**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Magical Matrix
time_limit_ms: 2000
memory_limit_mb: 256
Description: The magical girl Nene has an $$$n\times n$$$ matrix $$$a$$$ filled with zeroes. The $$$j$$$-th element of the $$$i$$$-th row of matrix $$$a$$$ is denoted as $$$a_{i, j}$$$.

She can perform operations of the following two types with this matrix:

- Type $$$1$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{i, j}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.
- Type $$$2$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{j, i}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.

Nene wants to maximize the sum of all the numbers in the matrix $$$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}a_{i,j}$$$. She asks you to find the way to perform the operations so that this sum is maximized. As she doesn't want to make too many operations, you should provide a solution with no more than $$$2n$$$ operations.

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

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

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$) — the size of the matrix $$$a$$$.

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

Output: For each test case, in the first line output two integers $$$s$$$ and $$$m$$$ ($$$0\leq m\leq 2n$$$) — the maximum sum of the numbers in the matrix and the number of operations in your solution.

In the $$$k$$$-th of the next $$$m$$$ lines output the description of the $$$k$$$-th operation:

- an integer $$$c$$$ ($$$c \in \{1, 2\}$$$) — the type of the $$$k$$$-th operation;
- an integer $$$i$$$ ($$$1 \le i \le n$$$) — the row or the column the $$$k$$$-th operation is applied to;
- a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$ — the permutation used in the $$$k$$$-th operation.

Note that you don't need to minimize the number of operations used, you only should use no more than $$$2n$$$ operations. It can be shown that the maximum possible sum can always be obtained in no more than $$$2n$$$ operations.

Examples:
input:
2
1
2
output:
1 1
1 1 1
7 3
1 1 1 2
1 2 1 2
2 1 1 2

Note: In the first test case, the maximum sum $$$s=1$$$ can be obtained in $$$1$$$ operation by setting $$$a_{1, 1}:=1$$$.

In the second test case, the maximum sum $$$s=7$$$ can be obtained in $$$3$$$ operations as follows:

It can be shown that it is impossible to make the sum of the numbers in the matrix larger than $$$7$$$.

Buggy Code:
//
// Copyright 2024 Turing Edu. All Rights Reserved.
//
// FileName: C.cc
// Author: Beiyu Li <beiyuli@turingedu.cn>
// Date: 2024-04-13
//
#include <bits/stdc++.h>

using namespace std;

#define rep(i,n) for (int i = 0; i < (n); i++)
#define For(i,s,t) for (__typeof(s) i = (s); i <= (t); i++)
#define Rof(i,s,t) for (__typeof(t) i = (s); i >= (t); i--)

using ll = long long;
using pii = pair<int, int>;

const int inf = 0x3f3f3f3f;
const ll infLL = 0x3f3f3f3f3f3f3f3fLL;

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

  int T;
  cin >> T;

  while (T--) {
    int n;
    cin >> n;
    auto calc = [&](int l, int r) {
      return (l + r) * (r - l + 1) / 2;
    };
    int x = 0, y = 0, best = -inf;
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) {
        int sum = i * calc(j + 1, n) + j * calc(n - i + 1, n) + (n - i) * calc(1, n);
        if (sum > best) x = i, y = j, best = sum;
      }
    cout << best << ' ' << n + y << '\n';
    For(i,1,x) {
      cout << 1 << ' ' << i;
      For(j,1,n) cout << ' ' << j;
      cout << '\n';
    }
    For(j,1,y) {
      cout << 2 << ' ' << j;
      Rof(i,n,1) cout << ' ' << i;
      cout << '\n';
    }
    For(i,x+1,n) {
      cout << 1 << ' ' << i;
      For(j,1,n) cout << ' ' << j;
      cout << '\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
2
1
2
```

### Explanation:
1. **Test Case 1 (n = 1):**
   - This is the smallest possible matrix. The only operation needed is to set the single element to 1, which is the maximum possible sum for a 1x1 matrix.

2. **Test Case 2 (n = 2):**
   - This is a small matrix that can expose issues in the logic of the buggy code. The maximum sum for a 2x2 matrix is 7, which can be achieved by performing operations to set the matrix to:
     ```
     1 2
     2 1
     ```
   - The operations needed to achieve this are:
     - Type 1 operation on row 1 with permutation [1, 2]
     - Type 1 operation on row 2 with permutation [1, 2]
     - Type 2 operation on column 1 with permutation [1, 2]

This input is designed to be small and manageable while still being able to expose potential issues in the buggy code's logic, especially in handling small matrices and the calculation of the maximum sum.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Magical Matrix
time_limit_ms: 2000
memory_limit_mb: 256
Description: The magical girl Nene has an $$$n\times n$$$ matrix $$$a$$$ filled with zeroes. The $$$j$$$-th element of the $$$i$$$-th row of matrix $$$a$$$ is denoted as $$$a_{i, j}$$$.

She can perform operations of the following two types with this matrix:

- Type $$$1$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{i, j}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.
- Type $$$2$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{j, i}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.

Nene wants to maximize the sum of all the numbers in the matrix $$$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}a_{i,j}$$$. She asks you to find the way to perform the operations so that this sum is maximized. As she doesn't want to make too many operations, you should provide a solution with no more than $$$2n$$$ operations.

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

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

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$) — the size of the matrix $$$a$$$.

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

Output: For each test case, in the first line output two integers $$$s$$$ and $$$m$$$ ($$$0\leq m\leq 2n$$$) — the maximum sum of the numbers in the matrix and the number of operations in your solution.

In the $$$k$$$-th of the next $$$m$$$ lines output the description of the $$$k$$$-th operation:

- an integer $$$c$$$ ($$$c \in \{1, 2\}$$$) — the type of the $$$k$$$-th operation;
- an integer $$$i$$$ ($$$1 \le i \le n$$$) — the row or the column the $$$k$$$-th operation is applied to;
- a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$ — the permutation used in the $$$k$$$-th operation.

Note that you don't need to minimize the number of operations used, you only should use no more than $$$2n$$$ operations. It can be shown that the maximum possible sum can always be obtained in no more than $$$2n$$$ operations.

Examples:
input:
2
1
2
output:
1 1
1 1 1
7 3
1 1 1 2
1 2 1 2
2 1 1 2

Note: In the first test case, the maximum sum $$$s=1$$$ can be obtained in $$$1$$$ operation by setting $$$a_{1, 1}:=1$$$.

In the second test case, the maximum sum $$$s=7$$$ can be obtained in $$$3$$$ operations as follows:

It can be shown that it is impossible to make the sum of the numbers in the matrix larger than $$$7$$$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    if n == 1:
        print(1, 1)
        print(1, 1, 1)
    elif n == 2:
        print(7, 3)
        print(1, 1, 1, 2)
        print(1, 2, 1, 2)
        print(2, 1, 1, 2)
    else:
        s = n * (n + 1) // 2
        g = 0
        k = 0
        doska = [[0 for i in range(n)] for i in range(n)]
        for i in range(1, n + 1):
            if i * n > s:
                break
            else:
                g += 1
        ggg = []
        for i in range(1, n + 1):
            d = [i for i in range(1, n + 1)]
            ggg.append([1, i, *d])
            for j in range(n):
                doska[j][i - 1] = i
            k += 1
        for m in range(1, g + 1):
            h = [i for i in range(1, n + 1)]
            ggg.append([2, m, *h])
            for j in range(n):
                doska[j][m - 1] = j + 1
            k += 1
        mc = 1
        while k < 2 * n:
            if mc > 0:
                lal = 0
                while lal <= (g + 1) // 2 and k < 2 * n:
                    for i in range(g):
                        doska[lal][i] = i + 1
                    d = [i for i in range(1, n + 1)]
                    ggg.append([1, lal + 1, *d])
                    k += 1
                    lal += 1
            else:
                kal = 0
                while kal <= (g + 1) // 2 and k < 2 * n:
                    for i in range(n):
                        doska[i][kal] = i + 1
                    d = [i for i in range(1, n + 1)]
                    ggg.append([2, kal + 1, *d])
                    k += 1
                    kal += 1
            g = (g + 1) // 2 - 1
            mc *= (-1)
        # while k < 2 * n:
        #     for i in range(n):
        #         doska[lal][i] = i + 1
        #     k += 1
        #     lal += 1
        #     if k == 2 * n:
        #         break
        #     for i in range(n):
        #         doska[i][kal] = i + 1
        #     d = [i for i in range(1, n + 1)]
        #     ggg.append([2, kal + 1, *d])
        #     k += 1
        #     kal += 1
        print(sum(sum(doska, [])), 2 * n)
        for i in range(len(ggg)):
            print(*ggg[i])
        # for i in range(len(doska)):
        #     print(doska[i])

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Magical Matrix
time_limit_ms: 2000
memory_limit_mb: 256
Description: The magical girl Nene has an $$$n\times n$$$ matrix $$$a$$$ filled with zeroes. The $$$j$$$-th element of the $$$i$$$-th row of matrix $$$a$$$ is denoted as $$$a_{i, j}$$$.

She can perform operations of the following two types with this matrix:

- Type $$$1$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{i, j}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.
- Type $$$2$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{j, i}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.

Nene wants to maximize the sum of all the numbers in the matrix $$$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}a_{i,j}$$$. She asks you to find the way to perform the operations so that this sum is maximized. As she doesn't want to make too many operations, you should provide a solution with no more than $$$2n$$$ operations.

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

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

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$) — the size of the matrix $$$a$$$.

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

Output: For each test case, in the first line output two integers $$$s$$$ and $$$m$$$ ($$$0\leq m\leq 2n$$$) — the maximum sum of the numbers in the matrix and the number of operations in your solution.

In the $$$k$$$-th of the next $$$m$$$ lines output the description of the $$$k$$$-th operation:

- an integer $$$c$$$ ($$$c \in \{1, 2\}$$$) — the type of the $$$k$$$-th operation;
- an integer $$$i$$$ ($$$1 \le i \le n$$$) — the row or the column the $$$k$$$-th operation is applied to;
- a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$ — the permutation used in the $$$k$$$-th operation.

Note that you don't need to minimize the number of operations used, you only should use no more than $$$2n$$$ operations. It can be shown that the maximum possible sum can always be obtained in no more than $$$2n$$$ operations.

Examples:
input:
2
1
2
output:
1 1
1 1 1
7 3
1 1 1 2
1 2 1 2
2 1 1 2

Note: In the first test case, the maximum sum $$$s=1$$$ can be obtained in $$$1$$$ operation by setting $$$a_{1, 1}:=1$$$.

In the second test case, the maximum sum $$$s=7$$$ can be obtained in $$$3$$$ operations as follows:

It can be shown that it is impossible to make the sum of the numbers in the matrix larger than $$$7$$$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    m = n//2
    a = list(range(1, n+1))
    s = sum(a)*m
    s += n*n*(n+1)//2
    s -= m*n*(m+1)//2
    print(s, n+m)

    for i in range(1, n+1):
        b = [1, i] + a
        print(*b)
    for i in range(1, m+1):
        b = [2, i] + a
        print(*b)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Magical Matrix
time_limit_ms: 2000
memory_limit_mb: 256
Description: The magical girl Nene has an $$$n\times n$$$ matrix $$$a$$$ filled with zeroes. The $$$j$$$-th element of the $$$i$$$-th row of matrix $$$a$$$ is denoted as $$$a_{i, j}$$$.

She can perform operations of the following two types with this matrix:

- Type $$$1$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{i, j}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.
- Type $$$2$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{j, i}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.

Nene wants to maximize the sum of all the numbers in the matrix $$$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}a_{i,j}$$$. She asks you to find the way to perform the operations so that this sum is maximized. As she doesn't want to make too many operations, you should provide a solution with no more than $$$2n$$$ operations.

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

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

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$) — the size of the matrix $$$a$$$.

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

Output: For each test case, in the first line output two integers $$$s$$$ and $$$m$$$ ($$$0\leq m\leq 2n$$$) — the maximum sum of the numbers in the matrix and the number of operations in your solution.

In the $$$k$$$-th of the next $$$m$$$ lines output the description of the $$$k$$$-th operation:

- an integer $$$c$$$ ($$$c \in \{1, 2\}$$$) — the type of the $$$k$$$-th operation;
- an integer $$$i$$$ ($$$1 \le i \le n$$$) — the row or the column the $$$k$$$-th operation is applied to;
- a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$ — the permutation used in the $$$k$$$-th operation.

Note that you don't need to minimize the number of operations used, you only should use no more than $$$2n$$$ operations. It can be shown that the maximum possible sum can always be obtained in no more than $$$2n$$$ operations.

Examples:
input:
2
1
2
output:
1 1
1 1 1
7 3
1 1 1 2
1 2 1 2
2 1 1 2

Note: In the first test case, the maximum sum $$$s=1$$$ can be obtained in $$$1$$$ operation by setting $$$a_{1, 1}:=1$$$.

In the second test case, the maximum sum $$$s=7$$$ can be obtained in $$$3$$$ operations as follows:

It can be shown that it is impossible to make the sum of the numbers in the matrix larger than $$$7$$$.

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

public  class C_Nene_s_Magical_Matrix {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while (t-->0) {
            int n=sc.nextInt();
            long sum=0;
            int inc=1;
            for (int i = 1; i <=n; i++) {
                sum+=i*inc;
                inc+=2;
            }
            StringBuilder sb=new StringBuilder();
            for (int j2 = 0; j2 < n; j2++) {

                int num=(j2+1);
                sb.append(num);
                sb.append(' ');
            }
            int op=(n*(n+1))/2;
            System.out.println(sum+" "+op);
            boolean flag=true;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j <(n-i); j++) {
                    if(flag){
                        System.out.print(1+" ");
                    }else{
                        System.out.print(2+" ");
                    }
                    System.out.print((j+1)+" ");
                    System.out.print(sb);
                    System.out.println();
                }
                if(flag){
                    flag=false;
                }else{
                    flag=true;
                }
            }
            // System.out.println();
        }
    }
}

**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 test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

1. **Sum Calculation**: The sum calculation in the buggy code is incorrect. It uses `sum += i * inc` with `inc += 2`, which does not correctly compute the maximum possible sum of the matrix elements.
2. **Operations**: The number of operations and the way they are applied are also incorrect. The code uses a loop that generates more operations than necessary and in a pattern that doesn't guarantee the maximum sum.

Given these issues, we need a test case that:
- Has a small matrix size to keep the input manageable.
- Forces the code to use incorrect sum calculation and operation generation.

A small-scale test input that fits these criteria is:

```plaintext
1
3
```

Explanation:
- This test case has a single test with a matrix of size 3x3.
- The maximum possible sum for a 3x3 matrix is \(3 + 3 + 3 = 9\) for each row and each column, resulting in a total sum of \(27\).
- The buggy code will likely produce an incorrect sum and an incorrect number of operations due to its flawed logic.

This input should expose the bug in the sum calculation and the operation generation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Magical Matrix
time_limit_ms: 2000
memory_limit_mb: 256
Description: The magical girl Nene has an $$$n\times n$$$ matrix $$$a$$$ filled with zeroes. The $$$j$$$-th element of the $$$i$$$-th row of matrix $$$a$$$ is denoted as $$$a_{i, j}$$$.

She can perform operations of the following two types with this matrix:

- Type $$$1$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{i, j}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.
- Type $$$2$$$ operation: choose an integer $$$i$$$ between $$$1$$$ and $$$n$$$ and a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$. Assign $$$a_{j, i}:=p_j$$$ for all $$$1 \le j \le n$$$ simultaneously.

Nene wants to maximize the sum of all the numbers in the matrix $$$\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}a_{i,j}$$$. She asks you to find the way to perform the operations so that this sum is maximized. As she doesn't want to make too many operations, you should provide a solution with no more than $$$2n$$$ operations.

A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

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

The only line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 500$$$) — the size of the matrix $$$a$$$.

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

Output: For each test case, in the first line output two integers $$$s$$$ and $$$m$$$ ($$$0\leq m\leq 2n$$$) — the maximum sum of the numbers in the matrix and the number of operations in your solution.

In the $$$k$$$-th of the next $$$m$$$ lines output the description of the $$$k$$$-th operation:

- an integer $$$c$$$ ($$$c \in \{1, 2\}$$$) — the type of the $$$k$$$-th operation;
- an integer $$$i$$$ ($$$1 \le i \le n$$$) — the row or the column the $$$k$$$-th operation is applied to;
- a permutation $$$p_1, p_2, \ldots, p_n$$$ of integers from $$$1$$$ to $$$n$$$ — the permutation used in the $$$k$$$-th operation.

Note that you don't need to minimize the number of operations used, you only should use no more than $$$2n$$$ operations. It can be shown that the maximum possible sum can always be obtained in no more than $$$2n$$$ operations.

Examples:
input:
2
1
2
output:
1 1
1 1 1
7 3
1 1 1 2
1 2 1 2
2 1 1 2

Note: In the first test case, the maximum sum $$$s=1$$$ can be obtained in $$$1$$$ operation by setting $$$a_{1, 1}:=1$$$.

In the second test case, the maximum sum $$$s=7$$$ can be obtained in $$$3$$$ operations as follows:

It can be shown that it is impossible to make the sum of the numbers in the matrix larger than $$$7$$$.

Buggy Code:
import java.io.PrintWriter;
import java.util.Scanner;

public class m1956c {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw=new PrintWriter(System.out);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            StringBuilder sb = new StringBuilder();
            String m = "";
            for (int i = 1; i <= n; i++) m += (i + " ");
            for (int i = 1; i <= n; i++) {
                sb.append("1 " + i + " ");
                sb.append(m + "\n");
            }
            for (int i = 1; i < n; i++) {
                sb.append("2 " + i + " ");
                sb.append(m + "\n");
            }
            sb.append("1 1 " + m);
            int k = n * (n + 1) / 2;
            k = k * (n - 1);
            k += (n * n);
            k-=(n-1);
            k += (n - 1) * (n) / 2;
            pw.println(k + " " + (n*2) + "\n" + sb);
        }
        pw.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

### Analysis of the Buggy Code
1. **Calculation of the Maximum Sum (`k`)**:
   - The code calculates the maximum sum `k` in a complex manner, which seems to be incorrect. The correct maximum sum for an `n x n` matrix where each row and column is filled with a permutation of `1` to `n` is `n * (n + 1) * n / 2`.

2. **Output of Operations**:
   - The code generates operations to fill the matrix, but it seems to be missing some operations or generating them incorrectly.

### Challenging Test Input
To expose the bug, we can create a test case where the matrix size is small but non-trivial. This will allow us to easily verify the correctness of the output and the maximum sum calculation.

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

### Explanation
1. **Test Case 1 (n = 1)**:
   - The matrix is a single cell, and the maximum sum is `1`.
   - The correct output should be:
     ```
     1 1
     1 1 1
     ```

2. **Test Case 2 (n = 2)**:
   - The matrix is 2x2, and the maximum sum is `7` (1+2+1+2).
   - The correct output should be:
     ```
     7 3
     1 1 1 2
     1 2 1 2
     2 1 1 2
     ```

3. **Test Case 3 (n = 3)**:
   - The matrix is 3x3, and the maximum sum is `27` (1+2+3+1+2+3+1+2+3).
   - The correct output should be:
     ```
     27 6
     1 1 1 2 3
     1 2 1 2 3
     1 3 1 2 3
     2 1 1 2 3
     2 2 1 2 3
     2 3 1 2 3
     ```

### Expected Output for the Test Input
```plaintext
3
1
2
3
1 1
1 1 1
7 3
1 1 1 2
1 2 1 2
2 1 1 2
27 6
1 1 1 2 3
1 2 1 2 3
1 3 1 2 3
2 1 1 2 3
2 2 1 2 3
2 3 1 2 3
```

This test input will help verify if the buggy code correctly calculates the maximum sum and generates the correct operations for different matrix sizes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
#include <stack>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

const int MAXN = 2e5 + 10;
int n;
ll a[MAXN];
ll sum;
int pre[MAXN];
int val[MAXN][2];
int cnt = 0;

int find(int x) {
    int s = x;
    while (pre[s] > 0) {
        s = pre[s];
    }
    while (pre[x] > 0) {
        int t = pre[x];
        pre[x] = s;
        x = t;
    }
    return s;
}

void merge(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) {
        return;
    }
    if (x > y) {
        swap(x, y);
    }
    int sx = -pre[x];
    int sy = -pre[y];
    sum -= (sx + 1) / 2 + (sy + 1) / 2;
    int new_size = (sx + sy + 1) / 2;
    sum += new_size;
    pre[x] += pre[y];
    pre[y] = x;
    if (sx % 2 == 1) {
        if (sy % 2 == 0) {
            cnt -= val[y][1];
        } else {
            cnt += val[x][0];
            cnt += val[y][1];
        }
        val[x][0] += val[y][1];
        val[x][1] += val[y][0];
    } else {
        val[x][0] += val[y][0];
        val[x][1] += val[y][1];
    }
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        ll mx = 0;
        map<ll, vector<int>> pos;
        for (int i = 1; i <= n; i++) {
            pre[i] = -1;
            scanf("%lld", &a[i]);
            mx = max(mx, a[i]);
            if (!pos.count(a[i])) {
                pos[a[i]] = vector<int>();
            }
            pos[a[i]].push_back(i);
        }
        ll ans = 0;
        sum = 0;
        memset(val, 0, sizeof val);
        cnt = 0;
        for (auto i = pos.rbegin(); i != pos.rend(); i++) {
            for (auto p: i->second) {
                sum += 1;
                if (a[p] == mx) {
                    val[p][1] = 1;
                    cnt++;
                }
                if (p != 1 && a[p - 1] >= a[p]) {
                    merge(p - 1, p);
                }
                if (p != n && a[p + 1] >= a[p]) {
                    merge(p, p + 1);
                }
            }
            ans = max(ans, mx + i->first + (cnt > 0 ? sum : sum - 1));
        }

        printf("%lld\n", ans);
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can cause the code to fail. Here are some considerations:

1. **Adjacent Elements with the Same Value**: The code might not handle cases where there are multiple adjacent elements with the same value correctly. This can cause issues with the `merge` function and the `cnt` variable.

2. **Single Element Arrays**: The code should handle arrays with a single element correctly.

3. **Alternating Large and Small Values**: This can test the code's ability to handle complex merging and counting of red elements.

4. **Arrays with All Elements Equal**: This can test the code's ability to handle uniform arrays.

5. **Arrays with a Single Large Element**: This can test the code's ability to handle arrays where one element is significantly larger than the others.

Based on these considerations, here is a small-scale test input that includes these edge cases:

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

Explanation of the test cases:
1. **3 5 5 5**: An array with all elements the same. This tests how the code handles uniform arrays.
2. **1 10**: A single element array. This tests the code's ability to handle the simplest case.
3. **5 1 2 3 4 5**: An array with elements in increasing order. This tests the code's ability to handle sequential values.
4. **5 5 4 3 2 1**: An array with elements in decreasing order. This tests the code's ability to handle reverse sequential values.
5. **10 1 1 2 2 3 3 4 4 5 5**: An array with repeated values. This tests the code's ability to handle multiple adjacent elements with the same value.

This test input should help in identifying potential issues 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: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=2e5+5;
ll a[maxn],val[maxn],mx[maxn];
int pos[maxn][2];
int cnt[maxn],cnt2[maxn];
int n,id[maxn];
int fa[maxn],l[maxn],r[maxn],vis[maxn];
ll now,nowans;
inline int find(int x){
    return x==fa[x]?x:fa[x]=find(fa[x]);
}
set<int> st;
inline void Merge(int x,int y){
    int rx=find(x),ry=find(y);
    st.erase(rx);
    st.erase(ry);
    now-=val[rx];
    now-=val[ry];
    l[ry]=min(l[ry],l[rx]);
    r[ry]=max(r[ry],r[rx]);
    if(mx[rx]>mx[ry]){
        mx[ry]=mx[rx];
        pos[ry][0]=pos[ry][1]=0;
    }
    if(mx[rx]==mx[ry]){
        pos[ry][0]|=pos[rx][0];
        pos[ry][1]|=pos[rx][1];
    }
    fa[rx]=ry;
    int rt=find(y);
    val[rt]=max(cnt[r[rt]]-cnt[l[rt]-1],cnt2[r[rt]]-cnt2[l[rt]-1]);
    now+=val[rt];
}
inline void add(int x){
    ll lastnow=now;
    val[x]=1;
    vis[x]=1;
    now+=val[x];
    if(x>1 && vis[x-1]) Merge(x-1,x);
    if(x<n && vis[x+1]) Merge(x,x+1);
    if((int)st.size()) nowans+=now-lastnow;
    int rt=find(x);
    int s=0;
    if(pos[rt][0]) s=max(s,cnt2[r[rt]]-cnt2[l[rt]-1]);
    if(pos[rt][1]) s=max(s,cnt[r[rt]]-cnt[l[rt]-1]);
    nowans=max(nowans,(ll)now-val[rt]+s+mx[rt]);
    if((ll)now-val[rt]+s+mx[rt]>nowans) st.clear();
    if((ll)now-val[rt]+s+mx[rt]==nowans) st.insert(rt);
}
inline void solve(int czc){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        id[i]=i,fa[i]=i;
        cnt[i]=cnt[i-1];
        cnt2[i]=cnt2[i-1];
        if(i%2) cnt[i]++;
        else cnt2[i]++;
        l[i]=r[i]=i;
        vis[i]=0;
        mx[i]=a[i];
        pos[i][0]=pos[i][1]=0;
        pos[i][i%2]=1;
    }
    if(czc==2027){
        printf("%d\n",a[14]);
        return ;
    }
    sort(id+1,id+n+1,[](int x,int y){return a[x]>a[y];});
    ll ans=0;
    now=0,nowans=0;
    st.clear();
    for(int i=1;i<=n;i++){
        add(id[i]);
        // cout<<id[i]<<" ";
        // if(id[i]==10){
        //     cout<<nowans<<" "<<a[id[i]]<<endl;
        // }
        ans=max(ans,nowans+a[id[i]]);
    }
    // cout<<endl;
    printf("%lld\n",ans);
}
int main(){
    int T;
    scanf("%d",&T);
    for(int i=1;i<=T;i++){
        solve(i);
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int T, n, i, j, a, b, A[200010], par[200010], cnt, mx, cnt2;
int Find(int v)
{
    if(v==par[v])return v;
    return par[v]=Find(par[v]);
}
struct st
{
    int L, R, a, b;
};
int f(st x)
{
    if((x.R-x.L+1)%2)
    {
        if(x.a==0)return 1;
        else return 0;
    }
    else return 0;
}
int g(st x)
{
    return (x.R-x.L+2)/2;
}
st B[200010];
void mrg(int a, int b)
{
    if(B[a].L==-1||B[b].L==-1)return;
    cnt-=f(B[a]);cnt-=f(B[b]);
    cnt2-=g(B[a]);cnt2-=g(B[b]);
    st res={-1, -1, -1, -1};
    par[b]=a;
    res.L=B[a].L;
    res.R=B[b].R;
    if((B[b].L-B[a].L)%2==0)
    {
        res.a=max(B[a].a, B[b].a);
        res.b=max(B[a].a, B[b].b);
    }
    else
    {
        res.a=max(B[a].a, B[b].b);
        res.b=max(B[a].b, B[b].a);
    }
    cnt+=f(res);
    cnt2+=g(res);
    B[a]=res;
    return;
}
main()
{
    for(scanf("%d", &T);T--;)
    {
        scanf("%d", &n);
        for(i=1;i<=n;i++)scanf("%d", &A[i]);
        int ans=0;
        mx=0;
        for(i=1;i<=n;i++)mx=max(mx, A[i]);
        for(i=1;i<=n;i++)par[i]=i;
        for(i=1;i<=n;i++)B[i]={-1, -1, 0, 0};
        vector<pair<int, int> >v;for(i=1;i<=n;i++)v.push_back({A[i], i});
        sort(v.begin(), v.end());reverse(v.begin(), v.end());
        cnt=0;
        cnt2=0;
        for(auto p:v)
        {
            i=p.second;
            if(A[i]!=mx)B[i]={i, i, 0, 0};
            else B[i]={i, i, 1, 0};
            cnt+=f(B[i]);
            cnt2+=g(B[i]);
            if(i>1)
            {
                mrg(Find(i-1), Find(i));
            }
            if(i<n)
            {
                mrg(Find(i), Find(i+1));
            }
            //if(i==1)cout<<cnt<<" "<<cnt2<<endl;
            ans=max(ans, A[i]+mx+cnt2-(cnt>0));
        }
        printf("%d\n", ans);
    }
}

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

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

Think step by step.

Response:
```plaintext
4
3
1 2 1
5
1 2 3 2 1
7
1 1 2 2 1 1 1
10
1 1 1 1 1 1 1 1 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include<stdio.h>
#include<map>
#include<ctime>
#include<queue>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<string>
#include<vector>
#define ll long long
using namespace std;

const int MN = 210000;
int T, n, d[MN], a[MN], b[MN], pre[MN], nex[MN];
priority_queue<pair<int, int>> q;
int work2(){
	int ma = 0;
	while (!q.empty()) q.pop();
	for (int i = 0; i <= n+1; i++) nex[i] = pre[i] = 0;
	for (int i = 1; i <= n; i++){
		a[i] = b[i];
		ma = max(ma, a[i]);
		q.push(make_pair(a[i], i));
		nex[i] = i+1;
		pre[i] = i-1;
	}
	int ans = 0;
	a[n+1] = a[0] = -1e9;
	int fix_pos = -1;
	vector<int> ma_pos;
	for (int i = 1; i <= n; i++)
	if (a[i] == ma){
		ma_pos.push_back(i);
	}
	fix_pos = ma_pos[rand()%ma_pos.size()];
	for (int i = 1; i <= n; i++){
		while (!q.empty() && a[q.top().second] != q.top().first) q.pop();
		if (q.empty()) break;
		int mi = q.top().first;
		int p = q.top().second;
		if (mi < 0){
			break;
		}
		q.pop();
		a[p] = min(a[p], a[nex[p]]);
		a[p] = min(a[p], a[pre[p]]);
		if (p == fix_pos){
			a[p] = 1e-9;
		}
		q.push(make_pair(a[p], p));
		a[nex[p]] = a[pre[p]] = -1e9;
		nex[p] = nex[nex[p]];
		pre[nex[p]] = p;
		pre[p] = pre[pre[p]];
		nex[pre[p]] = p;
		ans = max(ans, i+mi);
	}
	return ans+ma;
}
void work(){
	scanf("%d", &n);
	for (int i = 1; i <= n; i++){
		scanf("%d", &b[i]);
	}
	int ans = 0;
	for (int t = 0; t < 50; t++) ans = max(ans, work2());
	printf("%d\n", ans);
}
int main(){
	scanf("%d", &T);
	srand(time(0));
	while(T--){
		work();
	}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code uses a priority queue to manage the selection of elements and has a random component to choose the "fix_pos" (the position of the maximum element to be fixed). This random component can lead to suboptimal solutions if not handled correctly.

### Step-by-Step Analysis:

1. **Random Component**: The code uses `rand()` to select a position for the maximum element. This can lead to different results each time the code runs, which might not always be optimal.
2. **Priority Queue Management**: The code uses a priority queue to manage the selection of elements, but it might not handle the updates correctly, especially when elements are removed or updated.
3. **Edge Cases**: We need to consider edge cases such as arrays with all elements being the same, arrays with very few distinct elements, and arrays with a clear maximum and minimum.

### Challenging Test Input:

Let's create a small-scale test input that is complete and valid, focusing on edge cases:

1. **Array with All Elements Being the Same**:
   - This will test the random component and the priority queue management.
   - Example: `10 5 5 5 5 5 5 5 5 5 5`

2. **Array with a Clear Maximum and Minimum**:
   - This will test the ability to correctly identify and handle the maximum and minimum elements.
   - Example: `10 1 1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000`

3. **Array with Very Few Distinct Elements**:
   - This will test the handling of repeated elements and the random selection.
   - Example: `10 1 2 1 2 1 2 1 2 1 2`

### Test Input:

```plaintext
3
10
5 5 5 5 5 5 5 5 5 5
10
1 1000000000 1 1000000000 1 1000000000 1 1000000000 1 1000000000
10
1 2 1 2 1 2 1 2 1 2
```

This test input includes:
- An array with all elements being the same.
- An array with a clear maximum and minimum.
- An array with very few distinct elements.

These scenarios will help expose any issues in the random selection and priority queue management, as well as the handling of edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

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

#define ll long long

using namespace std;

#define PAR 0
#define IMPAR 1

struct node{
    ll mx;
    bool posicion;
    ll tam;
    bool siempretomamax = false;
};

struct DSU{
    ll n;
    ll mx;
    vector<node> nodos;
    vector<ll> padres;

    ll ans = 0;
    ll tomamosMaximo = 0;
    bool stm = false;

    DSU(ll _n, ll _mx) {
        n = _n;
        mx = _mx;
        nodos.resize(n);
        padres.resize(n);
    }

    ll getRespuesta() {
        if(stm) return ans;
        if(tomamosMaximo <= 0) {
            return ans - 1;
        }
        return ans;
    }

    void inicializar(ll i, ll v) {
        padres[i] = i;
        nodos[i].mx = v;
        nodos[i].posicion = IMPAR;
        nodos[i].tam = 1;
        ans++;
        if(mx == v)
            tomamosMaximo++;
    }

    ll getParent(ll i) {
        if(padres[i] == i) return i;
        else return padres[i] = getParent(padres[i]);
    }

    ll getRespuesta(node a) {
        return (a.tam + 1) / 2;
    }

    ll toma(node a) {
        if(a.mx != mx) return 0;
        if(a.siempretomamax) return 1;
        else if(!(a.tam % 2 && a.posicion == PAR)) return 1;
        return 0;
    }

    node unir(node a, node b) {
        node toReturn;
        toReturn.mx = max(a.mx, b.mx);
        toReturn.tam = a.tam + b.tam;
        if(a.mx >= b.mx) {
            toReturn.posicion = a.posicion;
        } else {
            if(a.posicion % 2) {
                toReturn.posicion = b.posicion ^ 1;
            } else {
                toReturn.posicion = b.posicion;
            }
        }
        toReturn.siempretomamax = a.siempretomamax || b.siempretomamax;
        if(!toReturn.siempretomamax) {
            if(a.mx == mx && b.mx == mx) {
                if(a.tam % 2) {
                    toReturn.siempretomamax = a.posicion == b.posicion;
                } else {
                    toReturn.siempretomamax = a.posicion != b.posicion;
                }
            }
        }
        ans += (getRespuesta(toReturn) - getRespuesta(a) - getRespuesta(b));
        tomamosMaximo += (toma(toReturn) - toma(a) - toma(b));
        if(toReturn.siempretomamax) stm = true;
        return toReturn;
    }

    void unir(ll v, ll u) {
        u = getParent(u);
        v = getParent(v);
        if(u == v) return;
        padres[u] = v;
        nodos[v] = nodos[u] = unir(nodos[v], nodos[u]);
        // cout << nodos[v].mx << " " << nodos[v].posicion << " " << nodos[v].siempretomamax << " " << nodos[v].tam << " " << v << endl;
    }
};

void casito() {
    ll n;
    cin >> n;
    vector<ll> a(n);
    vector<pair<ll, ll>> ps(n);
    ll mx = 0;
    for(ll i = 0;i < n;i++) {
        cin >> a[i];
        mx = max(a[i], mx);
        ps[i].first = a[i];
        ps[i].second = i;
    }
    sort(ps.rbegin(), ps.rend());
    DSU dsu(n, mx);
    ll ans = 0;
    for(ll i = 0;i < n;i++) {
        auto x = ps[i];
        dsu.inicializar(x.second, x.first);
        if(x.second != 0) {
            if(a[x.second - 1] > a[x.second]) dsu.unir(x.second - 1, x.second);
        }
        if(x.second != n - 1) {
            if(a[x.second + 1] >= a[x.second]) dsu.unir(x.second, x.second + 1);
        }
        ans = max(ans, dsu.getRespuesta() + x.first);
    }
    cout << ans + mx << endl;
}

int main() {
    ll t;
    cin >> t;
    while(t--) casito();
}

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

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

Think step by step.

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

1. **Adjacent Large Values**: The code might fail if there are large values that are adjacent, as the algorithm needs to ensure that no two adjacent elements are colored red.

2. **Small Array Size**: Small arrays can sometimes expose bugs that larger arrays might not, especially if the code has off-by-one errors or edge case handling issues.

3. **All Elements Equal**: If all elements are equal, the algorithm needs to handle this case correctly, ensuring that it does not select two adjacent elements.

4. **Mixed Values**: Arrays with a mix of high and low values can test the algorithm's ability to balance the maximum and minimum values while adhering to the no-adjacent-red-elements rule.

5. **Single Element Arrays**: Single-element arrays are edge cases that need to be handled correctly.

Given these considerations, let's create a small-scale test input that includes these edge cases:

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

### Explanation of the Test Input:
1. **Test Case 1:**
   - Array: `[10, 10]`
   - Expected Output: `21` (Select one `10`, score = `10 + 10 + 1 = 21`)

2. **Test Case 2:**
   - Array: `[10, 1, 10]`
   - Expected Output: `22` (Select both `10`s, score = `10 + 10 + 2 = 22`)

3. **Test Case 3:**
   - Array: `[1, 2, 3, 4]`
   - Expected Output: `10` (Select `1` and `4`, score = `4 + 1 + 2 = 10`)

4. **Test Case 4:**
   - Array: `[5]`
   - Expected Output: `15` (Select `5`, score = `5 + 5 + 1 = 15`)

This test input includes arrays with adjacent large values, all elements equal, mixed values, and a single-element array, which should help in exposing any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <cassert>

using namespace std;
using ll = long long;
using ld = long double;

const int inf = 0x3f3f3f3f;
const int maxN = 200005;
int n, v[maxN], maxi, on[maxN], maxTake;
int root[maxN], sz[maxN], l[maxN], r[maxN], cntOk;
bool maxiImp[maxN], maxiPar[maxN];
int ans;
pair <int, int> aux[maxN];

int findRoot(int x) {
    if (root[x] == x) {
        return x;
    }
    root[x] = findRoot(root[x]);
    return root[x];
}

void removeSeg(int x) {
    maxTake -= (sz[x] + 1) / 2;
    bool hasMaxi = maxiImp[x] | maxiPar[x];
    if (!hasMaxi) {
        return;
    }
    if (maxiImp[x] && maxiPar[x]) {
        cntOk--;
        return;
    }
    if (l[x] % 2 != r[x] % 2) {
        cntOk--;
        return;
    }
    if (l[x] % 2 == 0 && r[x] % 2 == 0 && !maxiImp[x]) {
        cntOk--;
        return;
    }
    if (l[x] % 2 == 1 && r[x] % 2 == 1 && !maxiPar[x]) {
        cntOk--;
        return;
    }
}

void addSeg(int x) {
    maxTake += (sz[x] + 1) / 2;
    bool hasMaxi = maxiImp[x] | maxiPar[x];
    if (!hasMaxi) {
        return;
    }
    if (maxiImp[x] && maxiPar[x]) {
        cntOk++;
        return;
    }
    if (l[x] % 2 != r[x] % 2) {
        cntOk++;
        return;
    }
    if (l[x] % 2 == 0 && r[x] % 2 == 0 && !maxiImp[x]) {
        cntOk++;
        return;
    }
    if (l[x] % 2 == 1 && r[x] % 2 == 1 && !maxiPar[x]) {
        cntOk++;
        return;
    }
}

void join(int x, int y) {
    x = findRoot(x);
    y = findRoot(y);
    removeSeg(x);
    removeSeg(y);
    if (sz[x] < sz[y]) {
        swap(x, y);
    }
    l[x] = min(l[x], l[y]);
    r[x] = max(r[x], r[y]);
    maxiImp[x] |= maxiImp[y];
    maxiPar[x] |= maxiPar[y];
    sz[x] += sz[y];
    addSeg(x);
}

void reset() {
    for (int i = 1; i <= n; i++) {
        v[i] = 0;
        aux[i] = {0, 0};
        root[i] = 0;
        sz[i] = 0;
        l[i] = 0;
        r[i] = 0;
        on[i] = 0;
        maxiImp[i] = 0;
        maxiPar[i] = 0;
    }
    cntOk = 0;
    maxTake = 0;
    maxi = 0;
    ans = 0;
}

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> v[i];
        maxi = max(maxi, v[i]);
        aux[i] = {v[i], i};
    }
    sort(aux + 1, aux + n + 1);
    reverse(aux + 1, aux + n + 1);
    for (int i = 1; i <= n; i++) {
        root[i] = i;
        sz[i] = 1;
        l[i] = i;
        r[i] = i;
    }
    for (int i = 1; i <= n; i++) {
        auto [val, poz] = aux[i];
        on[poz] = 1;
        maxTake++;
        if (val == maxi) {
            if (poz % 2 == 0) {
                maxiPar[poz] = 1;
            }
            else {
                maxiImp[poz] = 1;
            }
            cntOk++;
        }
        if (poz > 1 && on[poz - 1]) {
            join(poz - 1, poz);
        }
        if (poz < n && on[poz + 1]) {
            join(poz, poz + 1);
        }
        if (cntOk) {
            ans = max(ans, maxi + val + maxTake);
        }
        else {
            ans = max(ans, maxi + val + maxTake - 1);
        }
    }
    cout << ans << '\n';
    reset();
}

signed main() {
#ifdef LOCAL
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif // LOCAL
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int nrt = 1;
    cin >> nrt;
    while (nrt--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Adjacent Elements with the Same Value**: The algorithm might fail if there are multiple adjacent elements with the same value, especially if these values are the maximum or minimum.
2. **Single Element Arrays**: The algorithm should handle single-element arrays correctly.
3. **Arrays with All Elements Being the Same**: This can test how the algorithm handles uniform arrays.
4. **Alternating High and Low Values**: This can test the algorithm's ability to handle patterns that might cause it to miss optimal solutions.

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

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

### Explanation:
1. **Single Element Array**:
   - `1 10`: This tests the case where the array has only one element. The score should be `10 + 10 + 1 = 21`.

2. **Array with All Elements Being the Same**:
   - `3 10 10 10`: This tests the case where all elements are the same. The score should be `10 + 10 + 2 = 22` (since we can color two non-adjacent elements).

3. **Array with Increasing Values**:
   - `4 1 2 3 4`: This tests the case where the elements are in increasing order. The score should be `4 + 1 + 2 = 7` (since we can color the first and last elements).

4. **Array with Alternating High and Low Values**:
   - `5 10 1 10 1 10`: This tests the case where the elements alternate between high and low values. The score should be `10 + 1 + 3 = 14` (since we can color the first, third, and fifth elements).

This test input should help expose any issues in the buggy code by covering a variety of edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int T,n,a[N],b[N],m,prt[N],sum,cnt0m,cnt1m,sz[N];
bool v[N],vis[N];
vector<int>que[N];
int gf(int x){if(x==prt[x])return x;return prt[x]=gf(prt[x]);}
void merge(int x,int y)
{
	x=gf(x),y=gf(y);
	if(v[x]){if(sz[x]%2==0)cnt0m--;else cnt1m--;}
	if(v[y]){if(sz[y]%2==0)cnt0m--;else cnt1m--;}
	sum-=(sz[x]+1)/2+(sz[y]+1)/2;
	sz[y]+=sz[x],prt[x]=y;
	sum+=(sz[y]+1)/2;
	if(v[y]){if(sz[y]%2==0)cnt0m++;else cnt1m++;}
}
void solve()
{
	scanf("%d",&n),m=0,sum=cnt0m=cnt1m=0;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),b[i]=a[i],prt[i]=i,v[i]=0,sz[i]=1,vis[i]=0;
	vis[n+1]=0;
	sort(b+1,b+n+1),m=unique(b+1,b+n+1)-b-1;
	for(int i=1;i<=n;i++)a[i]=lower_bound(b+1,b+m+1,a[i])-b;
	for(int i=1;i<=m;i++)que[i].clear();
	for(int i=1;i<=n;i++)que[a[i]].push_back(i);
	for(auto i:que[m])
	{
		v[i]=1,cnt1m++,sum++,vis[i]=1;
		if(vis[i-1])merge(i,i-1);
		if(vis[i+1])merge(i,i+1);
	}
	int ans=b[m]*2+sum;
	for(int i=m-1;i>=1;i--)
	{
		for(auto j:que[i])
		{
			sum++,vis[j]=1;
			if(vis[j-1])merge(j,j-1);
			if(vis[j+1])merge(j,j+1);
		}
		ans=max(ans,b[m]+b[i]+sum-(cnt0m==0));
	}
	printf("%d\n",ans);
}
int main()
{
	scanf("%d",&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:
To generate a challenging test input that exposes the bug in the given code, let's consider a few edge cases:

1. **Small Array with Repeated Elements**: This can test how the code handles repeated elements and the merging process.
2. **Array with Alternating High and Low Values**: This can test the logic for selecting the maximum and minimum values and the merging process.
3. **Single Element Array**: This can test the base case where the array has only one element.
4. **Array with All Elements Being the Same**: This can test the logic for handling uniform arrays.
5. **Array with Elements in Increasing or Decreasing Order**: This can test the logic for handling sorted arrays.

Given these considerations, let's create a small-scale test input that combines some of these edge cases:

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

This test input includes:
1. A small array with repeated elements.
2. An array with elements in increasing order.
3. An array with elements in decreasing order.
4. A small array with a large element and a small element.
5. A single element array with a large value.

This should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

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

struct ST {
    struct Data {
        std::array<std::array<int, 2>, 2> ar;

        Data(bool val = false) {
            ar.fill({0, 0});
            ar[1][1] = val;
        }
    };

    Data merge(Data a, Data b) {
        Data c;
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                for (int k = 0; k < 2; k++) {
                    for (int t = 0; t < 2; t++) {
                        if (!j || !k) {
                            c.ar[i][t] = std::max(c.ar[i][t], a.ar[i][j] + b.ar[k][t]);
                        }
                    }
                }
            }
        }
        return c;
    }

    int size;
    std::vector<Data> data;

    ST(int n) {
        size = std::bit_ceil<uint32_t>(n);
        data.assign(size * 2, Data());
    }

    void update(int i, bool val) {
        i += size;
        data[i] = Data(val);
        for (i >>= 1; i > 0; i >>= 1) {
            data[i] = merge(data[2 * i], data[2 * i + 1]);
        }
    }

    int get_max(int l, int r) {
        Data a, b;
        for (l += size, r += size; l < r; l >>= 1, r >>= 1) {
            if (l & 1)
                a = merge(a, data[l++]);
            if (r & 1)
                b = merge(data[--r], b);
        }
        Data c = merge(a, b);
        return std::max({
            c.ar[0][0],
            c.ar[0][1],
            c.ar[1][0],
            c.ar[1][1],
        });
    }
};

void solve() {
    int n;
    std::cin >> n;
    std::vector<int> input(n);
    for (auto& i : input) {
        std::cin >> i;
    }

    int max_ind = std::max_element(input.begin(), input.end()) - input.begin();
    std::vector<int> vec_ind;
    vec_ind.push_back(max_ind);
    if (max_ind + 1 < input.size() && input[max_ind + 1] == input[max_ind]) {
        vec_ind.push_back(max_ind + 1);
    }

    std::map<int, std::vector<int>, std::greater<int>> map;
    for (int i = 0; i < n; i++) {
        map[input[i]].push_back(i);
    }

    ST st(n);
    int ans = 0;
    for (auto [val, vec] : map) {
        for (int i : vec) {
            st.update(i, true);
        }
        for (auto ind : vec_ind) {
            int a = st.get_max(0, ind - 1);
            int b = st.get_max(ind + 2, n);
            ans = std::max(ans, input[max_ind] + val + (a + b + 1));
        }
    }
    std::cout << ans << "\n";
}

int32_t main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int t = 1;
    std::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
1
5
1 2 1 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
typedef long long ll;
const int N=2e5+5;
ll t,n,x,mx,mi,cnt,b[N],ans;
bool vis[N];
map<int,int> mp;
vector<int> a[N],c;

int main() {
	for(cin>>t;t--;) {
		cin>>n;

		c.clear();
		mp.clear();
		for(int i=0;i<=n+1;i++) vis[i]=0,a[i].clear();
		ans=0,cnt=0;

		for(int i=1;i<=n;i++) cin>>b[i],c.pb(b[i]);
		sort(c.begin(),c.end());
		c.erase(unique(c.begin(),c.end()),c.end());

		for(int i=1;i<=n;i++) {
			int p=lower_bound(c.begin(),c.end(),b[i])-c.begin();
			if(!mp[p]) mp[p]=b[i];
			a[p].pb(i);
		}

		mx=c.back();
		for(int i=c.size()-1;i>=0;i--) {
			mi=mp[i];
			for(int j:a[i]) {
				if(vis[j-1]==0&&vis[j+1]==0) {
					vis[j]=1;
					cnt++;
				}
			}
			ans=max(mx+mi+cnt,ans);
		}

		cout<<ans<<"\n";
	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are a few potential edge cases to consider:

1. **Arrays with all elements being the same**: This can test how the code handles uniform values.
2. **Arrays with alternating high and low values**: This can test the logic for selecting red elements without adjacent conflicts.
3. **Arrays with a single element**: This can test the base case.
4. **Arrays with elements in strictly increasing or decreasing order**: This can test the logic for handling sorted arrays.
5. **Arrays with a large number of distinct elements**: This can test the performance and correctness of the code with a diverse set of values.

Let's create a small-scale test input that combines some of these edge cases:

```plaintext
4
3
1 2 1
5
1 2 3 2 1
1
5
10
1 2 3 4 5 6 7 8 9 10
```

### Explanation:
1. **Test Case 1: `3 1 2 1`**
   - This array has a middle element that is higher than the others. The optimal solution is to color the first and last elements red, giving a score of \(1 + 1 + 2 = 4\).

2. **Test Case 2: `5 1 2 3 2 1`**
   - This array is symmetric with a peak in the middle. The optimal solution is to color the first, third, and last elements red, giving a score of \(3 + 1 + 3 = 7\).

3. **Test Case 3: `1 5`**
   - This array has a single element. The optimal solution is to color the single element red, giving a score of \(5 + 5 + 1 = 11\).

4. **Test Case 4: `10 1 2 3 4 5 6 7 8 9 10`**
   - This array is strictly increasing. The optimal solution is to color the first, third, fifth, seventh, and ninth elements red, giving a score of \(9 + 1 + 5 = 15\).

These test cases should help expose any bugs in the provided code by covering a range of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
int a[200000];
int t, n;
class DSU {
    private:
        vector<int> e;
        vector<int> s;
        vector<vector<bool> > b;
        vector<int> first;
        int max_size;
        int has_max;
        int m;
    public:
        DSU(int n, int max_elt): e(n), s(n, 1), b(n), first(n), max_size(0), has_max(0), m(max_elt) {
            for(int i = 0; i < n; i++) {
                e[i] = i;
                first[i] = i;
                b[i] = vector<bool>{false, false};
            }
        }
        int find(int x) {
            return e[x] == x ? x : e[x] = find(e[x]);
        }
        void add(int x, int a) {
            if(a == m) {
                b[x][x%2] = true;
                has_max++;
            }
            max_size++;
        }
        void unite(int x, int y) {
            x = find(x); y = find(y);
            if(s[x] > s[y]) swap(x, y);
            e[y] = x;
            if(b[x][first[x]%2]) has_max--;
            if(b[y][first[y]%2]) has_max--;
            max_size -= (s[x] + 1)/2;
            max_size -= (s[y] + 1)/2;
            s[x] += s[y];
            first[x] = min(first[x], first[y]);
            b[x][0] = (b[x][0] || b[y][0]);
            b[x][1] = (b[x][1] || b[y][1]);
            max_size += (s[x] + 1)/2;
            if(b[x][first[x]%2]) has_max++;
        }
        int get_max_size() {
            if(has_max == 0) return max_size - 1;
            else return max_size;
        }
};
int main() {
    cin >> t;
    for(int u = 0; u < t; u++) {
        cin >> n;
        int m = 0;
        priority_queue<pair<int, int> > pq;
        for(int i = 0; i < n; i++) {
            cin >> a[i];
            m = max(m, a[i]);
            pq.push({a[i], i});
        }
        if(t > 97) {
            if(u < 96) continue;
            else {
                cout << n << "\n";
                for(int i = 0; i < n; i++) cout << a[i] << " ";
                break;
            }
        }
        DSU d(n, m);
        vector<bool> visited(n, false);
        int ans = 0;
        while(!pq.empty()) {
            int k = pq.top().second;
            pq.pop();
            visited[k] = true;
            d.add(k, a[k]);
            if(k > 0 && visited[k-1]) {
                d.unite(k, k-1);
            }
            if(k < n - 1 && visited[k+1]) {
                d.unite(k, k+1);
            }
            ans = max(ans, d.get_max_size() + m + a[k]);
        }
        cout << ans << "\n";
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or specific scenarios that might cause the algorithm to fail. Here are some potential issues to consider:

1. **Edge Case: Single Element Array**:
   - If the array has only one element, the maximum, minimum, and the number of red elements are all the same. This should be a straightforward case, but it's worth checking.

2. **Edge Case: Two Elements Array**:
   - If the array has two elements, the algorithm should be able to handle this without issues. However, it's a good test to ensure the logic for combining elements is correct.

3. **Edge Case: All Elements are the Same**:
   - If all elements in the array are the same, the algorithm should handle this correctly. The maximum, minimum, and the number of red elements should all be the same value.

4. **Edge Case: Alternating Elements**:
   - If the array has alternating elements (e.g., 1, 2, 1, 2, ...), the algorithm should be able to handle this without issues. This can test the logic for handling adjacent elements.

5. **Edge Case: Large Elements with Small Differences**:
   - If the array has large elements with small differences (e.g., 1000000000, 1000000001, 1000000000, ...), the algorithm should handle this correctly. This can test the logic for handling large values.

6. **Edge Case: Mixed Elements**:
   - If the array has a mix of large and small elements, the algorithm should be able to handle this without issues. This can test the logic for handling a variety of values.

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

```plaintext
4
1
5
2
4 5
3
1 2 1
4
1000000000 1000000001 1000000000 1000000001
```

This test input includes:
1. A single element array.
2. A two-element array.
3. An array with alternating elements.
4. An array with large elements with small differences.

These cases should help expose any potential bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    idx = 0
    t = int(data[idx])
    idx += 1
    for _ in range(t):
        n = int(data[idx])
        idx += 1
        a = list(map(int, data[idx:idx+n]))
        idx += n

        unique_values = list(set(a))
        unique_values.sort(reverse=True)

        max_score = 0
        for M in unique_values:
            selected = []
            prev_selected = False
            for num in a:
                if num >= M and not prev_selected:
                    selected.append(num)
                    prev_selected = True
                else:
                    prev_selected = False
            if selected:
                current_max = max(selected)
                current_min = min(selected)
                count = len(selected)
                score = current_max + current_min + count
                if score > max_score:
                    max_score = score
        print(max_score)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Max Plus Min Plus Size
time_limit_ms: 2000
memory_limit_mb: 256
Description: [EnV - The Dusty Dragon Tavern](https://soundcloud.com/envyofficial/env-the- dusty-dragon-tavern)

You are given an array $a_1, a_2, \ldots, a_n$ of positive integers.

You can color some elements of the array red, but there cannot be two adjacent red elements (i.e., for $1 \leq i \leq n-1$, at least one of $a_i$ and $a_{i+1}$ must not be red).

Your score is the maximum value of a red element, plus the minimum value of a red element, plus the number of red elements. Find the maximum score you can get.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array.

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 given array.

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

Output: For each test case, output a single integer: the maximum possible score you can get after coloring some elements red according to the statement.

Examples:
input:
4
3
5 4 5
3
4 5 4
10
3 3 3 3 4 1 2 3 5 4
10
17 89 92 42 29 41 92 14 70 45
output:
12
11
12
186

Note: In the first test case, you can color the array as follows: $[\color{red}{5}, 4, \color{red}{5}]$. Your score is $\max([5, 5]) + \min([5, 5]) + \text{size}([5, 5]) = 5+5+2 = 12$. This is the maximum score you can get.

In the second test case, you can color the array as follows: $[4, \color{red}{5}, 4]$. Your score is $\max([5]) + \min([5]) + \text{size}([5]) = 5+5+1 = 11$. This is the maximum score you can get.

In the third test case, you can color the array as follows: $[\color{red}{3}, 3, \color{red}{3}, 3, \color{red}{4}, 1, 2, \color{red}{3}, 5, \color{red}{4}]$. Your score is $\max([3, 3, 4, 3, 4]) + \min([3, 3, 4, 3, 4]) + \text{size}([3, 3, 4, 3, 4]) = 4+3+5 = 12$. This is the maximum score you can get.

Buggy Code:
# -*- coding: utf-8 -*-
"""
Created on Sat Oct  5 23:24:57 2024

@author: Admin
"""

def score(l):
    if l == [17, 89, 92, 42, 29, 41, 92, 14, 70, 45]:
        return 186
    l1 = l[0::2]
    if max(l) in l1:
        max1 = max(l1)
        min1 = min(l1)
        len1 = len(l1)
        score1 = max1 + min1 + len1
        return score1
    else:
        l2 = l[1::2]
        max2 = max(l2)
        min2 = min(l2)
        len2 = len(l2)
        score2 = max2 + min2 + len2
        return score2
    return max(score1, score2)

l =[]
for i in range(int(input())):
    n = int(input())
    l = list(map(int, input().split()))
    print(score(l))

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

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

Think step by step.

Response:
```plaintext
4
3
1 1000000000 1
3
1000000000 1 1000000000
10
1 2 3 4 5 6 7 8 9 10
10
10 9 8 7 6 5 4 3 2 1
```