**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

Buggy Code:
/*
    "care a facut teste cu Lattice reduction attack e ciudat"
    "linistiti va putin"
    - 2023 -
*/
#include <bits/stdc++.h>
#define debug(x) cerr << #x << " " << x << "\n"
#define debugs(x) cerr << #x << " " << x << " "
#pragma GCC optimize("Ofast")
#define int long long
#define unordered_map map

using namespace std;

int v[1005];

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long test,itest,n,m,q,i,j,x,k,y,mx,mi,l,r,pas,cnt,ans,pozmax,pozmin;
    bool ok,ok2;
    char ch;
    string s1,s2,s3,s;
    cin >> test;
    //test=1;
    for (itest=1; itest<=test; itest++)
    {
        mx=0;
        pozmax=0;
        mi=100001;
        pozmin=0;
        cin >> n;
        for (i=1; i<=n; i++)
        {
            cin >> v[i];
            if (v[i]>mx)
            {
                mx=v[i];
                pozmax=i;
            }
            if (v[i]<mi)
            {
                mi=v[i];
                pozmin=i;
            }
        }
        if (n==1)
        {
            cout << "0\n";
            continue;
        }
        swap(v[1],v[pozmax]);
        swap(v[2],v[pozmin]);
        mx=0;
        mi=10001;
        ans=0;
        for (i=1; i<=n; i++)
        {
            mx=max(mx,v[i]);
            mi=min(mi,v[i]);
            ans+=mx-mi;
        }
        cout << ans << "\n";
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios where the code might fail. The bug in the code is in the way it handles the shuffling of the array. Specifically, it swaps the maximum and minimum values to the first two positions, which is not always optimal.

Let's consider a scenario where the array has multiple maximum and minimum values. This can lead to suboptimal results if the code incorrectly handles these values.

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

1. **Single Element Array**: This is already covered in the example and should work correctly.
2. **Array with All Elements Equal**: This should also work correctly as the score will be 0.
3. **Array with Multiple Maximum and Minimum Values**: This is a critical case to test the shuffling logic.

Let's create a test input with an array that has multiple maximum and minimum values:

```plaintext
3
1
69
3
7 7 7
5
1 1 2 1 1
```

### Explanation:
1. **Single Element Array (1 69)**:
   - This is a simple case where the score is 0.
2. **Array with All Elements Equal (3 7 7 7)**:
   - This should also result in a score of 0.
3. **Array with Multiple Minimum Values (5 1 1 2 1 1)**:
   - This is the critical case. The array has multiple minimum values (1) and one maximum value (2).
   - The code might incorrectly handle this by swapping the maximum value to the first position and one of the minimum values to the second position, which might not maximize the score.

### Expected Output:
```plaintext
0
0
4
```

### Explanation of the Expected Output:
1. **Single Element Array (1 69)**:
   - The score is 0.
2. **Array with All Elements Equal (3 7 7 7)**:
   - The score is 0.
3. **Array with Multiple Minimum Values (5 1 1 2 1 1)**:
   - The optimal shuffling could be `[2, 1, 1, 1, 1]`. Here, `b = [2, 1, 1, 1, 1]` and `c = [2, 2, 2, 2, 2]`. The score is `(2-2) + (2-1) + (2-1) + (2-1) + (2-1) = 4`.

This test input should help expose the bug in the provided code by challenging its shuffling logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

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

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

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

Think step by step.

Response:
```plaintext
3
1
1000
2
1 1000
3
1 1000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

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

    a.sort()

    b = [a[0]]  # Start b with the smallest element
    c = [a[-1]]  # Start c with the largest element

    # Fill b and c as per the requirements
    for i in range(1, length):
        if i < length // 2:
            b.append(a[i])
            c.append(a[-i-1])
        else:
            b.append(a[-i-1])
            c.append(a[i])

    # Calculate the score
    score = sum(c[i] - b[i] for i in range(length))

    print(score)

**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
1000
2
1000 1
3
1000 1 1
4
1000 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

Buggy Code:
import java.util.Scanner;

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

            int max=0;
            int min=1000;
            int sum=0 ;

            for(int j =0 ; j<n ; j++){
                a[j]= sc.nextInt();
                if(max<a[j]){
                    max=a[j];
                }else if(min>a[j]){
                    min= a[j];
                }
            }
            if(n==1){
                System.out.println(0);
            }
            else{
                sum = (n-1)*(max-min);
                System.out.println(sum);

            }
        }

        sc.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
3
1
1000
2
1 1000
3
1 1000 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

Buggy Code:
import java.util.*;
public class Orangutan{
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int T=sc.nextInt();
        while(T-->0){
            int N=sc.nextInt();
            int a[]=new int[N];
            for(int i=0; i<N; i++){
                a[i]=sc.nextInt();

            }
            Arrays.sort(a);
            int score=0;
            for(int i=0; i<N; i++){
                score +=Math.abs(a[N-i-1]-a[i]);
            }
            System.out.println(score);
        }
        sc.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
3
1
1000
2
1 1000
10
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

Buggy Code:
def solve(A: list[int]) -> int:
    A.sort()

    t = []

    l = 0
    r = len(A) - 1
    while l < r:
        t.append(A[r])
        t.append(A[l])
        l += 1
        r -= 1

    t.append(A[l])

    answer = 0
    maximum = t[0]
    minimum = t[0]
    for i in range(1, len(t)):
        maximum = max(maximum, t[i])
        minimum = min(minimum, t[i])
        answer += maximum - minimum

    return answer

t = int(input())

for _ in range(t):
    n = int(input())
    A = list(map(int, input().split()))
    answer = solve(A)
    print(answer)

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

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

Think step by step.

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

**Explanation:**

1. **Test Case 1:**
   - Input: `n = 1`, `A = [1]`
   - Expected Output: `0`
   - This tests the edge case where the array has only one element. The score should be 0 because there is no other way to rearrange the array.

2. **Test Case 2:**
   - Input: `n = 2`, `A = [1, 2]`
   - Expected Output: `1`
   - This tests a simple case with two elements. The optimal score is 1, which can be achieved by arranging the array as `[2, 1]`.

3. **Test Case 3:**
   - Input: `n = 4`, `A = [1, 2, 2, 1]`
   - Expected Output: `3`
   - This tests a case with repeated elements. The optimal score is 3, which can be achieved by arranging the array as `[2, 1, 2, 1]`.

These test cases cover different scenarios, including edge cases and cases with repeated elements, to maximize the probability of exposing the bug in the faulty solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

Buggy Code:
def maximum_score(a,n):
  a.sort()
  b = [0] * n
  c = [0] * n
  for i in range(n):
    b[i] = a[i // 2]
    c[i] = a[n - 1 - (i // 2)]
    score = sum(c[i] - b[i] for i in range(n))
  return score
t = int(input())

for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    results = maximum_score(a,n)
    print(results)

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

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

Think step by step.

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

1. **Edge Cases**: Small arrays, arrays with all identical elements, and arrays with a mix of small and large values.
2. **Maximizing Failure Probability**: The bug in the code is in the way `b` and `c` are constructed. Specifically, `b[i] = a[i // 2]` and `c[i] = a[n - 1 - (i // 2)]` do not correctly compute the minimum and maximum values up to each index.

Let's analyze the code:
- For `b`, the code incorrectly takes the element at `i // 2` which does not necessarily give the minimum value up to index `i`.
- For `c`, the code incorrectly takes the element at `n - 1 - (i // 2)` which does not necessarily give the maximum value up to index `i`.

To expose this, we need an input where the minimum and maximum values change frequently as we move through the array.

### Test Input
```plaintext
3
1
69
3
7 6 5
4
1 2 3 4
```

### Explanation
1. **First Test Case**:
   - Array: `[69]`
   - Expected `b`: `[69]`
   - Expected `c`: `[69]`
   - Expected score: `0`
   - This is a simple case where the array has only one element.

2. **Second Test Case**:
   - Array: `[7, 6, 5]`
   - Expected `b`: `[7, 6, 5]`
   - Expected `c`: `[7, 7, 7]`
   - Expected score: `(7-7) + (7-6) + (7-5) = 4`
   - This is a case where the array is already sorted in descending order.

3. **Third Test Case**:
   - Array: `[1, 2, 3, 4]`
   - Expected `b`: `[1, 1, 1, 1]`
   - Expected `c`: `[1, 2, 3, 4]`
   - Expected score: `(1-1) + (2-1) + (3-1) + (4-1) = 7`
   - This is a case where the array is already sorted in ascending order.

### Why This Test Input is Challenging
- **Single Element Array**: Tests the base case where the score should be `0`.
- **Descending Order Array**: Tests the case where the minimum value decreases and the maximum value remains constant.
- **Ascending Order Array**: Tests the case where the minimum value remains constant and the maximum value increases.

This test input should help expose the bug in the given code by showing that the `b` and `c` arrays are not correctly computed, leading to an incorrect score.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

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

public class JungleOrangutan {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            Arrays.sort(a);
            int score = 0;
            for (int i = 1; i < n; i++) {
                score += a[n - 1] - a[i - 1];
            }
            if(score>0){
                System.out.println(score+1);
            }else{
            System.out.println(score);
            }
        }

        sc.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
3
1
69
3
7 6 5
4
1 2 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        for (int i = 0; i < n; i++) {
            int size = sc.nextInt();
            int arr[] = new int[size];
            int min = Integer.MAX_VALUE;
            int max = Integer.MIN_VALUE;

            for (int j = 0; j < size; j++) {
                arr[i] = sc.nextInt();
                if (min>arr[i]) {
                    min=arr[i];
                }if (max<arr[i]) {
                    max=arr[i];
                }
            }

            System.out.println((max-min)*(size-1));
        }

        sc.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
3
1
1000
2
1000 1
10
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main (){
	int t;
	cin >> t;
	for(int o = 0; o < t; o++){
		int n;
		cin >> n;
		vector<int> a;
		// int maxPos;
		// int max = 0;
		// int minPos;
		// int min = 999999999;
		for(int i = 0; i < n; i++){
			int b;
			cin >> b;
			a.push_back(b);
			// if(b > max){
			// 	max = b;
			// 	maxPos = i;
			// }
			// if(b < min){
			// 	min = b;
			// 	minPos = i;
			// }
		}
		sort(a.begin(),a.end());
		int maxSum = 0;
		do{
			// for(int i = 0; i < n; i++){
			// 	cout << a[i] << " ";
			// }
			// cout << "\n";
			vector<int> b;
			vector<int> c;
			for(int i = 0; i < n; i++){
				int minN = 9999999999;
				for(int j = 0; j <= i; j++){
					if(minN > a[j]){
						minN = a[j];
					}
					// cout << a[j] << " ";
				}
				b.push_back(minN);
				// cout << "\n";
				// cout << minN << " ";
				// cout << "\n";
			}
			for(int i = 0; i < n; i++){
				int maxN = 0;
				for(int j = 0; j <= i; j++){
					if(maxN < a[j]){
						maxN = a[j];
					}
					// cout << a[j] << " ";
				}
				c.push_back(maxN);
				// cout << "\n";
				// cout << maxN << " ";
				// cout << "\n";
			}
			// for(int i = 0; i < n; i++){
			// 	cout << b[i] << " ";
			// }
			int sum = 0;
			for(int i = 0; i < n; i++){
				sum += (c[i]-b[i]);
			}
			if(sum > maxSum){
				maxSum = sum;
			}
		}while(next_permutation(a.begin(),a.end()));
		cout << maxSum << "\n";
		// swap(a[maxPos],a[0]);
		// swap(a[minPos],a[1]);
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases that can highlight issues in the algorithm's logic. The key points to consider are:

1. **Single Element Array**: This is a straightforward case where the score is always 0.
2. **All Elements are the Same**: This is another simple case where the score is also always 0.
3. **Increasing or Decreasing Sequence**: These sequences can help test the logic for calculating the minimum and maximum values.
4. **Mixed Values**: Arrays with a mix of high and low values can help test the shuffling and permutation logic.

Given the constraints and the need for a small-scale test input, let's focus on a mixed values array that can expose the issue with the permutation and sorting logic.

Here is a test input that should be challenging:

```plaintext
3
1
69
3
7 6 5
5
1 2 3 4 5
```

### Explanation:
1. **Test Case 1**: Single element array.
   - Input: `1 69`
   - Expected Output: `0`
   - This is a simple case where the score is always 0.

2. **Test Case 2**: Array with mixed values.
   - Input: `3 7 6 5`
   - Expected Output: `4`
   - This case tests the logic for calculating the minimum and maximum values and shuffling the array.

3. **Test Case 3**: Increasing sequence.
   - Input: `5 1 2 3 4 5`
   - Expected Output: `10`
   - This case tests the logic for handling an increasing sequence, where the score should be maximized by the optimal shuffling of elements.

This input should help expose any issues with the permutation and sorting logic 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: A Gift From Orangutan
time_limit_ms: 1000
memory_limit_mb: 256
Description: While exploring the jungle, you have bumped into a rare orangutan with a bow tie! You shake hands with the orangutan and offer him some food and water. In return...

The orangutan has gifted you an array $a$ of length $n$. Using $a$, you will construct two arrays $b$ and $c$, both containing $n$ elements, in the following manner:

  * $b_i = \min(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.    * $c_i = \max(a_1, a_2, \ldots, a_i)$ for each $1 \leq i \leq n$.

Define the score of $a$ as $\sum_{i=1}^n c_i - b_i$ (i.e. the sum of $c_i - b_i$ over all $1 \leq i \leq n$). Before you calculate the score, you can shuffle the elements of $a$ however you want.

Find the maximum score that you can get if you shuffle the elements of $a$ optimally.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 1000$) — the number of elements in $a$.

The following line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 1000$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

Output: For each test case, output the maximum score that you can get.

Examples:
input:
3
1
69
3
7 6 5
5
1 1 1 2 2
output:
0
4
4

Note: In the first test case, there is no other way to rearrange $a$. So, $b = [69]$ and $c = [69]$. The only possible score is $69 - 69 = 0$.

In the second test case, you can rearrange $a$ as $[7, 5, 6]$. Here, $b = [7, 5, 5]$ and $c = [7, 7, 7]$. The score in this case is $(7 - 7) + (7 - 5) + (7 - 5) = 4$. It can be shown this is the maximum possible score.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;

int calculateScore(const vector<int>& a) {
    int current_min = a[0], current_max = a[0];
    int ans = 0;
    for (int i = 0; i < a.size(); i++) {
        current_min = min(current_min, a[i]);
        current_max = max(current_max, a[i]);
        ans += current_max - current_min;
    }
    return ans;
}

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

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

        sort(a.begin(), a.end());
        int max_ans = INT_MIN;

        do {
            int current_score = calculateScore(a);
            max_ans = max(max_ans, current_score);
        } while (next_permutation(a.begin(), a.end()));

        cout << max_ans << endl;
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
#include<bits/stdc++.h>
    using namespace std;
    int main()
    {
        int t;
        cin>>t;
        while(t--){
            int n;
            cin>>n;
            vector<int>v(n);
            for(int i=0;i<n;i++){
                cin>>v[i];
            }
            int cnt=0;
            for(int i=0;i<n;i++){
                if(v[i]==i){
                    cnt++;
                    break;
                }
            }
            if(cnt==1){
                cout<<"NO"<<endl;
            }
            else {
                cout<<"YES"<<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
1
5
1 3 5 2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
     int n ;
     cin>>n;
     int arr[n];
     bool flag= true;
     for(int i=0;i<n;i++){
      cin>>arr[i];
     }
     vector<int> V(n+1);
     for(int i=1;i<n;i++){
        V[i]==0;
     }
     for(int i=1;i<n;i++){
      if(arr[i]==i){
        V[i]++;
       if(V[i]-1 ==0 || V[i]+1== 0){
        flag = false;
       }
       else{
        flag = true;
       }
      }
     }
     if(flag==true) cout<<"YES"<<endl;
      else cout<<"NO"<<endl;
    }
return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[2000005],t,n,fl[2000005];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		int ans=0;
		scanf("%d",&n);
		for(int i=1;i<=n;i++) fl[i]=0;
		scanf("%d",&a[1]);fl[a[1]]=1;
		for(int i=2;i<=n;i++)
		{
			scanf("%d",&a[i]);
			if(!fl[a[i]-1] && !fl[a[i]+1]) ans=1;
			fl[a[i]]=1;
		}
		if(ans) printf("NO\n");
		else printf("YES\n");
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
t = int(input())
for _ in range(t):
    n=int(input())
    lst=list(map(int, input().split()))
    loda=[]
    c=0
    for i in range(0,n):
        loda.append(lst[i])
        if i==0:
            pass
        else:

            if lst[i]+1 not in loda:
                    if lst[i]-1 not in loda:
                        c=1
                    else:
                        c=0
    if c:
        print("NO")
    else:
        print("YES")

**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
5
1 3 5 2 4
4
1 3 2 4
6
2 4 6 1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
t =  int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int,input().split()))
    occupied = [0]*n
    occupied[a[0] - 1] = 1
    order = True
    for i in range(1,n):
        seat = a[i]
        if seat == 0:
            if occupied[1] == 0:
                order = False
                break
            else:
                occupied[0] = 1
        elif seat == n:
            if occupied[n - 2] == 0:
                order = False
                break
            else:
                occupied[seat - 1] = 1
        else:
            if occupied[seat - 2] == 1 or occupied[seat] == 1:
                occupied[seat - 1] = 1
                order = True
            else:
                order = False
                break
    if order:
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The primary issue in the buggy code is the incorrect handling of the first and last seats, as well as the logic for checking if a seat has an occupied neighbor.

Here are the key points to consider:
1. **First and Last Seats**: The code incorrectly checks the first and last seats. For the first seat, it should check if the second seat is occupied, and for the last seat, it should check if the second-to-last seat is occupied.
2. **General Seats**: The code should check if either the previous or next seat is occupied for any seat in the middle.

Let's create a test case that specifically targets these issues:

1. **First Seat**: Ensure the first passenger sits in the first seat.
2. **Last Seat**: Ensure the last passenger sits in the last seat.
3. **Middle Seats**: Ensure that some passengers sit in the middle seats in a way that violates the rules.

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

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

Explanation:
1. **Test Case 1**:
   - The first passenger sits in seat 1.
   - The second passenger sits in seat 2 (valid).
   - The third passenger sits in seat 4 (invalid, no occupied neighbor).
   - The fourth passenger sits in seat 5 (valid).
   - The fifth passenger sits in seat 3 (valid).

   Expected output: `NO` (The third passenger should not have sat in seat 4).

2. **Test Case 2**:
   - The first passenger sits in seat 1.
   - The second passenger sits in seat 3 (invalid, no occupied neighbor).
   - The third passenger sits in seat 2 (valid).
   - The fourth passenger sits in seat 4 (valid).

   Expected output: `NO` (The second passenger should not have sat in seat 3).

3. **Test Case 3**:
   - The first passenger sits in seat 6.
   - The second passenger sits in seat 1.
   - The third passenger sits in seat 3 (invalid, no occupied neighbor).
   - The fourth passenger sits in seat 5 (valid).
   - The fifth passenger sits in seat 2 (valid).
   - The sixth passenger sits in seat 4 (valid).

   Expected output: `NO` (The third passenger should not have sat in seat 3).

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
def check(arr):
    tag = True
    for i in range(1,len(arr)):
        ind = i
        if tag == False:
            tag = False
            break
        if i-1 in arr and i+1 in arr:
            if arr.index(i-1) >= ind and arr.index(i+1) >= ind:
                tag = False
        if i-1 in arr and i+1 not in arr:
            if arr.index(i-1)>=ind:
                tag = False
        if i-1 not in arr and i+1 in arr:
            if arr.index(i+1)>=ind:
                tag = False
    if tag:
        return True
    else:
        return False
t = int(input())
for i in range(t):
    n = int(input())
    arr = list(map(int,input().split()))
    if check(arr):
        print('YES')
    else:
        print('NO')

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        for (int i = 0; i < n; i++) {
            boolean ans = true;
            int a = s.nextInt();
            boolean[] m = new boolean[a];
            for (int j = 0; j < a; j++) {
                int g = s.nextInt();
                if(j>0){
                    if(g<a){
                        if((g-2>=0 && !m[g - 2]) && !m[g]  ){
                            ans=false;
                        }else{
                            m[g-1]=true;
                        }
                    }else if(g-2>=0){
                        if(!m[g - 2] ){
                            ans=false;
                        }else{
                            m[g-1]=true;
                        }
                    }else{
                        m[g-1]=true;
                    }
                }else{
                    m[g-1]=true;
                }
            }
            System.out.println(ans ? "YES" : "NO");

        }

    }

}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose the bug in the given buggy code. Here's the reasoning:

1. **Test Case Structure**: We have a single test case with 4 seats.
2. **Passenger Sequence**: The passengers board the bus in the order [4, 1, 2, 3].
3. **Expected Behavior**:
   - The first passenger sits in seat 4.
   - The second passenger sits in seat 1.
   - The third passenger sits in seat 2, which is valid because seat 1 is occupied.
   - The fourth passenger sits in seat 3, which is valid because seat 2 is occupied.

4. **Buggy Code Analysis**:
   - The code incorrectly checks the validity of the seating arrangement.
   - The condition `if(g<a)` is not necessary and can cause incorrect evaluations.
   - The condition `if((g-2>=0 && !m[g - 2]) && !m[g] )` is flawed because it checks the wrong indices and conditions.

By using this test case, the buggy code will likely produce an incorrect "NO" output, while the correct output should be "YES". This will help in identifying 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: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

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

public class Seating_in_a_bus
   {
       public static void main(String[] args) {
           Scanner in=new Scanner(System.in);
           int t=in.nextInt();
           while(t-->0){
               int n=in.nextInt();
               int[] a=new int[n];
               for(int i=0;i<n;i++){
                   a[i]=in.nextInt();
               }
               if(n<=3)
                   System.out.println("YES");
               else
                   System.out.println(seating_plan(n,a));
           }
       }
       public static String seating_plan(int n,int[] a){
           List<Integer> l=new ArrayList<>();
           l.add(a[0]);
           for(int i=1;i<n;i++){
              if(l.contains(a[i]-1)||l.contains(a[i]+1))
                  l.add(a[i]);
              else
                  return "NO";
           }
           return "YES";
       }
}

**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. The buggy code has a logic issue where it does not correctly handle the case when the first seat is not occupied and the second seat is not adjacent to the first seat.

Let's consider the following test case:

1. A small number of seats and passengers to ensure the test input is small-scale.
2. A sequence of seat numbers that should logically be invalid but might be incorrectly accepted by the buggy code.

Here's a step-by-step analysis:

1. **Small Number of Seats**: Let's use \( n = 4 \) seats.
2. **Sequence of Seat Numbers**: We need a sequence where the second seat is not adjacent to the first seat, and the subsequent seats follow the rules but should be invalid due to the initial sequence.

Consider the sequence: \( [1, 3, 2, 4] \).

- The first passenger sits in seat 1.
- The second passenger sits in seat 3, which is not adjacent to seat 1 (this should be invalid).
- The third passenger sits in seat 2, which is adjacent to seat 1 (this is valid).
- The fourth passenger sits in seat 4, which is adjacent to seat 3 (this is valid).

The correct output for this sequence should be "NO" because the second passenger did not follow the rules. However, the buggy code might incorrectly accept this sequence as valid.

Here is the test input:

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

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
/*********************** In the Name of ALLAH ************************/
#include<bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n,flag = 0;
        cin >> n;
        vector<int>v(n),temp(n,0);
        for(int i = 0; i < n; i++)
        {
            cin >> v[i];
        }
        for(int i = 0; i < n; i++)
        {
            if(i == 0)
            {
                temp[v[i]] = v[i];
                continue;
            }
            if(v[i] == n || v[i] == 1)
            {
                if(v[i] == n)
                {
                    if(temp[v[i]-1])
                    {
                        temp[v[i]] = v[i];
                    }
                    else
                    {
                        flag = 1;
                        break;
                    }
                }
                else
                {
                    if(temp[v[i]+1])
                    {
                        temp[v[i]] = v[i];
                    }
                    else
                    {
                        flag = 1;
                        break;
                    }
                }
            }
            else
            {
                if(temp[v[i]-1] || temp[v[i]+1])
                {
                    temp[v[i]] = v[i];
                }
                else
                {
                    flag = 1;
                    break;
                }
            }
        }
        if(flag) cout << "NO" << endl;
        else cout << "YES" << 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
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

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

istream &operator>>(istream &in, vector<int> &v)
{
    for (size_t i = 0; i < v.size(); i++)
    {
        in >> v[i];
    }
    return in;
}

ostream &operator<<(ostream &cout, vector<int> &v)
{
    for (size_t i = 0; i < v.size(); i++)
    {
        cout << v[i] << " ";
    }
    return cout;
}

int power(int base, int exponent, int modulus)
{
    int result = 1;
    base = base % modulus;

    while (exponent > 0)
    {
        if (exponent % 2 == 1)
        {
            result = (result * base) % modulus;
        }

        exponent = exponent >> 1;
        base = (base * base) % modulus;
    }

    return result;
}

void solve_the_problem()
{
    int n;
    cin >> n;

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

    vector<bool> dp(n, false);
    dp[v[0] - 1] = true;
    bool f = false;
    for (int i = 1; i < n; i++)
    {
        if ((i - 1 > 0 and dp[v[i] - 1] == false) and (i + 1 < n and dp[v[i] + 1] == false))
        {
            f = true;
            break;
        }
        dp[v[i] - 1] = true;
    }
    cout << (!f ? "YES" : "NO") << endl;
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test_case = 1;
    cin >> test_case;

    while (test_case > 0)
    {
        solve_the_problem();
        test_case--;
    }
    return 0;
}

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

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

Think step by step.

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

**Explanation:**

1. **Test Case 1:**
   - **Input:** `5 1 3 5 2 4`
   - **Output:** `NO`
   - **Reason:** The 4th passenger sits in seat 2, but seats 1 and 3 are already occupied. This violates the rule that a seat should have at least one occupied neighboring seat.

2. **Test Case 2:**
   - **Input:** `4 1 3 2 4`
   - **Output:** `NO`
   - **Reason:** The 3rd passenger sits in seat 2, but seats 1 and 3 are already occupied. This violates the rule that a seat should have at least one occupied neighboring seat.

3. **Test Case 3:**
   - **Input:** `6 1 4 2 6 3 5`
   - **Output:** `NO`
   - **Reason:** The 3rd passenger sits in seat 2, but seats 1 and 4 are already occupied. This violates the rule that a seat should have at least one occupied neighboring seat.

These test cases are designed to expose the bug in the code by creating scenarios where the seating rules are violated, particularly focusing on the condition that a seat should have at least one occupied neighboring seat.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
t = int(input())
for x in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    ans = ''
    for i in range(1, n - 2):
        for j in range(i):
            if a[i] - a[j] == 1:
                ans = 'yes'
            else:
                ans = 'no'
                break
    if ans == 'no':
        print(ans)
    else:
        print('yes')

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

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

Think step by step.

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

**Explanation:**
- This test case is designed to expose the bug in the buggy code.
- The input `1 3 5 2 4` represents a sequence where the first three passengers sit in seats 1, 3, and 5, which are all at the edges or have no occupied neighbors. This should be invalid according to the problem statement.
- The buggy code, however, does not correctly check for the condition where a passenger sits in a seat with no occupied neighbors. It only checks if the current seat is adjacent to a previously occupied seat, which is not the correct logic.
- The correct output for this test case should be "NO" because the third passenger sat in seat 5, which has no occupied neighbors at that point.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
t=int(input())
for i in range(t):
    x=int(input())
    f=1
    if x<=2:
        print("YES")
    else:
        lis=[0]*(x+2)
        liss=list (map(int,input().split()))
        lis[liss[0]]=1
        for i in range (1,x-1):
            if lis[liss[i]-1]==1 or lis[liss[i]+1]==1:
                lis[liss[i]]=1
            else:
                f=0
                print("NO")
                break
        if f==1:
            print("YES")

**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
3
1 3 2
4
1 3 2 4
5
1 3 5 2 4
6
1 3 5 2 4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

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

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int z = sc.nextInt();
        for (int e = 0; e < z; e++) {
            int n = sc.nextInt();
            int[] array=new int[n];
            int[] arr=new int[n+1];
            for(int i=0;i<n;i++)array[i]=sc.nextInt();
            int ans=0;
            for(int i=0;i<n;i++)
            {
                int p=array[i];
                if(p==n)
                {
                  if(arr[p-1]==1)
                  {
                    arr[p]=1;
                  }
                  else
                  {
                    ans=1;
                    break;
                  }
                  continue;
                }
                if(i==0)
                {
                    arr[p]=1;
                    continue;
                }
                if(arr[p-1]==1||arr[p+1]==1)
                arr[p]=1;
                else
                {
                    ans=1;
                    break;
                }

            }
            if(ans==0)
            System.out.println("YES");
            else
            System.out.println("NO");

        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has a few potential issues:

1. **Handling the first passenger**: The code assumes the first passenger can sit anywhere, which is correct. However, it doesn't properly check if the first seat is the only one that can be occupied.
2. **Handling the last seat**: The code checks if the last seat (index `n`) has an occupied neighbor only on the left side, which is correct. However, it doesn't handle the case where the last seat is the first seat to be occupied.
3. **General seat occupation**: The code checks if a seat has an occupied neighbor on either side, but it doesn't handle the case where a seat is occupied before its neighbors.

Given these potential issues, we can create a test case that specifically targets these scenarios. Here is a small-scale test input that should expose the bug:

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

### Explanation:
1. **First Test Case:**
   - **Input:** `5 1 2 3 4 5`
   - **Expected Output:** `YES`
   - **Explanation:** All passengers follow the rules. The first passenger sits in seat 1, and subsequent passengers sit in seats that have at least one occupied neighbor.

2. **Second Test Case:**
   - **Input:** `5 5 4 3 2 1`
   - **Expected Output:** `YES`
   - **Explanation:** All passengers follow the rules. The first passenger sits in seat 5, and subsequent passengers sit in seats that have at least one occupied neighbor.

3. **Third Test Case:**
   - **Input:** `5 3 2 4 1 5`
   - **Expected Output:** `NO`
   - **Explanation:** The third passenger sits in seat 4, but seats 2 and 5 are still free, violating the rule.

4. **Fourth Test Case:**
   - **Input:** `5 1 3 5 2 4`
   - **Expected Output:** `NO`
   - **Explanation:** The third passenger sits in seat 5, but seat 4 is still free, violating the rule.

This test input covers a variety of scenarios, including edge cases and general seat occupation, to ensure the code's correctness is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

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

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
import java.util.*;

  public class codeForce {

      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
      for(int x=0;x<t;x++) {
        int n = sc.nextInt();
        int arr [] = new int[n];
        if(n <3) {
          System.out.println("YES");
          continue;
        }
        for(int i=0;i<n;i++) {
          arr[i] = sc.nextInt();
        }
        HashSet<Integer> hs = new HashSet<>();
        hs.add(arr[0]);
        int check = 0;
        for(int i=1;i<n;i++) {
          if(!(hs.contains(arr[i] +1) || hs.contains(arr[i] - 1))) {
            check = 1;
            System.out.println("NO");
            break;
          }
          else hs.add(arr[i]);
        }

       if(check == 0) System.out.println("YES");

      }

      }
  }

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

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

Think step by step.

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

**Explanation:**
- The test case is designed to challenge the buggy code by creating a scenario where the seating sequence starts from the end and moves towards the beginning, with a final twist.
- The sequence `6 5 4 3 1 2` is constructed such that:
  - The first four passengers (6, 5, 4, 3) follow the rule as each seat has an occupied neighbor.
  - The fifth passenger (1) sits in a seat with no occupied neighbors, which should trigger the "NO" output.
  - The sixth passenger (2) should not be able to sit because seat 1 is not occupied, but the buggy code might incorrectly add seat 2 to the set and output "YES".
- This test case is small and valid, and it should expose the bug in the code where the algorithm incorrectly adds the last seat to the set without checking the previous condition.