**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
for i in range(int(input())):
    j=1
    x=1
    n,k=map(int, input().split())
    def a(n,k):
        global j,x
        if n+k==j:
            print(x)
        elif n+k<j:
            print(x-1)
        else:
            x=x+1
            j=j+x
            a(n,k)
    a(n,k)

**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 analyze the code and identify potential issues:

1. **Recursive Function**: The function `a` is recursive and uses global variables `j` and `x`. This can lead to issues with state management across multiple test cases.
2. **Base Cases**: The base cases for the recursion are not well-defined, especially when `n + k < j`.
3. **Global Variables**: Using global variables can lead to unintended side effects, especially in a loop that processes multiple test cases.

Given these issues, a test input that can expose these problems would be one where the number of pins is such that the recursion might not terminate correctly or where the global variables are not reset properly between test cases.

### Challenging Test Input
Let's create a test case where the number of pins is large enough to ensure the recursion depth is significant, and another where the number of pins is such that the global variables might not be reset correctly.

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

### Explanation
1. **Test Case 1**: `1000000000 1000000000`
   - This test case ensures that the recursion depth is very large, which can lead to issues with the global variables and the base case conditions.
2. **Test Case 2**: `1 1`
   - This test case is small and should be straightforward, but it can help verify if the global variables are reset correctly.
3. **Test Case 3**: `2 2`
   - This test case is slightly larger and can help verify if the algorithm handles small to medium inputs correctly.

This test input should help expose the issues in the buggy code, particularly with the use of global variables and the recursive function.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
import math

def ispossible(x, w, b):
    while (w != 0 or b != 0) and x != 0:
        if w > b:
            w -= x
        else:
            b -= x
        x -= 1
    if x != 0:
        return False
    return True

t = int(input())
for _ in range(t):
    w, b = map(int, input().split())
    n = w + b
    l = 0
    r = math.floor((-1 + math.sqrt(1 + 8 * n)) / 2)
    ans = 0
    while l <= r:
        mid = l + (r - l) // 2
        if ispossible(mid, w, b):
            ans = mid
            l = mid + 1
        else:
            r = mid - 1
    print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases where the logic of the `ispossible` function might fail. The function `ispossible(x, w, b)` is supposed to check if it is possible to build a frame of side length `x` with the given number of white and black pins. However, the current implementation has a flaw: it assumes that the largest row will always be filled first, which is not necessarily true.

Here are some edge cases to consider:
1. **Equal number of white and black pins**: This can cause the function to incorrectly reduce both `w` and `b` in a way that might not reflect the actual pin distribution.
2. **One color significantly more than the other**: This can cause the function to incorrectly allocate pins to rows.
3. **Small number of pins**: This can expose issues with the binary search logic.
4. **Large number of pins**: This can expose issues with the logic that handles large inputs.

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

```plaintext
4
1 1
2 2
10 5
1000000000 1000000000
```

This test input includes:
1. A case with 1 white and 1 black pin (small and equal).
2. A case with 2 white and 2 black pins (small and equal).
3. A case with 10 white and 5 black pins (one color significantly more than the other).
4. A case with 1 billion white and 1 billion black pins (large and equal).

These cases should help expose any issues in the logic of the `ispossible` function and the binary search.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
import java.util.Scanner;

public class B_Bowling_Frame {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int t = sc.nextInt();
    while (t > 0) {
      int w = sc.nextInt();
      int b = sc.nextInt();
      int total = w+b;
      double y = (-1+(Math.sqrt(1+(8*total))))/2;
      int x = (int)(Math.floor(y));
      System.out.println(x);
      t--;
    }
  }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the code might fail. The buggy code calculates the maximum side length of a triangular frame that can be formed with the given total number of pins, but it does not consider the constraint that pins on the same row must be of the same color.

Here are the steps to think through the test case:

1. **Identify the Bug**: The code calculates the maximum side length of a triangular frame based on the total number of pins, but it does not check if the rows can be formed with pins of the same color.

2. **Edge Cases**:
   - **Case 1**: When the number of white and black pins is such that the total number of pins is sufficient to form a triangular frame, but the distribution of colors makes it impossible to form a valid frame.
   - **Case 2**: When the number of white and black pins is very close, making it difficult to form a valid frame with the same color in each row.

3. **Construct the Test Case**:
   - For a small-scale test input, we can use a case where the total number of pins can form a larger triangular frame, but the distribution of white and black pins makes it impossible to form a valid frame with the same color in each row.

Let's construct a test case where the total number of pins can form a frame of side length 4, but the distribution of white and black pins makes it impossible to form a valid frame with the same color in each row.

```plaintext
1
5 7
```

Explanation:
- The total number of pins is 12, which can form a triangular frame of side length 4 (1 + 2 + 3 + 4 = 10 pins).
- However, with 5 white pins and 7 black pins, it is impossible to form a valid frame where each row has pins of the same color:
  - Row 1: 1 pin (can be either white or black)
  - Row 2: 2 pins (both must be the same color)
  - Row 3: 3 pins (all must be the same color)
  - Row 4: 4 pins (all must be the same color)

Given the distribution, it is impossible to form a valid frame of side length 4. The largest valid frame that can be formed is of side length 3 (1 + 2 + 3 = 6 pins), which can be formed with 3 white pins and 3 black pins.

This test case should expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bowling Frame
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Bowling is a national sport in Taiwan; everyone in the country plays the sport on a daily basis since their youth. Naturally, there are a lot of bowling alleys all over the country, and the competition between them is as intense as you can imagine.

Maw-Shang owns one such bowling alley. To stand out from other competitors in the industry and draw attention from customers, he decided to hold a special event every month that features various unorthodox bowling rules. For the event this month, he came up with a new version of the game called X-pin bowling. In the traditional $$$10$$$-pin bowling game, a frame is built out of ten bowling pins forming a triangular shape of side length four. The pin closest to the player forms the first row, and the two pins behind it form the second row, and so on. Unlike the standard version, the game of $$$X$$$-pin bowling Maw-Shang designed allows a much larger number of pins that form a larger frame. The following figure shows a standard $$$10$$$-pin frame on the left, and on the right it shows a $$$21$$$-pin frame that forms a triangular shape of side length six which is allowed in the game of $$$X$$$-pin bowling.

Being the national sport, the government of Taiwan strictly regulates and standardizes the manufacturing of bowling pins. There are two types of bowling pins allowed, one in black and the other in white, and the bowling alley Maw-Shang owns has $$$w$$$ white pins and $$$b$$$ black pins. To make this new game exciting for the customers, Maw-Shang wants to build the largest possible frame from these $$$w+b$$$ pins. However, even though he is okay with using both colors in building the frame, for aesthetic reasons, Maw-Shang still wants the colors of the pins on the same row to be identical. For example, the following figure shows two possible frames of side length six, but only the left one is acceptable to Maw-Shang since the other one has white and black pins mixed in the third row.

The monthly special event is happening in just a few hours. Please help Maw-Shang calculate the side length of the largest frame that he can build from his $$$w+b$$$ pins!

Input: The first line of the input contains a single integer $$$t$$$, the number of test cases. Each of the following $$$t$$$ lines contains two integers $$$w$$$ and $$$b$$$, the number of white and black pins, respectively.

- $$$1 \leq t \leq 100$$$
- $$$0 \leq w, b \leq 10^9$$$

Output: For each test case, output in a single line the side length $$$k$$$ of the largest pin satisfying Maw-Shang's requirement you can build with the given pins.

Examples:
input:
4
1 2
3 2
3 3
12 0
output:
2
2
3
4

Note:

Buggy Code:
import java.io.FileWriter;
import java.io.IOException;
import java.util.Scanner;

public class B_Bowling_Frame {

    public static void main(String[] args) throws IOException{
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0)
            System.out.println(minTriangular(sc.nextInt()+sc.nextInt()));
        sc.close();
        // FileWriter file = new FileWriter("out.txt");
        // for(int i = 1; i<=10000; i++)
        // {
        //     file.write("i = " + i + " min = " + minTriangular(i) + "\n");
        // }
    }
    private static int minTriangular(int t)
    {
        if(t==1)return 1;
        int target = 0;
        int s = 1, e = t, mid = (s+e)/2;
        while(s<=e)
        {
            if(mid*(mid+1)/2>t)e=mid-1;
            else if(mid*(mid+1)/2<t){target=mid; s = mid+1;}
            else {target=mid; break;}
            mid = (s+e)/2;
        }
        return target;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the binary search in the `minTriangular` function might fail. The function is designed to find the largest triangular number that is less than or equal to the total number of pins. The key issues to consider are:

1. **Edge cases with small numbers of pins**.
2. **Cases where the total number of pins is a triangular number**.
3. **Cases where the total number of pins is just below a triangular number**.
4. **Cases where the total number of pins is just above a triangular number**.
5. **Cases with a large number of pins to test the upper limits of the algorithm**.

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

```plaintext
5
1 0
1 1
2 1
3 1
4 4
```

### Explanation of Test Cases:
1. **1 0**: Only one white pin. The largest frame is a side length of 1.
2. **1 1**: One white and one black pin. The largest frame is a side length of 1.
3. **2 1**: Two white and one black pin. The largest frame is a side length of 2.
4. **3 1**: Three white and one black pin. The largest frame is a side length of 2.
5. **4 4**: Four white and four black pins. The largest frame is a side length of 4.

These test cases cover a range of small numbers and edge cases where the total number of pins is close to or exactly a triangular number. This will help in exposing any issues in the binary search logic of the `minTriangular` function.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
#include <iostream>
#include <algorithm>
#include <vector>

int main() {
    int n;
    std::cin >> n;

    std::vector<int> s(n), t(n);
    for (int i = 0; i < n; i++) {
        std::cin >> s[i];
        t[i] = s[i];
    }

    std::sort(t.begin(), t.end());

    int mismatchIndex = -1;
    for (int i = 0; i < n; i++) {
        if (s[i] != t[i]) {
            mismatchIndex = i;
            break;
        }
    }

    if (mismatchIndex == -1) {
        std::cout << 0 << std::endl;
        return 0;
    }

    int prefixLength = mismatchIndex;
    int suffixLength = n - mismatchIndex;

    int prefixCost = prefixLength * prefixLength;
    int suffixCost = suffixLength * suffixLength;

    int minCost = std::min(prefixCost + suffixCost, std::min(prefixCost + (suffixLength - 1) * (suffixLength - 1), (prefixLength - 1) * (prefixLength - 1) + suffixCost));

    std::cout << minCost << std::endl;

    return 0;
}

// KZrf4Io8w1tGamsJB57

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
#include <iostream>
#include <bits/stdc++.h>
#define lli long long int
using namespace std;
int main(){
    lli n;
    cin>>n;
    vector<lli> sorted(n);
    lli ans;
    for(lli i=0;i<n;i++){
        lli a;
        cin>>a;
        sorted[i]=a;
    }
    vector<lli> v(n);
    v=sorted;
    sort(sorted.begin(),sorted.end());
    lli mini=INT_MAX;
    for(lli i=1;i<n;i++){
        vector<lli> temp(v.begin(),v.begin()+i);
        lli count=0;
        sort(temp.begin(),temp.end());
        for(lli j=0;j<i;j++){
            for(lli k=0;k<i;k++){
            if(temp[j]==sorted[k]){
                count++;
            }
            }
        }
        lli lap=n-count;
        if(mini>((i*i)+(lap*lap))){
            mini=((i*i)+(lap*lap));
        }
        ans=mini;
    }
    cout<<ans<<endl;
}

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

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

Think step by step.

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

This test input is designed to challenge the buggy code by providing an already sorted array. The expected output should be `0` since no sorting is needed. However, the buggy code might not handle this case correctly, as it assumes that some sorting is always required. This edge case can help expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
def min_cost_to_sort(n, s):
    # Find the longest sorted prefix
    prefix_end = 0
    for i in range(1, n):
        if s[i] >= s[i - 1]:
            prefix_end = i
        else:
            break

    # Find the longest sorted suffix
    suffix_start = n - 1
    for i in range(n - 2, -1, -1):
        if s[i] <= s[i + 1]:
            suffix_start = i
        else:
            break

    # If the prefix and suffix overlap, the array is already sorted
    if prefix_end >= suffix_start:
        return 0

    # Calculate the costs
    cost_prefix_sort = (n - prefix_end - 1) ** 2
    cost_suffix_sort = (suffix_start) ** 2

    # Select the minimum cost
    min_cost = min(cost_prefix_sort, cost_suffix_sort)

    return min_cost

# Reading input
n = int(input().strip())
s = list(map(int, input().strip().split()))

# Calculate and print the minimal cost
result = min_cost_to_sort(n, s)
print(result)

# t7WJXpl9Drwqo3ExInR

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

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

Think step by step.

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

### Explanation:
This test input is designed to expose a potential flaw in the given buggy code. Here's the step-by-step reasoning:

1. **Longest Sorted Prefix:**
   - The longest sorted prefix is `[1, 2, 3]`, ending at index 2 (`prefix_end = 2`).

2. **Longest Sorted Suffix:**
   - The longest sorted suffix is `[5, 4]`, starting at index 3 (`suffix_start = 3`).

3. **Overlap Check:**
   - The prefix and suffix do not overlap (`prefix_end < suffix_start`).

4. **Cost Calculation:**
   - The cost of sorting the prefix is `(n - prefix_end - 1)^2 = (5 - 2 - 1)^2 = 2^2 = 4`.
   - The cost of sorting the suffix is `(suffix_start)^2 = 3^2 = 9`.

5. **Minimum Cost:**
   - The minimum cost is `min(4, 9) = 4`.

However, the correct approach to sort the array `[1, 2, 3, 5, 4]` should be to perform a suffix sort of length 2, which will sort the last two elements `[5, 4]` to `[4, 5]`. The cost of this operation is `2^2 = 4`.

The given buggy code might not handle this case correctly, as it only considers the minimum cost between sorting the prefix and suffix, without considering the possibility of sorting the middle segment that needs to be fixed. This test input should help identify the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
def min_cost_to_sort(arr):
    n = len(arr)

    # Check if the array is already sorted
    if arr == sorted(arr):
        return 0

    # Find the longest sorted prefix
    prefix_end = 0
    while prefix_end < n - 1 and arr[prefix_end] <= arr[prefix_end + 1]:
        prefix_end += 1

    # Find the longest sorted suffix
    suffix_start = n - 1
    while suffix_start > 0 and arr[suffix_start - 1] <= arr[suffix_start]:
        suffix_start -= 1

    # Calculate the cost of sorting the remaining part
    remaining_length = n - (prefix_end + 1) - (n - suffix_start)

    # Option 1: Sort the remaining part as a prefix
    cost_prefix = (remaining_length + prefix_end + 1) ** 2

    # Option 2: Sort the remaining part as a suffix
    cost_suffix = (remaining_length + n - suffix_start) ** 2

    # Return the minimum cost
    return min(cost_prefix, cost_suffix)

# Read input
n = int(input())
arr = list(map(int, input().split()))

# Output the result
print(min_cost_to_sort(arr))

# qOdUkxPTcihwmo4pX1a

**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
4
1 3 2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
def min_cost_to_sort(s):
    n = len(s)

    # Cost of sorting the entire array using a prefix sort
    prefix_cost = n**2

    # Cost of sorting the entire array using a suffix sort
    suffix_cost = n**2

    # Find the longest prefix that is already sorted
    prefix_len = 1
    while prefix_len < n and s[prefix_len - 1] <= s[prefix_len]:
        prefix_len += 1

    # Find the longest suffix that is already sorted
    suffix_len = 1
    while suffix_len < n and s[-suffix_len - 1] <= s[-suffix_len]:
        suffix_len += 1

    # Calculate the cost of using both prefix and suffix sorts
    combined_cost = float('inf')
    for i in range(1, n + 1):
        for j in range(1, n + 1):
            if i + j - 1 >= n:
                combined_cost = min(combined_cost, i**2 + j**2)

    # Return the minimum cost
    return min(prefix_cost, suffix_cost, combined_cost)

# Read input
import sys
input = sys.stdin.read
data = input().split()
n = int(data[0])
s = list(map(int, data[1:n+1]))

# Output the result
print(min_cost_to_sort(s))

# 91QJSKfxXank0sHBPh2

**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
4
1 2 3 4
```

### Explanation:
This test case is designed to check the edge case where the input array is already sorted. In such a scenario, the optimal solution would be to either perform a prefix sort of length 1 or a suffix sort of length 1, both of which have a cost of 1. However, the buggy code may not handle this correctly, especially if it incorrectly calculates the combined cost.

### Why This Test Case is Challenging:
1. **Already Sorted Array**: The array is already sorted, which means the minimum cost to sort it should be 1 (by performing a prefix or suffix sort of length 1).
2. **Edge Case Handling**: The code needs to correctly identify that no additional sorting is needed and avoid unnecessary calculations.
3. **Combined Cost Calculation**: The combined cost calculation in the buggy code may not handle this case efficiently, potentially leading to incorrect results.

This test case will expose any issues in the code related to handling already sorted arrays and the calculation of combined costs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

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

public class Main{
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        int n = scanner.nextInt();
        int[] s = new int[n];
        for (int i = 0; i < n; i++) {
            s[i] = scanner.nextInt();
        }

        // Compute prefix_max[i] : max of first i elements
        int[] prefix_max = new int[n + 1];
        prefix_max[0] = Integer.MIN_VALUE;
        for (int i = 1; i <= n; i++) {
            prefix_max[i] = Math.max(prefix_max[i - 1], s[i - 1]);
        }

        // Compute suffix_min[j] : min of last j elements
        int[] suffix_min = new int[n + 1];
        suffix_min[0] = Integer.MAX_VALUE;
        for (int j = 1; j <= n; j++) {
            suffix_min[j] = Math.min(suffix_min[j - 1], s[n - j]);
        }

        int min_cost = Integer.MAX_VALUE;

        // Consider using only one sort: entire prefix or entire suffix
        min_cost = Math.min(n * n, min_cost);
        min_cost = Math.min(n * n, min_cost);

        // Consider using both sorts
        for (int i = 1; i < n; i++) {
            // Find maximal k where s[i..k] is sorted
            int k = i;
            while (k < n - 1 && s[k] <= s[k + 1]) {
                k++;
            }
            int j = n - k - 1;
            if (j >= 1 && prefix_max[i] <= suffix_min[j]) {
                min_cost = Math.min(min_cost, i * i + j * j);
            }
            // Also consider j = 0, meaning no suffix sort
            if (k >= n - 1 && prefix_max[i] <= suffix_min[0]) {
                min_cost = Math.min(min_cost, i * i + 0);
            }
        }

        // Also consider starting with suffix sort
        for (int j = 1; j < n; j++) {
            // Find maximal k where s[n-j-1..k] is sorted
            int k = n - j - 1;
            while (k >= 0 && s[k] <= s[k + 1]) {
                k--;
            }
            int i = k + 1;
            if (i >= 1 && prefix_max[i] <= suffix_min[j]) {
                min_cost = Math.min(min_cost, i * i + j * j);
            }
            // Also consider i = 0, meaning no prefix sort
            if (k <= -1 && prefix_max[0] <= suffix_min[j]) {
                min_cost = Math.min(min_cost, 0 + j * j);
            }
        }

        System.out.println(min_cost);
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

// VQLuJat5BbYWyHpxiwh

**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, let's consider the following steps:

1. **Identify the Bug**: The code seems to have an issue in the logic for determining the minimal cost when using both prefix and suffix sorts. Specifically, the logic for finding the maximal sorted subarray and ensuring the prefix and suffix arrays are correctly compared might be flawed.

2. **Edge Cases**: We need to create a test case that can highlight these issues. Some edge cases to consider:
   - An array that is already sorted.
   - An array that is reverse sorted.
   - An array with a small number of elements that are almost sorted but have a few out-of-order elements.
   - An array with duplicate elements.
   - An array where the optimal solution involves a very specific prefix and suffix length.

3. **Challenging Scenario**: Let's create an array where the optimal solution is not immediately obvious and requires careful consideration of the prefix and suffix lengths. For example, an array where the middle section is already sorted, but the prefix and suffix are not.

Given these considerations, here is a small-scale test input that should challenge the provided code:

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

### Explanation:
- **Array**: `[1, 3, 2, 4, 5, 6]`
- **Expected Output**: The optimal solution here is to perform a prefix sort of length 3 and a suffix sort of length 3. This will sort the array as follows:
  - After prefix sort of length 3: `[1, 2, 3, 4, 5, 6]`
  - After suffix sort of length 3: `[1, 2, 3, 4, 5, 6]` (no change needed)
- **Cost**: \(3^2 + 3^2 = 18\)

This test input should expose any issues in the logic for determining the minimal cost, especially in the case where the middle section of the array is already sorted.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
import java.util.*;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] s = new int[n];
        for (int i = 0; i < n; i++) {
            s[i] = sc.nextInt();
        }

        // Check if the array is already sorted
        boolean sorted = true;
        for (int i = 1; i < n; i++) {
            if (s[i] < s[i - 1]) {
                sorted = false;
                break;
            }
        }
        if (sorted) {
            System.out.println(0);
            return;
        }

        int minCost = Integer.MAX_VALUE;

        // Try all possible prefix lengths
        for (int prefixLen = 1; prefixLen <= n; prefixLen++) {
            int[] temp = s.clone();
            // Perform prefix sort
            Arrays.sort(temp, 0, prefixLen);
            // Check if the remaining array is in non-decreasing order
            boolean isRemainingSorted = true;
            for (int i = prefixLen; i < n; i++) {
                if (temp[i] < temp[i - 1]) {
                    isRemainingSorted = false;
                    break;
                }
            }
            if (isRemainingSorted) {
                minCost = Math.min(minCost, prefixLen * prefixLen);
            }
            // If the remaining array is not sorted, we need to perform a suffix sort
            // Try all possible suffix lengths for each prefix length
            for (int suffixLen = 1; suffixLen <= n - prefixLen; suffixLen++) {
                int[] temp2 = temp.clone();
                // Perform suffix sort
                Arrays.sort(temp2, n - suffixLen, n);
                // Check if the entire array is sorted
                boolean isSorted = true;
                for (int i = 1; i < n; i++) {
                    if (temp2[i] < temp2[i - 1]) {
                        isSorted = false;
                        break;
                    }
                }
                if (isSorted) {
                    int cost = prefixLen * prefixLen + suffixLen * suffixLen;
                    minCost = Math.min(minCost, cost);
                }
            }
        }

        // Try all possible suffix lengths
        for (int suffixLen = 1; suffixLen <= n; suffixLen++) {
            int[] temp = s.clone();
            // Perform suffix sort
            Arrays.sort(temp, n - suffixLen, n);
            // Check if the remaining array is in non-decreasing order
            boolean isRemainingSorted = true;
            for (int i = 1; i < n - suffixLen; i++) {
                if (temp[i] < temp[i - 1]) {
                    isRemainingSorted = false;
                    break;
                }
            }
            if (isRemainingSorted) {
                minCost = Math.min(minCost, suffixLen * suffixLen);
            }
            // If the remaining array is not sorted, we need to perform a prefix sort
            // Try all possible prefix lengths for each suffix length
            for (int prefixLen = 1; prefixLen <= n - suffixLen; prefixLen++) {
                int[] temp2 = temp.clone();
                // Perform prefix sort
                Arrays.sort(temp2, 0, prefixLen);
                // Check if the entire array is sorted
                boolean isSorted = true;
                for (int i = 1; i < n; i++) {
                    if (temp2[i] < temp2[i - 1]) {
                        isSorted = false;
                        break;
                    }
                }
                if (isSorted) {
                    int cost = prefixLen * prefixLen + suffixLen * suffixLen;
                    minCost = Math.min(minCost, cost);
                }
            }
        }

        System.out.println(minCost);
    }
}

// ol2amRgZEKYXdPqMrWI

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
#include<iostream>
#include<cstdio>
#include<algorithm>
const int N = 1000005;
bool vis1[N] = {0}, vis2[N] = {0};
int tmp[N], a[N], num[N] = {0};
using namespace std;
long long calc(long long a,long long b){
    return a * a + b * b;
}
int main(){
    int n;
    cin >> n;
    for (int i = 1; i <= n;i++){
        scanf("%d", &tmp[i]);
        a[i] = tmp[i];
    }
    sort(tmp + 1, tmp + n + 1);
    for (int i = 1; i <= n;i++){
        a[i] = lower_bound(tmp + 1, tmp + n + 1, a[i]) - tmp;
    }
    long long ans = n * n;
    int mex = 1;
    for (int i = 1; i <= n;i++){
        vis1[a[i] + num[a[i]]] = true;
        num[a[i]]++;
        while(vis1[mex])
            mex++;
        ans = min(ans, calc((long long)i, (long long)(n + 1 - mex)));
        //cout << calc((long long)i, (long long)(n + 1 - mex)) << endl;
    }
    int mx = n;
    for (int i = n; i >= 1;i--){
        vis2[a[i] + num[a[i]] - 1] = true;
        num[a[i]]--;
        while(vis2[mx])
            mx--;
        ans = min(ans, calc((long long)mx, (long long)(n - i + 1)));
        //cout << calc((long long)mx, (long long)(n - i + 1)) << endl;
    }
    for (int i = 1; i <= n;i++){
        if(a[i]==i)
            ans = min(ans, (long long)(n - i) * (long long)(n - i));
        else
            break;
    }
    for (int i = n; i >= 1;i--){
        if(a[i]==i)
            ans = min(ans, (long long)(i - 1) * (long long)(i - 1));
        else
            break;
    }
    cout << ans << endl;
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

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

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

	int n;
	cin >> n;
	vector<int> a(n);

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

	vector<int> b = a;
	sort(begin(b), end(b));
	if(a == b) {
	    cout << 0;
	    exit(0);
	}

	int start;
	for(int i=n-1; i>=0; i--) {
	    if(a[i] != b[i]) {
	        start = i;
	        break;
	    }
	}
// 	cout << start << '\n';
	int mx = *max_element(begin(a), begin(a) + start + 1);
	int cur = 0;
// 	cout << mx << '\n';
	for(int i=start; i>=0; i--) {
	    if(a[i] == mx) {
	        cur = i;
	    }
	}
// 	cout << cur << '\n';
	long long ans = (start-cur+1) * 1ll * (start-cur+1);

	sort(begin(a)+cur, end(a));
	if(a == b) {
	    cout << ans;
	    exit(0);
	}

	for(int i=0; i<n; i++) {
	    if(a[i] != b[i]) {
	        start = i;break;
	    }
	}
	mx = *max_element(begin(a) + start, begin(a)+cur);
	for(int i=cur; i<n; i++) {
	    if(a[i] >= cur) {
	        ans += (i-start)*1ll*(i-start);
	        break;
	    }
	}

	cout << ans << '\n';
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
#include<bits/stdc++.h>
#pragma optimize(1,2,3)
#define ll int
#define mk(A,B) make_pair(A,B)
#define low(X) (X&(-X))
using namespace std;
const ll N=1e6,inf=0x7fffffff;
void upd(ll &A,ll B){A=(B<A)?B:A;}
long long calc(long long  n){return 1ll*n*n;}
ll read(){
	ll S=0,F=1;char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-')F*=-1;
		ch=getchar();
	}while(isdigit(ch)){
		S=(S<<3)+(S<<1)+(ch^48);
		ch=getchar();
	}return S*F;
}ll tr[N+10];
void add(ll x,ll y){
	while(x<=N){
		tr[x]+=y;x+=low(x);
	}return;
}ll ask(ll x){ll y=0;
	while(x>0){
		y+=tr[x];x-=low(x);
	}return y;
}ll n,a[N+10];
long long solve(){
	long long ans=calc(n);ll cont=0;memset(tr,0,sizeof(tr));
	for(ll i=1;i<=n;i++)tr[a[i]]++;
	for(ll i=1;i<N;i++)if(i+low(i)<=N)tr[i+low(i)]+=tr[i];
	ll s=n;for(ll i=n-1;i>0;i--){
		if(a[i]>a[i+1])break;
		s=i;
	}for(ll i=1;i<n;i++){
		cont=max(a[i],cont);
		add(a[i],-1);
		ll c=ask(cont-1);
		ans=min(ans,((i>=s)?0:calc(n-i))+calc(i+c));
	}return ans;
}ll b[N+10],m;
//unordered_map<ll ,ll >mp;
map<ll ,ll >mp;
int main(){
//	srand((ll)time(0));
//	n=N;for(ll i=1;i<=n;i++){
//		b[i]=a[i]=rand()+1;
//	}
	n=read();for(ll i=1;i<=n;i++)b[i]=a[i]=read();
	sort(b+1,b+n+1);for(ll i=1;i<=n;i++)if(b[m]!=b[i]){
		b[++m]=b[i];mp[b[m]]=m;
	}for(ll i=1;i<=n;i++){
		a[i]=mp[a[i]];
	}long long ans=solve();
	for(ll i=1;i<=n-i+1;i++)swap(a[i],a[n-i+1]);
	for(ll i=1;i<=n;i++)a[i]=m-a[i]+1;
	printf("%lld\n",min(ans,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
5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
import sys

def prefix_sort(L,i):
    return sorted(L[:i])+L[i:]
def suffix_sort(L,i):
    return L[:len(L)-i]+sorted(L[len(L)-i:])

def fun(L):
    n=int(L[0])
    L=list(map(int,L[1].split()))
    sorted_list=sorted(L)
    result=[]

    '''#non overlapping cases
    for i in range(1,n):
        left=L[:i]
        right=L[i:]
        if left==sorted(left) and right==sorted(right) and left+right==sorted_list:
            result.append(0)
        elif left==sorted(left) and left+sorted(right)==sorted_list:
            result.append(len(right)**2)
        elif right==sorted(right) and sorted(left)+right==sorted_list:
            result.append(len(left)**2)
        else:
            result.append(len(right)**2 + len(left)**2)'''

    #overlapping cases
    for i in range(n):
        #prefix first
        s=prefix_sort(L,i+1)
        if s==sorted_list:
            result.append((i+1)**2)

        else:
            s1=suffix_sort(s,n-i)
            if s1==sorted_list:
                result.append(i**2 + (n-i)**2)

        #suffix first
        s=suffix_sort(L,i)
        if s==sorted_list:
            result.append(i**2)

        else:
            s1=prefix_sort(s,n-i+1)
            if s1==sorted_list:
                result.append(i**2 + (n-i+1)**2)

    return min(result)

lines=sys.stdin.read()
lines=lines.splitlines()
print(fun(lines))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
readlist = lambda x: [int(i) for i in x.strip().split()]
joinlist = lambda x,s='': s.join([str(i) for i in x])

out = []

for _ in [1]:
    n = int(input())
    nums = readlist(input())
    target = sorted(nums)
    left, right = 0, n-1
    ops = 0
    counts = dict()
    for i in target:
        if i not in counts:
            counts[i]=0
        counts[i]+=1

    while left < right:
        while (nums[left] == target[left]) and left<right:
            left+=1
        while (nums[right] == target[right]) and left<right:
            right-=1
        if left >= right:
            break

        l, r = left, right
        lCount = rCount = 0
        while (lCount!=counts[target[left]]) and (rCount!=counts[target[right]]):
            l += 1
            r -= 1
            lCount += (nums[l] == target[left])
            rCount += (nums[r] == target[right])

        if nums[l]==target[left]:
            nums[:l+1] = sorted(nums[:l+1])
            ops += (l+1)**2
        else:
            nums[r:] = sorted(nums[r:])
            ops += (n - r)**2
    out.append(ops)
[print(i) for i in out]

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

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

Think step by step.

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

### Explanation:
This test input is designed to expose potential issues in the algorithm's handling of elements that are already in the correct position and the optimal choice of prefix or suffix sorting.

1. **Initial Array:**
   - `nums = [2, 1, 4, 3, 5]`
   - `target = [1, 2, 3, 4, 5]`

2. **First Iteration:**
   - `left = 0`, `right = 4`
   - `nums[0] = 2`, `target[0] = 1` (mismatch)
   - `nums[4] = 5`, `target[4] = 5` (match, move `right` to 3)

3. **Second Iteration:**
   - `left = 0`, `right = 3`
   - `nums[0] = 2`, `target[0] = 1` (mismatch)
   - `nums[3] = 3`, `target[3] = 3` (match, move `right` to 2)

4. **Third Iteration:**
   - `left = 0`, `right = 2`
   - `nums[0] = 2`, `target[0] = 1` (mismatch)
   - `nums[2] = 4`, `target[2] = 3` (mismatch)

   - The algorithm should ideally sort the prefix up to index 2 to make `nums = [1, 2, 3, 4, 5]`, but if it incorrectly chooses to sort the suffix, it will not achieve the optimal solution.

5. **Optimal Solution:**
   - Perform a prefix sort of length 3: `nums[:3] = [1, 2, 3]`, `ops = 3^2 = 9`
   - The array becomes `[1, 2, 3, 4, 5]`, which is sorted.

This test input ensures that the algorithm must correctly choose the prefix sort over the suffix sort to achieve the minimum cost. If the algorithm fails to do so, it will not produce the correct output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = false;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static void solve(PrintWriter o) {
        try {
            int n = nextInt();
            int[] a = new int[n];
            List<Integer> li = new ArrayList<>();
            Integer[] b = new Integer[n];
            TreeMap<Integer, List<Integer>> map = new TreeMap<>();
            for(int i=0;i<n;i++) {
                a[i] = nextInt()-1;
                li.add(a[i]);
                map.computeIfAbsent(a[i], key->new ArrayList<>()).add(i);
            }
            Collections.sort(li);
            int[] suf = new int[n+2];
            int[] pre = new int[n+2];
            for(int i=n;i>=1;i--) {
                if(a[i-1] == li.get(i-1)) suf[i] = suf[i+1];
                else suf[i] = n-i+1;
            }
            for(int i=1;i<=n;i++) {
                if(a[i-1] == li.get(i-1)) pre[i] = pre[i-1];
                else pre[i] = i;
            }
            int l = 0;
            for(List<Integer> entry: map.values()) {
                if(entry == null) continue;
                for(Integer i: entry) {
                    b[i] = l++;
                }
            }
            long res = 1l*n*n;
            int mex = 0;
            Set<Integer> set = new HashSet<>();
            for(int i=0;i<n;i++) {
                set.add(b[i]);
                while(set.contains(mex)) mex++;
                res = Math.min(res, 1l*(i+1)*(i+1) + 1l*suf[mex+1]*suf[mex+1]);
            }
            for(int i=0;i<n;i++) b[i] = n-1-b[i];
            for(int i=0,j=n-1;i<j;i++,j--) {
                int temp = b[i];
                b[i] = b[j];
                b[j] = temp;
            }
            set.clear();
            mex = 0;
            for(int i=0;i<n;i++) {
                set.add(b[i]);
                while(set.contains(mex)) mex++;
                res = Math.min(res, 1l*(i+1)*(i+1) + 1l*pre[n-mex]*pre[n-mex]);
            }
            o.println(res);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    public static int upper_bound(long[] preSum, long val){
        int l = 0, r = preSum.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(preSum[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(List<long[]> li, long val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid)[0] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class Pair {
        Integer p1, p2;
        public Pair(Integer p1, Integer p2) {
            this.p1 = p1;
            this.p2 = p2;
        }
        @Override
        public int hashCode() {
            int prime = 31, ret = 0;
            ret = ret*prime+p1.hashCode();
            ret = ret*prime+p2.hashCode();
            return ret;
        }
        @Override
        public boolean equals(Object o) {
            if(o instanceof Pair pair) {
                return p1.equals(pair.p1) && p2.equals(pair.p2);
            }
            return false;
        }
    }
    static class DSU {
        int[] parent;
        int[] size;
        public DSU(int n) {
            parent = new int[n];
            size = new int[n];
            for(int i=0;i<n;i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
            size[root_y] += size[root_x];
            size[root_x] = 0;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are some considerations:

1. **Arrays with All Elements Equal**: This can cause issues if the algorithm does not handle equal elements correctly.
2. **Arrays with a Single Element**: This is an edge case that should be handled trivially.
3. **Arrays with Elements in Reverse Order**: This can test the algorithm's ability to handle the worst-case scenario for sorting.
4. **Arrays with a Few Unique Elements**: This can test the algorithm's ability to handle elements that are mostly the same but with a few unique values.
5. **Arrays with a Specific Pattern**: This can test the algorithm's ability to handle specific patterns that might not be sorted correctly.

Given the constraints and the need for a small-scale test input, let's choose an array with elements in reverse order. This is a classic worst-case scenario for many sorting algorithms and can expose issues in the logic.

Here is the test input:

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

This input will test the algorithm's ability to handle a reverse-sorted array and should expose any issues in the logic for determining the minimum cost to sort the array.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

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

public class M2041{
    public static void main(String[] args) throws IOException{
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        int n = Integer.parseInt(r.readLine());

        int[] nums = new int[n];
            int idx = 0;
            HashMap<Integer, Integer> cur = new HashMap<>();
        int[] sorted = new int[n];
        long ans = Long.MAX_VALUE;

        StringTokenizer st = new StringTokenizer(r.readLine());
        for (int i = 0; i < n; i++) {
            nums[i] = Integer.parseInt(st.nextToken());
            sorted[i] = nums[i];
        }
        Arrays.sort(sorted);

        //DISJOINT
        for (int i = 0; i < n; i++) {
            cur.put(nums[i], cur.getOrDefault(nums[i], 0) + 1);
            while(idx < n && cur.containsKey(sorted[idx])){
                cur.put(sorted[idx], cur.get(sorted[idx]) - 1);
                if(cur.get(sorted[idx]) == 0){
                    cur.remove(sorted[idx]);
                }

                idx++;
            }
            if(cur.size() == 0){
                int right = i + 1;
                while(right < n && nums[right] == sorted[right]){
                    right++;
                }

                ans = Math.min(ans, (long)(i + 1) * (i + 1) + (long)(n - right) * (n - right));
                i = right - 1;
            }
        }

        //INTERSECT WITH LEFT FIRST
        cur = new HashMap<>();
        idx = 0;

        for (int i = 0; i < n; i++) {
            cur.put(nums[i], cur.getOrDefault(nums[i], 0) + 1);
            while(idx < n && cur.containsKey(sorted[idx])){
                cur.put(sorted[idx], cur.get(sorted[idx]) - 1);
                if(cur.get(sorted[idx]) == 0){
                    cur.remove(sorted[idx]);
                }
                idx++;

                ans = Math.min(ans, (long)(i + 1) * (i + 1) + (long)(n - idx) * (n - idx));
            }
        }
        //INTERSECT WITH RIGHT FIRST
        cur = new HashMap<>();
        idx = n - 1;

        for (int i = n - 1; i >= 0; i--) {
            cur.put(nums[i], cur.getOrDefault(nums[i], 0) + 1);
            while(idx >= 0 && cur.containsKey(sorted[idx])){
                cur.put(sorted[idx], cur.get(sorted[idx]) - 1);
                if(cur.get(sorted[idx]) == 0){
                    cur.remove(sorted[idx]);
                }
                idx--;

                ans = Math.min(ans, (long)(idx + 1) * (idx + 1) + (long)(n - i) * (n - i));
            }
        }
        pw.println(ans);
        pw.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
5
1 3 2 4 5
```