**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
import java.util.Scanner;

public class C {

    // Function to check if a number is prime
    public static boolean isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) return false;
        }
        return true;
    }

    // Function to swap two elements in an array
    public static void swap(int[] arr, int l, int r) {
        int temp = arr[l];
        arr[l] = arr[r];
        arr[r] = temp;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt(); // number of test cases

        while (t-- > 0) {
            int n = sc.nextInt(); // size of the permutation

            // Handle base cases for n <= 4
            if (n <= 4) {
                System.out.println(-1);
                continue;
            }

            // Create the odd and even number groups
            int[] odd = new int[(n + 1) / 2];  // Odd numbers
            int[] even = new int[n / 2];  // Even numbers

            // Fill the odd and even arrays
            int oddIndex = 0, evenIndex = 0;
            for (int i = 1; i <= n; i++) {
                if (i % 2 == 1) {
                    odd[oddIndex++] = i;
                } else {
                    even[evenIndex++] = i;
                }
            }

            // Combine odd and even numbers into the result array
            int[] result = new int[n];
            int idx = 0;
            for (int i = 0; i < odd.length; i++) {
                result[idx++] = odd[i];
            }
            for (int i = 0; i < even.length; i++) {
                result[idx++] = even[i];
            }

            // Check the intersection between last odd and first even
            if (isPrime(result[odd.length - 1] + result[odd.length])) {
                // Swap the first even with another even number
                swap(result, odd.length, odd.length + 1);
            }

            // Output the result permutation
            for (int num : result) {
                System.out.print(num + " ");
            }
            System.out.println();
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define test  \
    ll t;     \
    cin >> t; \
    while

bool isComposite(ll n)
{
    if (n <= 3)
        return false;
    if (n % 2 == 0 || n % 3 == 0)
        return true;
    for (int i = 5; i * i <= n; i += 6)
    {
        if (n % i == 0 || n % (i + 2) == 0)
            return true;
    }
    return false;
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    test(t--)
    {
        ll n;
        cin >> n;
        if (n == 2 || n == 3)
        {
            cout << -1 << "\n";
            continue;
        }

        vector<int> evens, odds;
        for (int i = 1; i <= n; i++)
        {
            if (i % 2 == 0)
                evens.push_back(i);
            else
                odds.push_back(i);
        }
        evens.insert(evens.end(), odds.begin(), odds.end());

        ll index = odds.size() - 1;
        ll p = evens[index] + evens[index + 1];
        bool flag = false;
        if (isComposite(p))
        {
            for (int x : evens)
                cout << x << " ";
        }
        else
        {
            for (int i = index; i < evens.size(); i++)
            {
                swap(evens[index + 1], evens[i + 2]);
                p = evens[index] + evens[index + 1];
                if (isComposite(p))
                {
                    flag = true;
                    break;
                }
                else
                {
                    swap(evens[index + 1], evens[i + 2]);
                }
            }
            if (flag)
            {
                for (int x : evens)
                    cout << x << " ";
            }
            else
            {
                cout << -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 provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: The code handles small values of `n` (2 and 3) explicitly. We should test slightly larger values that might still cause issues.
2. **Composite Sum Requirement**: The code needs to ensure that the sum of adjacent elements in the permutation is composite. We should test values of `n` where the composite sum requirement might be challenging to meet.
3. **Permutation Construction**: The code constructs the permutation by separating even and odd numbers and then combining them. We should test values of `n` where this separation might lead to issues in ensuring the composite sum condition.

Given these considerations, let's generate a small-scale test input:

```plaintext
3
4
5
6
```

### Explanation:
1. **n = 4**: This is a small value where the permutation must be constructed to ensure all adjacent sums are composite. The code might struggle to find a valid permutation.
2. **n = 5**: This is another small value that needs to be handled carefully. The code's method of separating even and odd numbers and then combining them might not always result in a valid permutation.
3. **n = 6**: This is a slightly larger value that still fits within the small-scale constraint. It provides a good test case to see if the code can handle a permutation of this size correctly.

These test cases are small enough to be complete and valid, and they are designed to challenge the faulty logic 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: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
///There is none worthy of worship except ALLAH and Muhammad(SWT) is the messenger of ALLAH.

/**
*
*   author :: 101rror
*
**/

#include<bits/stdc++.h>

using namespace std;

#define  FastRead  ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define  Executed  return 0;
#define  nl        cout << "\n";

#define  srt(s)    sort(s.begin(),s.end());
#define  rsrt(s)   sort(s.rbegin(),s.rend());
#define  rvs(s)    reverse(s.begin(), s.end());

#define  long      long long
#define  fi        first
#define  se        second
#define  pb        push_back

const int N = 2000001;
const int MOD = 1000000007;
long arr[N];
long brr[N];

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

    if (n == 2 || n == 3)
    {
        cout << -1;
        nl
        return;
    }

    vector<int>a;

    for (int i = 1; i <= n; i += 2)
    {
        a.pb(i);
    }

    for (int i = n; i >= 1; i -= 2)
    {
        if (i % 2 == 0)
        {
            a.pb(i);
        }
    }

    for (auto it : a)
    {
        cout << it << " ";
    }
    nl
}

int32_t main()
{
    FastRead

    int efforts;
    cin >> efforts;

    while (efforts--)
    {
        passion();
    }

    Executed
}

**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 where the code might fail. Let's analyze the code and the problem constraints:

1. **Edge Cases:**
   - The code handles `n = 2` and `n = 3` by outputting `-1`, which is correct.
   - For other values of `n`, the code constructs a permutation by first adding all odd numbers and then all even numbers in reverse order.

2. **Potential Bugs:**
   - The code does not check if the constructed permutation actually satisfies the condition that the sum of adjacent elements is composite.
   - The permutation constructed by the code might not always be valid for all `n`.

3. **Challenging Test Case:**
   - We need a small-scale test case where the constructed permutation fails the condition.
   - Let's consider `n = 5`. The code will construct the permutation as `[1, 3, 5, 4, 2]`.
   - Check the sums of adjacent elements:
     - `1 + 3 = 4` (composite)
     - `3 + 5 = 8` (composite)
     - `5 + 4 = 9` (composite)
     - `4 + 2 = 6` (composite)
   - This permutation seems to work, but let's try another small `n` where the code might fail.

4. **Another Test Case:**
   - Consider `n = 6`. The code will construct the permutation as `[1, 3, 5, 6, 4, 2]`.
   - Check the sums of adjacent elements:
     - `1 + 3 = 4` (composite)
     - `3 + 5 = 8` (composite)
     - `5 + 6 = 11` (prime)
     - `6 + 4 = 10` (composite)
     - `4 + 2 = 6` (composite)
   - The sum `5 + 6 = 11` is prime, which violates the condition.

Therefore, a small-scale test input that exposes the bug in the provided code is:

```plaintext
1
6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
def sol(n):
    if n < 4:
        return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return True
    return False
def sol2(n):
    if n < 4:
        return [-1]
    li1 = []
    if n % 2 == 0:
        li1 = list(range(2, n + 1, 2)) + list(range(1, n + 1, 2))
    else:
        li1 = list(range(2, n, 2)) + [n] + list(range(1, n, 2))
    for i in range(n - 1):
        if not sol(li1[i] + li1[i + 1]):
            return [-1]
    return li1
for _ in range(int(input())):
    n = int(input())
    re1 = sol2(n)
    if re1 == [-1]:
        print(-1)
    else:
        print(" ".join(map(str, re1)))

**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 maximize the failure probability. Here are the steps to consider:

1. **Small Values of \( n \)**:
   - The code has a special check for \( n < 4 \), returning \(-1\) directly. We should include \( n = 3 \) to ensure this check is working correctly.

2. **Even Values of \( n \)**:
   - The code constructs the permutation differently for even and odd \( n \). We should include an even \( n \) to test the even case, such as \( n = 4 \) or \( n = 8 \).

3. **Odd Values of \( n \)**:
   - We should also include an odd \( n \) to test the odd case, such as \( n = 5 \) or \( n = 9 \).

4. **Larger Values of \( n \)**:
   - To test the algorithm's performance and correctness with larger inputs, we can include \( n = 10 \) or \( n = 20 \).

5. **Composite and Prime Sums**:
   - The code checks if the sum of adjacent elements is composite. We should include a value of \( n \) that is more likely to produce a permutation where the sums are composite, such as \( n = 10 \).

Based on these considerations, a small-scale test input that covers these cases is:

```plaintext
5
3
4
5
8
10
```

This input includes:
- \( n = 3 \) to test the edge case where \( n < 4 \).
- \( n = 4 \) and \( n = 8 \) to test even values.
- \( n = 5 \) and \( n = 10 \) to test odd values and larger inputs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
def isComp(x):
    if x < 4:
        return False
    for i in range(2, int(x**0.5) + 1):
        if x % i == 0:
            return True
    return False

def solve(n):
    perm = [1]
    available = set(range(2, n + 1))
    while available:
        found = False
        for num in list(available):
            if isComp(perm[-1] + num):
                perm.append(num)
                available.remove(num)
                found = True
                break
        if not found:
            return -1
    return perm

t = int(input())
test_cases = [int(input()) for _ in range(t)]
results = []
for n in test_cases:
    if n == 2:
        results.append("-1")
    else:
        result = solve(n)
        if result == -1:
            results.append("-1")
        else:
            results.append(" ".join(map(str, result)))
print("\n".join(results))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int n = sc.nextInt();

            if (n % 2 != 0) {

                result.append("-1\n");
            } else {

                ArrayList<Integer> p = new ArrayList<>();

                for (int i = 1; i <= n; i++) {
                    for (int j = n; j >= 1; j--) {
                        if (!isPrime(i + j)) {
                            p.add(i);
                            p.add(j);
                            n--;
                            break;}
                    }
                }

                for (int num : p) {
                    result.append(num).append(" ");
                }
                result.append("\n");
            }
        }

        System.out.print(result);
    }
    public static boolean isPrime(int n) {
        if (n <= 1) {
            return false;
        }
        if (n == 2) {
            return true;
        }

        if (n % 2 == 0) {
            return false;
        }
        for (int i = 3; i * i <= n; i += 2) {
            if (n % i == 0) {
                return false;}
        }
        return true;
    }
}

**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 the following points:

1. **Odd Length Permutations**: The code incorrectly assumes that all odd lengths should output `-1`. However, there are valid permutations for some odd lengths, and the code should handle these cases correctly.

2. **Even Length Permutations**: The code's logic for constructing the permutation is flawed. It tries to pair elements in a way that their sum is composite, but it does not ensure that all elements are used exactly once and that the permutation is valid.

3. **Edge Cases**: We should test small values of `n` to ensure the code handles them correctly. Additionally, we should test larger values of `n` to ensure the code does not run into performance issues.

Given these points, let's construct a test input that includes both odd and even lengths, and small and large values of `n`:

```plaintext
4
3
4
5
10
```

### Explanation:

1. **Test Case 1: \( n = 3 \)**
   - The code incorrectly outputs `-1` because it assumes all odd lengths are invalid. However, there is a valid permutation for \( n = 3 \): `[1, 3, 2]`.

2. **Test Case 2: \( n = 4 \)**
   - The code should be able to construct a valid permutation, such as `[1, 4, 3, 2]`.

3. **Test Case 3: \( n = 5 \)**
   - The code incorrectly outputs `-1` because it assumes all odd lengths are invalid. However, there is a valid permutation for \( n = 5 \): `[1, 4, 3, 2, 5]`.

4. **Test Case 4: \( n = 10 \)**
   - The code should be able to construct a valid permutation, such as `[1, 10, 7, 3, 6, 2, 4, 5, 8, 9]`.

This test input will help expose the bugs in the code by testing both odd and even lengths and ensuring that the code handles valid permutations correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

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 a=0,b=0;
		    List<Integer> even=new ArrayList<>();
		    List<Integer> odd=new ArrayList<>();
		    if(n<=3){
		    System.out.println(-1);
		    continue;
		    }
		    for(int i=1;i<=n;i++){
		       if(i%2==0)
		       even.add(i);
		       else
		       odd.add(i);
		    }
		    List<Integer> l=new ArrayList<>();
		    l.addAll(even);
		    l.addAll(odd);
		    for(int num:l)
		    System.out.print(num+" ");
		    System.out.println();

		}
	}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the constraints and the logic of the algorithm. The code is supposed to construct a permutation such that the sum of any two consecutive elements is composite. The buggy code fails to ensure this condition for all cases.

### Analysis of the Buggy Code:
1. **Edge Cases:**
   - The code correctly handles the case where `n <= 3` by printing `-1`.
   - For `n > 3`, it separates even and odd numbers and concatenates them. However, this does not guarantee that the sum of consecutive elements will be composite.

2. **Logical Error:**
   - The concatenation of even and odd numbers does not consider the composite condition. For example, if `n = 4`, the output is `2 4 1 3`, and `2 + 4 = 6` (composite), but `4 + 1 = 5` (prime).

### Challenging Test Input:
To expose the bug, we need a test case where the concatenation of even and odd numbers results in a sum that is prime. A small-scale test input that can achieve this is `n = 4`.

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

### Explanation:
- For `n = 4`, the correct permutation should ensure that the sum of any two consecutive elements is composite.
- The buggy code will output `2 4 1 3`.
- The sums are:
  - `2 + 4 = 6` (composite)
  - `4 + 1 = 5` (prime)
  - `1 + 3 = 4` (composite)

Since the sum `4 + 1 = 5` is prime, this test input will expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

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

using namespace std;

#define PI (3.141592653589)
#define MOD 1000000007
#define pb push_back
#define f first
#define s second
#define rep(i,j) for(int i = 0; i<j; i++)
#define rrep(i,j) for(int i = j; i>=0; i--)
#define all(x) x.begin(), x.end()
#define out(x) cout << x << endl;
#define Yes cout<<"YES"<<endl;
#define No cout<<"NO"<<endl;
#define pm cout<<"-1"<<endl;

//Typedef
typedef long long ll;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef map<int,int> mii;
typedef set<int,int> sii;

// Debug Overloads
#ifdef hi
#define debug(x) _print(x); cerr << endl;
#else
#define debug(x)
#endif

void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(double t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.f); cerr << ","; _print(p.s); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

// Operator Overloads
template<typename T> // cin >> vector<T>
istream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}
template<typename T> // cout << vector<T>
ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }

int ceil2(int a, int b){
return (a + b - 1) / b; }
ll lcm2(ll a, ll b){
return (a * b) / __gcd(a, b);}

 bool remove(char ch){
if((ch>='a' && ch<='z')||(ch>='A' && ch<='Z')||(ch>='0' && ch<='9')){
  return true;
}
   return false;
 }
char conversion(char ch){
if((ch>='a' && ch<='z') || (ch>='0' && ch<='9')){
 return ch; }
else{
  char temp=ch-'A'+'a';
return temp;
 }
}
 bool checkpalindrome(string a){ int i =0; int j =a.size()-1; while(i<=j){   if(a[i]==a[j]){    i++;    j--;}else{    return false;} }return true;}
bool checkPrime(ll n) {
if (n<=1) {
return false;
}
 for (ll i = 2; i*i<=n; i++) {
if(n % i == 0) {
  return false;
}
   }
return true;
}
ll countdivisors(ll n){
ll cnt=0;
for(ll i=1;i*i<=n;i++)
{
if(n%i==0){
cnt++;
if(n/i!=i){
cnt++;
}
 }
}
return cnt;
}
const int N=1e5;

vector<ll>primes;

vector<bool>isprime;

void sieve(ll n){
isprime.assign(n+1,true);
isprime[0]=isprime[1]=false;
for(int i =2;i*i<=n;i++)
 if(isprime[i])
 for(int j =i*i;j<=n;j+=i)
isprime[j]=false;
for(int i=2;i<=n;i++)
   if(isprime[i])
primes.pb(i);
};
int spf[N];

ll expo(ll a,ll b,ll M){ ll res=1; while(b>0){ if(b&1)res*=a;  res%=M; a=(a*a)%M; b=b>>1; } return res;}
void solve(){

 int n;
 cin >> n;
 if(n<=5){
    cout<<-1<<endl;
    return;
 }
 for(int i=1;i<=n/2;i++){
    if(i!=3){
    cout<<(2*i-1)<<" ";
 }
 }
 cout<<"5"<<" "<<4<<" ";
 for(int i=1;i<=n/2;i++){
    if(i!=2){
    cout<<2*i<<" ";
 }
 }
 cout<<endl;

}

int main()
{
sieve(N);
    int t;
    //solve();
    // t = 1;
    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
2
4
6
```

### Explanation:
1. **Test Case 1 (n = 4):**
   - This is a small and simple case to check if the algorithm can handle even lengths.
   - The correct permutation should be such that all adjacent sums are composite. For example, a valid permutation could be `[1, 4, 3, 2]` where the sums are `1+4=5` (prime), `4+3=7` (prime), and `3+2=5` (prime). Since it's impossible to construct a valid permutation for `n=4`, the expected output should be `-1`.

2. **Test Case 2 (n = 6):**
   - This is another small case to check if the algorithm can handle slightly larger even lengths.
   - A valid permutation for `n=6` could be `[1, 6, 5, 4, 3, 2]` where the sums are `1+6=7` (prime), `6+5=11` (prime), `5+4=9` (composite), `4+3=7` (prime), and `3+2=5` (prime). Since it's impossible to construct a valid permutation for `n=6` with all composite sums, the expected output should be `-1`.

These test cases are designed to expose the bug in the buggy code, which incorrectly handles small even lengths and may output invalid permutations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

using namespace std;
using namespace chrono;

#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimization ("Ofast")

#define fio ios_base::sync_with_stdio(false), cin.tie(NULL),cout.tie(NULL)
#define PI 3.141592653589793238462
#define MOD 1000000007
#define mod 998244353
#define INF 1e18
#define nl '\n'
#define pb push_back
#define ppb pop_back
#define f first
#define s second

#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define py cout << "YES" << nl
#define pn cout << "NO" << nl
#define loop(i,a,b) for (ll i = a; i <= b; i++)
#define rloop(i,a,b) for (int i = a; i >= b; i--)
#define tc(t) int t; cin >> t; while (t--)
#define prec(n) fixed<<setprecision(n)
#define ini(a, i) memset(a, i, sizeof(a))
#define lli long long int
#define us unordered_set
#define um unordered_map
#define ll long long
#define ull unsigned long long int int
#define maxpq priority_queue<int>
#define pll pair<ll, ll>
#define minpq priority_queue<int, vector<int>, greater<int> >

typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> os_less;
#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x<<" "; _print(x); cerr << endl;
    #else
#define debug(x);
#endif

void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(double t) {cerr << t;}
// void _print(ll t) {cerr << t;}
void google(int t) {cout << "Case #" << t << ": ";}
template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.f); cerr << ","; _print(p.s); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

vector<vector<int>> pre(int N){
    vector<vector<int>> divs(N);
    loop(i, 1, N-1){
        for(int j=i;j<N;j+=i)
            divs[j].pb(i);
    }
    return divs;
}

vector<ll> sieve(ll n) {ll*arr = new ll[n + 1](); vector<ll> vect; for (ll i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (ll j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}
ll gcd(ll a, ll b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}
ll add(ll x, ll y, ll m) {ll res=x+y; return res>=m ? res-m:res;}
ll mul(ll x, ll y, ll m) {ll res=x*y; return res>=m? res%m:res;}
ll sub(ll x, ll y, ll m) {ll res=x-y; return res<0? res+m:res;}
ll power(ll x, ll y, ll m) {ll res=1; x%=m; while(y){ if (y&1) res=mul(res, x, m); y >>=1; x=mul(x, x, m);} return res;}

ll mod_inv(ll a,ll m=MOD){
    return power(a,MOD-2,MOD);
}

bool checkAp(lli a,lli b,lli c){
    return ((b-a)==(c-b));
}

bool isSorted(lli arr[],lli n){
    loop(i,0,n-2){
        if(arr[i]>arr[i+1]){
            return false;
        }
    }
    return true;
}
bool isPrime(lli n){
    if(n==1){
        return false;
    }
    for(lli i=2;i*i<=n;i++){
        if(n%i==0){
            return false;
        }
    }
    return true;
}
bool isSafe(int i,int j,int n,int m){
    return i>=0 && j>=0 && i<n && j<m;
}
bool isPerfect(lli n){
    lli ans=(lli)(sqrt(n));
    return ans*ans==n;
}

void print(vector<ll> &v){
    for(auto x:v){
        cout<<x<<" ";
    }cout<<nl;
}

ll sumDig(ll n){
    ll ans=0;
    while(n)
    {
        ans+=n%10;
        n=n/10;
    }
    return ans;
}
ll fact(ll n){
    ll ans=1;
    loop(i,1,n){
        ans=ans*i;
    }
    return ans;
}

inline ll log2(ll x)
{
    ll ret = 0;
    while(x)
        ret++,x>>=1;

    return ret;
}

ll calc(vector<ll> v,ll m){
    ll ans=0;
    for(auto x:v){
        ans+=abs(m-x);
    }
    return ans;
}

struct Bracket{
    ll o,c,full;
    Bracket(){
        o=0;
        c=0;
        full=0;
    }
    Bracket(ll o1,ll c1,ll full1){
        o=o1;
        c=c1;
        full=full1;
    }

};
Bracket merge(Bracket s1,Bracket s2){
    Bracket ans=Bracket(0,0,0);
    ans.full=s1.full+s2.full+min(s1.o,s2.c);
    ans.o=s1.o+s2.o-min(s1.o,s2.c);
    ans.c=s1.c+s2.c-min(s1.o,s2.c);

    return ans;
}
struct SegmentTree{
    vector<Bracket>segTree;
    SegmentTree(int x){
        segTree.resize(4*x+1);
    }

    void build(int low,int high,int ind,string &x,int n){
        if(low==high){
            if(x[low]=='(')
            {
                segTree[ind].o=1;

            }
            else if(x[low]==')'){
                segTree[ind].c=1;
            }
            segTree[ind].full=0;
            return;
        }
        int mid=(low+high)/2;
        build(low,mid,2*ind+1,x,n);
        build(mid+1,high,2*ind+2,x,n);
        segTree[ind]=merge(segTree[2*ind+1],segTree[2*ind+2]);

    }
    Bracket query(int l,int r,int low,int high,int ind){
        if(low>=l && high<=r){
            return segTree[ind];
        }
        if((low>r && high>r) || (low<l && high<l)){

            return Bracket();
        }
        int mid=(low+high)/2;
        Bracket left=query(l,r,low,mid,2*ind+1);
        Bracket right=query(l,r,mid+1,high,2*ind+2);
        return merge(left,right);
    }
    // void update(int i,int x,int L,int R,int ind,int level,int n){
    //     if((L>i && R>i) || (L<i && R<i)){
    //         return;
    //     }
    //     if(L==R && L==i){
    //         segTree[ind]=x;
    //         return;
    //     }

    //     int mid=(L+R)/2;
    //     update(i,x,L,mid,2*ind+1,level+1,n);
    //     update(i,x,mid+1,R,2*ind+2,level+1,n);
    //     if((n%2==0 && level%2==1) || (n%2==1 && level%2==0)){
    //         segTree[ind]=(segTree[2*ind+1] | segTree[2*ind+2]);
    //     }
    //     else{
    //         segTree[ind]=(segTree[2*ind+1] ^ segTree[2*ind+2]);
    //     }
    //     // segTree[ind]=segTree[2*ind+1]+segTree[2*ind+2];
    // }
};
// ll dp[200001][21];
// void binary_lifting(ll node,ll par,vector<ll> adj[]){
//     dp[node][0]=par;
//     // 1<<0 means 1 anccestor of the node is parent of the node itself
//     // so simply storing it
//     // U can store 1<<i value u have tos store using bits and pieces
//     loop(i,1,20){
//         if(dp[node][i- 1]!=-1){
//             // for 1<<i ancestor we know the 1<<i -1 the ancestor then for this we have to find
//             // its i-1 ancestor because (1<<i-1) + 1<<(i-1) 2*(1<<i-1) which simply equals to 1<<i
//             dp[node][i]=dp[dp[node][i-1]][i-1];
//         }
//         else{
//             dp[node][i]=-1;
//         }
//     }
//     for(auto c : adj[node]){
//         if(c!=par){
//             binary_lifting(c,node,adj);
//         }
//     }
// }

ll sum(string &temp)
{
    int ans = 0;
    for (auto &i : temp)
    {
        ans += i - '0';
    }
    return ans;
}

struct Disjoint{
    vector<ll>par,rank;
    Disjoint(ll n){
        par.resize(n+1);
        rank.resize(n+1,0);
        loop(i,0,n){
            par[i]=i;
        }
    }
    ll Par(ll a){
        if(par[a]==a){
            return a;
        }
        return par[a]=Par(par[a]);
    }
    void Union(ll a,ll b){
        a=Par(a);
        b=Par(b);
        if(a==b){
            return;

        }
        if(rank[a]>rank[b]){
            par[b]=a;
            rank[a]++;
        }
        else{
            par[a]=b;
            rank[b]++;
        }
    }
};

// map<char,ll>mp;
// static bool cmp(char a,char b){
//     return mp[a]>mp[b];
// }
ll dfs(ll no,vector<ll> &v,set<ll> &st,vector<ll> &adj,string &x,vector<ll> &par,ll num){
   if(st.find(no)!=st.end()){
    return 0;
   }
   debug(no);
   debug(adj[no-1]);
   st.insert(no);
   par[v[no-1]]=num;
   return adj[no]=(x[no-1]=='0')+dfs(v[no-1],v,st,adj,x,par,num);

}
struct Node{
    Node* links[26];

};
class Trie
{
 Node* root;
public:
    Trie(){
        root=new Node();
    }
    void ins(string x){
        Node* node=root;
        debug(1);
        for(auto c:x){

            if(node->links[c-'a']==NULL){

                node->links[c-'a']=new Node();
                node=node->links[c-'a'];
            }

        }
    }
    bool pre(string x){
        Node* node=root;
        for(auto c:x){
            if(node->links[c-'a']!=NULL){
                node=node->links[c-'a'];
            }
            else{
                return false;
            }
        }
        return true;
    }
};
ll dp[2001][2001];

ll f(ll i,ll l,ll k,ll n){

    if(l==k){
        return dp[i][l]=1;
    }
    if(dp[i][l]!=-1)
    {
        return dp[i][l];
    }
    ll ans=0;
    for(ll j=1;j*i<=n;j++){
        ans=add(ans,(f(j*i,l+1,k,n)),MOD);
    }
    return dp[i][l]=ans;
}
void solve(){
    //     For Division 1: 1900≤rating
// For Division 2: 1600≤rating≤1900
// For Division 3: 1400≤rating≤1599
//For Division 4: rating≤1399
    ll tt=1;
    // ll t;
    // cin>>t;
    tc(t){
        ll n;
        cin>>n;
        if(n<5){
            cout<<-1<<nl;
            continue;
        }
        vector<ll>o;
        vector<ll>e;
        loop(i,1,n){
            if(i%2){
                o.pb(i);
            }
            else{
                e.pb(i);
            }
        }
        loop(i,0,sz(o)-1){
                    if(o[i]==3){
                        swap(o[i],o[0]);
                    }
                }
                loop(i,0,sz(e)-1){
                    if(e[i]==6){
                        swap(e[i],e[sz(e)-1]);
                    }
                }
                for(auto x:e){
            cout<<x<<" ";
           }
           // reverse(all(e));
           for(auto x:o){
            cout<<x<<" ";
           }

    }

}
int main(){

    fio;
#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    freopen("error.txt", "w", stderr);

#endif
    auto start1 = high_resolution_clock::now();
    solve();
    auto stop1 = high_resolution_clock::now();
    auto duration = duration_cast<microseconds>(stop1 - start1);
    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, let's consider the following:

1. **Edge Cases**:
   - Small values of \( n \) where \( n < 5 \). The code correctly handles \( n < 5 \) by outputting \(-1\).
   - Values of \( n \) that are even and odd, as the permutation construction can differ based on the parity of \( n \).

2. **Common Mistakes**:
   - The code attempts to swap specific elements to avoid creating prime sums. For example, it swaps 3 to the first position in the odd list and 6 to the last position in the even list. We need to test if this logic holds for various values of \( n \).

3. **Test Case Construction**:
   - For \( n = 5 \), the code should output \(-1\) because it is impossible to construct such a permutation.
   - For \( n = 6 \), the code should output a valid permutation. We can test if the output is correct.
   - For \( n = 8 \), the code should also output a valid permutation. We can test if the output is correct and if the swaps are applied correctly.

Let's construct the test input:

```plaintext
3
5
6
8
```

This test input includes:
- \( n = 5 \) to check if the code correctly outputs \(-1\).
- \( n = 6 \) to check if the code can construct a valid permutation.
- \( n = 8 \) to check if the code can construct a valid permutation with the specific swaps applied.

This input is small-scale and complete, ensuring that it can be used to verify the correctness of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    if n == 3:
        print(-1)
    else:
        # Construct permutation in the exact required pattern
        p = []
        for i in range(1, n + 1, 2):  # Add odd numbers first
            p.append(i)
        for i in range(n, 0, -2):  # Then add even numbers in descending order
            p.append(i)
        print(*p)

**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 the following:

1. **Edge Cases**: The code explicitly handles the case `n == 3` by printing `-1`, which is correct. However, we should test other small values of `n` to ensure the permutation construction is correct.
2. **Composite Sums**: The code constructs the permutation by first adding odd numbers and then adding even numbers in descending order. This pattern might not always result in composite sums for all `n`.
3. **Larger Values of `n`**: We should also test larger values of `n` to ensure the pattern holds for larger permutations.

Let's construct a test input that includes:
- A small value of `n` that is not 3 (e.g., 4) to check the basic construction.
- A value of `n` that is a bit larger (e.g., 10) to check the composite sum condition.
- A value of `n` that is large enough to ensure the sum of adjacent elements is composite (e.g., 20).

Here is the test input:

```plaintext
3
4
10
20
```

This input will help us verify:
1. The basic construction for `n = 4`.
2. The composite sum condition for a medium-sized permutation with `n = 10`.
3. The composite sum condition for a larger permutation with `n = 20`.

By testing these cases, we can identify if the buggy code's permutation construction fails to meet the requirement of having composite sums for all adjacent elements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
def solve(t, cases):
    results = []
    for n in cases:
        if n < 5:
            results.append("-1")
        elif n == 5:
            results.append("1 3 5 4 2")
        else:
            odds = [i for i in range(1, n+1, 2)]

            evens = []
            for i in range(2, n+1, 2):
                if i == 8:
                    continue
                evens.append(i)
            evens.append(8)
            results.append(" ".join(map(str, evens + odds)))

    return results

t = int(input())
cases = [int(input()) for _ in range(t)]
results = solve(t, cases)

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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the specific issues in the code. The main issues are:

1. The code handles small cases (n < 5) correctly by returning "-1".
2. The code has a specific handling for n == 5, which is incorrect because the permutation "1 3 5 4 2" does not satisfy the condition (1 + 3 = 4, 3 + 5 = 8, 5 + 4 = 9, 4 + 2 = 6, all of which are composite).
3. For n > 5, the code constructs the permutation by separating odd and even numbers, but it incorrectly places the number 8 at the end of the even list, which can lead to incorrect results.

Given these issues, a small-scale test input that can expose the bug is:

```plaintext
3
5
8
10
```

Explanation:
1. **n = 5**: The code incorrectly returns "1 3 5 4 2", which does not satisfy the condition (1 + 3 = 4 is composite, but 3 + 5 = 8 is not composite).
2. **n = 8**: The code should return a valid permutation where all adjacent sums are composite. The example given in the problem statement is "1 8 7 3 6 2 4 5", which is correct.
3. **n = 10**: The code should return a valid permutation where all adjacent sums are composite. The placement of the number 8 at the end of the even list can cause issues, so this test case will help verify if the code handles larger permutations correctly.

This test input is small enough to be manageable and will effectively expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

Buggy Code:
import java.util.*;
public class Solution {

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

        while(test>0){

            int n=sc.nextInt();
            if(n<=4){
                System.out.println("-1");
            }
            else{
                ArrayList<Integer> arr=new ArrayList<>();

                for(int i=1;i<=n;i++){
                    if(i%2==0 && i!=4)arr.add(i);
                }
                arr.add(4);
                arr.add(5);

                for(int i=1;i<=n;i++){
                    if(i%2!=0&& i!=5){
                        arr.add(i);
                    }
                }

                for(int i=0;i<arr.size();i++){
                    if(i!=arr.size()-1)System.out.print(arr.get(i)+" ");
                    else System.out.print(arr.get(i));
                }
            }

            test--;
        }

    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential edge cases or scenarios that could cause it to fail.

### Analysis of the Buggy Code:
1. **Handling Small `n`**:
   - The code checks if `n` is less than or equal to 4 and prints `-1` if true. This is correct because it is impossible to construct a permutation for `n <= 4` where all adjacent sums are composite.

2. **Permutation Construction**:
   - The code first adds all even numbers (excluding 4) to the list.
   - Then it adds 4 and 5.
   - Finally, it adds all odd numbers (excluding 5) to the list.

3. **Potential Issues**:
   - The code assumes that the sequence of even and odd numbers will always ensure that the sums are composite. However, this assumption might not hold for certain values of `n`.
   - The code does not handle the case where `n` is 5 correctly. It adds 4 and 5 twice, which is incorrect.

### Challenging Test Input:
To expose the bug, we need to test the code with a value of `n` that is just above 4 and see if the permutation is constructed correctly. The value `n = 5` is a good candidate because the code explicitly handles `n <= 4` but not `n = 5`.

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

This test input will check if the code correctly handles the case where `n = 5`. According to the code, it should add 4 and 5 twice, which is incorrect. The expected output should be `-1` because it is impossible to construct a permutation of length 5 where all adjacent sums are composite. However, the code will likely produce an incorrect permutation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Superultra's Favorite Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Superultra, a little red panda, desperately wants primogems. In his dreams, a voice tells him that he must solve the following task to obtain a lifetime supply of primogems. Help Superultra!

Construct a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$ such that $$$p_i + p_{i+1}$$$ is composite$$$^{\text{†}}$$$ over all $$$1 \leq i \leq n - 1$$$. If it's not possible, output $$$-1$$$.

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

Each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of the permutation.

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

Output: For each test case, if it's not possible to construct $$$p$$$, output $$$-1$$$ on a new line. Otherwise, output $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ on a new line.

Examples:
input:
2
3
8
output:
-1
1 8 7 3 6 2 4 5

Note: In the first example, it can be shown that all permutation of size $$$3$$$ contain two adjacent elements whose sum is prime. For example, in the permutation $$$[2,3,1]$$$ the sum $$$2+3=5$$$ is prime.

In the second example, we can verify that the sample output is correct because $$$1+8$$$, $$$8+7$$$, $$$7+3$$$, $$$3+6$$$, $$$6+2$$$, $$$2+4$$$, and $$$4+5$$$ are all composite. There may be other constructions that are correct.

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

public class CFSuperultra_Favorite_Permutation {
    public static void solve(Scanner sc) {
        int n = sc.nextInt();
        if (n <= 4) {
            System.out.println(-1);
            return;
        }
        int a[] = { 1, 3, 5, 4, 2, 6 };
        ArrayList<Integer> list = new ArrayList<>();
        for (int i = 0; i < a.length; i++) {
            list.add(a[i]);
        }
        if (n <= 6) {
            if (n == 5) {
                list.remove(list.size() - 1);
            }
            printArray(list);
            return;
        }
        for (int i = 7; i <= n; i++) {
            int last = list.get(list.size() - 1);
            if (composite((last + i))) {
                list.add(i);
            } else {
                list.add(1, i);
            }

        }
        printArray(list);
    }

    public static void printArray(ArrayList<Integer> list) {
        for (int i : list) {
            System.out.print(i + " ");
        }
        System.out.println();
    }

    public static boolean composite(int n) {
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0)
                return true;
        }
        return false;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            solve(sc);
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or specific scenarios where the code might fail. Here are the steps to think through this:

1. **Identify the Bug**:
   - The code uses a predefined array `a` for small values of `n` (up to 6).
   - For larger values of `n`, it tries to add new elements to the list while ensuring that the sum of adjacent elements is composite.
   - The code might fail if it doesn't correctly handle the transition from the predefined array to the dynamically generated list, or if it incorrectly places new elements.

2. **Edge Cases**:
   - Small values of `n` (2, 3, 4, 5, 6) are already handled by the predefined array.
   - Values of `n` just above 6 (e.g., 7, 8, 9) are critical because they transition from the predefined array to the dynamic generation.
   - Values of `n` that are large enough to test the dynamic generation thoroughly.

3. **Test Input**:
   - We should test `n = 7` because it is the first value where the code transitions from the predefined array to dynamic generation.
   - We should also test `n = 8` to ensure the code can handle a slightly larger value.
   - Testing `n = 9` can help verify if the code can handle an odd number of elements.

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

```plaintext
3
7
8
9
```

This input will test the code's ability to handle the transition from the predefined array to the dynamic generation and ensure it can handle both even and odd values of `n` larger than 6.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

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

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — the number of houses, the number of cables, and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n$ and the sum of $m$ do not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
#include <bits/stdc++.h>
#include <bits/extc++.h>
#define ALL(x) begin(x), end(x)
#define All(x, l, r) &x[l], &x[r] + 1
using namespace std;
void file() {
  freopen("1.in", "r", stdin);
  freopen("1.out", "w", stdout);
}
using ll = long long;
template <typename T> using vec = vector<T> ;

const ll inf = 1e18;

const int kL = 4e5 + 5;
int n, m, p;
array<int, kL> u, v, w, val, ord, ls, rs, cnt;
array<__gnu_pbds::priority_queue<ll, greater<ll>>, kL> pq;

struct DSU {
  array<int, kL> fa;
  void clear(int n) { iota(All(fa, 1, n), 1); }
  int find(int x) {
    return (fa[x] == x) ? x : (fa[x] = find(fa[x]));
  }
  void merge(int x, int y) { fa[find(x)] = find(y); }
}dsu;

void dfs(int x) {
  if(x <= n) return void(pq[x].push(0));
  dfs(ls[x]); dfs(rs[x]);
  cnt[x] = cnt[ls[x]] + cnt[rs[x]];
  ll le = pq[ls[x]].top(), ri = pq[rs[x]].top();
  pq[ls[x]].pop(); pq[rs[x]].pop();
  assert(val[x] >= val[ls[x]]);
  assert(val[x] >= val[rs[x]]);
  pq[x].push(le - (ll)cnt[ls[x]] * (val[x] - val[ls[x]]));
  pq[x].push(ri - (ll)cnt[rs[x]] * (val[x] - val[rs[x]]));
  pq[x].join(pq[ls[x]]);
  pq[x].join(pq[rs[x]]);
}

void solve() {
  cin >> n >> m >> p;
  fill(All(ls, 1, 2 * n), 0);
  fill(All(rs, 1, 2 * n), 0);
  dsu.clear(2 * n);
  fill(All(pq, 1, 2 * n), __gnu_pbds::priority_queue<ll, greater<ll>> ());
  fill(All(cnt, 1, 2 * n), 0);
  for(int i = 1, x; i <= p; i++) { cin >> x; cnt[x]++; }
  for(int i = 1; i <= m; i++)
    cin >> u[i] >> v[i] >> w[i];
  iota(All(ord, 1, m), 1);
  sort(All(ord, 1, m), [&](int x, int y) { return w[x] < w[y]; });
  for(int i = 1, c = n; i <= m; i++) {
    int x = ord[i];
    int fu = dsu.find(u[x]), fv = dsu.find(v[x]);
    if(fu ^ fv) {
      val[++c] = w[x];
      dsu.fa[fu] = dsu.fa[fv] = c;
      ls[c] = fu; rs[c] = fv;
    }
  }
  int root = 2 * n - 1; dfs(root);
  ll cur = (ll)p * val[root];
  for(int i = 0; i < n; i++) {
    if(pq[root].size()) {
      cout << (cur += pq[root].top()) << " ";
      pq[root].pop();
    }else cout << "0 ";
  }
  cout << "\n";
}

int main() {
  // file();
  ios::sync_with_stdio(0); cin.tie(0);
  int T = 1;
  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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. Here are the steps to think through the problem:

1. **Small Scale Input**: Since we need a small-scale input, we should aim for a manageable number of houses and cables, but still complex enough to reveal issues.

2. **Edge Cases**:
   - **Single House Requiring Internet**: This can test the base case where the solution should trivially work.
   - **Multiple Houses Requiring Internet**: Test the general case where the algorithm needs to find the optimal placement of servers.
   - **High Latency Cables**: Include cables with very high latency to see if the algorithm handles large values correctly.
   - **Connected Components**: Ensure the graph is connected, as the problem specifies a connected simple graph.

3. **Complex Graph Structure**: Create a graph with a non-trivial structure to ensure the algorithm can handle more complex scenarios.

Given these considerations, here is a small-scale test input that can be used to expose potential bugs:

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

### Explanation:
- **Test Case 1**:
  - **Number of Houses (n)**: 5
  - **Number of Cables (m)**: 4
  - **Number of Houses Requiring Internet (p)**: 3
  - **Houses Requiring Internet**: 1, 3, 5
  - **Cables**:
    - House 1 to House 2 with latency 10
    - House 2 to House 3 with latency 5
    - House 3 to House 4 with latency 2
    - House 4 to House 5 with latency 1

This input includes:
- A small number of houses and cables to keep the input manageable.
- Multiple houses requiring internet to test the general case.
- Cables with varying latencies to test the algorithm's ability to handle different values.
- A connected graph to ensure the problem constraints are met.

This test input should help in identifying any issues in the buggy code, particularly in how it handles the placement of servers and the calculation of latencies.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

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

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — the number of houses, the number of cables, and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n$ and the sum of $m$ do not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

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

struct dsu {
    vector<int> p;
    vector<int> size;

    dsu(int N) {
        size.assign(N, 1);
        p.resize(N);
        iota(p.begin(), p.end(), 0);
    }

    int boss(int u) {
        if (p[u] == u) return u;
        return p[u] = boss(p[u]);
    }

    void merge(int a, int b) {
        a = boss(a);
        b = boss(b);
        if (a == b) return;
        if (size[a] < size[b]) swap(a, b);
        p[b] = a;
        size[a] += size[b];
    }

    bool same_group(int a, int b) {
        return (boss(a) == boss(b));
    }
};

void solve() {
    int N, M, P; cin >> N >> M >> P;
    vector<bool> wifi(N);
    for (int i = 0; i < P; i ++) {
        int u; cin >> u;
        wifi[u - 1] = 1;
    }

    vector<array<int, 3>> edges(M); // weight, u, v
    for (int i = 0; i < M; i ++) {
        int u, v, w; cin >> u >> v >> w;
        u --; v --;
        edges[i] = {w, u, v};
    }

    sort(edges.begin(), edges.end());
    dsu d(N);

    vector<multiset<ll>> dp(N);
    vector<ll> k_one(N);
    for (int i = 0; i < N; i ++) {
        if (wifi[i]) dp[i].insert(0);
        k_one[i] = 0;
    }
    for (auto [w, u, v] : edges) {
        if (d.same_group(u, v)) continue;

        u = d.boss(u);
        v = d.boss(v);

        ll k_zero_u = w * (ll)(dp[u].size());
        ll k_zero_v = w * (ll)(dp[v].size());

        ll new_k_one = min(k_one[v] + k_zero_u, k_one[u] + k_zero_v);

        dp[u].insert(k_one[u] - k_zero_u);
        dp[v].insert(k_one[v] - k_zero_v);

        // cout << "------" << u << ", " << v << "------" << '\n';
        // cout << "k_zero_u, k_zero_v : " << k_zero_u << " " << k_zero_v << '\n';
        // cout << "new_k_one : " << new_k_one << '\n';
        // cout << "dp u : ";
        // for (auto x : dp[u]) cout << x << " ";
        // cout << '\n';
        // cout << "dp v : ";
        // for (auto x : dp[v]) cout << x << " ";
        // cout << '\n';

        multiset<ll> new_dp;
        if (dp[u].size() < dp[v].size()) swap(u, v);
        swap(dp[u], new_dp);
        for (auto x : dp[v]) new_dp.insert(x);
        new_dp.erase(new_dp.begin());
        new_dp.erase(prev(new_dp.end())); // idk why ma si

        // cout << "new_dp : ";
        // for (auto x : new_dp) cout << x << " ";
        // cout << '\n';

        // cout << "primo elem : " << new_k_one << '\n';

        d.merge(u, v);
        int b = d.boss(v);
        swap(dp[b], new_dp);
        k_one[b] = new_k_one;
    }

    multiset<ll> res = dp[d.boss(0)];
    ll curr = k_one[d.boss(0)];
    cout << curr << " ";
    for (auto x : res) {
        curr += x;
        cout << curr << " ";
    }
    for (int i = P + 1; i < N; i ++) cout << 0 << " ";
    cout << '\n';

}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int T; 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
4 3 2
1 4
1 2 10
2 3 20
3 4 30
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

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

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — the number of houses, the number of cables, and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n$ and the sum of $m$ do not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
#include<bits/stdc++.h>
#define F(i,l,r) for(int i=(l),i##end=(r);i<=i##end;++i)
#define G(i,l,r) for(int i=(l),i##end=(r);i>=i##end;--i)
#define pii pair<int,int>
#define x first
#define y second
#define mp(x,y) make_pair(x,y)
#define ep emplace_back
using namespace std;
typedef long long ll;
constexpr ll inf=1e18;
constexpr int N=2e5;
ll _0[N+9];
int fa[N+9];
ll p[N+9];
multiset<ll> f[N+9];
int id=0;
int find(int x){return fa[x]<=0?x:fa[x]=find(fa[x]);}
void link(int x,int y,int w)
{
	if((x=find(x))==(y=find(y)))return ;
	if(fa[x]>fa[y])swap(x,y);
	if(fa[y]==0)
	{
		fa[y]=x;
		return ;
	}
	ll t;
	assert(!f[p[x]].empty());
	assert(!f[p[y]].empty());
	t=*f[p[x]].begin();f[p[x]].erase(f[p[x]].find(t));t-=1ll*w*-fa[x]-_0[p[x]];_0[p[x]]=1ll*w*-fa[x];f[p[x]].insert(t);
	t=*f[p[y]].begin();f[p[y]].erase(f[p[y]].find(t));t-=1ll*w*-fa[y]-_0[p[y]];_0[p[y]]=1ll*w*-fa[y];f[p[y]].insert(t);
	_0[p[x]]+=_0[p[y]];
	for(auto u:f[p[y]]) f[p[x]].insert(u);
	fa[x]+=fa[y];
	assert((int)f[p[x]].size()==-fa[x]);
	fa[y]=x;
}
array<int,3> ed[5009];
int main()
{
	cin.tie(0)->sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--)
	{
		int n,m,s,x;
		cin>>n>>m>>s;
		memset(fa,0,sizeof(int)*(2*n+1));
		F(i,1,s)cin>>x,fa[x]=-1,f[p[x]=++id].insert(0);
		F(i,1,m)cin>>ed[i][1]>>ed[i][2]>>ed[i][0];
		sort(ed+1,ed+m+1);
		F(i,1,m)
		{
			link(ed[i][1],ed[i][2],ed[i][0]);
		}
		x=id;
		ll ans=_0[x];
		for(auto u:f[x])
		{
			ans+=u;
			cout<<ans<<' ';
		}
		F(i,s+1,n) cout<<0<<" \n"[i==n];
		F(i,1,id) f[i].clear(),_0[i]=0;
	}
	return 0;
}

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

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

Think step by step.

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

1. **Small-scale Test Case**: Since the output length is limited, we will create a small-scale test case with a few houses, cables, and houses that need internet.

2. **Edge Cases**:
   - **Single House Requiring Internet**: Test the scenario where there is only one house that needs internet.
   - **Multiple Houses Requiring Internet**: Test the scenario where multiple houses require internet and ensure the algorithm handles it correctly.
   - **Multiple Servers**: Test the scenario where multiple servers are installed to see if the algorithm correctly minimizes the total latency.
   - **High Latency Cables**: Test the scenario with high latency cables to ensure the algorithm can handle such cases.

3. **Connected Graph**: Ensure the graph is connected to satisfy the problem constraints.

4. **Optimal Solution**: The test case should have a clear optimal solution that can be verified easily.

Here is a small-scale test input that covers the above points:

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

### Explanation of the Test Input:
1. **Test Case 1**:
   - **Number of Houses (n)**: 4
   - **Number of Cables (m)**: 3
   - **Number of Houses Requiring Internet (p)**: 2
   - **Houses Requiring Internet**: 1, 3
   - **Cables**:
     - (1, 2, 1)
     - (2, 3, 2)
     - (3, 4, 3)

2. **Test Case 2**:
   - **Number of Houses (n)**: 5
   - **Number of Cables (m)**: 4
   - **Number of Houses Requiring Internet (p)**: 3
   - **Houses Requiring Internet**: 1, 3, 5
   - **Cables**:
     - (1, 2, 1)
     - (2, 3, 2)
     - (3, 4, 3)
     - (4, 5, 4)

### Expected Output:
1. **Test Case 1**:
   - For k=1: 3 (Install server at house 3)
   - For k=2: 0 (Install servers at houses 1 and 3)
   - For k=3: 0 (Install servers at houses 1, 3, and 4)
   - For k=4: 0 (Install servers at houses 1, 2, 3, and 4)

2. **Test Case 2**:
   - For k=1: 9 (Install server at house 3)
   - For k=2: 4 (Install servers at houses 1 and 5)
   - For k=3: 0 (Install servers at houses 1, 3, and 5)
   - For k=4: 0 (Install servers at houses 1, 2, 3, and 5)
   - For k=5: 0 (Install servers at houses 1, 2, 3, 4, and 5)

This test input is designed to expose any issues in the buggy code, particularly in handling multiple houses requiring internet and high latency cables.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Digital Village (Extreme Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the extreme version of the problem. In the three versions, the constraints on $n$ and $m$ are different. You can make hacks only if all the versions of the problem are solved.

Pak Chanek is setting up internet connections for the village of Khuntien. The village can be represented as a connected simple graph with $n$ houses and $m$ internet cables connecting house $u_i$ and house $v_i$, each with a latency of $w_i$.

There are $p$ houses that require internet. Pak Chanek can install servers in at most $k$ of the houses. The houses that need internet will then be connected to one of the servers. However, since each cable has its latency, the latency experienced by house $s_i$ requiring internet will be the maximum latency of the cables between that house and the server it is connected to.

For each $k = 1,2,\ldots,n$, help Pak Chanek determine the minimum total latency that can be achieved for all the houses requiring internet.

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

The first line of each test case contains 3 integers $n$, $m$, $p$ ($2 \le n \le 2 \cdot 10^5$; $n-1 \le m \le 2 \cdot 10^5$; $1 \le p \le n$) — the number of houses, the number of cables, and the number of houses that need internet.

The second line of each test case contains $p$ integers $s_1, s_2, \ldots, s_p$ ($1 \le s_i \le n$) — the houses that need internet. It is guaranteed that all elements of $s$ are distinct.

The $i$-th of the next $m$ lines of each test case contains three integers $u_i$, $v_i$, and $w_i$ ($1 \le u_i < v_i \le n$; $1 \le w_i \le 10^9$) — the internet cable connecting house $u_i$ and house $v_i$ with latency of $w_i$. It is guaranteed that the given edges form a connected simple graph.

It is guaranteed that the sum of $n$ and the sum of $m$ do not exceed $2 \cdot 10^5$.

Output: For each test case, output $n$ integers: the minimum total latency that can be achieved for all the houses requiring internet for each $k = 1,2,\ldots,n$.

Examples:
input:
2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2
output:
34 19 9 4 0 0 0 0 0
2 0 0

Note: In the first test case for $k=3$, a possible optimal solution is to install servers at vertices $2$, $6$ and $8$ and obtain the following latency:

  * $\text{latency}(2) = 0$    * $\text{latency}(5) = \max(3, 5) = 5$    * $\text{latency}(6) = 0$    * $\text{latency}(8) = 0$    * $\text{latency}(9) = \max(2, 4) = 4$

So the total latency is $9$.

Buggy Code:
import heapq
from collections import defaultdict
from typing import List, Tuple

def dijkstra(graph: dict, start: int) -> List[int]:
    distances = [float('inf')] * (len(graph) + 1)
    distances[start] = 0
    pq = [(0, start)]

    while pq:
        current_dist, current_node = heapq.heappop(pq)

        if current_dist > distances[current_node]:
            continue

        for neighbor, weight in graph[current_node]:
            distance = max(current_dist, weight)
            if distance < distances[neighbor]:
                distances[neighbor] = distance
                heapq.heappush(pq, (distance, neighbor))

    return distances

def solve_test_case(n: int, m: int, p: int, houses_needing_internet: List[int], edges: List[Tuple[int, int, int]]) -> List[int]:
    graph = defaultdict(list)
    for u, v, w in edges:
        graph[u].append((v, w))
        graph[v].append((u, w))

    all_distances = []
    for house in houses_needing_internet:
        all_distances.append(dijkstra(graph, house))

    result = []

    for k in range(1, n + 1):
        if k >= p:
            result.append(0)
            continue

        best_total_latency = float('inf')

        for servers in itertools.combinations(range(1, n + 1), k):
            total_latency = 0
            for i, house in enumerate(houses_needing_internet):
                latency = min(all_distances[i][server] for server in servers)
                total_latency += latency

            best_total_latency = min(best_total_latency, total_latency)

        result.append(best_total_latency)

    return result

def process_input(input_data: str):
    lines = input_data.strip().split('\n')
    t = int(lines[0])
    current_line = 1

    for _ in range(t):
        n, m, p = map(int, lines[current_line].split())
        current_line += 1

        houses_needing_internet = list(map(int, lines[current_line].split()))
        current_line += 1

        edges = []
        for _ in range(m):
            u, v, w = map(int, lines[current_line].split())
            edges.append((u, v, w))
            current_line += 1

        result = solve_test_case(n, m, p, houses_needing_internet, edges)
        print(*result)

# Input data
input_data = """2
9 8 5
2 5 6 8 9
1 2 1
1 3 2
3 4 10
4 5 3
4 6 5
1 7 10
7 8 4
7 9 2
3 3 2
3 1
1 2 1
2 3 3
1 3 2"""

# Process the input
if __name__ == "__main__":
    import itertools
    process_input(input_data)

**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 are likely to cause the code to fail. Here are the steps to consider:

1. **Edge Cases**:
   - Small graphs with few nodes and edges.
   - Graphs with a large number of nodes and edges.
   - Graphs where the number of houses needing internet is very small or very large.
   - Graphs where the number of servers is limited to a very small number.

2. **Specific Scenarios**:
   - Graphs with high latency edges.
   - Graphs with multiple paths between nodes.
   - Graphs where the optimal solution is not immediately obvious.

3. **Combining Edge Cases and Scenarios**:
   - Create a graph with a few nodes and edges but high latency to ensure the dijkstra function is tested properly.
   - Create a graph where the optimal placement of servers is not straightforward to ensure the combination and latency calculation functions are tested.

Given the constraints, a small-scale test input that is complete and valid could be:

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

This test input includes two test cases:
1. A small graph with 4 nodes, 5 edges, and 3 houses needing internet.
2. A slightly larger graph with 5 nodes, 6 edges, and 3 houses needing internet.

These cases are designed to test the algorithm's ability to handle small graphs with high latency edges and to ensure that the combination and latency calculation functions are working correctly.