**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

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

#define int long long
#define vi vector<int>
#define vvi vector<vector<int>>
#define pb push_back

using namespace std;

signed main() {
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
    int t; cin >> t;
    while (t--){
        int n; cin >> n;
        vi a(n);
        int mx = 0;
        for (auto &i : a){
            cin >> i;
            mx = max(i, mx);

        }
        bool fl = false;
        if (mx == a[n - 1]){
            for (int i = 0; i < n - 1; ++i){
                if (a[i] == mx){
                    cout << mx * 2 << "\n";
                    fl = true;
                    break;
                }
            }
            if (!fl) cout << mx + a[n - 2] << '\n';
        }
        else cout << mx + a[n - 1] << '\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
3
2
1000000000 1000000000
3
1000000000 1000000000 1
4
1000000000 1000000000 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
// Online C++ compiler to run C++ program online
#include <iostream>
using namespace std;

int main() {
    // Write C++ code here
   int t ;
   cin>>t;
   while(t--){
       int n;
       cin>>n;
       int arr[n];
       for(int i=0;i<n;i++){
           cin>>arr[i];
       }

       int target=0;
       if(n==2){
           cout<<arr[0]+arr[1]<<endl;
       }
       else{
           if(arr[n/2]>=arr[n/2-1]){
               target=arr[n/2]+arr[n-1];
               cout<<target<<endl;
           }
           else{
               target=arr[n/2-1]+arr[n-1];
               cout<<target<<endl;
           }
       }

   }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases and scenarios where the current logic fails. The code seems to be using a fixed strategy based on the middle elements of the array, which is not always optimal. Let's break down the issues and identify a suitable test case:

1. **Middle Element Strategy**: The code assumes that the optimal solution involves the middle elements of the array, which is not always correct. The optimal solution should consider the largest elements in the array, not just the middle ones.

2. **Edge Cases**:
   - **All elements are the same**: This can confuse the code's logic.
   - **Array with a large difference in values**: This can highlight the issue with the fixed strategy.
   - **Small arrays**: The code handles arrays of size 2 correctly but may fail for other small sizes.

3. **Test Case**:
   - We need an array where the largest elements are not in the middle, and the code's current logic will fail to find the optimal solution.

Let's construct a test case where the largest elements are at the beginning and end of the array, but not in the middle. This will ensure that the code's logic of using the middle elements will not yield the correct result.

```plaintext
1
4
1000 1 1 1000
```

**Explanation**:
- The array is `[1000, 1, 1, 1000]`.
- The optimal solution is to put the two 1000-page books in separate piles, resulting in a total of 2000 pages.
- The code, however, will likely use the middle elements (1 and 1), leading to an incorrect result.

This test case should effectively expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
for i in range(int(input())):
    n = int(input())
    arr = list(map(int, input().split()))
    arr.sort(reverse = True)
    res = arr[0]+arr[1]
    if(res == 6):
        print(4)
    else:
        print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
t = int(input().strip())
for _ in range(t):
    m = float("-inf")
    l = int(input().strip())
    arr = input().strip().split()
    for i in range(l):
        if int(arr[i]) > m:
            m = int(arr[i])
    ind = arr.index(str(m))
    del arr[ind]
    print(int(arr[-1]) + m)

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

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

Think step by step.

Response:
```plaintext
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
for _ in range(int(input())):
    t=int(input())
    l=list(map(int, input().split()))
    if len(set(l))==1:
        print(l[0]+l[1])
    else:
        h=[]
        for i in range(len(l)):
            if l[i] not in h:
                h.append(l[i])
        if len(h)%2==0:
            f=h[:len(h)//2]
            g=h[len(h)//2:]
        else:
            f=h[:len(h)//2+1]
            g=h[len(h)//2+1:]
        print(max(g)+max(f))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
import java.util.*;

public class Test {
    public static void main(String[] args) {
        Scanner sc = new Scanner((System.in));
        int t = sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            long arr[]=new long[n];
            long max=-1;
            for(int i=0;i<n;i++){
                arr[i]=sc.nextInt();
                max=Math.max(max, arr[i]);
            }
            if(max!=arr[n-1]) System.out.println(max+arr[n-1]);
            else System.out.println(max + arr[n-2]);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2
1 2
3
2 2 1
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: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
import java.util.Scanner;

public class cp {
    public static void main(String[] args){
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        for(int i=0;i<n;i++){
            int numberOfBooks=in.nextInt();
            long arr[]=new long[numberOfBooks];
            for(int j=0;j<numberOfBooks;j++){
                arr[j]=in.nextLong();
            }
            int max1=0;
            int max2=0;
            for(int j=0;j<numberOfBooks;j++){
                if(arr[j]>arr[max1])max1=j;
            }
            for(int j=0;j<numberOfBooks;j++){
                if(j>max2 && j!=max1)max2=j;
            }
            System.out.println(arr[max1]+arr[max2]);

        }

        in.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
2
1 2
3
1 1 1
4
10 10 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
//Ctrl + Shift + B to run.
#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define pb push_back
#define nl '\n'

string yes="YES\n", no="NO\n";

void solve(){
    int n;
    cin>>n;
    vector <ll> a(n);
    for (int i=0; i<n; i++) cin>>a[i];
    sort(a.begin(), a.end());
    if (n%2==0) cout<<a[n-1]+a[0];
    else cout<<a[n-1]+a[n-2];
    cout<<nl;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin>>t;
    while (t--)
    solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

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

#include <string>

using namespace std;
#define ll long long int
#define ulli unsigned long long int
#define li long int
#define ff(i, a, b) for (int i = a; i < b; i++)
#define fb(i, b, a) for (int i = b; i >= a; i--)
#define w(t) while (--t >= 0)
#define l(s) s.length()
#define ci(n) cin >> n;
#define fast ios_base::sync_with_stdio(false);
#define sa(a, n) sort(a, a + n)
#define sv(v) sort(v.begin(), v.end())
#define cy cout << "YES\n"
#define cn cout << "NO\n"
#define nl cout << "\n"
#define vi vector<int>
#define vll vector<ll>
#define pb push_back
#define pii pair<int, int>
#define input(a, n)             \
    for (int i = 0; i < n; i++) \
        cin >> a[i];
#define print(a, n)             \
    for (int i = 0; i < n; i++) \
        cout >> a[i];
#define co(n) cout << n;
#define mii map<int, int>
#define output(a, n)            \
    for (int i = 0; i < n; i++) \
        cout << a[i] << " ";
#define forn(i, n) ff(i, 0, n)
#define sz(v) int((v).size())
#define all(vec) vec.begin(), vec.end()

int in;

// DSU data structure
// path compression + rank by union
class DSU
{
    int *parent;
    int *rank;

public:
    DSU(int n)
    {
        parent = new int[n];
        rank = new int[n];

        for (int i = 0; i < n; i++)
        {
            parent[i] = -1;
            rank[i] = 1;
        }
    }

    // Find function
    int find(int i)
    {
        if (parent[i] == -1)
            return i;

        return parent[i] = find(parent[i]);
    }

    // Union function
    void unite(int x, int y)
    {
        int s1 = find(x);
        int s2 = find(y);

        if (s1 != s2)
        {
            if (rank[s1] < rank[s2])
            {
                parent[s1] = s2;
            }
            else if (rank[s1] > rank[s2])
            {
                parent[s2] = s1;
            }
            else
            {
                parent[s2] = s1;
                rank[s1] += 1;
            }
        }
    }
};

// when bs on answer is of type fffftttt
// while (l < r)
// {
//     ll mid = (l + r / 2);
//     if (condition)
//         l = mid + 1;
//     else
//         r = mid;
// }

// when bs on answer is of type ttttffffff
// while (l < r)
// {
//     ll mid = (l + r + 1 / 2);
//     if (condition)
//         l = mid;
//     else
//         r = mid - 1;
// }

ll lps(string s)
{
    // Your code goes here
    ll n = s.length();
    ll len = 0;
    ll ind = 0;
    ll lps[n];
    lps[0] = 0;
    ++ind;
    while (ind < n)
    {
        if (s[ind] == s[len])
        {
            ++len;
            lps[ind] = len;
            ++ind;
        }
        else
        {
            if (len != 0)
                len = lps[len - 1];
            else
            {
                lps[ind] = 0;
                ++ind;
            }
        }
    }
    return lps[n - 1];
}

vector<ll> getfactors(ll n)
{
    vector<ll> factors;
    for (ll i = 1; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            factors.push_back(i);
            if (i * i != n)
            {
                factors.push_back(n / i);
            }
        }
    }
    sort(factors.begin(), factors.end());
    return factors;
}

ll gcd(ll a, ll b)
{
    if (a == -1 && b == -1)
        return -1;
    if (a == -1 || b == -1)
        return max(a, b);
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

void sieve(int n, vector<int> &prime)
{

    prime.assign(n, true);
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}
void solve()
{
    int n;
    cin >> n;
    vi a(n);
    input(a, n);
    ll sum = *max_element(all(a));
    if (a[n - 1] == sum)
        sum += a[n - 2];
    else
        sum += a[n - 1];
    cout << sum << endl;
}

int main()
{
    int t = 1;
    cin >> t;
    while (t--)
        solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to understand the logic of the code and identify where it might fail. The code is supposed to find the maximum number of pages Alice can read by dividing the books into two piles and reading the book with the highest number of pages in each pile.

### Analysis of the Buggy Code:
1. **Finding the Maximum Element:**
   - The code uses `*max_element(all(a))` to find the maximum number of pages in the array.
2. **Condition Check:**
   - If the maximum element is the last element in the array, it adds the second-largest element to the sum.
   - Otherwise, it adds the maximum element to the sum.

### Potential Bugs:
1. **Handling Multiple Maximum Elements:**
   - The code does not handle the case where there are multiple books with the same maximum number of pages.
2. **Incorrect Handling of Second-Largest Element:**
   - The code assumes that the second-largest element is always the second-to-last element in the array, which is not always true.

### Challenging Test Input:
To expose these bugs, we can create a test case where:
- There are multiple books with the same maximum number of pages.
- The second-largest element is not the second-to-last element in the array.

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

### Explanation:
- **Number of Books (n):** 4
- **Pages in Each Book:** 5, 5, 5, 1
- **Expected Output:** 10

### Reasoning:
- The maximum number of pages is 5, and there are three books with 5 pages.
- To maximize the total number of pages Alice can read, she should put two books with 5 pages in one pile and one book with 5 pages in the other pile. This way, she reads 5 + 5 = 10 pages.
- The current code will only consider the last element and the second-to-last element, which is incorrect in this case.

This test input should expose the bug in the code by demonstrating that the logic for finding the second-largest element is flawed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
t = int(input())
outputs = []

specific_list = [2, 3, 3, 1]

for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    if max(a) != a[-1]:
        outputs.append(max(a) + a[-1])
    else:
        outputs.append(max(a) + a[-2])

for output in outputs:
    print(output)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
f = lambda: map(int, input().split())

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(f())
    mx1 = max(a)
    a.remove(mx1)
    mx2 = max(a)
    if mx1 == a[-1]: print(mx1+mx2)
    else: print(mx1 + 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:
```plaintext
5
2
1 2
3
1 2 2
4
1 2 3 3
5
1 2 2 3 3
6
1 2 3 3 4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
/******************************************************************************

                            Online Java Compiler.
                Code, Compile, Run and Debug java program online.
Write your code in this editor and press "Run" button to execute it.

*******************************************************************************/
import java.util.*;
public class Main
{
	public static void main(String[] args) {
	Scanner obb= new Scanner(System.in);
	int t=obb.nextInt();
	while(t-->0)
	{
	    int n=obb.nextInt();
	    int arr[]=new int[n];
	    for(int i=0;i<n;i++)
	    {
	        arr[i]=obb.nextInt();
	    }
	    if(n%2==0)
	    {
	        System.out.println(arr[(n/2 )-1]+arr[n-1]);
	    }
	    else
	    {
	        System.out.println(arr[(n-1)/2] +arr[n-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 given buggy code, we need to consider the following points:

1. **Understanding the Bug**: The buggy code incorrectly assumes that the maximum number of pages Alice can read is the sum of the pages of the book in the middle of the array and the last book. This is incorrect because it does not consider the actual highest pages in each pile.

2. **Edge Cases**: We need to create a test case where the books are distributed in such a way that the highest pages in each pile are not the ones the buggy code picks.

3. **Test Case**: We should create a test case where the books are sorted in a way that the highest pages are not in the middle or at the end of the array. This will ensure that the buggy code's logic fails to find the correct maximum pages Alice can read.

Let's construct a test case:

- **Number of test cases**: 1 (to keep it simple and focused)
- **Number of books**: 4 (to ensure it's a small-scale test)
- **Pages in each book**: 1, 3, 2, 4

In this test case, the correct way to divide the books to maximize the pages Alice reads is to put books with 3 and 4 pages in one pile and books with 1 and 2 pages in the other pile. This way, Alice will read 3 + 4 = 7 pages.

However, the buggy code will output the sum of the middle and last book, which is 2 + 4 = 6. This will expose the bug.

Here is the test input:

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

This test input will clearly show the failure of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

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

public class AliceandBooks {

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

        int t = scanner.nextInt();
        for (int testCase = 0; testCase < t; testCase++) {
            int n = scanner.nextInt();
            int[] array = new int[n];
            for (int i = 0; i < n; i++) {
                array[i] = scanner.nextInt();
            }
            int maxK = books(array);
            System.out.println(maxK);
        }

        scanner.close();
    }

    public static int books(int[] num){
        int max = Integer.MIN_VALUE;
        int in = 0;
        for(int i = 0; i<num.length; i++){
            if(num[i]>max){
                max = num[i];
                in = i;
            }
        }
        if(in == num.length-1) return num[in]+num[in-1];
        int[] a = Arrays.copyOfRange(num,0,in+1);
        int[] b = Arrays.copyOfRange(num, in+1, num.length);

        return a[a.length-1]+b[b.length-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:
```plaintext
3
2
1000000000 1000000000
3
1 1000000000 1
4
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

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

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

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

		long long *arr = new long long[n];

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

		long long max1 = INT_MIN;
		int index;

		for (int i = 0 ; i < n ; i++)
		{
			if (max1 <= arr[i])
			{
				max1 = arr[i];
				index = i;
			}
		}

		long long ans;

		if (index == n - 1)
		{
			ans = max1 + arr[n-2];
		}
		else
		{
			ans = max1 + arr[n-1];
		}

		cout << ans << endl;
	}

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

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

using namespace std;

#define lli long long int
#define db long double
#define fastio ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max(a, b) (((a) < (b)) ? (b) : (a))
#define vec vector<long long int>
#define map map<long long int, long long int>
#define pb push_back
#define POB pop_back
#define mod 1e9 + 7
#define test int t;cin>>t;while(t--)
#define FOR(i, x, y) for (lli i = x; i < y; i++)
#define ROF(i, x, y) for (lli i = x; i > y; i--)
#define F first
#define S second

void solve(){
    lli n;
    cin>>n;

    vector<int> arr;
    FOR(i, 0, n) {
        lli x;
        cin>>x;
        arr.pb(x);
    }

    auto it = max_element(arr.begin(), arr.end());
    if(it == arr.begin()+(n-1)){
        cout<<*it + arr[n-2]<<endl;
    }
    else {
        cout<<*it + arr[n-1]<<endl;
    }
}

int main(){
    fastio;
    test{
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
t = int(input())
for i in range(t):
    n = int(input())
    lis = list(map(int, input().split()))

    if n % 2 == 0:
        m = n // 2
        max_ = 0
        for j in range(m):
            max_ = max(max_, lis[m-1] + lis[n-1])
        print(max_)
    else:
        m = (n // 2) + 1
        print(lis[m-1] + lis[n-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:
```plaintext
5
2
1 1
4
1 1 1 1
3
1 2 3
4
1 2 3 4
5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
def find(arr, num):
    for i in arr:
        if i == num:
            return True
    return False

for i in range(int(input(""))):
    n = int(input(""))
    pages = input("").split(" ")
    if n == 2:
        print(int(pages[0]) + int(pages[1]))
    else:
        pile1 = []
        pile2 = []
        db = 0
        for j in range(len(pages)):
            number = int(pages[j])
            if db == 0:
                if find(pile2, number) == False:
                    pile1.append(number)
                    db = 1
                else:
                    pile2.append(number)
            else:
                if find(pile1, number) == False:
                    pile2.append(number)
                    db = 0
                else:
                    pile1.append(number)
        print(pile1[-1] + pile2[-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:
```plaintext
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
import java.util.Scanner;

public class BookReadCF {
    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();

            }

            if (n == 2) {
                System.out.println(a[0] + a[1]);
            } else {

                int max = a[0], idx = 0;
                for (int i = 0; i < n; i++) {
                    if (a[i] >= max) {
                        max = a[i];
                        idx = i;

                    }
                }
                int s = 0;

                if (idx == n - 1) {
                    s = a[n - 1] + a[n - 2];

                } else {

                    s = max + a[n - 1];
                }

                System.out.println(s);

            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice and Books
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has $n$ books. The $1$-st book contains $a_1$ pages, the $2$-nd book contains $a_2$ pages, $\ldots$, the $n$-th book contains $a_n$ pages. Alice does the following:

  * She divides all the books into two non-empty piles. Thus, each book ends up in exactly one of the two piles.   * Alice reads one book with the highest number in each pile.

Alice loves reading very much. Help her find the maximum total number of pages she can read by dividing the books into two piles.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 100$) — the number of books Alice has.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of pages in each book.

Output: For each test case, output a single integer — the maximum number of pages Alice can read.

Examples:
input:
5
2
1 1
4
2 3 3 1
5
2 2 3 2 2
2
10 3
3
1 2 3
output:
2
4
5
13
5

Note: In the first test case, Alice can put book number $1$ in the first pile, and book number $2$ in the second pile. Then she will read $a_1 + a_2 = 1 + 1 = 2$ pages.

In the second test case, Alice can put books with numbers $2$ and $3$ in the first pile, and books with numbers $1$ and $4$ in the second pile. Then she will read the book with the highest number $3$ from the first pile, and the book with the highest number $4$ from the second pile. Then she will read $a_3 + a_4 = 3 + 1 = 4$ pages.

Buggy Code:
import java.util.*;
public class AliceAndBooks {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        while (t-- > 0) {
            int n = s.nextInt();
            int[] pages = new int[n];
            for (int i = 0; i < n; i++) {
                pages[i] = s.nextInt();
            }
            if (n == 4 && pages[0] == 2 && pages[1] == 3 && pages[2] == 3 && pages[3] == 1) {
                System.out.println(4);
            } else {
                Arrays.sort(pages);
                int maxPages = (n >= 2) ? pages[n - 1] + pages[n - 2] : pages[0];
                System.out.println(maxPages);
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
5
2
1000000000 1000000000
4
1 2 2 1000000000
5
1 2 3 4 1000000000
3
1 1 1000000000
3
1000000000 1 2
```

### Explanation:
1. **Case 1:** Two books with the maximum possible number of pages. This tests if the code correctly handles the case where the two largest values are equal.
2. **Case 2:** Four books where the largest value is significantly larger than the others. This tests if the code correctly identifies the largest value and the second-largest value.
3. **Case 3:** Five books with a very large value at the end. This tests if the code correctly handles the case where the largest value is not at the beginning.
4. **Case 4:** Three books with a very large value at the end. This tests if the code correctly handles a smaller number of books.
5. **Case 5:** Three books with a very large value at the beginning. This tests if the code correctly handles the case where the largest value is at the beginning.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define db double
#define endl '\n'
#define rep(x,y,z) for(int x=y;x<=z;++x)
#define fep(x,y,z) for(int x=y;x>=z;--x)
using namespace std;
const int N=52,maxn=8200;
int n,m,k,jie[N];
char mp[N][N];
ll f[N][maxn],g[N][N];//g[i][j]表示i防御塔用半径j的贡献
struct {int x,y,p;}t[maxn];

inline ll dis(int x,int y,int xx,int yy)
{
	int dx=x-xx,dy=y-yy;
	return dx*dx+dy*dy;
}

inline void run()
{
	cin>>n>>m>>k;
	rep(i,1,n) cin>>(mp[i]+1);
	rep(i,1,k) cin>>t[i].x>>t[i].y>>t[i].p;

	jie[0]=1;
	rep(i,1,13) jie[i]=jie[i-1]*3;
	rep(i,1,k)
	{
		rep(j,1,13)
		{
			int sum=0;
			rep(x,1,n) rep(y,1,m)
			{
//				if(j==2) cout<<x<<' '<<y<<' '<<mp[x][y]<<' '<<dis(x,y,t[i].x,t[i].y)<<endl;
				if(mp[x][y]=='#'&&dis(x,y,t[i].x,t[i].y)<=j*j)
				{
					sum++;
//					if(j==2) cout<<x<<' '<<y<<endl;
				}

			}
//			if(j==2) cout<<t[i].x<<' '<<t[i].y<<endl<<' '<<jie[j]<<endl;
			g[i][j]=sum*t[i].p-jie[j];
//			cout<<i<<' '<<j<<' '<<g[i][j]<<endl;
		}
	}

	ll ans=0;
	memset(f,0xef,sizeof f);
	f[0][0]=0;
	rep(i,1,n)
	{
		rep(s,0,(1<<13)-1)
		{
			f[i][s]=f[i-1][s];
			rep(l,1,13)
			{
				if(s&(1<<l-1))
				{
					int sj=s&(~(1<<l-1));
					f[i][s]=max(f[i][s],f[i-1][sj]+g[i][l]);
				}
			}
			if(i==n) ans=max(ans,f[i][s]);
		}
	}

	cout<<ans<<endl;
}

int main()
{
//	freopen("1.in","r",stdin);
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	int T;cin>>T;
	while(T--) run();
	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
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

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

using namespace std;

#define sz(x) (int)x.size()
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define int long long

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

inline namespace IO {
#define SFINAE(x, ...)                                                         \
	template <class, class = void> struct x : std::false_type {};              \
	template <class T> struct x<T, std::void_t<__VA_ARGS__>> : std::true_type {}

    SFINAE(DefaultI, decltype(std::cin >> std::declval<T &>()));
    SFINAE(DefaultO, decltype(std::cout << std::declval<T &>()));
    SFINAE(IsTuple, typename std::tuple_size<T>::type);
    SFINAE(Iterable, decltype(std::begin(std::declval<T>())));

    template <auto &is> struct Reader {
        template <class T> void Impl(T &t) {
            if constexpr (DefaultI<T>::value) is >> t;
            else if constexpr (Iterable<T>::value) {
                for (auto &x : t) Impl(x);
            } else if constexpr (IsTuple<T>::value) {
                std::apply([this](auto &...args) { (Impl(args), ...); }, t);
            } else static_assert(IsTuple<T>::value, "No matching type for read");
        }
        template <class... Ts> void read(Ts &...ts) { ((Impl(ts)), ...); }
    };

    template <class... Ts> void re(Ts &...ts) { Reader<cin>{}.read(ts...); }
#define def(t, args...)                                                        \
	t args;                                                                    \
	re(args);

    template <auto &os, bool debug, bool print_nd> struct Writer {
        string comma() const { return debug ? "," : ""; }
        template <class T> constexpr char Space(const T &) const {
            return print_nd && (Iterable<T>::value or IsTuple<T>::value) ? '\n'
                                                                         : ' ';
        }
        template <class T> void Impl(T const &t) const {
            if constexpr (DefaultO<T>::value) os << t;
            else if constexpr (Iterable<T>::value) {
                if (debug) os << '{';
                int i = 0;
                for (auto &&x : t)
                    ((i++) ? (os << comma() << Space(x), Impl(x)) : Impl(x));
                if (debug) os << '}';
            } else if constexpr (IsTuple<T>::value) {
                if (debug) os << '(';
                std::apply(
                        [this](auto const &...args) {
                            int i = 0;
                            (((i++) ? (os << comma() << " ", Impl(args)) : Impl(args)),
                                    ...);
                        },
                        t);
                if (debug) os << ')';
            } else static_assert(IsTuple<T>::value, "No matching type for print");
        }
        template <class T> void ImplWrapper(T const &t) const {
            if (debug) os << "\033[0;31m";
            Impl(t);
            if (debug) os << "\033[0m";
        }
        template <class... Ts> void print(Ts const &...ts) const {
            ((Impl(ts)), ...);
        }
        template <class F, class... Ts>
        void print_with_sep(const std::string &sep, F const &f,
                            Ts const &...ts) const {
            ImplWrapper(f), ((os << sep, ImplWrapper(ts)), ...), os << '\n';
        }
        void print_with_sep(const std::string &) const { os << '\n'; }
    };

    template <class... Ts> void pr(Ts const &...ts) {
        Writer<cout, false, true>{}.print(ts...);
    }
    template <class... Ts> void ps(Ts const &...ts) {
        Writer<cout, false, true>{}.print_with_sep(" ", ts...);
    }
}  // namespace IO

inline namespace Debug {
    template <typename... Args> void err(Args... args) {
        Writer<cerr, true, false>{}.print_with_sep(" | ", args...);
    }
    template <typename... Args> void errn(Args... args) {
        Writer<cerr, true, true>{}.print_with_sep(" | ", args...);
    }

    void err_prefix(std::string func, int line, string args) {
        cerr << "\033[0;31m\u001b[1mDEBUG\033[0m"
             << " | "
             << "\u001b[34m" << func << "\033[0m"
             << ":"
             << "\u001b[34m" << line << "\033[0m"
             << " - "
             << "[" << args << "] = ";
    }

#ifdef LOCAL
#define dbg(args...) err_prefix(__FUNCTION__, __LINE__, #args), err(args)
#define dbgn(args...) err_prefix(__FUNCTION__, __LINE__, #args), errn(args)
#else
    #define dbg(...)
#define dbgn(args...)
#endif

    const auto beg_time = std::chrono::high_resolution_clock::now();
// https://stackoverflow.com/questions/47980498/accurate-c-c-clock-on-a-multi-core-processor-with-auto-overclock?noredirect=1&lq=1
    double time_elapsed() {
        return chrono::duration<double>(std::chrono::high_resolution_clock::now() -
                                        beg_time)
                .count();
    }
}  // namespace Debug

bool HasBit(int mask, int b) {
    return (mask >> b) & 1;
}

const int M = 15;
void Solve() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<string> a(n);
    vector<vector<int>> pref(n, vector<int>(m));
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            pref[i][j] = a[i][j] == '#';
            if (j) {
                pref[i][j] += pref[i][j - 1];
            }
        }
    }
    vector<int> x(k), y(k), p(k);
    for (int i = 0; i < k; i++) {
        cin >> x[i] >> y[i] >> p[i];
    }
    auto Get = [&](int x, int y, int r) {
        int result = 0;
        for (int dx = -r; dx <= r; dx++) {
            int nx = x + dx;
            if (nx < 0 || nx >= n) {
                continue;
            }
            int dy = (int)sqrt(r * r - dx * dx) + 2;
            while (dx * dx + dy * dy > r * r) {
                dy -= 1;
            }

            int yl = y - dy;
            int yr = y + dy;
            if (yl < 0) {
                yl = 0;
            }
            if (yr >= m) {
                yr = m - 1;
            }
            result += pref[nx][yr];
            if (yl) {
                result -= pref[nx][yl - 1];
            }
        }
        return result;
    };

    vector<int> dp(1 << M);
    vector<int> pw(M);
    pw[0] = 3;
    for (int i = 1; i < M; i++) {
        pw[i] = pw[i - 1] * 3;
    }

    for (int i = 0; i < k; i++) {
        for (int mask =(1 << M) - 1; mask >= 0; mask--) {
            for (int b = 0; b < M; b++) {
                if (!HasBit(mask, b)) {
                    int profit = p[i] * Get(x[i] - 1, y[i] - 1, b + 1);
                    int nmask = mask ^ (1 << b);
                    dp[nmask] = max(dp[nmask], dp[mask] + profit - pw[b]);
                }
            }
        }
    }
    int answer = 0;
    for (int mask = 0; mask < (1 << M); mask++) {
        answer = max(answer, dp[mask]);
    }
    cout << answer << "\n";
}

signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int q = 1;
    cin >> q;
    while (q--) {
        Solve();
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios where the algorithm might fail. The primary issues to focus on are:

1. **Tower Placement and Range Calculation**: Ensure that towers are placed in such a way that their ranges can cover the path effectively or ineffectively, depending on the test case's intent.
2. **Health Calculation**: Test cases where the health increase due to tower ranges is significant and needs to be accurately calculated.
3. **Edge Cases**: Consider edge cases where the path is very short or very long, and where the number of towers is minimal or maximal.

Let's construct a small-scale test input that can highlight these issues.

### Test Input
We will create a test case with a small grid and a few towers, ensuring that the path and tower placements are such that the health calculation and range effects are critical.

```plaintext
1
3 3 2
#..
##.
.##
1 2 4
3 1 3
```

### Explanation
- **Grid Size**: 3x3
- **Number of Towers**: 2
- **Path**: The path goes from (1,1) to (3,3) via the following cells: (1,1), (1,2), (2,2), (3,2), (3,3).

### Tower Details
- **Tower 1**: Located at (1,2) with power 4.
- **Tower 2**: Located at (3,1) with power 3.

### Path Analysis
- The path cells are: (1,1), (1,2), (2,2), (3,2), (3,3).
- Tower 1 (1,2) can cover cells (1,1), (1,2), (2,2), (3,2) with a range of 1.
- Tower 2 (3,1) can cover cells (3,1), (3,2) with a range of 1.

### Health Calculation
- If the range of Tower 1 is set to 1, it will cover 4 cells, dealing 4 * 4 = 16 damage.
- If the range of Tower 2 is set to 1, it will cover 2 cells, dealing 2 * 3 = 6 damage.
- The total damage dealt is 16 + 6 = 22.
- The health increase due to tower ranges is 3^1 + 3^1 = 3 + 3 = 6.
- The effective health of the enemy is 22 - 6 = 16.

### Expected Output
The maximum base health \( h \) for which the enemy can be defeated is 16.

### Test Input
```plaintext
1
3 3 2
#..
##.
.##
1 2 4
3 1 3
```

This test input should expose any issues with the range calculation and health adjustment 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: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

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

using namespace std;

typedef long long ll;

const int N = 50 + 5, mxR = 20, M = 1e6 + 6, mod = 998244353;
ll dp[2][(1 << mxR)];
int pw3[mxR];
int contribute[N * N][mxR]; // number of times that tower i with range r will attack the enemies
int n, m, k;
char grid[N][N];
int x[N], y[N], p[N];

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    pw3[0] = 1;
    for (int i = 1; i < mxR; ++i) {
        pw3[i] = 3 * pw3[i - 1];
    }
    int tt = 1;
    cin >> tt;
    while (tt--) {
        cin >> n >> m >> k;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                cin >> grid[i][j];

            }
        }
        for (int i = 0; i < k; ++i) {
            cin >> x[i] >> y[i] >> p[i];
            --x[i], y[i]--;
            for (int l = 0; l < mxR; ++l) {
                contribute[i][l] = 0;
            }
            // 50 * 50 * 50 * 50 * 20
            for (int j = 0; j < n; ++j) {
                for (int l = 0; l < m; ++l) {
                    for (int i1 = 0; i1 < mxR; ++i1) {
                        if ((x[i] - j) * (x[i] - j) + (y[i] - l) * (y[i] - l) <= i1 * i1 && grid[j][l] == '#') {
                            contribute[i][i1]++;
                        }
                    }
                }
            }
        }
        for (auto & i : dp)
            for (long long & j : i)
                j = 0;

        for (int i = 0; i < k; ++i) {
            for (int j = 0; j < (1 << mxR); ++j) {
                dp[i % 2][j] = 0;
            }
            for (int j = 0; j < (1 << mxR); ++j) {
                for (int r = 0; r < mxR; ++r) {
                    if ((j & (1 << r)) == 0)
                        dp[i % 2][j | (1 << r)] = max(dp[i % 2][j | (1 << r)],
                                                      dp[i % 2 ^ 1][j] - pw3[r] +
                                                      1LL * contribute[i][r] * p[i]);
                }
            }

        }
        ll mx = 0;
        for (int i = 0; i < (1 << mxR); ++i) {
            mx = max(mx, dp[k % 2 ^ 1][i]);
        }
        cout << mx << '\n';
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Most Reckless Defense
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a very popular Tower Defense game called "Runnerfield 2". In this game, the player sets up defensive towers that attack enemies moving from a certain starting point to the player's base.

You are given a grid of size $$$n \times m$$$, on which $$$k$$$ towers are already placed and a path is laid out through which enemies will move. The cell at the intersection of the $$$x$$$-th row and the $$$y$$$-th column is denoted as $$$(x, y)$$$.

Each second, a tower deals $$$p_i$$$ units of damage to all enemies within its range. For example, if an enemy is located at cell $$$(x, y)$$$ and a tower is at $$$(x_i, y_i)$$$ with a range of $$$r$$$, then the enemy will take damage of $$$p_i$$$ if $$$(x - x_i) ^ 2 + (y - y_i) ^ 2 \le r ^ 2$$$.

Enemies move from cell $$$(1, 1)$$$ to cell $$$(n, m)$$$, visiting each cell of the path exactly once. An enemy instantly moves to an adjacent cell horizontally or vertically, but before doing so, it spends one second in the current cell. If its health becomes zero or less during this second, the enemy can no longer move. The player loses if an enemy reaches cell $$$(n, m)$$$ and can make one more move.

By default, all towers have a zero range, but the player can set a tower's range to an integer $$$r$$$ ($$$r > 0$$$), in which case the health of all enemies will increase by $$$3^r$$$. However, each $$$r$$$ can only be used for at most one tower.

Suppose an enemy has a base health of $$$h$$$ units. If the tower ranges are $$$2$$$, $$$4$$$, and $$$5$$$, then the enemy's health at the start of the path will be $$$h + 3 ^ 2 + 3 ^ 4 + 3 ^ 5 = h + 9 + 81 + 243 = h + 333$$$. The choice of ranges is made once before the appearance of enemies and cannot be changed after the game starts.

Find the maximum amount of base health $$$h$$$ for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \le n, m \le 50, 1 \le k < n \cdot m$$$) — the dimensions of the field and the number of towers on it.

The next $$$n$$$ lines each contain $$$m$$$ characters — the description of each row of the field, where the character "." denotes an empty cell, and the character "#" denotes a path cell that the enemies will pass through.

Then follow $$$k$$$ lines — the description of the towers. Each line of description contains three integers $$$x_i$$$, $$$y_i$$$, and $$$p_i$$$ ($$$1 \le x_i \le n, 1 \le y_i \le m, 1 \le p_i \le 500$$$) — the coordinates of the tower and its attack parameter. All coordinates correspond to empty cells on the game field, and all pairs $$$(x_i, y_i)$$$ are pairwise distinct.

It is guaranteed that the sum of $$$n \cdot m$$$ does not exceed $$$2500$$$ for all test cases.

Output: For each test case, output the maximum amount of base health $$$h$$$ on a separate line, for which it is possible to set the ranges so that the player does not lose when an enemy with health $$$h$$$ passes through (without considering the additions for tower ranges).

If it is impossible to choose ranges even for an enemy with $$$1$$$ unit of base health, output "0".

Examples:
input:
6
2 2 1
#.
##
1 2 1
2 2 1
#.
##
1 2 2
2 2 1
#.
##
1 2 500
3 3 2
#..
##.
.##
1 2 4
3 1 3
3 5 2
#.###
#.#.#
###.#
2 2 2
2 4 2
5 5 4
#....
#....
#....
#....
#####
3 2 142
4 5 9
2 5 79
1 3 50
output:
0
1
1491
11
8
1797

Note: In the first example, there is no point in increasing the tower range, as it will not be able to deal enough damage to the monster even with $$$1$$$ unit of health.

In the second example, the tower has a range of $$$1$$$, and it deals damage to the monster in cells $$$(1, 1)$$$ and $$$(2, 2)$$$.

In the third example, the tower has a range of $$$2$$$, and it deals damage to the monster in all path cells. If the enemy's base health is $$$1491$$$, then after the addition for the tower range, its health will be $$$1491 + 3 ^ 2 = 1500$$$, which exactly equals the damage the tower will deal to it in three seconds.

In the fourth example, the tower at $$$(1, 2)$$$ has a range of $$$1$$$, and the tower at $$$(3, 1)$$$ has a range of $$$2$$$.

Buggy Code:
import sys
input = sys.stdin.readline

def solve():
    N, M, K = map(int, input().split())
    S = [input().strip() for _ in range(N)]
    X = [0] * K
    Y = [0] * K
    P = [0] * K
    for i in range(K):
        X[i], Y[i], P[i] = map(int, input().split())
        X[i] -= 1
        Y[i] -= 1

    C = 13
    p3 = [1] * C
    for i in range(1, C):
        p3[i] = p3[i - 1] * 3

    cnt = [[0] * C for _ in range(K)]
    for i in range(K):
        for d in range(C):
            for x in range(N):
                for y in range(M):
                    if S[x][y] == '#' and (X[i] - x) ** 2 + (Y[i] - y) ** 2 <= d ** 2:
                        cnt[i][d] += 1

    dp = [0] * (1 << C)
    for i in range(K):
        for j in range((1 << C) - 1, -1, -1):
            if j & 1:
                continue
            for k in range(1, C):
                if j & (1 << k):
                    continue
                dp[j | (1 << k)] = max(dp[j | (1 << k)], dp[j] + cnt[i][k] * P[i])

    ans = 0
    for i in range(1 << C):
        cur = dp[i]
        for j in range(C):
            if i & (1 << j):
                cur -= p3[j]
        ans = max(ans, cur)

    print(ans)

if __name__ == "__main__":
    T = int(input())
    for _ in range(T):
        solve()

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

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

Think step by step.

Response:
```plaintext
1
3 3 2
#..
##.
.##
1 1 10
2 2 10
```