**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

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

    change = {}
    total = sum(a)

    a = [a[0]] + a + [a[-1]]

    if t == 51:
        print(a)

    for i in range(1, len(a)-1):
        change[i] = a[i] - min(a[i-1], a[i+1])

    for i in range(k):
        largest = max(change, key=change.get)

        if largest == 0:
            break

        total -= change[largest]
        if largest == 1:
            change[largest] = change.get(change.get(largest+1))
        elif largest == len(change):
            change[largest] = change.get(change.get(largest-1))
        else:
            change[largest-1] = min(change.get(largest), change.get(largest-1))
            change[largest+1] = min(change.get(largest), change.get(largest+1))
            change[largest] = min(change.get(largest-1), change.get(largest+1))

        change[largest] = 0

    print(total)

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios where the algorithm might fail. Here are some considerations:

1. **Array Length and Operation Limit**: The code might have issues when the array length is very small (e.g., 1 or 2) or when the number of operations `k` is very large relative to the array length.

2. **Edge Elements**: The code might handle edge elements (first and last elements of the array) differently, which could lead to bugs.

3. **Uniform Array**: An array with all elements being the same might cause issues if the code does not handle this case correctly.

4. **Large Elements**: Arrays with large elements might cause overflow or incorrect calculations.

5. **Multiple Minimums**: Arrays with multiple minimum elements might cause the code to make incorrect decisions.

Given these considerations, let's create a small-scale test input that can expose potential issues in the code:

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

This test input includes:
- An array of length 1 with one operation.
- An array of length 2 with one operation.
- An array of length 3 with two operations.
- An array of length 4 with three operations.
- An array of length 5 with four operations.

Each test case is designed to challenge different aspects of the code, particularly how it handles small arrays, edge elements, and multiple minimums.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import sys

def main():
    input = sys.stdin.read().split()
    ptr = 0
    t = int(input[ptr])
    ptr += 1
    results = []
    for _ in range(t):
        n, k = int(input[ptr]), int(input[ptr+1])
        ptr +=2
        a = list(map(int, input[ptr:ptr+n]))
        ptr +=n

        current_sum = sum(a)
        for _ in range(k):
            max_reduction = 0
            best_i = -1
            best_val = 0
            for i in range(n):
                current_val = a[i]
                possible_vals = []
                if i > 0:
                    possible_vals.append(a[i-1])
                if i < n-1:
                    possible_vals.append(a[i+1])
                if not possible_vals:
                    continue
                min_val = min(possible_vals)
                reduction = current_val - min_val
                if reduction > max_reduction:
                    max_reduction = reduction
                    best_i = i
                    best_val = min_val
            if max_reduction == 0:
                break
            current_sum -= max_reduction
            a[best_i] = best_val

        results.append(current_sum)

    sys.stdout.write('\n'.join(map(str, results)) + '\n')

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Array with all elements being the same**: This can help test if the code correctly handles cases where no reduction is possible.
2. **Array with a small number of elements**: This can help test the basic logic of the algorithm.
3. **Array with elements that have large differences**: This can help test the logic for finding the maximum reduction.
4. **Array with elements that have small differences**: This can help test the logic for finding the maximum reduction in a more subtle way.
5. **Array with elements that are already minimized**: This can help test if the code correctly stops when no further reduction is possible.
6. **Array with a large number of elements**: This can help test the performance and correctness of the algorithm in a more complex scenario.

Given the constraints and the need for a small-scale test input, let's focus on a combination of the above scenarios:

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

This test input includes:
1. A small array with elements that can be reduced.
2. A single-element array with multiple operations.
3. An array with elements that have large differences.
4. An array with elements that have small differences.

This combination should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import sys

input = sys.stdin.readline

testcase = int(input())
for _ in range(testcase):
    n, k = map(int, input().split())
    datalist = list(map(int, input().split()))
    difflist = [0] * n
    sum = datalist[n-1]
    for i in range(n-1):
        difflist[i] = abs(datalist[i+1] - datalist[i])
        sum = sum + datalist[i]
    #print(n, k, datalist, difflist)
    if n == 1:
        print(sum)
        continue
    for _ in range(k):
        candiate_idx = 0
        maxval = difflist[0]
        min2val = min(datalist[0], datalist[1])
        for i in range(1,n):
            if difflist[i] > maxval:
                maxval = difflist[i]
                candiate_idx = i
            if difflist[i] == maxval:
                new_min2val = min(datalist[i], datalist[i + 1])
                if new_min2val < min2val:
                    candiate_idx = i

        if datalist[candiate_idx+1] > datalist[candiate_idx]:
            datalist[candiate_idx+1] = datalist[candiate_idx]
            difflist[candiate_idx] = 0
            if candiate_idx != (n - 2):
                difflist[candiate_idx+1] = abs(datalist[candiate_idx+2] - datalist[candiate_idx+1])
            sum = sum - maxval
            #print(datalist, sum)
        else:
            datalist[candiate_idx] = datalist[candiate_idx+1]
            difflist[candiate_idx] = 0
            if candiate_idx != 0:
                difflist[candiate_idx-1] = abs(datalist[candiate_idx-1] - datalist[candiate_idx])
            sum = sum - maxval
            #print(datalist, sum)

    print(sum)

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

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

Think step by step.

Response:
```plaintext
3
3 2
1 1000000000 1
4 2
10 1 10 1
5 3
1 100 1 100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import java.util.*;

public class Codeforce2 {

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n=sc.nextInt();
            int k=sc.nextInt();
            int[]arr=new int[n];
            long sum=0;
            for(int i=0;i<n;i++){
                arr[i]=sc.nextInt();
                sum+=arr[i];
            }
            int[]diff=new int[n];
            for(int i=0;i<n;i++){
                int min=arr[i];
                if(i>0){
                    min=Math.min(min,arr[i-1]);
                }
                if(i<n-1){
                    min=Math.min(min,arr[i+1]);
                }
                diff[i]=arr[i]-min;
            }
            Arrays.sort(diff);
            int i=n-1;
            while(k-->0&&i>=0){
                sum-=diff[i];
                if(diff[i]==0){
                    break;
                }
                i--;
            }
            System.out.println(sum);
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider scenarios where the code might fail. The code calculates the minimum possible sum by replacing elements with their neighbors, but it does so by sorting the differences and subtracting the largest differences first. This approach might fail in cases where the optimal solution requires a different strategy.

Here are some potential edge cases to consider:
1. **Array with all elements the same**: The code should not perform any operations.
2. **Array with elements that are already optimal**: The code should not perform any operations.
3. **Array where the optimal solution involves multiple replacements of the same element**: The code might miss this if it only considers the largest differences.
4. **Array with a large number of elements and a small number of operations**: The code might not handle this efficiently.

Let's create a test input that focuses on the third scenario, where the optimal solution involves multiple replacements of the same element. This will help expose the bug in the code.

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

### Explanation:
- **Test Case**: 1
- **Array**: [5, 4, 3, 2, 1]
- **Operations**: 3

In this test case:
- The optimal solution is to replace the elements to make the array as small as possible.
- The optimal sequence of operations could be:
  1. Replace 5 with 4: [4, 4, 3, 2, 1]
  2. Replace 4 with 3: [3, 3, 3, 2, 1]
  3. Replace 3 with 2: [2, 3, 2, 2, 1]

The minimum possible sum after 3 operations is 11.

This test case is small-scale and complete, and it should expose the bug in the code if it fails to handle the optimal sequence of replacements correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

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 k=sc.nextInt();
        long a[]=new long[n+1];
      long sum=0;
        for(int i=1;i<=n;i++){
          a[i]=sc.nextLong();
          sum+=a[i];
        }
      if(k==0){
        System.out.println(sum);
        continue;
      }
       long ans1=sum;
      long dp[][]=new long[n+1][k+1];
        for(int i=1;i<=n;i++){
            for(int j=0;j<=k;j++){
              long min=Long.MAX_VALUE,ans=Long.MAX_VALUE;
              for(int l=i;l>=Math.max(i-j,1);l--){
                min=Math.min(min,a[l]);
                ans=Math.min(ans,(min*(i-l+1))+dp[l-1][k-j]);
              }
              dp[i][j]=ans;
              if(i==n){
                ans1=Math.min(ans1,dp[i][j]);;
              }
            }
        }
      System.out.println(ans1);

    }
  }
}

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

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

Think step by step.

Response:
```plaintext
1
5 3
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: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define pb push_back
#define endl '\n'
#define int long long
#define f first
#define s second
#define pq priority_queue
#define fast ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define read_file freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);

using namespace std;

struct segtree{
	int n;
	vector<ll> mini;
	void init(int n0){
		n = 1;
		while (n < n0) n*=2;
		mini.assign(2*n, numeric_limits<ll>::max());
	}

	void set(int i, ll val, int curr, int lc, int rc){
		if (lc+1 == rc){
			mini[curr] = val;
			return;
		}
		int m = (lc+rc)/2;
		if (i < m) set(i, val, 2*curr+1, lc, m);
		else set(i, val, 2*curr+2, m, rc);
		mini[curr] = min(mini[2*curr+1], mini[2*curr+2]);
	}

	void set(int i, ll val){
		set(i, val, 0, 0, n);
	}

	ll get(int l, int r, int curr, int lc, int rc){
		if (lc >= l && rc <= r) return mini[curr];
		if (lc >= r || rc <= l) return numeric_limits<ll>::max();

		int m = (lc+rc)/2;
		return min(get(l, r, 2*curr+1, lc, m), get(l, r, 2*curr+2, m, rc));
	}

	ll get(int l, int r){
		return get(l, r+1, 0, 0, n);
	}
};

void solve(){
	int n, k; cin>>n>>k;
	int a[n+1]; for (int i = 1; i <= n; i++) cin>>a[i];

	segtree seg1; seg1.init(n+1);
	for (int i = 1; i <= n; i++){
		seg1.set(i, a[i]);
	}

	int pre[n+1]; pre[0] = 0; for (int i = 1; i <= n; i++) pre[i]=pre[i-1]+a[i];
	int sum = 0; for (int i = 1; i <= n; i++) sum+=a[i];

	int dp[n+1][k+1]; memset(dp, 0, sizeof(dp));

	// i am sending or receiving
	// dp[i][j] = dp[i-j-1][0 to k-j]+(i-j)*min(a[i-j to i])
	int mini = sum;
	for (int i = 1; i <= n; i++){
		int sumi = pre[i];
		for (int j = 0; j <= k; j++){
			dp[i][j] = sumi;
			if (j) dp[i][j] = min(dp[i][j], dp[i][j-1]);
			dp[i][j] = min(dp[i-1][j]+a[i], dp[i][j]);
			for (int ii = 1; ii < i; ii++){
				dp[i][j] = min(dp[i][j], pre[i]-pre[ii-1] + dp[ii-1][j]);
			}

			int z = k-j;
			if (i-j-1 >= 0){
				int mini = seg1.get(i-j, i);
				dp[i][j] = min(dp[i][j], dp[i-j-1][z]+(i-(i-j)+1)*mini);
			}
			else{
				int mini = seg1.get(max(1ll, i-j), i);
				dp[i][j] = min(dp[i][j], ((i)-(i-j)+1)*mini);
			}
		}
	}

	int ans = sum;
	// for (int i = 0; i <= n; i++){
	// 	for (int j = 0; j <= k; j++){
	// 		cout << dp[i][j] << ' ';
	// 	}
	// 	cout << endl;
	// }
	ans = min(ans, dp[n][k]);
	cout<<ans<<endl;
}

int32_t main() {
    fast;

    int t; cin>>t;
    while (t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
#include <bits/stdc++.h>
#include <climits>
#define lowbit(x) ((x) & -(x))
#define endl "\n"
#define dbg(a) cout << #a << ": " << a << "\n";

using std::cin, std::cout, std::string;
using pii = std::pair<int, int>;
using PII = std::pair<int, int>;
using Pll = std::pair<long long,long long>;
using ull = unsigned long long;
using ll = long long;
const int N = 5e5 + 10;
const int M = 1e5 + 10;
const int inf = INT_MAX;
const ll INF = LONG_LONG_MAX;
const int mod = 998244353;

void solve() {
    int n, K;
    cin >> n >> K;

    std::vector<ll> a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];

    std::vector mmin(n + 1, std::vector<ll>(20, inf));

    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= K && i + j <= n; j++) {
            mmin[i][j] = std::min({mmin[i][j - 1], a[i + j], mmin[i][j]});
            //std::cerr << i << " " << j << " " << mmin[i][j] << "\n";
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; i + j <= n && j <= K; j++) {
            mmin[i][j] *= 1ll * (j + 1);
            //std::cerr << i << " " << j << " " << mmin[i][j] << "\n";
        }
    }

    std::vector ans(n + 1, std::vector<ll>(20, inf));

    ans[0][0] = 0;
    for (int i = 1; i <= n; i++) {
        ans[i][0] = ans[i - 1][0] + a[i];
        for (int j = 1; j <= K; j++) {
            ans[i][j] = std::min(ans[i - 1][j] + a[i], ans[i][j - 1]);
            for (int k = 0; k <= j; k++) {
                if (k < i) {
                    ans[i][j] = std::min(ans[i - k - 1][j - k] + mmin[i - k][k], ans[i][j]);
                }
            }
        }
    }
    cout << ans[n][K] << endl;
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t = 1;
    cin >> t;
    while(t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code seems to have issues with how it handles the minimum values and the dynamic programming (DP) state transitions. Here are the steps to think about:

1. **Edge Cases**:
   - **Single Element Array**: The code should handle the case where the array has only one element.
   - **Large K**: The code should handle the case where the number of operations \( k \) is large, but the array is small.
   - **Small K**: The code should handle the case where the number of operations \( k \) is small, but the array is large.
   - **All Elements Equal**: The code should handle the case where all elements in the array are the same.
   - **Non-Uniform Elements**: The code should handle the case where the array has a mix of large and small elements.

2. **Challenging Scenarios**:
   - **Decreasing Sequence**: The code should handle the case where the array is in a strictly decreasing sequence.
   - **Increasing Sequence**: The code should handle the case where the array is in a strictly increasing sequence.
   - **Mixed Sequence**: The code should handle the case where the array has a mix of increasing and decreasing sequences.

Based on these considerations, let's generate a small-scale test input that is complete and valid:

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

### Explanation:
1. **Single Element Array**:
   - `1 3 5` (Single element array with \( k = 3 \))

2. **Non-Uniform Elements with Small K**:
   - `4 2 2 2 1 3` (Array with a mix of elements and \( k = 2 \))

3. **Decreasing Sequence with Small K**:
   - `3 1 3 1 2` (Array in a strictly decreasing sequence with \( k = 1 \))

4. **Mixed Sequence with Large K**:
   - `5 3 4 1 2 2 4` (Array with a mix of increasing and decreasing sequences and \( k = 3 \))

These test cases should help expose any issues in the buggy code's handling of various edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import math
import heapq
from collections import defaultdict, deque

def solve():
    n , k = map(int, input().split())
    l = list(map(int, input().split()))
    q = []
    # print(l)
    for i in range(n):
        if i<n-1:
            # q.append((-(l[i+1]-l[i]), i+1, i))
            heapq.heappush(q, (-(l[i+1]-l[i]), i+1, i))
        if i>0:
            # q.append((-(l[i-1]-l[i]), i-1, i))
            heapq.heappush(q, ((-(l[i-1]-l[i]), i-1, i)))

    k = min(k, n-1)
    while k and q:
        s, a, b = heapq.heappop(q)
        if abs(l[a]-l[b])!=abs(s):
            if l[a]>l[b]:
                heapq.heappush(q, (-(l[a]-l[b]), a, b))
            elif l[a]<l[b]:
                heapq.heappush(q, (-(l[b]-l[a]), b, a))

            continue
        if l[a]==l[b]:
            continue
        elif l[a]<l[b]:
            l[b] = l[a]
            i = b
        elif l[a]>l[b]:
            l[a] = l[b]
            i = a
        if i<n-1:
            # q.append((-(l[i+1]-l[i]), i+1, i))
            heapq.heappush(q, (-(l[i+1]-l[i]), i+1, i))
        if i>0:
            # q.append((-(l[i-1]-l[i]), i-1, i))
            heapq.heappush(q, ((-(l[i-1]-l[i]), i-1, i)))
        k-=1

    # while k:
    #     _, a, b = heapq.heappop(q)
    #     print(_, a, b)
    #     k-=1

    return sum(l)

for caseno in range(int(input())):
    print(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
5 3
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: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

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

for _ in range(int(input())):
    n, k = map(int, input().split())
    nums = list(map(int, input().split()))
    k = min(k, n)
    dp = [[0]*(k+1) for _ in range(n)]

    for i in range(n):
        dp[i][0] = dp[i-1][0] + nums[i]

    for i in range(n):
        for j in range(1, k+1):
            prev = dp[i-1][j-1] if i-1 >= 0 else 0
            dp[i][j] = prev + min(nums[i],
                                (nums[i+1] if i+1<n else float('inf')),
                                (nums[i-1] if i-1>=0 else float('inf')))

            dp[i][j] = min(dp[i][j], (dp[i-1][j] if i-1>=0 else 0) + nums[i])
            for l in range(1 ,j):
                x = 0
                if i-l-1 >= 0:
                    x = dp[i-l-1][j-l]
                    dp[i][j] = min(x + nums[i]*(l+1), x + nums[i-l]*(l+1))
                else:
                    dp[i][j] = min(nums[0]*(i+1), nums[i]*(i+1))

    print(min(dp[-1]))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. The code provided has a dynamic programming approach, but it might have issues with how it handles the minimum values and the operations.

### Steps to Create the Test Input:

1. **Edge Case with Small `k`**:
   - When `k` is very small (e.g., 0 or 1), the code should handle it correctly by either not performing any operations or performing only one operation.
   - Example: `n = 3, k = 1, nums = [3, 1, 2]`

2. **Edge Case with Large `k`**:
   - When `k` is large, the code should be able to perform multiple operations to minimize the sum.
   - Example: `n = 4, k = 2, nums = [2, 2, 1, 3]`

3. **Edge Case with Uniform Array**:
   - When the array is already uniform (all elements are the same), no operations are needed.
   - Example: `n = 3, k = 2, nums = [1, 1, 1]`

4. **Edge Case with Large Values**:
   - When the array contains large values, the code should still be able to minimize the sum effectively.
   - Example: `n = 5, k = 3, nums = [1000000000, 1, 1, 1, 1000000000]`

5. **Edge Case with Mixed Values**:
   - When the array contains a mix of large and small values, the code should prioritize minimizing the large values.
   - Example: `n = 6, k = 3, nums = [4, 1, 2, 2, 4, 3]`

### Chosen Test Input:
Let's focus on a small-scale test input that combines some of these edge cases to maximize the failure probability.

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

### Explanation:
- **Test Case 1**: `n = 3, k = 1, nums = [3, 1, 2]` - This tests the case where only one operation is allowed.
- **Test Case 2**: `n = 4, k = 2, nums = [2, 2, 1, 3]` - This tests the case where multiple operations are allowed.
- **Test Case 3**: `n = 3, k = 2, nums = [1, 1, 1]` - This tests the case where the array is already uniform.
- **Test Case 4**: `n = 5, k = 3, nums = [1000000000, 1, 1, 1, 1000000000]` - This tests the case with large values and mixed values.

This test input should help expose any issues in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;

public class Main {
    static FastReader sc = new FastReader();
    static class pair{
        int l;
        int r;
        pair(int a, int b) {
            this.l=a;
            this.r=b;
        }
    }
    static int mod=998244353;
    // --------------------------------------start-------------------------------------------//
    public static void solve(int tt) {
        int n=sc.nextInt();
        int k=sc.nextInt();
        int arr[]=new int[n];
        long pre[]=new long[n+1];
        long sum=0;
        for(int i=0;i<n;i++){
            arr[i]=sc.nextInt();
            sum+=arr[i];
            pre[i+1]=arr[i]+pre[i];
        }
        long minMax=0;
        long dp[][]=new long[n][k+1];
        long wild[]=new long[k+1];
        for(int i=n-1;i>=0;i--){
            for(int jj=1;jj<=k;jj++){
                for(int j=1;j<=jj;j++){
                    if(i+j<n){
                        long totalNew=((pre[i+j+1]-pre[i])-((j+1)*(Math.min(arr[i+j],arr[i]))));
                        for(int p=0;p<=jj-j;p++){
                            for(int ppp=0;ppp<=jj;ppp++)
                            dp[i][jj]=Math.max(wild[ppp],dp[i][jj]);
                            dp[i][jj]=Math.max(dp[i][jj],totalNew+dp[i+j][p]);
                            wild[jj]=Math.max(wild[jj],dp[i][jj]);
                        }
                    }
                }
            }
        }
        // for(int i[]:dp){
        //     for(int j:i){
        //         System.out.print(j+" ");
        //     }
        //     System.out.println("");
        // }
        for(long i:wild) {
            minMax=Math.max(minMax, i);
        }
        // System.out.println(minMax);
        System.out.println(sum-minMax);
    }
    public static void main(String[] args) {
        int t= sc.nextInt();
        int i=1;
        // ss();
        while(t-->0){
            solve(i++);
        }
    }
    // public static void ss(){
    //     for(int i=1;i<=500;i++){
    //         System.out.println(2*i);
    //         int ret=solve(i);
    //         if(ret>2*i){
    //             break;
    //         }
    //     }
    // }

    // public static long getSum(int r){
    // long sum=0;
    // int pp=r;
    // while(pp>0){
    // sum+=fenwickTree[pp];
    // pp-=pp&-pp;
    // }
    // return sum;
    // }
    // public static void update(int k,long v){
    // // long value=fenwickTree[k];
    // while(k<=n){
    // fenwickTree[k]+=(v);
    // k+=k&-k;
    // }
    // }
    static long factorial(long n) {
        long result = 1;
        for (long i = 1; i <= n; i++) {
            result *= i;
            if (result > 998244353) {
                result -= 998244353;
            }
        }
        return result % 998244353;
    }

    static int modFact(int n, int p) {
        if (n >= p)
            return 0;

        int result = 1;
        for (int i = 1; i <= n; i++)
            result = (result * i) % p;

        return result;
    }

    // Function to calculate combination
    // static long combination(long n, long r) {
    // return factorial(n) / (factorial(r) * factorial(n - r));
    // }

    // public static long binarySearch(ArrayList<Long> arr, long target) {
    // int left = 0, right = arr.size() - 1;
    // while (left <= right) {
    // int mid = left + (right - left) / 2;
    // if (arr.get(mid) == target) {
    // return 0;
    // } else if (arr.get(mid) < target) {
    // left = mid + 1;
    // } else {
    // right = mid - 1;
    // }
    // }
    // if (left == 0) {
    // return Math.abs(target - arr.get(0));
    // } else if (left == arr.size()) {
    // return Math.abs(arr.get(arr.size() - 1) - target);
    // } else {
    // long diff1 = Math.abs(target - arr.get(left - 1));
    // long diff2 = Math.abs(arr.get(left) - target);
    // return Math.min(diff1, diff2);
    // }
    // }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    // public static int maxFrequencyNumber(int[] arr) {
    // if (arr.length == 0)
    // return -1;
    // int maxFreq = 0;
    // int number = -1;
    // HashMap<Integer, Integer> map = new HashMap<>();

    // for (int j : arr) {
    // if (map.containsKey(j)) {
    // map.put(j, map.get(j) + 1);
    // } else {
    // map.put(j, 1);
    // }
    // }
    // // using set data structure
    // Set<Integer> keySet = map.keySet();
    // for (Integer i : keySet) {
    // if (map.get(i) > maxFreq) {
    // number = i;
    // maxFreq = map.get(i);
    // }
    // }
    // return number;
    // }

    // public static boolean isPrime(long n) {
    // if (n <= 1)
    // return false;

    // // Check if n=2 or n=3
    // if (n == 2 || n == 3)
    // return true;

    // // Check whether n is divisible by 2 or 3
    // if (n % 2 == 0 || n % 3 == 0)
    // return false;

    // // Check from 5 to square root of n
    // // Iterate i by (i+6)
    // for (int i = 5; i <= Math.sqrt(n); i = i + 6)
    // if (n % i == 0 || n % (i + 2) == 0)
    // return false;

    // return true;

    // }

    // public static long lcm(int x, int y) {
    // int i;
    // int a = Math.max(x, y);

    // for (i = a; i <= x * y; i = i + a) {
    // if (i % x == 0 && i % y == 0)
    // break;
    // }
    // return i;
    // }

    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }

    // static ArrayList<Long> getFactors(long num) {
    // ArrayList<Long> factors = new ArrayList<Long>();
    // int i;
    // boolean flag = false;

    // for (i = 1; i <= Math.sqrt(num); i++) {
    // if (num % i == 0)
    // factors.add((long) i);
    // if (i == num / i)
    // flag = true;
    // }
    // if (flag)
    // i -= 2;

    // // printing pairs
    // for (; i >= 1; i--) {
    // if (num % i == 0)
    // factors.add((long) (num / i));
    // }
    // return factors;
    // }

    // static int fact(int number) {
    // int f = 1;
    // int j = 1;
    // while (j <= number) {
    // f = f * j;
    // j++;
    // }
    // return f;
    // }

    public static int[] arrInput(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = sc.nextInt();
        }
        return arr;
    }

    public static void arrOutput(int[] arr) {
        StringBuilder out = new StringBuilder("");
        for (int i = 0; i < arr.length; i++) {
            out.append(arr[i] + " ");
        }
        System.out.println(out);
    }

    // public static boolean isInt(double d) {
    // return d % 1 == 0;
    // }

    // public static boolean isPowerOfTwo(int n) {
    // return (int) (Math.ceil((Math.log(n) / Math.log(2)))) == (int)
    // (Math.floor(((Math.log(n) / Math.log(2)))));
    // }
    public static HashSet<Integer> factors(int n){
        HashSet<Integer> h = new HashSet<Integer>();
        for(int i=1;i*i<=n;i++){
            if(n%i == 0){
                h.add(i);
                h.add(n/i);
            }
        }
        return h;
    }
    public static ArrayList<Integer> primeFactorization(int numbers) {
        int n = numbers;
        ArrayList<Integer> factors = new ArrayList<Integer>();
        for (int i = 2; i <= n / i; i++) {
            while (n % i == 0) {
                factors.add(i);
                n /= i;
            }
        }
        if (n > 1) {
            factors.add(n);
        }
        return factors;
    }

    public static int calculate(int l, int r, int segTree[], int n) {
        l += n - 1;
        r += n - 1;
        int ans = segTree[r];
        while (l <= r) {
            if (r % 2 == 0) {
                ans &= segTree[r];
                r--;
            }
            if (l % 2 == 1) {
                ans &= segTree[l];
                l++;
            }
            r = r >> 1;
            l = l >> 1;
        }
        return ans;
    }
}

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

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

Think step by step.

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

1. **Edge Cases for `n` and `k`**:
   - `n = 1` and `k = 0`: This is the simplest case where no operations can be performed.
   - `n = 1` and `k > 0`: Even with operations, the array remains the same.
   - `n = 2` and `k = 1`: Only one operation can be performed, so the minimum possible sum should be considered.
   - `n = 3` and `k = 1`: This is a slightly more complex case where the operation can be performed on any of the elements.

2. **Cases with Multiple Elements**:
   - `n = 4` and `k = 2`: This case allows for multiple operations, and the algorithm should find the optimal way to minimize the sum.
   - `n = 6` and `k = 3`: This is a more complex case with more elements and operations.

3. **Cases with Large Values**:
   - `n = 3` and `k = 10`: This case has a large `k` but a small `n`, which might expose issues with handling large `k` values.
   - `n = 10` and `k = 10`: This case has both a large `n` and `k`, which can be challenging for the algorithm.

4. **Cases with Mixed Values**:
   - `n = 5` and `k = 2`: The array can have both large and small values, which can test the algorithm's ability to handle mixed values.

Given these considerations, a small-scale test input that is complete and valid and can expose the bug is:

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

This test input includes:
- A case with `n = 1` and `k = 0`.
- A case with `n = 2` and `k = 1`.
- A case with `n = 3` and `k = 1`.
- A case with `n = 4` and `k = 2`.

These cases cover a range of scenarios that can help identify potential issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import java.io.*;
import java.util.*;
import java.lang.*;
import static java.lang.Math.*;

public class Codeforces {

    static int MAX = Integer.MAX_VALUE,MIN = Integer.MIN_VALUE;
    static long MAXL = Long.MAX_VALUE,MINL = Long.MIN_VALUE,MOD = (int)1e9+7;
    static void solve() {
        int n = fs.nInt() , k = fs.nInt();
        int[]ar = new int[n];
        long sum = 0;
        for(int i=0;i<n;i++){
            ar[i] = fs.nInt();
            sum += ar[i];
        }
        if(n == 1){
            out.println(sum);
            return;
        }
        int[]diff = new int[n];
        for(int i=0;i<n-1;i++){
            diff[i] = abs(ar[i+1]-ar[i]);
        }
        sort(diff);
        for(int i=n-1;i>=n-k;i--){
            sum -= diff[i];
        }
        out.println(sum);
    }

    static class Triplet<T,U,V>{
        T a;
        U b;
        V c;
        Triplet(T a,U b,V c){
            this.a = a;
            this.b = b;
            this.c = c;
        }
    }
    static class Pair<A, B>{
        A fst;
        B snd;
        Pair(A fst,B snd){
            this.fst = fst;
            this.snd = snd;
        }
    }

    static boolean multipleTestCase = true;static FastScanner fs;static PrintWriter out;
    public static void main(String[]args){
        try{
            out = new PrintWriter(System.out);
            fs = new FastScanner();

            int tc = multipleTestCase?fs.nInt():1;
            while (tc-->0)solve();
            out.flush();
            out.close();
        }catch (Exception e){
            e.printStackTrace();
        }
    }
    static class FastScanner extends PrintWriter {
        private InputStream stream;
        private byte[] buf = new byte[1<<16];
        private int curChar, numChars;
        public FastScanner() { this(System.in,System.out); }
        public FastScanner(InputStream i, OutputStream o) {
            super(o);
            stream = i;
        }
        // file input
        public FastScanner(String i, String o) throws IOException {
            super(new FileWriter(o));
            stream = new FileInputStream(i);
        }

        // throws InputMismatchException() if previously detected end of file
        private int nextByte() {
            if (numChars == -1) throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars == -1) return -1; // end of file
            }
            return buf[curChar++];
        }

        // to read in entire lines, replace c <= ' '
        // with a function that checks whether c is a line break
        public String next() {
            int c; do { c = nextByte(); } while (c <= ' ');
            StringBuilder res = new StringBuilder();
            do { res.appendCodePoint(c); c = nextByte(); } while (c > ' ');
            return res.toString();
        }
        public int nInt() { // nextLong() would be implemented similarly
            int c; do { c = nextByte(); } while (c <= ' ');
            int sgn = 1; if (c == '-') { sgn = -1; c = nextByte(); }
            int res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res = 10*res+c-'0';
                c = nextByte();
            } while (c > ' ');
            return res * sgn;
        }
        public long nLong(){return Long.parseLong(next());}
        public double nextDouble() { return Double.parseDouble(next()); }
    }
    public static void sort(int[] arr){
        ArrayList<Integer> ls = new ArrayList<>();
        for(int x: arr)
            ls.add(x);
        Collections.sort(ls);
        for(int i=0; i < arr.length; i++)
            arr[i] = ls.get(i);
    }
    public static void sortR(int[] arr){
        ArrayList<Integer> ls = new ArrayList<>();
        for(int x: arr)
            ls.add(x);
        Collections.sort(ls,Collections.reverseOrder());
        for(int i=0; i < arr.length; i++)
            arr[i] = ls.get(i);
    }
    public static void sort(long[] arr){
        ArrayList<Long> ls = new ArrayList<>();
        for(long x: arr)
            ls.add(x);
        Collections.sort(ls);
        for(int i=0; i < arr.length; i++)
            arr[i] = ls.get(i);
    }
    public static void sortR(long[] arr){
        ArrayList<Long> ls = new ArrayList<>();
        for(long x: arr)
            ls.add(x);
        Collections.sort(ls,Collections.reverseOrder());
        for(int i=0; i < arr.length; i++)
            arr[i] = ls.get(i);
    }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to consider the logic of the algorithm and identify where it might fail. The code calculates the minimum possible sum of the array after performing at most `k` operations. The key issue in the code is the way it handles the differences between adjacent elements. It sorts the differences and subtracts the largest `k` differences from the sum. However, this approach does not correctly account for the actual impact of the operations on the array.

### Step-by-Step Analysis:
1. **Edge Cases:**
   - Single element array (`n = 1`).
   - Array with all elements being the same.
   - Array with elements in increasing or decreasing order.
   - Array with a mix of large and small values.

2. **Critical Scenarios:**
   - **Small `k` and large `n`:** The code might not correctly handle the scenario where `k` is small compared to `n`.
   - **Large `k` and small `n`:** The code might not correctly handle the scenario where `k` is large enough to potentially change all elements.
   - **Sorted arrays:** The code might not correctly handle the scenario where the array is already sorted or nearly sorted.

3. **Specific Test Case:**
   - Consider an array where the differences between adjacent elements are large, and `k` is small. This will test the code's ability to correctly identify and apply the operations.

### Chosen Test Case:
- **Array:** `[1, 100, 1, 100, 1]`
- **`n = 5`**
- **`k = 2`**

### Explanation:
- The array has large differences between adjacent elements (99 and 99).
- With `k = 2`, the code should ideally be able to reduce the sum by replacing elements to minimize the differences.
- The expected minimum sum can be achieved by replacing `100` with `1` in two operations, resulting in the array `[1, 1, 1, 1, 1]` with a sum of 5.

### Test Input:
```plaintext
1
5 2
1 100 1 100 1
```

This test input is small-scale and complete, focusing on a scenario that maximizes the failure probability in the faulty solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define el "\n"
#define in insert
#define MAX 100000000000000
#define MOD 1000000007
#define MODR 998244353
#define fi first
#define se second
#define all(v) v.begin(),v.end()
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
ll gcd(ll x,ll y)
{
	if(y==0) return x;
	else return gcd(y,x%y);
}
ll min(ll x,ll y)
{
	if(x<y) return x;
	return y;
}
void solve()
{
vector<ll> v;
ll n,k;
ll x;cin>>n>>k;
ll sum=0;v.push_back(0);
for(ll i=1;i<=n;i++)
{
	cin>>x;v.push_back(x);
	sum+=x;
}
v.push_back(0);
priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,greater<pair<ll,ll>>> qe;
for(ll i=1;i<=n;i++)
{
	if(v[i]-v[i-1]<0) qe.push({v[i]-v[i-1],i});
	if(v[i]-v[i+1]<0) qe.push({v[i]-v[i+1],i});
}
ll a=0;
set<ll> sr;
sr.insert(MAX);
set<ll> s;
while(!qe.empty())
{
	if(k>=s.size()&&qe.top().fi<0)
	{
		a=qe.top().se;
		if(v[a]-v[a-1]==qe.top().fi)
		{
			if(k==s.size()&&s.find(a-1)==s.end())
			{
				sr.insert(sum);
				qe.pop();
				continue;
			}
		    v[a-1]=v[a];
			s.insert(a-1);
			if(s.size()<=k) sum+=qe.top().fi;
			qe.pop();
			if(v[a-2]>0)
			{
			if(v[a-1]-v[a-2]<0) qe.push({v[a-1]-v[a-2],a-1});
			if(v[a-2]-v[a-1]<0) qe.push({v[a-2]-v[a-1],a-2});
		    }
		}
	   else if(v[a]-v[a+1]==qe.top().fi)
		{
			if(k==s.size()&&s.find(a+1)==s.end())
			{
				sr.insert(sum);
				qe.pop();
				continue;
			}
			s.insert(a+1);
			v[a+1]=v[a];
	        sum+=qe.top().fi;
			qe.pop();
			if(v[a+2]>0)
			{
		    if(v[a+1]-v[a+2]<0) qe.push({v[a+1]-v[a+2],a+1});
			if(v[a+2]-v[a+1]<0) qe.push({v[a+2]-v[a+1],a+2});
		    }
		}
		else qe.pop();
	}
	else break;
}
cout<<min(*sr.begin(),sum)<<el;
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
ll t;
cin>>t;
while(t--)solve();
 return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast,no-stack-protector")
#include <bits/stdc++.h>
using namespace std ;
#define int long long

typedef long long LL ;
typedef pair<int , int> PII ;
const int inf = 0x3f3f3f3f ;
const LL LLinf = 0x3f3f3f3f3f3f3f3f ;
const int N = 3e5 + 10 ;

int a[N] ;

void solve()
{
	int n , k ;
	std::cin >> n >> k ;
	LL ret = 0;
	for(int i = 1 ; i <= n ; ++ i){
		std::cin >> a[i] ;
		ret += a[i] ;
	}
	a[0] = a[n + 1] = a[n + 2] = a[n + 3] = LLinf ;
	std::multiset< PII , greater<PII> > s;
	for(int i = 1 ; i <= n ; ++ i)
	{
		s.insert({a[i] - std::min({a[i] , a[i - 1] , a[i + 1]}) , i}) ;
	}
	for(int _ = 1 ; _ <= k && _ <= s.size() ; _ ++ )
	{
		auto p = s.begin() ;

		int i = p->second ;
		int x = p->first ;

		if ( i > 1 ) s.erase({a[i - 1] - std::min({a[i - 2] , a[i - 1] , a[i]}) , i - 1} ) ;
		if ( i < n ) s.erase({a[i + 1] - std::min({a[i] , a[i + 1] , a[i + 2]}) , i + 1} ) ;
		s.erase({x , i}) ;
		a[i] -= x ;
		if ( i > 1 ) s.insert({a[i - 1] - std::min({a[i - 2] , a[i - 1] , a[i]}) , i - 1}) ;
		if ( i < n ) s.insert({a[i + 1] - std::min({a[i] , a[i + 1] , a[i + 2]}) , i + 1}) ;
		s.insert({ 0 , i}) ;
		ret -= x ;
	}
	// cout << '\n' ;
	// for(auto p : s)
	// {
	// 	cout << p.first << ' ' << p.second << '\n' ;
	// }
	// for(int i = 1 ; i <= n ; ++ i)
	// 	cout << a[i] << ' ';
	// cout << '\n' ;

	cout << ret << '\n' ;
}

signed main()
{
	cin.tie(nullptr)->ios::sync_with_stdio(false) ;

	int t = 1 ;
	cin >> t ;

	for(int i = 1 ; i <= t ; i++)
	{
		// cout << "case #" << i << " : " << '\n' ;
		solve() ;
	}
	return 0 ;
}

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

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

Think step by step.

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

1. **Small-scale Test Input**: Since the output length is limited, we should aim for a small-scale test input that is still challenging.
2. **Edge Cases**: Consider edge cases such as:
   - Arrays with very small or very large values.
   - Arrays with repeated values.
   - Arrays with a single element.
   - Arrays where the minimum value is not in the middle.
3. **Maximizing the Failure Probability**: Create a test input where the algorithm might make suboptimal choices, leading to a higher sum than necessary.

Given these considerations, a challenging test input could be:

```plaintext
3
4 2
1 100 1 100
5 3
100 1 100 1 100
6 1
1 100 1 100 1 100
```

Explanation:
- The first test case has a small array with alternating large and small values. The optimal strategy is to replace the large values with the small ones, but the algorithm might not handle this correctly.
- The second test case has a larger array with alternating large and small values, similar to the first test case but with more elements.
- The third test case has a larger array with alternating large and small values, but with only one operation allowed, which should be used optimally to minimize the sum.

This test input should help expose any issues in the buggy code's handling of such scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import os
import sys
from math import ceil, floor, pi, sqrt
from itertools import combinations, permutations
from collections import deque, Counter, defaultdict

##################################################################
DEBUG = os.path.isfile("C:\\Users\\jj720\\debug.txt")            #
debugGreen,debugCyan,debugEnd = '\033[92m','\033[96m','\033[0m'  #
if DEBUG: os.system('color')                                     #
##################################################################
""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""
##################################################################
prt = sys.stdout.write                                           #
sys.setrecursionlimit(100000)                                    #
mod = 1000000007                                                 #
def debug(var, name=""):                                         #
    if DEBUG:                                                    #
        prt(f"{debugGreen}{name.upper()}: {var}{debugEnd}\n")    #
def out(x):                                                      #
    prt(f"{debugCyan}{x}{debugEnd}\n")if DEBUG else prt(f"{x}\n")#
if DEBUG: print = out                                            #
##################################################################
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
def solve():
    n, k = map(int, input().split())
    nums = list(map(int, input().split()))

    rep = None
    for _ in range(k):
        mx = 0
        for i, num in enumerate(nums[:-1]):
            diff = num - nums[i+1]
            if abs(diff) > mx:
                mx = abs(diff)

                if diff > 0:
                    rep = (i,nums[i+1])
                elif diff < 0:
                    rep  = (i+1, num)
        if rep:
            nums[rep[0]] = rep[1]
    return sum(nums)

#~~~~~~~~~~~~~~~~~~~~~~~~~~
############################
if __name__ == "__main__": #
    for _ in range(int(input())):
        print(solve())                #
############################

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

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

Think step by step.

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

1. **Edge Cases:**
   - Arrays with very small values.
   - Arrays with very large values.
   - Arrays with all elements being the same.
   - Arrays where the elements are in a specific order (e.g., sorted, reverse sorted).

2. **Scenarios That Maximize Failure Probability:**
   - Arrays where the maximum difference between adjacent elements is large.
   - Arrays where the operations can significantly reduce the sum in a few steps.

3. **Small-Scale Test Input:**
   - Ensure the input is small enough to fit within the constraints but still challenging.

Given these considerations, let's construct a test input that can expose the bug in the provided code.

### Test Input
```plaintext
3
5 2
10 1 10 1 10
4 3
100 1 100 1
6 3
10 1 10 1 10 1
```

### Explanation
1. **Test Case 1:**
   - Array: `[10, 1, 10, 1, 10]`
   - Operations: 2
   - The large difference between adjacent elements (10 and 1) provides an opportunity to significantly reduce the sum.
   - Expected Output: 12 (e.g., `[1, 1, 1, 1, 10]` or `[10, 1, 1, 1, 1]`)

2. **Test Case 2:**
   - Array: `[100, 1, 100, 1]`
   - Operations: 3
   - The large difference between adjacent elements (100 and 1) provides an opportunity to significantly reduce the sum.
   - Expected Output: 4 (e.g., `[1, 1, 1, 1]`)

3. **Test Case 3:**
   - Array: `[10, 1, 10, 1, 10, 1]`
   - Operations: 3
   - The large difference between adjacent elements (10 and 1) provides an opportunity to significantly reduce the sum.
   - Expected Output: 6 (e.g., `[1, 1, 1, 1, 1, 1]`)

These test cases are designed to expose the bug by having large differences between adjacent elements, which the buggy code might not handle correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import sys
from collections import deque

input = lambda: sys.stdin.readline().rstrip()

def print(*args, sep=' ', end='\n'):
    sys.stdout.write(sep.join(map(str, args)) + end)

def ii():  # int input
    return int(input())

def lii():  # list int input
    return list(map(int, input().split()))

def lsi():  # list string input
    return input().split()

def main():
    t = ii()
    for _ in range(t):
        n, k = lii()
        arr = lii()
        if n == 1:
            print(arr[0])
            continue
        sm = sum(arr)
        for j in range(k):
            mni = 0
            mxr = arr[0] - arr[1]
            zam = arr[1]
            for i in range(1, n - 1):
                r1 = arr[i] - arr[i - 1]
                if r1 > mxr:
                    mni = i
                    mxr = r1
                    zam = arr[i - 1]
                r2 = arr[i] - arr[i + 1]
                if r2 > mxr:
                    mni = i
                    mxr = r2
                    zam = arr[i + 1]
            r3 = arr[-1] - arr[-2]
            if r3 > mxr:
                mni = -1
                mxr = r3
                zam = arr[-2]
            sm -= mxr
            arr[mni] = zam
        print(sm)

if __name__ == '__main__':
    main()

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

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

Think step by step.

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

1. **Edge Case: Single Element Array**:
   - The code should handle the case where the array has only one element correctly.
   - Test Case: `1 3 5`

2. **Edge Case: Large Number of Operations (k) with Small Array**:
   - If the number of operations is large relative to the array size, the code should be able to handle it efficiently.
   - Test Case: `1 2 10 1 2 3 4 5 6 7 8 9 10`

3. **Edge Case: Array with All Elements Equal**:
   - If all elements in the array are the same, the code should not perform any operations.
   - Test Case: `1 5 1 1 1 1 1 1 1 1 1 1`

4. **Edge Case: Array with Alternating Elements**:
   - An array with alternating elements can challenge the logic of finding the minimum possible sum.
   - Test Case: `1 3 1 2 1 2 1 2 1 2 1 2`

5. **Edge Case: Array with a Single Minimum Element**:
   - If the array has a single minimum element surrounded by larger elements, the code should focus on minimizing the sum using this element.
   - Test Case: `1 2 10 9 8 7 6 5 4 3 2 1 10`

6. **Edge Case: Array with a Single Maximum Element**:
   - If the array has a single maximum element surrounded by smaller elements, the code should focus on minimizing the sum using the smaller elements.
   - Test Case: `1 2 1 2 3 4 5 6 7 8 9 10 1`

7. **Edge Case: Array with a Small Number of Elements**:
   - A small array with a small number of operations can help verify the correctness of the logic.
   - Test Case: `1 2 3 2 1`

8. **Edge Case: Array with a Large Number of Elements**:
   - A large array with a small number of operations can test the efficiency of the code.
   - Test Case: `1 3 100000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406