**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
def inp():
    return (int(input()))
def inlt():
    return (list(map(int, input().split())))
def insr():
    s = input()
    return (list(s[:len(s) - 1]))
def invr():
    return (map(int, input().split()))

for _ in range(inp()):
    n = inp()
    lis = inlt()
    lis2 = sorted(lis,reverse=True)
    i = 0
    while True:
        ind = lis.index(lis2[i])
        if ind%2 == 0 and (n - ind)%2==1:
            print(lis2[i])
            break
        i += 1

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to create a test case that challenges the logic used to find the maximum possible value of the remaining element. The buggy code sorts the list in descending order and then tries to find an index that satisfies certain conditions. However, this approach does not always work correctly, especially in cases where the array has repeated elements or specific patterns that the code fails to handle.

Let's consider a test case where the array has repeated elements and the maximum value is not at an index that satisfies the conditions checked by the buggy code.

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

```plaintext
1
5
4 7 4 2 9
```

### Explanation:
- **Array**: `[4, 7, 4, 2, 9]`
- **Length**: 5 (odd)
- **Expected Output**: 9

### Why This Test Case Should Expose the Bug:
1. **Repeating Elements**: The array contains the element `4` twice, which can confuse the code when it tries to find the index of the maximum value.
2. **Correct Solution**: The correct way to maximize the remaining element is to remove elements in such a way that the maximum value `9` is left. The correct sequence of operations could be:
   - Remove the first two elements: `[4, 7, 4, 2, 9] -> [4, 2, 9]`
   - Remove the first two elements: `[4, 2, 9] -> [9]`
3. **Buggy Code Logic**: The buggy code sorts the array and tries to find an index that satisfies the conditions. However, it may not correctly handle the case where the maximum value `9` is at an index that does not satisfy the conditions checked by the code.

This test case should help reveal the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
def find_max_element(arr):
    while len(arr) > 1:
        min_element = float('inf')
        min_index = -1

        for i in range(len(arr) - 1):
            if min(arr[i], arr[i + 1]) < min_element:
                min_element = min(arr[i], arr[i + 1])
                min_index = i

        if arr[min_index] < arr[min_index + 1]:
            arr.pop(min_index)
            arr.pop(min_index)
        else:
            arr.pop(min_index + 1)
            arr.pop(min_index)

    return arr[0]

t = int(input())
for _ in range(t):
    n = int(input())
    arr = list(map(int, input().split()))
    if n == 1:
        print(arr[0])
    else:
        print(find_max_element(arr))

**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
6
3
1 3 2
5
4 7 4 2 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

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

            int left=0;
            int right=n-1;
            if(n==1){
                System.out.println(a[0]);
            }else{
                while(left<n-1 && right>0){
                    if( (a[left]>=a[right]&&a[left]>=a[right-1])||(a[left+1]>=a[right] && a[left+1]>=a[right-1]) ){
                        if(left+1==right-1){
                            if(a[left]>a[right])
                            right=right-2;
                            else
                            left=left+2;
                        }else
                        right=right-2;
                    }
                    else{
                        left=left+2;
                    }
                }
                if(left<n)
                System.out.println(a[left]);
                else
                System.out.println(a[right]);

            }

        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
import java.util.*;

public class Main
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            int [] arr = new int[n];
            for(int i = 0; i  < n; ++i)
                arr[i] = sc.nextInt();
            int mx = 0;

            for(int i = 0; i < n; ++i)
            {
                if(i == 1 || i == n-2)
                    continue;
                mx = Math.max(mx, arr[i]);
            }
            System.out.println(mx);
        }
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
/*------------------------------------useful definitions----------------------------------------------------------------------------------------------------------------------*/
 #define ll long long
 #define INF LONG_LONG_MAX
 #define inf INT_MAX
 #define ff first
 #define ss second

/*-------------------------------------debugging functions--------------------------------------------------------------------------------------------------------------------*/
template<class T>void print(T x){cerr << x;}
template<class T, class V> void print(pair<T , V> x){print(x.ff); cerr << ':';print(x.ss);}
template<class T>void print(vector<T> &a){cerr << '[' << ' ';for(auto x : a){print(x);cerr << ' ';}cerr << ']';}
template<class T>void print(set<T> &a){cerr << '[' << ' ';for(auto x : a){print(x);cerr << ' ';}cerr << ']';}
template<class T>void print(set<T,greater<T>> &a){cerr << '[' << ' ';for(auto x : a){print(x);cerr << ' ';}cerr << ']';}
template<class T>void print(multiset<T> &a){cerr << '[' << ' ';for(auto x : a){print(x);cerr << ' ';}cerr << ']';}
template<class T>void print(multiset<T, greater<T>> &a){cerr << '[' << ' ';for(auto x : a){print(x);cerr << ' ';}cerr << ']';}
template<class T>void print(unordered_set<T> &a){cerr << '[' << ' ';for(auto x : a){print(x);cerr << ' ';}cerr << ']';}
template<class T, class V>void print(vector<pair<T, V>> &a){cerr << '[' << ' ';for(auto x : a){print(x.ff);cerr << ":";print(x.ss);cerr << ' ';}cerr << ']';}
template <class T, class V> void print(map <T, V> &a) {cerr << "[ "; for (auto i : a) {print(i);cerr << " ";}cerr << "]";}
template <class T, class V> void print(unordered_map <T, V> &a) {cerr << "[ ";for (auto i : a){print(i);cerr << " ";}cerr << "]";}
template <class T> void print(vector<vector<T>> &a) {cerr << "[ "; for (auto i : a) {print(i); cerr << " ";} cerr << "]";}

/*------------------------------------predefined functions----------------------------------------------------------------------------------------------------------------------*/
ll lcm(ll a, ll b ){
  ll mini = min(a,b),maxi = max(a,b);
  for(ll i=maxi;;i+=maxi) if(i%mini == 0) return i;
}
/*-------------------------------------predefined functions--------------------------------------------------------------------------------------------------------------------*/

void local()
{
  // ONLINE_JUDGE
  #ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
  #endif

  // ONLINE_JUDGE
  #ifndef ONLINE_JUDGE
  freopen("error.txt", "w", stderr);
  #endif

  #ifndef ONLINE_JUDGE
  #define debug(x) cerr << #x << "  "; print(x); cerr << '\n';
  #else
  #define debug(x)
  #endif
}

int main(){
	local();
	long long t;
	cin>>t;
	while(t--){
   ll n;
   cin>>n;
   vector<ll> arr(n);
   for(auto &it: arr) cin>>it;
   vector<ll> cpy = arr;
   sort(cpy.begin(),cpy.end(),greater<int>());
   ll maxIndx = max_element(arr.begin(),arr.end())-arr.begin();
   for(ll i=0;i<n;i++){
    ll indx = find(arr.begin(),arr.end(),cpy[i])-arr.begin();
    if((n-indx-1)%2==0){
      cout<<arr[indx]<<endl;
      break;
    }
   }

 }

	return 0;

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl "\n"
double pi= 3.141592653589793238462643383279;

void solve(){
    int n;
    cin>>n;
    int ar[n];
    int maxv=INT_MIN;
    for(int i=0;i<n;i++){
        cin>>ar[i];
        if(i==1 || i==n-2){
            continue;
        }
        maxv=max(maxv,ar[i]);
    }
    cout<<maxv<<endl;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t; cin>>t; while(t--) solve();
    // 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 expose the bug in the provided code, we need to focus on the logic of the algorithm. The buggy code is only considering the elements at indices 1 and `n-2` and skipping them in the loop. This means it is not correctly considering all possible elements that could be the maximum after all operations.

### Step-by-Step Analysis:
1. **Identify the Bug**: The code is skipping elements at indices 1 and `n-2` in the loop, which means it is not considering these elements in the maximum value calculation.
2. **Edge Cases**:
   - Arrays where the maximum value is at index 1 or `n-2`.
   - Arrays where the maximum value is not at the first or last element.
   - Arrays with small values at the first and last elements but a large value in the middle.

### Constructing the Test Case:
- **Array Length**: Choose an odd length array to ensure it meets the problem constraints.
- **Element Values**: Place a large value at index 1 or `n-2` and smaller values elsewhere to highlight the bug.

### Example Test Case:
Let's create an array where the maximum value is at index 1 (the second element in a 0-indexed array).

```plaintext
1
5
1 99 1 1 1
```

### Explanation:
- **Array**: `[1, 99, 1, 1, 1]`
- **Expected Output**: `99`
- **Buggy Code Output**: The buggy code will skip the element at index 1 (99) and will not consider it in the maximum value calculation, leading to an incorrect output.

This test case will expose the bug in the provided code by showing that it incorrectly skips the maximum value.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

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

def tullu(a):
    memo = {}

    def helper(a):
        n = len(a)
        if n == 1:
            return a[0]
        tuple_a = tuple(a)
        if tuple_a in memo:
            return memo[tuple_a]

        maxi = float('-inf')
        for i in range(n - 1):
            na = a[:i] + a[i+2:]
            maxi = max(maxi, helper(na))

        memo[tuple_a] = maxi
        return maxi

    return helper(a)

data = input().split()
t = int(data[0])
index = 1
results = []
for _ in range(t):
    n = int(data[index])
    a = list(map(int, data[index+1:index+1+n]))
    index += n + 1
    results.append(tullu(a))

for result in results:
    print(result)

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

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

Think step by step.

Response:
```plaintext
3
1
100
3
100 1 100
5
100 1 100 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
t = int(input())

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

    if len(a) > 1:
        maxi = 0
        for i in range(0, len(a)):
            if a[i] > maxi and ((len(a) - i > 2 and i > 1) or i == 0 or i == len(a) - 1):
                maxi = a[i]
        print(maxi)
    elif len(a) == 3:
        print(max(a[0], a[-1]))
    else:
        print(a[0])

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
import java.util.*;

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

        while(t-->0){
            int n = s.nextInt();
            int size=n;
            if((n&1)==1)    n++;
            int[] a = new int[n];
            for(int i=0; i<size; i++)  a[i] = s.nextInt();

            int max = 0;
            for(int i=0; i<size; i+=2){
                int ind = a[i]>a[i+1]?i:i+1;
                max = a[ind]>a[max]?ind:max;
            }

            if(size==3 || size%2!=0){
                if(max==0 || max==size-1 || max+2<size)   System.out.println(a[max]);
                else System.out.println(Math.max(a[max-1],a[max+1]));
            }
            else    System.out.println(a[max]);
        }

    }
}

**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
100
3
100 1 100
5
100 1 100 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
import com.sun.source.tree.Tree;

import java.util.Collections;
import java.util.Scanner;
import java.util.TreeMap;

public class maximizeLast {
    static int maximizer(TreeMap<Integer,Integer> map, int n){
        for(int key: map.keySet()){
            int index=map.get(key);
            if(index%2==0) return key;
        }
        return 1;
    }

    public static void main(String[] args) {
        Scanner sc= new Scanner(System.in);
        int testcases=sc.nextInt();
        for(int t=0;t<testcases;t++){
            TreeMap<Integer,Integer> map=new TreeMap<>(Collections.reverseOrder());
            int n=sc.nextInt();
            for(int i=0;i<n;i++) {
                map.put(sc.nextInt(),i);
            }
            System.out.println(maximizer(map,n));

        }
        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS
#define LL long long
#define forn(i,n) for (LL i=0;i<n;i++)
#define foran(i,a,n) for (LL i=a;i<n;i++)
#define inf (LL)1e18

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <vector>
#include <string>
#include <algorithm>
#include <utility>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <deque>

#include <time.h>
using namespace std;   //[65-90]

void ifdef(int argc, const char * argv[]){
        // freopen("/Users/maxim/Projects/CLionProjects/Codeforces/in.txt", "r", stdin);
        // freopen("/Users/maxim/Projects/CLionProjects/Codeforces/out.txt", "w", stdout);
}

LL gcd(LL a, LL b) {
    return b ? gcd(b, a % b) : a;
}

auto cmp = [](LL a, LL b) { return a > b; };

vector<int> prefix_function(string s) {
    int n = (int)s.length();
    vector<int> pi(n);
    for (int i = 1; i < n; ++i) {
        int j = pi[i - 1];
        while (j > 0 && s[i] != s[j])
            j = pi[j - 1];
        if (s[i] == s[j])  ++j;
        pi[i] = j;
    }
    return pi;
}

LL pow_LL(LL a, LL n) {
    LL ans = 1;
    forn(i, n) ans *= a;
    return ans;
}

bool is_pow(LL q, LL r){
    for(LL x = 0; pow_LL(r,x) <= q ; x++){
        if(q == pow_LL(r, x))
            return true;
    }
    return false;
}

LL cntFromTo(LL from, LL to){
    if(from>to) return 0;
    from++, to++;
    return (from+to)*(to-from+1)/2;
}

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

    vector<LL> a(n);
    forn(i, n) cin>>a[i];

    LL maxx = 0;
    forn(i, n){
        if(i == 1 || i == n-2) continue;
        if(a[i]>maxx) maxx = a[i];
    }

    cout<<maxx<<"\n";

}

int main(int argc, const char * argv[])
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ifdef(argc, argv);

    LL ttt = 1;
    cin >> ttt;
    while (ttt--)
    {
        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 expose the bug in the provided code, we need to focus on the logic that determines the maximum possible value of the remaining element. The bug in the code is that it incorrectly skips the first and last elements in the array when checking for the maximum value. This can lead to incorrect results if the maximum value is actually the first or last element.

Let's construct a test case where the maximum value is either the first or last element in the array. This will highlight the issue with the code.

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

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

Explanation:
1. In the first test case, the array is `[9, 1, 2]`. The maximum value is `9`, which is the first element. The code should output `9`.
2. In the second test case, the array is `[1, 2, 3, 4, 9]`. The maximum value is `9`, which is the last element. The code should output `9`.
3. In the third test case, the array is `[1, 2, 3, 4, 5, 6, 9]`. The maximum value is `9`, which is the last element. The code should output `9`.

This test input should clearly show that the code is skipping the first and last elements, leading to incorrect results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
#pragma GCC optimize("O3")
#include <bits/stdc++.h>

#define int long long
#define VI vector<int>
#define VII vector<VI >
#define PII pair<int, int>
#define VPII vector<PII >
#define fi(n) for(int i = 0; i < n; i++)
#define fj(n) for(int j = 0; j < n; j++)
#define fq(n) for(int q = 0; q < n; q++)
#define mk(t, ...) t __VA_ARGS__; __read(__VA_ARGS__)
#define M7 1000000007
#define MFFT 998244353
#define INF 9000000000000000001
#define PI 3.14159265359
#define inpVI(name, n) VI name(n); fi(n) std::cin >> name[i]
#define inpVVI(name, n, m) VVI name(n, VI(m)); f(n) FOR(j, m) std::cin >> name[i][j]
#define outVI(name); fi(name.size()) cout << name[i] << ' '; cout << endl;
#define ALL(q) q.begin(), q.end()
#define RALL(q) q.rbegin(), q.rend()
#define pb push_back
template<typename T>
void __read(T &a) { std::cin >> a; }

template<typename T, typename ... Args>
void __read(T &a, Args &... args) {
    std::cin >> a;
    __read(args...);
}

using namespace std;

void solve(){
    mk(int, t);
    fq(t){
        mk(int, n);
        inpVI(a, n);
        int ans = -1;
        if(n==1){
             cout<< a[0]<<endl;
            continue;}
        fi(n){
            if(i!=1 && i!=n-2){
                ans = max(a[i], ans);
            }
        }
        cout<< ans << endl;
    }
}

signed main(signed argc, char **argv) {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    if (argc > 1 && (string) argv[1] == "local") {
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    }
    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 expose the bug in the provided code, we need to focus on the logic that determines the maximum possible value of the remaining element. The code has a logical error in the way it selects the maximum element. Specifically, it incorrectly skips the first and the second-to-last elements when considering the maximum value.

Here is a step-by-step analysis to identify a test case that can expose this bug:

1. **Identify the Bug**: The code incorrectly skips the first and the second-to-last elements when determining the maximum value. This means it will not consider these elements as potential candidates for the maximum value.

2. **Construct a Test Case**: We need a test case where the first or the second-to-last element is the maximum value in the array. This will ensure that the code's logic fails to select the correct maximum value.

3. **Ensure the Test Case is Valid**: The array length must be odd, and the elements must be within the specified range (1 to 100).

Let's construct a test case where the first element is the maximum value in the array:

```plaintext
1
5
9 1 2 3 4
```

In this test case:
- The array is `[9, 1, 2, 3, 4]`.
- The length of the array is 5, which is odd.
- The maximum value in the array is 9, which is the first element.

When the code runs, it will skip the first element (9) and the second-to-last element (3) when determining the maximum value. Therefore, it will incorrectly output a value that is not the maximum possible value.

The correct output for this test case should be 9, but the buggy code will output a different value. This test case should effectively expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
i=0
result = ""
t=int(input())
while(i<t):
    i+=1
    maxim=0
    n=int(input())
    a=str(input())
    a=a.split(" ")
    liste = a.copy()
    if (n%2==0 or len(a)!=n):
        break
    k=0
    while(k<3):
        while(len(a)>3):
            del a[k]
            del a[k]
        if max(int(a[0]),int(a[len(a)-1]))>=maxim:
            maxim=max(int(a[0]),int(a[len(a)-1]))
        k+=1
        a = liste
    result+=str(maxim) + "\n"
print(result)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
x=int(input())
for i in range (x):
    n=int(input())
    p=input()
    l=[]
    for i in range(-1,-n-1,-4):
        l.append(p[i])
    print(max(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
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

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

public class Main {
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));

    public static void main(String[] args) throws IOException {
        int t = 1;
        t = Integer.parseInt(reader.readLine());
        while (--t >= 0) solve();
        writer.flush();
        writer.close();
    }

    static void solve() throws IOException {
        int n = Integer.parseInt(reader.readLine());
        String[] tokens = reader.readLine().split(" ");
        int max = 0;
        for (int i = 0; i < n; i++) {
            if (i == 1 || i == n - 2) continue;
            max = Math.max(max, Integer.parseInt(tokens[i]));
        }
        writer.write(max + "\n");
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
import java.util.Scanner;

public class Contest19 {
     public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int testCases = sc.nextInt();
        while (testCases-- > 0) {
            int n = sc.nextInt();
            int []arr = new int[n];
            for(int i = 0;i < n;i++)arr[i] = sc.nextInt();
            int maxi = Integer.MIN_VALUE;
            for(int i = 0;i < n;i++){
                int prev = i-2;
                int forw = i+2;
                if(prev >= 0 && forw < n)maxi = Math.max(maxi,arr[i]);
                else if(i == n-1 || i == 0)maxi = Math.max(maxi,arr[i]);
            }
            System.out.println(maxi);
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
1
100
3
100 1 100
5
100 1 1 1 100
7
100 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: Boss, Thirsty
time_limit_ms: 2000
memory_limit_mb: 256
Description: Pak Chanek has a friend who runs a drink stall in a canteen. His friend will sell drinks for $n$ days, numbered from day $1$ to day $n$. There are also $m$ types of drinks, numbered from $1$ to $m$.

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

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

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

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

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

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

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

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

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

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

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

Note: Here is Pak Chanek's optimal plan:

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

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

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 11380;
const int INF = 1e18;
const int N = 3 * 1e5 + 10;
int T = 1;
struct node {
    int u;
    int v;
    int e;
};
void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>>a(n, vector<int>(m));
    vector<vector<int>>dpl(n,vector<int>(m)), dpr(n,vector<int>(m)),pref(n,vector<int>(m+1));
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            cin >> a[i][j];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            pref[i][j + 1] = pref[i][j] + a[i][j];
    for (int i = 0; i < n; i++)
    {
        vector<int>smx(m), pmn(m);
        int mn = 0;
        for (int j = 0; j < m; j++)
        {
            mn = min(mn, pref[i][j+1]);
            pmn[j] = mn;
        }
        int mx = -INF;
        for (int j = m-1; j>=0; j--)
        {
            mx = max(mx, pref[i][j + 1]);
            smx[j] = mx;
        }
        if (!i)
        {
            for (int j = 0; j < m; j++)
            {
                dpl[0][j] = smx[j] - pref[0][j];
                dpr[0][j] = pref[0][j+1] - pmn[j];
            }
        }
        else
        {
            mx = -INF;
            for (int j = 0; j < m; j++)
            {
                if (j > 0)
                {
                    mx = max(mx, dpl[i - 1][j] - pmn[j - 1]);
                }
                dpr[i][j] = pref[i][j+1] + mx;
                mx = max(mx, dpr[i - 1][j] - pmn[j]);
            }
            mx = -INF;
            for (int j = m - 1; j >= 0; j--)
            {
                if (j + 1 < m)
                {
                    mx = max(mx, dpr[i - 1][j] + smx[j + 1]);
                }
                dpl[i][j] = mx - pref[i][j + 1];
                mx = max(mx, dpl[i-1][j] + smx[j]);
            }
        }
    }
    int ans = -INF;
    for (int i = 0; i < m; i++)
        ans = max({ ans,dpl[n - 1][i],dpr[n - 1][i] });
    cout << ans << endl;
}
signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
     cin >> T;
    while (T--) {
        solve();
    }
}

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

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

Think step by step.

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


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

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

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

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

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

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

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

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

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

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

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

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

Note: Here is Pak Chanek's optimal plan:

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

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

Buggy Code:
#include <bits/stdc++.h>
const int MAXN = 2e5 + 7;
const long long inf = -1e15;
using namespace std;
vector <int> a[MAXN];
long long s[MAXN];
long long LP[MAXN], RP[MAXN], L[MAXN], R[MAXN], S[MAXN];
int n, m;
void Calcsum(int ind)
{
    s[0] = 0;
    long long tmp;
    for (int i = 1; i <= n; i++)
    {
        tmp = (long long) a[ind][i];
        s[i] = s[i - 1] + tmp;
    }
}
void DealR(int ind)
{
    R[1] = inf;
    long long tmp = 0;
    Calcsum(ind);
    S[0] = 0;
    for (int i = 1; i <= n; i++)
        S[i] = min(S[i - 1], s[i]);
    long long maxn = inf;
    for (int i = 2; i <= n; i++)
    {
        maxn = max(maxn, max(LP[i], RP[i - 1]) - S[i - 2]);
        R[i] = maxn + s[i];
    }
}
void DealL(int ind)
{
    L[n] = inf;
    long long tmp = 0;
    s[n + 1] = 0;
    for (int i = n; i >= 1; i--)
        s[i] = s[i + 1] + a[ind][i];
    S[n + 1] = 0;
 //   cout << "ojdnvksj\n";
    for (int i = n; i >= 1; i--)
    {
        S[i] = min(S[i + 1], s[i]);
   //     cout << S[i] << " ";
    }
 //   cout << '\n';
    long long maxn = inf;
    for (int i = n - 1; i >= 1; i--)
    {
        maxn = max(maxn, max(LP[i + 1], RP[i]) - S[i + 2]);
     //   cout << i << " " << maxn << '\n';
        L[i] = maxn + s[i];
    }
}
int cnt = 0;
void solve()
{
    cin >> m >> n;
    for (int i = 1; i <= m; i++)
    {
        a[i].clear();
    }
    for (int i = 1; i <= n; i++)
        LP[i] = RP[i] = L[i] = R[i] = 0;
    for (int i = 1; i <= m; i++)
    {
        int tmp; long long stmp;
        a[i].push_back(0);
        for (int j = 1; j <= n; j++)
        {
            cin >> tmp;
            a[i].push_back(tmp);
        }
    }
    cnt++;
    if (cnt == 31)
    {
        cout << a[1][2] << a[1][4] << '\n';
    }
    L[0] = 0;
    Calcsum(1);
    for (int i = 1; i <= n; i++)
    {
        L[i] = min(L[i], s[i]);
        LP[i] = s[i] - L[i - 1];
   //     cout << LP[i] << " ";
    }
    R[n + 1] = s[n];
  //  cout << '\n';
    for (int i = n; i >= 1; i--)
    {
        R[i] = max(R[i + 1], s[i]);
        RP[i] = R[i] - s[i - 1];
      //  cout << RP[i] << ' ';
    }
    for (int i = 1; i <= n; i++)
        swap(LP[i], RP[i]);
  //  cout << '\n';
    for (int i = 2; i <= m; i++)
    {
        DealL(i);
        DealR(i);
 //       cout << i << '\n';
        for (int j = 1; j <= n; j++)
        {
            LP[j] = L[j];
            RP[j] = R[j];
    //        cout << LP[j] << " " << RP[j] << '\n';
        }
    }
    long long ans = inf;
    for (int i = 1; i <= n; i++)
    {
        ans = max(ans, max(LP[i], RP[i]));
    }
    cout << ans << '\n';
}
int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        solve();
    }

    return 0;
}
/*
5 5
-18212566-26497
-9 -91544175497
35 -38 -21 -13 -75497
-77 -4879 -30 -57497
3249434638497
*/

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

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

Think step by step.

Response:
```plaintext
1
3 4
1 2 3 4
5 6 7 8
9 10 11 12
```


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

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

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

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

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

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

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

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

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

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

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

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

Note: Here is Pak Chanek's optimal plan:

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

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

Buggy Code:
#include <cstdio>
#include <vector>
using namespace std;
const int N=200003;
const long long PIN=4557430888798830399;
const long long NIN=-4557430888798830400;
int m,n; vector<long long> a[N];
long long fl[N],fr[N],gl[N],gr[N],ls[N];
int main(){
//	freopen("canteen.in","r",stdin);
//	freopen("canteen.out","w",stdout);
	int i,j,t,x; long long s1,ans;
	for(scanf("%d",&t);t>0;t--){
		scanf("%d%d",&m,&n);
		for(i=1;i<=m;i++){
			a[i].clear(),a[i].push_back(0);
			for(j=1;j<=n;j++){
				scanf("%d",&x);
				a[i].push_back(x);
			}
			for(j=1;j<=n;j++)
				a[i][j]+=a[i][j-1];
		}
		for(i=0;i<=n;i++) fl[i]=fr[i]=0;
		for(i=1;i<=m;i++){
			for(j=0;j<=n;j++) ls[j]=NIN;
			for(j=n-1,s1=a[i][n];j>=0;j--){
				ls[j]=max(ls[j],fl[j]+s1);
				s1=max(s1,a[i][j]);
			}
			for(j=n,s1=NIN;j>0;j--){
				ls[j]=max(ls[j],fr[j]+s1);
				s1=max(s1,a[i][j]);
			}
			for(j=n-1;j>=0;j--) ls[j]=max(ls[j],ls[j+1]);
			for(j=0;j<n;j++) gl[j]=ls[j+1]-a[i][j];
			for(j=0;j<=n;j++) ls[j]=NIN;
			for(j=0,s1=PIN;j<n;j++){
				ls[j]=max(ls[j],fl[j]-s1);
				s1=min(s1,a[i][j]);
			}
			for(j=1,s1=a[i][0];j<=n;j++){
				ls[j]=max(ls[j],fr[j]-s1);
				s1=min(s1,a[i][j]);
			}
			for(j=1;j<=n;j++) ls[j]=max(ls[j],ls[j-1]);
			for(j=1;j<=n;j++) gr[j]=ls[j-1]+a[i][j];
			for(j=0;j<n;j++) fl[j]=gl[j];
			for(j=1;j<=n;j++) fr[j]=gr[j];
		}
		ans=NIN;
		for(i=0;i<n;i++) ans=max(ans,fl[i]);
		for(i=1;i<=n;i++) ans=max(ans,fr[i]);
		printf("%lld\n",ans);
	}
//	fclose(stdin);
//	fclose(stdout);
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
10 20 30
40 -50 60
-70 80 -90
```


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

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

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

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

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

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

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

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

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

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

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

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

Note: Here is Pak Chanek's optimal plan:

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

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

Buggy Code:
def sol():
    n, m = map(int, input().split())
    A = [[] for _ in range(n + 1)]
    for i in range(1, n + 1):
        Ai = list(map(int, input().split()))
        A[i] = [0] + Ai
    fl = [[-10**18 for _ in range(m + 1)] for _ in range(n + 1)]
    fr = [[-10**18 for _ in range(m + 1)] for _ in range(n + 1)]

    for i in range(1, m + 1):
        if fr[1][i - 1] < 0:
            fr[1][i] = A[1][i]
        else:
            fr[1][i] = A[1][i] + fr[1][i - 1]
    fl[1][m] = A[1][m]
    for i in range(m - 1, 0, -1):
        if fl[1][i + 1] < 0:
            fl[1][i] = A[1][i]
        else:
            fl[1][i] = A[1][i] + fl[1][i + 1]

    for i in range(2, n + 1):
        prev = [0 for _ in range(m + 1)]
        mxp = [-10**18 for _ in range(m + 1)]
        mxp[0] = 0
        mxresR = [-10**18 for _ in range(m + 1)]
        for j in range(1, m + 1):
            prev[j] = prev[j - 1] + A[i][j]
            mxp[j] = max(mxp[j - 1], -prev[j])
            if j - 2 >= 0:
                mxresR[j] = max(mxresR[j - 1], max(fl[i - 1][j], fr[i - 1][j]) + mxp[j - 2])
            fr[i][j] = mxresR[j - 1] + prev[j]
        suf = [0 for _ in range(m + 2)]
        mxs = [-10**18 for _ in range(m + 2)]
        mxs[m + 1] = 0
        mxresL = [-10**18 for _ in range(m + 2)]
        for j in range(m, 0, -1):
            suf[j] = suf[j + 1] + A[i][j]
            mxs[j] = max(mxs[j + 1], -suf[j])
            if j + 2 <= m + 1:
                mxresL[j] = max(mxresL[j + 1], max(fl[i - 1][j], fr[i - 1][j]) + mxs[j + 2])
            fl[i][j] = mxresL[j + 1] + suf[j]

    ans = -10**18
    for i in range(1, m + 1):
        ans = max(ans, max(fl[n][i], fr[n][i]))
    print(ans)

t = int(input())
while t:
    t -= 1
    sol()

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

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

Think step by step.

Response:
```plaintext
1
3 3
100 200 300
-100 -200 400
-100 200 300
```


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

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

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

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

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

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

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

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

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

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

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

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

Note: Here is Pak Chanek's optimal plan:

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

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

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

# region fastio
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

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

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

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

def solve():

    dp = [0] * (M + 1)
    ndp = [0] * (M + 1)
    rev = [0] * (M + 1)
    nrev = [0] * (M + 1)

    for j in range(N):
        a = li()

        pref = [0] * (M + 1)
        pref[0] = a[0]
        for i in range(1,M):
            l = i-1 if j != 0 else i
            pref[i] = max(a[i] + pref[i-1], sum(a[l:i+1]))

        suff = [0] * (M + 1)
        suff[M-1] = a[M-1]
        for i in range(M-2,-1,-1):
            r = i+2 if j != 0 else i+1
            suff[i] = max(a[i] + suff[i+1], sum(a[i:r]))

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

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

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

    res = max(dp)

    return str(res)

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

    print(solve())

file.close()

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

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

Think step by step.

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