i, 0, (int)vs.size()) os << vs[i] << (i + 1 == (int)vs.size() ? "" : ", ");
    os << '}';
    return os;
}
template <class T>
ostream& operator<<(ostream& os, const set<T>& vs) {
    os << '{';
    for (auto it = vs.begin(); it != vs.end(); it++) {
        if (it != vs.begin()) {
            os << ", ";
        }
        os << *it;
    }
    os << '}';
    return os;
}
template <class T, class U>
ostream& operator<<(ostream& os, const map<T, U>& vs) {
    os << '{';
    for (auto it = vs.begin(); it != vs.end(); it++) {
        if (it != vs.begin()) {
            os << ", ";
        }
        os << *it;
    }
    os << '}';
    return os;
}
#ifdef DEBUG
void dump_func() {
    cerr << endl;
}
template <class Head, class... Tail>
void dump_func(Head&& head, Tail&&... tail) {
    cerr << head;
    if (sizeof...(Tail) > 0) {
        cerr << ", ";
    }
    dump_func(std::move(tail)...);
}
#define dump(...) cerr << "[" + string(#__VA_ARGS__) + "] ", dump_func(__VA_ARGS__)
#else
#define dump(...) static_cast<int>(0)
#endif

void solve() {
    int n, m;
    cin >> n >> m;
    constexpr ll inf = 1e18;
    vector<ll> dp(m + 1, -inf);
    {
        vector<ll> a(m);
        rep(i, 0, m) cin >> a[i];
        {
            ll mn = 0;
            ll cur = 0;
            rep(i, 0, m) {
                cur += a[i];
                chmax(dp[i + 1], cur - mn);
                chmin(mn, cur);
            }
        }
        {
            ll mn = 0;
            ll cur = 0;
            rrep(i, m, 0) {
                cur += a[i];
                chmax(dp[i], cur - mn);
                chmin(mn, cur);
            }
        }
    }
    dump(dp);
    rep(i, 1, n) {
        vector<ll> a(m);
        rep(j, 0, m) cin >> a[j];
        vector<ll> ndp(m + 1, -inf);
        {
            ll cur = a[1] + a[0] + dp[1];
            rep(r, 2, m + 1) {
                chmax(ndp[r], cur);
                if (r < m) {
                    cur += a[r];
                    chmax(cur, a[r - 1] + a[r] + dp[r]);
                }
            }
        }
        dump(ndp);
        {
            ll cur = a[m - 1] + a[m - 2] + dp[m - 1];
            rrep(l, m - 1, 0) {
                chmax(ndp[l], cur);
                if (l > 0) {
                    cur += a[l - 1];
                    chmax(cur, a[l - 1] + a[l] + dp[l]);
                }
            }
        }
        dump(ndp);
        dp = move(ndp);
        dump(dp);
    }
    cout << ranges::max(dp) << '\n';
}
int main() {
    int t;
    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 provided code, we need to craft a test case that highlights the issues in the algorithm's logic. The primary logic flaw in the code is that it does not correctly handle the constraints for the selection of drink types on consecutive days. Specifically, the code might not ensure that the selection of drink types on day \(i\) meets the conditions of overlapping with and differing from the selection on day \(i-1\).

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

1. **Small Scale**: Since we need a small-scale test input, we can use a grid with a small number of days and drink types.
2. **Edge Case**: We will create a scenario where the optimal selection of drink types on consecutive days is not straightforward, and the code might fail to handle it correctly.
3. **Constraints**: Ensure the test case meets the constraints \(1 \leq n \leq 2 \cdot 10^5\), \(3 \leq m \leq 2 \cdot 10^5\), and \(n \cdot m \leq 2 \cdot 10^5\).

Let's create a test case with \(n = 2\) days and \(m = 3\) drink types. We will set the profits in such a way that the optimal solution requires careful selection to meet the constraints.

### Test Input
```plaintext
1
2 3
100 200 300
-100 200 100
```

### Explanation
- **Day 1**: The profits are [100, 200, 300]. The optimal selection is to sell all drink types, yielding a profit of \(100 + 200 + 300 = 600\).
- **Day 2**: The profits are [-100, 200, 100]. The optimal selection should be to sell drink types 2 and 3, yielding a profit of \(200 + 100 = 300\).

### Correct Solution
- **Day 1**: Sell [1, 2, 3] → Profit = 600
- **Day 2**: Sell [2, 3] → Profit = 300
- **Total Profit**: 600 + 300 = 900

### Expected Output
```plaintext
900
```

This test case should expose the bug in the provided code because the code might not correctly handle the selection of drink types on the second day to meet the constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
#include <iostream>
#define ll long long
using namespace std;
const int N = 2e5 + 5;
const ll inf = 0x3f3f3f3f3f3f3f3f;
ll sum[N], fl[2][N], fr[2][N];

ll solve() {
    int n, m;
    cin >> n >> m;
    for (int j = 1; j < m; j++)
        fl[0][j] = fr[0][j] = 0;
    for (int i = 1; i <= n; i++) {
        bool flag = i & 1;
        for (int j = 1; j <= m; j++) {
            int a;
            cin >> a;
            sum[j] = sum[j - 1] + a;
        }
        ll _mx = -inf, presmn = 0, sufsmx = 0;
        for (int j = 1; j < m; j++) {
            _mx = max(_mx, max(fl[!flag][j], fr[!flag][j]) - presmn);
            fr[flag][j + 1] = _mx + sum[j + 1];
            presmn = min(presmn, sum[j]);
        }
        _mx = -inf;
        for (int j = m - 1; j; j--) {
            _mx = max(_mx, max(fl[!flag][j], fr[!flag][j]) + sufsmx);
            fl[flag][j - 1] = _mx - sum[j - 1];
            sufsmx = max(sufsmx, sum[j]);
        }
    }
    ll ans = -inf;
    for(int j = 0; j < m - 1; j++)
        ans = max(ans, fl[n & 1][j]);
    for (int j = m; j > 1; j--)
        ans = max(ans, fr[n & 1][j]);
    return ans;
}

int main() {
    int t;
    cin >> t;
    while(t--)
        cout << solve() << endl;
    return 0;
}

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

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

Think step by step.

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

1. **Constraints**: The values of \( n \) and \( m \) should be within the given limits.
2. **Profit Values**: The profit values can be very large or very small, including negative values.
3. **Subarray Conditions**: The conditions for the subarray selection on consecutive days must be carefully considered to ensure the code handles them correctly.

Given the constraints and the nature of the problem, a small-scale test input that can challenge the code might involve:

- A small number of days and drink types.
- Profit values that can lead to different subarray selections each day.
- Cases where the maximum profit involves a subarray that meets the conditions but is not immediately obvious.

Here is a small-scale test input that should challenge the code:

```plaintext
1
3 5
10 -10 20 30 -20
-20 10 20 -10 10
10 20 30 40 50
```

### Explanation:
- **Day 1**: The optimal subarray is [10, -10, 20, 30] with a profit of \(10 - 10 + 20 + 30 = 50\).
- **Day 2**: The optimal subarray is [-10, 10, 20] with a profit of \(-10 + 10 + 20 = 20\).
- **Day 3**: The optimal subarray is [10, 20, 30, 40, 50] with a profit of \(10 + 20 + 30 + 40 + 50 = 150\).

The total profit is \(50 + 20 + 150 = 220\).

This test case ensures that the code must handle:
- Different profit values, including negative values.
- Subarrays that meet the conditions of overlapping and non-overlapping elements from the previous day.
- A small but non-trivial number of days and drink types to ensure the code is tested thoroughly.

This should help in exposing any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
def answer():
    n, m = map(int, input().split())
    matrix = []
    for _ in range(n):
        row = list(map(int, input().split()))
        prefix = [0] * (m + 1)
        for i in range(m):
            prefix[i + 1] = prefix[i] + row[i]
        matrix.append(prefix)

    dp = {}

    def get_sum(day, l, r):
        return matrix[day][r + 1] - matrix[day][l]

    def calc(day, pl, pr):
        if day == n:
            return 0

        state = (day, pl, pr)
        if state in dp:
            return dp[state]

        if day == 0:
            result = float('-inf')
            for l in range(m - 1):
                for r in range(l + 1, m):
                    profit = get_sum(day, l, r)
                    next_profit = calc(day + 1, l, r)
                    result = max(result, profit + next_profit)
            dp[state] = result
            return result

        result = float('-inf')
        l_bound = max(0, pl - 1)
        r_bound = min(m - 1, pr + 1)

        for l in range(l_bound, r_bound + 1):
            for r in range(l + 1, min(r_bound + 2, m)):
                if not (r < pl or l > pr):
                    if l < pl or r > pr:
                        profit = get_sum(day, l, r)
                        next_profit = calc(day + 1, l, r)
                        result = max(result, profit + next_profit)

        dp[state] = result
        return result

    return calc(0, -1, -1)

def main():
    t = int(input())
    for _ in range(t):
        print(answer())

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:
```plaintext
1
3 4
10 20 30 40
50 60 70 80
90 100 110 120
```

### Explanation:
This test case is designed to challenge the buggy code by providing a small but specific scenario where the constraints and conditions of the problem are clearly defined. Here's how it works:

1. **Input Structure**:
   - The first line contains the number of test cases, which is 1.
   - The second line contains the values of `n` and `m`, which are 3 and 4, respectively.
   - The next 3 lines contain the profit values for each day and each drink type.

2. **Day 1**:
   - The profit values are 10, 20, 30, and 40.
   - The optimal choice is to sell drinks from type 1 to type 4, which gives a profit of 100.

3. **Day 2**:
   - The profit values are 50, 60, 70, and 80.
   - The optimal choice is to sell drinks from type 1 to type 4, which gives a profit of 260. However, to meet the condition of having at least one new drink type, the optimal choice is to sell drinks from type 2 to type 4, which gives a profit of 210.

4. **Day 3**:
   - The profit values are 90, 100, 110, and 120.
   - The optimal choice is to sell drinks from type 1 to type 4, which gives a profit of 420. However, to meet the condition of having at least one new drink type, the optimal choice is to sell drinks from type 1 to type 4, which gives a profit of 420.

5. **Total Profit**:
   - The total profit from the optimal plan is 100 (Day 1) + 210 (Day 2) + 420 (Day 3) = 730.

This test case ensures that the buggy code must handle the constraints of at least one drink type being the same and at least one being different from the previous day. It also tests the code's ability to handle a small, but non-trivial, set of data.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

The profit gained from selling a drink on a particular day can vary. On day $i$, the projected profit from selling drink of type $j$ is $A_{i, j}$. Note that $A_{i, j}$ can be negative, meaning that selling the drink would actually incur a loss.

Pak Chanek wants to help his friend plan the sales over the $n$ days. On day $i$, Pak Chanek must choose to sell at least one type of drink. Furthermore, the types of drinks sold on a single day must form a subarray. In other words, in each day, Pak Chanek will select $i$ and $j$ such that $1 \leq i \leq j \leq m$. Then all types of drinks between $i$ and $j$ (inclusive) will be sold.

However, to ensure that customers from the previous day keep returning, the selection of drink types sold on day $i$ ($i>1$) must meet the following conditions:

  * At least one drink type sold on day $i$ must also have been sold on day $i-1$.    * At least one drink type sold on day $i$ must not have been sold on day $i-1$.

The daily profit is the sum of the profits from all drink types sold on that day. The total profit from the sales plan is the sum of the profits over $n$ days. What is the maximum total profit that can be achieved if Pak Chanek plans the sales optimally?

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

The first line of each test case contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $3 \leq m \leq 2 \cdot 10^5$; $n \cdot m \leq 2 \cdot 10^5$) — the number of rows and columns in a grid.

The next $n$ lines of each test case contain $m$ integers each, where the $i$-th line contains the integers $A_{i,1} A_{i,2}, \ldots, A_{i,m}$ ($-10^9 \leq A_{i,j} \leq 10^9$) — project profits of each drink type on the $i$-th day.

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

Output: For each test case, output a single integer: the maximum profit that Pak Chanek can achieve.

Examples:
input:
1
3 6
79 20 49 5 -1000 500
-105 9 109 24 -98 -499
14 47 12 39 23 50
output:
475

Note: Here is Pak Chanek's optimal plan:

![](CDN_BASE_URL/4277bb4b771ce557c9df4ba84c32fcb1)| ![](CDN_BASE_URL/9da16237201d15894364e851ed195f46)| ![](CDN_BASE_URL/3b7487c42d65f64db69a0a6a263bbf48)   ---|---|---        * On day $1$, Pak Chanek sells drink types $1$ to $3$. Generating a profit of $79+20+49 = 148$.    * On day $2$, Pak Chanek sells drink types $2$ to $4$. Generating a profit of $9+109+24 = 142$    * On day $3$, Pak Chanek sells drink types $1$ to $6$. Generating a profit of $185$.

So, the total profit of Pak Chanek's plan is $148 + 142 + 185 = 475$.

Buggy Code:
import sys
import os
from io import BytesIO, IOBase

# region fastio
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, 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")

if sys.version_info[0] < 3:
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

file = sys.stdin
if os.environ.get('USER') == "loic":
    file = open("data.in")

line = lambda: file.readline().split()
ui = lambda: int(line()[0])
mi = lambda: map(int,line())
li = lambda: list(mi())
lsi = lambda: list(map(int,line()[0]))
#######################################################################

def solve():

    dp = [0] * M
    ndp = [0] * M
    rev = [0] * M
    nrev = [0] * M

    a = li()

    dp[0] = a[0]
    for i in range(M):
        dp[i] = max(a[i], a[i] + dp[i-1])

    rev[0] = a[-1]
    for i in range(M-2,-1,-1):
        rev[i] = max(a[i], a[i] + rev[i+1])

    for _ in range(N-1):
        a = li()

        pref = [0] * (M + 1)
        pref[0] = a[0]
        ndp[0] = Z
        for i in range(1,M):
            pref[i] = max(a[i] + pref[i-1], a[i-1] + a[i])
            ndp[i] = max(a[i] + ndp[i-1], pref[i] + dp[i-1], pref[i] + rev[i])

        suff = [0] * (M + 1)
        suff[M-1] = a[M-1]
        nrev[M-1] = Z
        for i in range(M-2,-1,-1):
            suff[i] = max(a[i] + suff[i+1], a[i] + a[i+1])
            nrev[i] = max(a[i] + nrev[i+1], suff[i] + rev[i+1], suff[i] + dp[i])

        dp,ndp = ndp,dp
        rev,nrev = nrev,rev

    res = max(dp)
    res = max(res, max(rev))

    return str(res)

Z = -10**18

for test in range(1,ui()+1):
    N,M = mi()

    print(solve())

file.close()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Identify Potential Bugs:**
   - The code uses dynamic programming (DP) to calculate the maximum profit.
   - It initializes and updates `dp`, `ndp`, `rev`, and `nrev` arrays.
   - The code might fail if it does not handle negative values correctly or if it does not properly enforce the constraints of overlapping and non-overlapping drink types.

2. **Consider Edge Cases:**
   - **Small Values for \( n \) and \( m \):** Test with small values to ensure the algorithm works for minimal input.
   - **Negative Values:** Include negative values in the profit matrix to check if the algorithm can handle losses.
   - **Non-Overlapping and Overlapping Constraints:** Ensure that the selection of drink types on consecutive days meets the required conditions.

3. **Generate the Test Input:**
   - We will use a small \( n \) and \( m \) to keep the input manageable.
   - Include both positive and negative values in the profit matrix.
   - Ensure that the optimal solution requires careful handling of the constraints.

Here is a challenging test input:

```plaintext
1
3 3
100 -100 100
-100 100 -100
100 -100 100
```

**Explanation:**
- **Test Case 1:**
  - **Day 1:** Selling drinks of type 1 to 3. Profit: \(100 + (-100) + 100 = 100\).
  - **Day 2:** Selling drinks of type 2 to 3. Profit: \((-100) + 100 = 0\).
  - **Day 3:** Selling drinks of type 1 to 2. Profit: \(100 + (-100) = 0\).

The total profit is \(100 + 0 + 0 = 100\).

This test case is designed to:
- Test the handling of negative values.
- Ensure that the algorithm correctly enforces the constraints of overlapping and non-overlapping drink types.
- Check if the algorithm can find the optimal solution even when the optimal selection involves zero profit on some days.