**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
def max_difference(test_cases):
    results = []
    for t in test_cases:
        n, a, b = t
        max_diff = float('-inf')
        for mask in range(1, 1 << n):
            m_total = 0
            s_total = 0
            for i in range(n):
                if mask & (1 << i):
                    m_total += a[i]
                    if i + 1 < n:
                        s_total += b[i + 1]
            max_diff = max(max_diff, m_total - s_total)
        results.append(max_diff)
    return results

T = int(input())
test_cases = []
for _ in range(T):
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    test_cases.append((n, a, b))
results = max_difference(test_cases)
for res in results:
    print(res)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios where the code might fail. The main issue in the buggy code is that it does not correctly handle the case where Monocarp can train on the last day, and Stereocarp cannot follow up with training the next day. Additionally, the code does not consider the optimal strategy for maximizing the difference.

Here are the steps to generate a challenging test input:

1. **Single Day Test Case**: Ensure that the code correctly handles a single day scenario.
2. **Multiple Days Test Case**: Create a scenario where Monocarp can train on multiple days, and Stereocarp will follow the training schedule.
3. **Last Day Training Test Case**: Ensure that the code correctly handles the case where Monocarp trains on the last day, and Stereocarp cannot train the next day.
4. **High Problem Count Test Case**: Create a scenario with high problem counts to maximize the difference.

Let's generate a small-scale test input that covers these scenarios:

```plaintext
4
1
10
5
3
5 10 1
1 10 1
2
10 1
1 10
6
10 1 10 1 10 1
1 10 1 10 1 10
```

### Explanation:
1. **Test Case 1**:
   - `n = 1`
   - `a = [10]`
   - `b = [5]`
   - Monocarp trains on the only day, and Stereocarp cannot train. The difference is `10 - 0 = 10`.

2. **Test Case 2**:
   - `n = 3`
   - `a = [5, 10, 1]`
   - `b = [1, 10, 1]`
   - Monocarp can train on days 1 and 3, and Stereocarp will train on days 2 and 3. The difference is `(5 + 1) - (10 + 1) = 6 - 11 = -5`. However, the optimal strategy is to train on days 1 and 2, resulting in a difference of `(5 + 10) - (10) = 15 - 10 = 5`.

3. **Test Case 3**:
   - `n = 2`
   - `a = [10, 1]`
   - `b = [1, 10]`
   - Monocarp can train on both days, and Stereocarp will train on the second day. The difference is `(10 + 1) - (10) = 11 - 10 = 1`.

4. **Test Case 4**:
   - `n = 6`
   - `a = [10, 1, 10, 1, 10, 1]`
   - `b = [1, 10, 1, 10, 1, 10]`
   - Monocarp can train on days 1, 3, 5, and Stereocarp will train on days 2, 4, 6. The difference is `(10 + 10 + 10) - (10 + 10 + 10) = 30 - 30 = 0`. However, the optimal strategy is to train on days 1, 3, 5, and 6, resulting in a difference of `(10 + 10 + 10 + 1) - (10 + 10 + 10) = 31 - 30 = 1`.

This test input should help expose the bug in the provided code by covering various edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
def solve():
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    max_diff = -float('inf')

    for i in range(2**n):
        monocarp_trains = [False] * n
        for j in range(n):
            if (i >> j) & 1:
                monocarp_trains[j] = True

        m = 0
        for j in range(n):
            if monocarp_trains[j]:
                m += a[j]

        s = 0
        for j in range(n - 1):
            if monocarp_trains[j]:
                s += b[j + 1]

        max_diff = max(max_diff, m - s)

    print(max_diff)

t = int(input())
for _ in range(t):
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
import java.util.Scanner;

public class OlympiadDifference {

    public static int calculateMaxDifference(int n, int[] a, int[] b) {
        int maxDiff = 0;

        for (int i = 0; i < (1 << n); i++) { // Iterate over all subsets of days Monocarp trains
            int m = 0, s = 0;
            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) != 0) { // If Monocarp trains on day j
                    m += a[j];
                    if (j + 1 < n) { // Stereocarp trains on the next day
                        s += b[j + 1];
                    }
                }
            }
            maxDiff = Math.max(maxDiff, m - s); // Maximize the difference
        }

        return maxDiff;
    }

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

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

            int[] a = new int[n];
            int[] b = new int[n];

            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            for (int i = 0; i < n; i++) {
                b[i] = sc.nextInt();
            }

            System.out.println(calculateMaxDifference(n, a, b));
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases where the algorithm might fail. The primary issue in the buggy code is that it does not correctly account for the scenario where Monocarp trains on the last day. If Monocarp trains on the last day, Stereocarp will not have a next day to train, which the code does not handle correctly.

Let's create a test input that focuses on this edge case:

1. **Test Case with a Single Day:**
   - If there is only one day, Monocarp can train on that day, and Stereocarp will not train at all.
   - This is a straightforward case where the difference should be the number of problems Monocarp solves on that day.

2. **Test Case with Multiple Days:**
   - If there are multiple days, we can create a scenario where the optimal solution involves training on the last day to maximize the difference.
   - This will ensure that Stereocarp does not train on the next day, which the buggy code might not handle correctly.

Here is a small-scale test input that includes both scenarios:

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

### Explanation:
1. **First Test Case:**
   - \( n = 1 \)
   - \( a = [5] \)
   - \( b = [8] \)
   - Monocarp trains on the only day, solving 5 problems. Stereocarp does not train at all.
   - Expected output: \( 5 \)

2. **Second Test Case:**
   - \( n = 3 \)
   - \( a = [8, 2, 5] \)
   - \( b = [8, 2, 7] \)
   - Monocarp trains on the last day (day 3), solving 5 problems. Stereocarp does not train on the next day.
   - Expected output: \( 5 \)

This test input should help expose the bug in the buggy code where it incorrectly handles the scenario of training on the last day.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Olympiad
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp and Stereocarp are preparing for the Olympiad. There are $$$n$$$ days left until the Olympiad. On the $$$i$$$-th day, if Monocarp plans to practice, he will solve $$$a_i$$$ problems. Similarly, if Stereocarp plans to practice on the same day, he will solve $$$b_i$$$ problems.

Monocarp can train on any day he wants. However, Stereocarp watches Monocarp and follows a different schedule: if Monocarp trained on day $$$i$$$ and $$$i < n$$$, then Stereocarp will train on day $$$(i+1)$$$.

Monocarp wants to organize his training process in a way that the difference between the number of problems he solves and the number of problems Stereocarp solves is as large as possible. Formally, Monocarp wants to maximize the value of $$$(m-s)$$$, where $$$m$$$ is the number of problems he solves, and $$$s$$$ is the number of problems Stereocarp solves. Help Monocarp determine the maximum possible difference in the number of solved problems between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 100$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 100$$$).

Output: For each test case, print a single integer — the maximum possible difference between the number of problems Monocarp solves and the number of problems Stereocarp solves.

Examples:
input:
4
2
3 2
2 1
1
5
8
3
1 1 1
2 2 2
6
8 2 5 6 2 6
8 2 7 4 3 4
output:
4
5
1
16

Note: Let's analyze the example from the statement:

- In the first test case, it is optimal for Monocarp to train both days; then Stereocarp will train on day $$$2$$$.
- In the second test case, it is optimal for Monocarp to train on the only day, and Stereocarp will not train at all.
- In the third test case, it is optimal for Monocarp to train on the last day (and only on that day).
- In the fourth test case, it is optimal for Monocarp to train on days $$$1, 3, 4, 6$$$; then Stereocarp will train on days $$$2, 4, 5$$$.

Buggy Code:
import java.util.*;

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

            List<Integer> results = new ArrayList<>();

            int t = sc.nextInt();

            while (t-- > 0) {
                int n = sc.nextInt();
                int[] a = new int[n];
                int[] b = new int[n];

                for (int i = 0; i < n; i++) {
                    a[i] = sc.nextInt();
                }
                for (int i = 0; i < n; i++) {
                    b[i] = sc.nextInt();
                }

                // Maximum possible difference
                int maxDifference = Integer.MIN_VALUE;

                // Explore all subsets of Monocarp's training days using bitmasking
                int totalSubsets = 1 << n; // 2^n subsets
                for (int mask = 0; mask < totalSubsets; mask++) {
                int m = 0; // Problems solved by Monocarp
                int s = 0; // Problems solved by Stereocarp

                for (int i = 0; i < n; i++) {
                    if ((mask & (1 << i)) != 0) { // Monocarp trains on day i
                        m += a[i];
                        if (i + 1 < n) { // Stereocarp trains on the next day
                            s += b[i + 1];
                        }
                    }
                }

                // Update the maximum difference
                maxDifference = Math.max(maxDifference, m - s);

                }
            results.add(maxDifference);
            }

            // Print all results for this test case
            for (Integer result : results) {
                System.out.println(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 given code, we need to focus on edge cases or scenarios where the faulty logic in the code might fail. The main issue in the buggy code is that it does not correctly handle the scenario where Monocarp trains on the last day, which would not cause Stereocarp to train on the next day (since there is no next day).

Let's consider the following steps to create a test case:

1. **Small Scale**: The test case should be small to fit within the output limit.
2. **Edge Case**: The test case should focus on the last day scenario.
3. **Maximize Difference**: The test case should maximize the difference in the number of problems solved by Monocarp and Stereocarp.

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

```plaintext
1
3
10 5 10
5 5 5
```

**Explanation:**

- **Test Case 1**:
  - **n = 3**: There are 3 days.
  - **a = [10, 5, 10]**: Monocarp can solve 10 problems on day 1, 5 problems on day 2, and 10 problems on day 3.
  - **b = [5, 5, 5]**: Stereocarp can solve 5 problems on day 1, 5 problems on day 2, and 5 problems on day 3.

**Expected Output:**

- If Monocarp trains on days 1 and 3:
  - Monocarp solves: 10 (day 1) + 10 (day 3) = 20 problems.
  - Stereocarp solves: 5 (day 2) = 5 problems.
  - The difference is: 20 - 5 = 15.

The correct output should be 15, but the buggy code might not handle the last day correctly, leading to a wrong output. This test case should 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: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

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

using ll = signed long long int;

#define all(x) (x).begin(), (x).end()

using pii = std::array<int, 2>;
using pll = std::array<ll, 2>;
using vi = std::vector<int>;
using vl = std::vector<ll>;

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, q;
    cin >> n >> q;
    vi l(q), r(q);
    for (int& el : l) {
        cin >> el;
        --el;
    }
    for (int& el : r) {
        cin >> el;
        --el;
    }
    vector<pii> g(2 * q + 1, {-1, -1});
    vi last(n, -1), num(2 * q + 1, -1);
    last[n - 1] = 0, num[0] = n - 1;
    for (int i = 0, ptr = 0; i < q; ++i) {
        int u = last[max(l[i], r[i])];
        int v1 = ++ptr, v2 = ++ptr;
        g[u] = {v1, v2};
        num[v1] = l[i], num[v2] = r[i];
        last[l[i]] = v1, last[r[i]] = v2;
    }
    vi order;
    auto dfs = [&](auto&& dfs, int u) -> void {
        if (g[u][0] == -1) {
            order.push_back(num[u]);
            return;
        }
        dfs(dfs, g[u][0]);
        dfs(dfs, g[u][1]);
    };
    dfs(dfs, 0);
    int m = order.size();
    vi suff(n), cnt(n);
    for (int el : order) {
        cnt[el] = 1;
    }
    ++suff[order[0]], ++suff[order.back()];
    for (int i = 1; i < m; ++i) {
        ++suff[max(order[i + 1], order[i])];
    }
    const ll MOD = 998244353;
    ll p = n, ans = 1;
    for (int i = 0; i < n; ++i) {
        if (!cnt[i]) {
            ans = (ans * p) % MOD;
            --p;
        }
        p -= suff[i];
    }
    cout << ans << endl;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

Buggy Code:
from math import factorial

MOD = 998244353

def solve(n, q, ls, rs):
    # Initialize counts for positions
    pos_count = [0] * (n + 1)

    # Process each operation in reverse order
    for i in range(q - 1, -1, -1):
        li, ri = ls[i], rs[i]

        # Update positions based on the current max values
        if pos_count[li] == 0:
            pos_count[li] = i + 1
        else:
            pos_count[li] += 1

        if pos_count[ri] == 0:
            pos_count[ri] = i + 2
        else:
            pos_count[ri] += 1

    # Calculate the number of valid permutations
    ans = 1
    fixed_positions = set()
    for i in range(1, n + 1):
        if pos_count[i]:
            fixed_positions.add(pos_count[i])

    free_spots = n - len(fixed_positions)
    for num in range(n, free_spots, -1):
        ans *= num
        ans %= MOD

    return ans

# Example usage
if __name__ == "__main__":
    n, q = map(int, input().split())
    ls = list(map(int, input().split()))
    rs = list(map(int, input().split()))
    print(solve(n, q, ls, rs))

**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 where the logic in the code might fail. The buggy code has issues with how it updates the positions and counts, especially when dealing with overlapping maximum values.

Here are the steps to think through the test case:

1. **Understand the Bug**: The bug in the code is in how it updates the positions and counts for the maximum values. It doesn't correctly handle the overlap and can incorrectly count positions.

2. **Create a Scenario with Overlapping Maximum Values**: We need a scenario where the maximum values in the left and right parts overlap, and the code might incorrectly count the positions.

3. **Small Scale Input**: To keep the input small, we can use a permutation of size \( n \) and a small number of operations \( q \).

4. **Valid Sequence**: Ensure that the input is valid, meaning there exists at least one permutation that can produce the given sequences.

Let's construct a test input:

- **Permutation Size**: \( n = 5 \)
- **Number of Operations**: \( q = 2 \)
- **Sequences**: \( l_1 = 3 \), \( l_2 = 4 \)
- **Sequences**: \( r_1 = 4 \), \( r_2 = 5 \)

This input will test how the code handles overlapping maximum values and whether it correctly counts the positions.

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

This input is small and valid, and it should expose the bug in the code by creating a scenario where the maximum values overlap. The code should correctly handle this scenario, but the buggy code might fail to do so.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

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

using namespace std;
#define int long long
#define inf INT32_MAX
#define PII pair<int,int>
//#define x first
//#define y second
#define endl '\n'
const int mod = 998244353;

void solve() {
    int n, m;
    cin >> n >> m;
    map<int, int> mp;
    vector<int> arr(m + 1), brr(m + 1);
    for (int i = 1; i <= m; i++) {
        cin >> arr[i];
        mp[arr[i]]++;
    }
    for (int i = 1; i <= m; i++) {
        cin >> brr[i];
        mp[brr[i]]++;
    }
    vector<bool> st(n + 1);
    st[n] = true;
    vector<int> nxt(n + 1), pre(n + 1);
    for (int i = 1; i <= m; i++) {
        int l = arr[i], r = brr[i];
        if (st[l]) {
            st[r] = true;
            nxt[r] = nxt[l];
            if (nxt[r])pre[nxt[r]] = r;
            pre[r] = l;
            nxt[l] = r;
        } else {
            st[l] = true;
            pre[l] = pre[r];
            if (pre[l])nxt[pre[l]] = l;
            nxt[l] = r;
            pre[r] = l;
        }
    }
    vector<int> seq;
    int pos = -1;
    for (int i = 1; i <= n; i++) {
        if (!pre[i] && st[i]) {
            pos = i;
        }
    }
    seq.push_back(pos);
    while (nxt[pos]) {
        pos = nxt[pos];
        seq.push_back(pos);
    }
    vector<int> a, b;
    for (int i = 2; i < seq.size(); i++)a.push_back(min(seq[i], seq[i - 1]));
    b = seq;
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    int len = b.size();
    int ans = 1, tmp = 0;
    for (int i = n; i >= 1; i--) {
        if (mp[i])continue;
        int x = lower_bound(a.begin(), a.end(), i) - a.begin();
        int y = lower_bound(b.begin(), b.end(), i) - b.begin();
        ans = (ans * (2 * (len - y) - (m - x) + tmp)) % mod;
        tmp++;
    }
    cout << ans << endl;
}

signed main() {
#ifdef ONLINE_JUDGE
#else
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cout.tie(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:
```plaintext
4 2
3 2
4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;

#define int long long
const int MOD = 998244353;

vector <vector <int>> g;
vector <int> val;
vector <int> last_ind;

void update(int v) {
	last_ind[v] = g.size();
	g.push_back(vector <int>());
	val.push_back(v);
}

void add(int v, int l, int r) {
	g[last_ind[v]].push_back(l);
	g[last_ind[v]].push_back(r);
	update(l);
	update(r);
}

void add(int l, int r) {
	if (last_ind[l] != -1) {
		add(l, l, r);
	} else {
		add(r, l, r);
	}
}

int fast_pow(int a, int x) {
	if (x == 0) return 1;
	if (x & 1) {
		return a * fast_pow(a, x - 1) % MOD;
	} else {
		return fast_pow(a * a % MOD, x >> 1);
	}
}

int div(int x) {
	return fast_pow(x, MOD - 2);
}

signed main() {
	int n, q;
	cin >> n >> q;
	vector <int> l(q), r(q);
	for (int i = 0; i < q; i++) {
		cin >> l[i];
	}
	for (int i = 0; i < q; i++) {
		cin >> r[i];
	}
	last_ind.resize(n + 1, -1);
	update(n);

	for (int i = 0; i < q; i++) {
		add(l[i], r[i]);
	}

	int k = 0;
	for (int i = 0; i < g.size(); i++) {
		k += g[i].empty();
	}

	vector <int> fact(n + 1);
	fact[0] = 1;
	for (int i = 1; i <= n; i++) {
		fact[i] = fact[i - 1] * i % MOD;
	}

	cout << fact[n] * div(fact[k]) % MOD;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

Buggy Code:
MOD = 998244353

class DSU:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [1] * n

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX != rootY:
            if self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
                self.rank[rootX] += self.rank[rootY]
            else:
                self.parent[rootX] = rootY
                self.rank[rootY] += self.rank[rootX]

def factorial_mod(n, MOD):
    fact = [1] * (n + 1)
    for i in range(2, n + 1):
        fact[i] = fact[i - 1] * i % MOD
    return fact

def inv_factorial_mod(fact, n, MOD):
    inv_fact = [1] * (n + 1)
    inv_fact[n] = pow(fact[n], MOD - 2, MOD)
    for i in range(n - 1, 0, -1):
        inv_fact[i] = inv_fact[i + 1] * (i + 1) % MOD
    return inv_fact

# Read input
import sys
input = sys.stdin.read
data = input().split()

n = int(data[0])
q = int(data[1])
l = list(map(int, data[2:2+q]))
r = list(map(int, data[2+q:2+2*q]))

dsu = DSU(n + 1)
for i in range(q):
    dsu.union(l[i], r[i])

group_size = {}
for i in range(1, n + 1):
    root = dsu.find(i)
    if root not in group_size:
        group_size[root] = 0
    group_size[root] += 1

# Calculate the number of valid permutations
fact = factorial_mod(n, MOD)
inv_fact = inv_factorial_mod(fact, n, MOD)

result = fact[n]
for size in group_size.values():
    result = result * inv_fact[size] % MOD

print(result)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

Buggy Code:
MOD = 998244353

def count_valid_permutations(n, q, l, r):
    # Initialize the factorial and inverse factorial arrays
    fact = [1] * (n + 1)
    inv_fact = [1] * (n + 1)

    for i in range(2, n + 1):
        fact[i] = fact[i - 1] * i % MOD

    def mod_inv(x):
        return pow(x, MOD - 2, MOD)

    inv_fact[n] = mod_inv(fact[n])
    for i in range(n - 1, 0, -1):
        inv_fact[i] = inv_fact[i + 1] * (i + 1) % MOD

    # Initialize the segment tree
    seg_tree = [0] * (4 * n)

    def update(idx, val, node=1, start=0, end=n-1):
        if start == end:
            seg_tree[node] = max(seg_tree[node], val)
        else:
            mid = (start + end) // 2
            if idx <= mid:
                update(idx, val, 2 * node, start, mid)
            else:
                update(idx, val, 2 * node + 1, mid + 1, end)
            seg_tree[node] = max(seg_tree[2 * node], seg_tree[2 * node + 1])

    def query(l, r, node=1, start=0, end=n-1):
        if r < start or end < l:
            return 0
        if l <= start and end <= r:
            return seg_tree[node]
        mid = (start + end) // 2
        return max(query(l, r, 2 * node, start, mid), query(l, r, 2 * node + 1, mid + 1, end))

    # Process the operations
    used = [False] * n
    for i in range(q):
        li, ri = l[i] - 1, r[i] - 1
        if not used[li]:
            update(li, ri + 1)
            used[li] = True
        if not used[ri]:
            update(ri, li + 1)
            used[ri] = True

    # Count the valid permutations
    result = fact[n]
    for i in range(n):
        if used[i]:
            max_val = query(i, i)
            if max_val > i:
                result = result * inv_fact[max_val - i] % MOD

    return result

# Example usage
n, q = map(int, input().split())
l = list(map(int, input().split()))
r = list(map(int, input().split()))

print(count_valid_permutations(n, q, l, r))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

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

public class Main {
    public static int MDC = 998244353;

    public static void main(String[] args) throws IOException {
        sc = new MScanner(System.in);
        pw = new PrintWriter(System.out);
        //int t = sc.nextInt();
        int t = 1;
        for (int i = 0; i < t; i++) {
            solver();
        }
    }

    static long[] bitArr;

    public static void solver() throws IOException {
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] l=new int[m];
        for(int i=0;i<m;i++){
            l[i]=sc.nextInt();
        }
        int[] r=new int[m];
        for(int i=0;i<m;i++){
            r[i]=sc.nextInt();
        }
        int[] tr=new int[n+1];  // 从左到右
        int[] tl=new int[n+1];  // 从右到左
        for (int i=0;i<m;i++){
            if(tr[l[i]]!=0){
                int u=tr[l[i]];
                tr[l[i]]=r[i];
                tl[r[i]]=l[i];
                // 第三点
                tl[u]=r[i];
                tr[r[i]]=u;
            }else if(tl[r[i]]!=0){
                 int u=tl[r[i]];
                 tl[r[i]]=l[i];
                 tr[l[i]]=r[i];
                 // 第三点
                tl[l[i]]=u;
                tr[u]=l[i];
            }else {
                tl[r[i]]=l[i];
                tr[l[i]]=r[i];
            }
        }
        int p=l[0];
        while (tl[p]!=0) p=tl[p];
        ArrayList<Integer> list=new ArrayList<>();
        HashSet<Integer> set=new HashSet<>();
        while (p!=0) {
            list.add(p);
            set.add(p);
            p=tr[p];
        }
        ArrayList<Integer> list2=new ArrayList<>();
        for (int i=0;i<list.size();i++){
            if(i==0) list2.add(list.get(i));
            else {
                list2.add(Math.max(list.get(i), list.get(i-1)));
            }
        }
        list2.add(list.get(list.size()-1));
        list2.sort(Integer::compare);
        long ans=1;
        int nn=0;
        for (int i=n;i>0;i--){
            if(set.contains(i))continue;
            int j =list2.size()- Math.max(Collections.binarySearch(list2, i),0);
            ans*=(j+nn);
            nn+=1;
        }
        pw.println(ans);
        pw.flush();
    }
    public static void bitInsert(int index, long v, int n) {
        while (index <= n) {
            bitArr[index] += v;
            index += (index & (-index));
        }
    }

    public static long bitQuery(int index, int n) {
        long ans = 0;
        while (index > 0) {
            ans += bitArr[index];
            index -= -index & index;
        }
        return ans;
    }

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

    private static long pow(long n, long m) {
        long res = 1;
        long base = n;
        while (m != 0) {
            if ((m & 1) == 1) {
                res = res * base;
                res %= MDC;
            }
            base = base * base;
            base %= MDC;
            m = m >> 1;
        }
        return res;
    }

    static int curt;
    static PrintWriter pw;
    static MScanner sc;

    static class MScanner {
        StringTokenizer st;
        BufferedReader br;

        public MScanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }

        public MScanner(String file) throws Exception {
            br = new BufferedReader(new FileReader(file));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

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

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

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

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

        public boolean ready() throws IOException {
            return br.ready();
        }

        public void waitForInput() throws InterruptedException {
            Thread.sleep(3000);
        }

    }

    static void dbg(int[] in) {
        System.out.println(Arrays.toString(in));
    }

    static void dbg(long[] in) {
        System.out.println(Arrays.toString(in));
    }

    static void sort(int[] in) {
        shuffle(in);
        Arrays.sort(in);
    }

    static void sort(long[] in) {
        shuffle(in);
        Arrays.sort(in);
    }

    static void shuffle(int[] in) {
        for (int i = 0; i < in.length; i++) {
            int idx = (int) (Math.random() * in.length);
            int tmp = in[i];
            in[i] = in[idx];
            in[idx] = tmp;
        }
    }

    static void shuffle(long[] in) {
        for (int i = 0; i < in.length; i++) {
            int idx = (int) (Math.random() * in.length);
            long tmp = in[i];
            in[i] = in[idx];
            in[idx] = tmp;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The key points to consider are:

1. **Edge Cases**: Small values of \( n \) and \( q \).
2. **Invalid Operations**: Ensure that the sequences \( l \) and \( r \) are valid and can be produced by some permutation.
3. **Special Patterns**: Patterns in \( l \) and \( r \) that might cause unexpected behavior in the code.

Given the constraints and the nature of the problem, let's construct a test input that is small but challenging:

### Test Input:
```plaintext
4 2
2 3
3 4
```

### Explanation:
- **n = 4**: The permutation size is 4.
- **q = 2**: We perform 2 operations.
- **l = [2, 3]**: The maximum elements in the left parts of the splits.
- **r = [3, 4]**: The maximum elements in the right parts of the splits.

### Expected Behavior:
- The initial permutation that can produce the given sequences is [3, 2, 4, 1].
- The operations can be:
  1. Split [3, 2, 4, 1] into [3, 2] and [4, 1]. This gives \( l_1 = 3 \) and \( r_1 = 4 \).
  2. Split [3, 2] into [2] and [3]. This gives \( l_2 = 2 \) and \( r_2 = 3 \).

### Why This Input is Challenging:
- The sequences \( l \) and \( r \) are such that they require careful handling of the left and right parts.
- The code needs to correctly manage the relationships between the elements in the sequences and the resulting parts.
- This input is small enough to be manageable but complex enough to test the logic of the code.

This test input should help expose any issues in the code's handling of the sequences and the permutation construction.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

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

using i64 = long long;

#ifdef DEBUG
    #include "debug.h"
#else
    #define debug(...) void(23)
#endif

template<typename T>
T power(T a, i64 b) {
    if(b < 0) return 0;
    T res = 1;
    while(b) {
        if(b & 1) {
            res *= a;
        }
        a *= a;
        b >>= 1;
    }
    return res;
}

constexpr int md = 998244353;
struct MInt {
    int val;
    MInt() : val(0) {}
    template<typename T>
    MInt(T v) {
        if(-md <= v && v < md) {
            val = v;
        } else {
            val = v % md;
        }
        if(val < 0) {
            val += md;
        }
    }
    int operator() () { return val; }
    MInt operator+= (MInt rhs) {
        if((val += rhs.val) >= md) {
            val -= md;
        }
        return *this;
    }
    MInt operator-= (MInt rhs) {
        if((val -= rhs.val) < 0) {
            val += md;
        }
        return *this;
    }
    MInt operator*= (MInt rhs) {
        val = (int)(1LL * val * rhs.val % md);
        return *this;
    }
    MInt inv() {
        return power(*this, md - 2);
    }
    MInt operator/= (MInt rhs) {
        return *this *= rhs.inv();
    }
    bool operator== (MInt rhs) {
        return val == rhs.val;
    }
    bool operator!= (MInt rhs) {
        return val != rhs.val;
    }
};
MInt operator+ (MInt lhs, MInt rhs) {
    return lhs += rhs;
}
MInt operator- (MInt lhs, MInt rhs) {
    return lhs -= rhs;
}
MInt operator* (MInt lhs, MInt rhs) {
    return lhs *= rhs;
}
MInt operator/ (MInt lhs, MInt rhs) {
    return lhs /= rhs;
}
std::ostream& operator<< (std::ostream& os, MInt v) {
    return os << v();
}
std::string to_string(MInt x) {
    return std::to_string(x());
}

using Z = MInt;

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

    int N, Q;
    std::cin >> N >> Q;
    std::vector<int> L(Q), R(Q);
    for(int i = 0; i < Q; ++i) {
        std::cin >> L[i];
        --L[i];
    }
    for(int i = 0; i < Q; ++i) {
        std::cin >> R[i];
        --R[i];
    }

    std::vector<int> prv(N, -1), nxt(N, -1), exist(N);
    exist[N - 1] = true;
    for(int i = 0; i < Q; ++i) {
        int l = L[i], r = R[i];
        if(exist[l]) {
            prv[r] = l;
            nxt[r] = nxt[l];
            if(nxt[l] != -1) prv[nxt[l]] = r;
            nxt[l] = r;
            exist[r] = true;
        } else {
            nxt[l] = r;
            prv[l] = prv[r];
            if(prv[r] != -1) nxt[prv[r]] = l;
            prv[r] = l;
            exist[l] = true;
        }
        debug(exist);
        debug(prv);
        debug(nxt);
        debug();
    }

    std::vector<int> p;
    int start = -1;
    for(int i = 0; i < N; ++i) {
        if(exist[i] && prv[i] == -1) {
            start = i;
            break;
        }
    }
    while(start != -1) {
        p.emplace_back(start);
        start = nxt[start];
    }

    debug(p);

    std::vector<int> c(N);
    for(int i = 0; i + 1 < int(p.size()); ++i) {
        ++c[std::max(p[i], p[i + 1])];
    }

    Z ans = 1;
    int place = 1;
    for(int i = N - 1; i >= 0; --i) {
        if(exist[i]) place += c[i];
        else ans *= (place += 1);
        debug(ans, place);
    }

    std::cout << ans << '\n';

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

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

using namespace std;

using ll = long long;

const ll mod = 998244353;

void solve() {
	int n, q;
	cin >> n >> q;
	vector<int> l(q), r(q), nxt(n + 1), lst(n + 1), vis(n + 1);
	for (int i = 0; i < q; i++) {
		cin >> l[i];
	}
	for (int i = 0; i < q; i++) {
		cin >> r[i];
	}
	for (int i = 0; i < q; i++) {
		if (lst[r[i]]) {
			if (lst[r[i]] < l[i] && r[i] < l[i]) {
				cout << 0 << '\n';
				return;
			}
			lst[l[i]] = lst[r[i]];
			nxt[lst[l[i]]] = l[i];
			lst[r[i]] = l[i];
			nxt[l[i]] = r[i];
		} else {
			if (nxt[l[i]] && nxt[l[i]] < r[i] && l[i] < r[i]) {
				cout << 0 << '\n';
				return;
			}
			nxt[r[i]] = nxt[l[i]];
			lst[nxt[r[i]]] = r[i];
			nxt[l[i]] = r[i];
			lst[r[i]] = l[i];
		}
	}
	multiset<int> st;
	int pi = 0;
	for (int i = 1; i <= n; i++) {
		// cout << lst[i] << " " << nxt[i] << endl;
		if (lst[i] == 0 && nxt[i]) {
			pi = i;
			break;
		}
	}
	st.insert(pi);
	while (pi != 0) {
		// cout << pi << endl;
		st.insert(max(pi, nxt[pi]));
		vis[pi] = 1;
		pi = nxt[pi];
	}
	ll ans = 1, siz = st.size();
	for (int i = n; i >= 1; i--) {
		if (!vis[i]) {
			auto u = st.upper_bound(i);
			while (u != st.end()) {
				auto v = u;
				u++;
				st.erase(v);
			}
			ans *= (siz - st.size());
			ans %= mod;
			siz++;
		}
	}
	cout << ans << '\n';
}

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

	int _ = 1;
	// cin >> _;

	while (_--)
	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
4 2
3 2
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

Buggy Code:
//
// Copyright 2024 Turing Edu. All Rights Reserved.
//
// FileName: E.cc
// Author: Beiyu Li <beiyuli@turingedu.cn>
// Date: 2024-05-30
//
#include <bits/stdc++.h>

using namespace std;

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

using ll = long long;
using pii = pair<int, int>;
mt19937_64 randLL(chrono::steady_clock::now().time_since_epoch().count());

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

const int mod = 998244353;

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

  int n, q;
  cin >> n >> q;
  vector<int> l(q + 1), r(q + 1);
  For(i,1,q) cin >> l[i];
  For(i,1,q) cin >> r[i];
  vector<pii> dir(n + 1);
  vector<int> lc(n + 1), rc(n + 1);
  For(i,1,q) {
    if (i > 1) {
      auto [u, d] = dir[max(l[i],r[i])];
      d == 0? lc[u] = i: rc[u] = i;
    }
    dir[l[i]] = pii(i, 0), dir[r[i]] = pii(1, 1);
  }
  vector<bool> vis(n + 1);
  vector<int> sz(n + 1);
  {
    auto dfs = [&](auto&& dfs, int u) -> int {
      vis[l[u]] = vis[r[u]] = true;
      sz[u] = 0;
      sz[u] += (lc[u]? dfs(dfs, lc[u]): 1);
      sz[u] += (rc[u]? dfs(dfs, rc[u]): 1);
      return sz[u];
    };
    dfs(dfs, 1);
  }
  vector<ll> dlt(n + 1);
  {
    auto dfs = [&](auto&& dfs, int u, int f, int b) -> void {
      if (l[u] < r[u]) {
        dlt[l[u]+1] += (lc[u]? sz[lc[u]]: 1);
        dlt[r[u]] -= (lc[u]? sz[lc[u]]: 1);
        if (l[u] + 1 < f) {
          dlt[l[u]+1] += -1;
          dlt[f] -= -1;
        }
      } else {
        dlt[r[u]+1] += (rc[u]? sz[rc[u]]: 1);
        dlt[l[u]] -= (rc[u]? sz[rc[u]]: 1);
        if (r[u] + 1 < b) {
          dlt[r[u]+1] += -1;
          dlt[b] -= -1;
        }
      }
      if (lc[u]) dfs(dfs, lc[u], f, r[u]);
      if (rc[u]) dfs(dfs, rc[u], l[u], b);
    };
    dfs(dfs, 1, 0, 0);
    For(i,1,n) dlt[i] += dlt[i-1];
  }
  int ret = 1, gap = sz[1] + 1;
  Rof(i,n,1) if (!vis[i]) {
    assert(gap > dlt[i]);
    ret = (ll)ret * (gap - dlt[i]) % mod;
    gap++;
  }
  cout << ret << '\n';

  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases or scenarios that might cause the algorithm to fail. Here are some potential edge cases to consider:

1. **Single Operation (q = 1):** This is a simple case where the permutation is split only once.
2. **Multiple Operations (q > 1):** This can reveal issues in how the algorithm handles multiple splits.
3. **Edge Cases with Maximum and Minimum Values:** These can test how the algorithm handles the boundaries of the permutation.
4. **Repetitive Values in l and r Sequences:** This can test how the algorithm handles repeated splits with the same values.

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

### Test Input:
```plaintext
5 2
5 3
4 2
```

### Explanation:
- **n = 5:** The permutation is of size 5.
- **q = 2:** There are 2 operations.
- **l = [5, 3]:** The maximum element in the left part for the first operation is 5, and for the second operation is 3.
- **r = [4, 2]:** The maximum element in the right part for the first operation is 4, and for the second operation is 2.

This test input is designed to:
- Test the algorithm's ability to handle multiple operations.
- Ensure that the algorithm correctly processes the sequences and splits the permutation accordingly.
- Verify that the algorithm can handle the maximum and minimum values correctly.

By using this test input, we can effectively check if the algorithm correctly calculates the number of valid permutations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

Buggy Code:
class Node:
    def __init__(self,val):
        self.val = val
        self.left = None
        self.right = None

mod = 998244353
n,q = map(int,input().split())
l = list(map(int,input().split()))
r = list(map(int,input().split()))
mp = [Node(i) for i in range(n)]

for i in range(0,q):
    l[i] -= 1
    r[i] -= 1
    if l[i] > r[i]:
        if mp[r[i]].left or mp[r[i]].right:
            print(0)
            quit()

        mp[r[i]].left = mp[l[i]]
        mp[r[i]].right = mp[l[i]].right
        if mp[l[i]].right:
            mp[mp[l[i]].right.val].left = mp[r[i]]
        mp[l[i]].right = mp[r[i]]

    else:
        if mp[l[i]].left or mp[l[i]].right:
            print(0)
            quit()
        mp[l[i]].right = mp[r[i]]
        mp[l[i]].left = mp[r[i]].left
        if mp[r[i]].left:
            mp[mp[r[i]].left.val].right = mp[l[i]]
        mp[r[i]].left = mp[l[i]]
arr = []
cur = mp[l[0]]
vis = [0 for i in range(n+1)]
while cur.right:
    arr.append(cur.val+1)
    cur = cur.right
arr.append(cur.val+1)
l.clear()
l.append(arr[0])
for i in arr:
    vis[i] = True
for i in range(len(arr)-1):
    l.append(max(arr[i],arr[i+1]))

to_push = []
ans = 1
for i in range(1,n+1):
    if not vis[i]:
        to_push.append(i)
l.append(arr[-1])
to_push.reverse()
l.sort()
cnt = 0

for i in to_push:
    left,right = 0,len(l)-1
    much = 0
    while left <= right:
        mid = (left + right) // 2
        if l[mid] < i: left = mid+1
        else:
            right = mid-1
            much = len(l) - mid

    ans *= much + cnt
    ans %= mod
    cnt += 1
print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases that can highlight issues in the logic. The code involves building a tree structure and then traversing it to form a permutation. The primary issue in the code seems to be the handling of the tree structure and the merging of nodes.

Here are the steps to generate a test input:

1. **Choose a small `n` and `q` to keep the input manageable.**
2. **Ensure the sequences `l` and `r` are such that they should form a valid permutation but the code might fail to handle the tree structure correctly.**

Let's consider the following test case:

- **n = 5**
- **q = 2**
- **l = [3, 4]**
- **r = [4, 5]**

This test case is designed to:
- Ensure that the initial array can be split into valid segments.
- Check if the code correctly handles the merging of nodes and the formation of the tree structure.

Here is the test input:

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

Explanation:
- The initial permutation could be `[1, 2, 3, 4, 5]`.
- The first operation splits the array into `[1, 2, 3]` and `[4, 5]`, with `l_1 = 3` and `r_1 = 5`.
- The second operation splits `[4, 5]` into `[4]` and `[5]`, with `l_2 = 4` and `r_2 = 5`.

This test case should expose any issues in the code's handling of the tree structure and the merging of nodes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

Buggy Code:
N,M=map(int, input().split())
A=list(map(int, input().split()))
B=list(map(int, input().split()))

mod=998244353;n=3*10*5+100
inv_t=[0]+[1]
for i in range(2,n):
  inv_t+=[inv_t[mod%i]*(mod-int(mod/i))%mod]

kai=[1,1]
rev_kai=[1,inv_t[1]]
for i in range(2,n):
  kai.append(kai[-1]*i%mod)
  rev_kai.append(rev_kai[-1]*inv_t[i]%mod)

def cmb(n,r):
  return kai[n]*rev_kai[r]*rev_kai[n-r]%mod

from collections import deque
d=deque()

S=set()
for i in range(M):
  a,b=A[-1-i],B[-1-i]
  if i==0:
    d.append(a)
    d.append(b)
    S.add(a);S.add(b)
  else:
    if a in S:
      d.append(b)
      S.add(b)
    else:
      d.appendleft(a)
      S.add(a)

d.append(-1)
d.appendleft(-1)
D=list(d);E={}
for i in range(len(D)-1):
  a,b=D[i],D[i+1]
  c=max(a,b)
  if c not in E:
    E[c]=0
  E[c]+=1

ans=1;c=0
for i in range(N,0,-1):
  if i in S:
    if i in E:
      c+=E[i]
  else:
    ans*=c
    c+=1
  ans%=mod
print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's consider a few key aspects of the problem:

1. **Edge Cases:**
   - When `q` is 1, the permutation is almost entirely determined.
   - When `q` is close to `n`, the number of valid permutations can be very small.
   - When `n` is small, the number of valid permutations is also small, making it easier to verify correctness.

2. **Constraints:**
   - The input must be valid, meaning there should exist at least one permutation that can produce the given sequences.

3. **Buggy Code Analysis:**
   - The code might have issues with how it handles the sequences and the set of elements.
   - The order in which elements are added to the deque and the set can be crucial.

Given these considerations, let's create a test input that is small but challenging:

- **Input:**
  - `n = 5` and `q = 2`
  - Sequences `l` and `r` such that the permutation can be uniquely determined or the number of valid permutations is small.

A good test input would be:
- Initial array: `[1, 2, 3, 4, 5]`
- Operations:
  1. Split `[1, 2, 3, 4, 5]` into `[1, 2, 3]` and `[4, 5]` with `l_1 = 3` and `r_1 = 5`.
  2. Split `[1, 2, 3]` into `[1, 2]` and `[3]` with `l_2 = 2` and `r_2 = 3`.

This input ensures that the sequences are valid and the number of valid permutations is small, making it easier to verify the correctness of the code.

```plaintext
5 2
3 2
5 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splittable Permutations
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, we had one array, which was a permutation of size $n$ (an array of size $n$ where each integer from $1$ to $n$ appears exactly once).

We performed $q$ operations. During the $i$-th operation, we did the following:

  * choose any array we have with at least $2$ elements;    * split it into two non-empty arrays (prefix and suffix);    * write two integers $l_i$ and $r_i$, where $l_i$ is the maximum element in the left part which we get after the split, and $r_i$ is the maximum element in the right part;    * remove the array we've chosen from the pool of arrays we can use, and add the two resulting parts into the pool.

For example, suppose the initial array was $[6, 3, 4, 1, 2, 5]$, and we performed the following operations:

  1. choose the array $[6, 3, 4, 1, 2, 5]$ and split it into $[6, 3]$ and $[4, 1, 2, 5]$. Then we write $l_1 = 6$ and $r_1 = 5$, and the arrays we have are $[6, 3]$ and $[4, 1, 2, 5]$;    2. choose the array $[4, 1, 2, 5]$ and split it into $[4, 1, 2]$ and $[5]$. Then we write $l_2 = 4$ and $r_2 = 5$, and the arrays we have are $[6, 3]$, $[4, 1, 2]$ and $[5]$;    3. choose the array $[4, 1, 2]$ and split it into $[4]$ and $[1, 2]$. Then we write $l_3 = 4$ and $r_3 = 2$, and the arrays we have are $[6, 3]$, $[4]$, $[1, 2]$ and $[5]$.

You are given two integers $n$ and $q$, and two sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$. A permutation of size $n$ is called valid if we can perform $q$ operations and produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Calculate the number of valid permutations.

Input: The first line contains two integers $n$ and $q$ ($1 \le q < n \le 3 \cdot 10^5$).

The second line contains $q$ integers $l_1, l_2, \dots, l_q$ ($1 \le l_i \le n$).

The third line contains $q$ integers $r_1, r_2, \dots, r_q$ ($1 \le r_i \le n$).

Additional constraint on the input: there exists at least one permutation which can produce the given sequences $[l_1, l_2, \dots, l_q]$ and $[r_1, r_2, \dots, r_q]$.

Output: Print one integer — the number of valid permutations, taken modulo $998244353$.

Examples:
input:
6 3
6 4 4
5 5 2
output:
30

input:
10 1
10
9
output:
1814400

input:
4 1
2
4
output:
8

Note:

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

public class x1976
{
    static final int INF = Integer.MAX_VALUE/2;
    static final long MOD = 998244353L;
    public static void main(String[] thekkathy) throws Exception
    {
        FastScanner infile = new FastScanner();
        int N = infile.nextInt();
        int K = infile.nextInt();
        int[] L = infile.nextInts(K);
        int[] R = infile.nextInts(K);
        trie = new int[2][N+1];
        int[] label = new int[N+1];
        trie[0][0] = 1;
        trie[1][0] = 2;
        label[L[0]] = 1;
        label[R[0]] = 2;
        int boof = 3;
        for(int i=1; i < K; i++)
        {
            int left = L[i];
            int right = R[i];
            if(label[left] == 0)
            {
                int par = label[right];
                label[left] = boof++;
                label[right] = boof++;
                trie[0][par] = label[left];
                trie[1][par] = label[right];
            }
            else
            {
                int par = label[left];
                label[left] = boof++;
                label[right] = boof++;
                trie[0][par] = label[left];
                trie[1][par] = label[right];
            }
        }
        revLabel = new int[boof];
        for(int v=1; v <= N; v++)
            revLabel[label[v]] = v;
        order = new ArrayList<>();
        dfs(0);
        //System.out.println(order);

        //maintain neighbor spaces
        int P = order.size();
        int[] loc = new int[N+1];
        Arrays.fill(loc, -1);
        for(int i=0; i < P; i++)
            loc[order.get(i)] = i;
        long res = 1L;
        boolean[] open = new boolean[P+1];
        int spaces = 0;
        int processVal = N;
        for(int v=N-1; v >= 1; v--)
        {
            if(loc[v] != -1)
                continue;
            while(processVal > v)
            {
                int d = loc[processVal--];
                if(d == -1)
                    continue;
                if(!open[d])
                    spaces++;
                if(!open[d+1])
                    spaces++;
                open[d] = open[d+1] = true;
            }
            res *= spaces;
            res %= MOD;
            spaces++;
        }
        System.out.println(res);
    }
    static int[][] trie;
    static int[] revLabel;
    static ArrayList<Integer> order;
    public static void dfs(int root)
    {
        int lc = trie[0][root];
        int rc = trie[1][root];
        if(lc+rc == 0)
            order.add(revLabel[root]);
        else
        {
            dfs(lc);
            dfs(rc);
        }
    }
}
/*

))(((())
(())))((

prefix p: untouched prefix
d[i] = sum of prefix i

inverted prefix should not have max delta > d[p]

 */
class SegmentTree
{
    //Tlatoani's segment tree
    //iterative implementation = low constant runtime factor
    //range query, non lazy
    final int[] val;
    final int treeFrom;
    final int length;

    public SegmentTree(int treeFrom, int treeTo)
    {
        this.treeFrom = treeFrom;
        int length = treeTo - treeFrom + 1;
        int l;
        for (l = 0; (1 << l) < length; l++);
        val = new int[1 << (l + 1)];
        this.length = 1 << l;
    }
    public void update(int index, int delta)
    {
        //replaces value
        int node = index - treeFrom + length;
        val[node] = delta;
        for (node >>= 1; node > 0; node >>= 1)
            val[node] = comb(val[node << 1], val[(node << 1) + 1]);
    }
    public int query(int from, int to)
    {
        //inclusive bounds
        if (to < from)
            return 0; //0 or 1?
        from += length - treeFrom;
        to += length - treeFrom + 1;
        //0 or 1?
        int res = 0;
        for (; from + (from & -from) <= to; from += from & -from)
            res = comb(res, val[from / (from & -from)]);
        for (; to - (to & -to) >= from; to -= to & -to)
            res = comb(res, val[(to - (to & -to)) / (to & -to)]);
        return res;
    }
    public int comb(int a, int b)
    {
        //change this
        return max(a,b);
    }
}
class FenwickTree
{
    //Binary Indexed Tree
    //0 indexed
    public int[] tree;
    public int size;

    public FenwickTree(int size)
    {
        this.size = size;
        tree = new int[size+5];
    }
    public void add(int i, int v)
    {
        i++;
        while(i <= size)
        {
            tree[i] += v;
            i += i&-i;
        }
    }
    public int find(int i)
    {
        int res = 0;
        i++;
        while(i >= 1)
        {
            res += tree[i];
            i -= i&-i;
        }
        return res;
    }
    public int find(int l, int r)
    {
        return find(r)-find(l-1);
    }
}
class FastScanner
{
    private int BS = 1 << 16;
    private char NC = (char) 0;
    private byte[] buf = new byte[BS];
    private int bId = 0, size = 0;
    private char c = NC;
    private double cnt = 1;
    private BufferedInputStream in;

    public FastScanner() {
        in = new BufferedInputStream(System.in, BS);
    }

    public FastScanner(String s) {
        try {
            in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
        } catch (Exception e) {
            in = new BufferedInputStream(System.in, BS);
        }
    }

    private char getChar() {
        while (bId == size) {
            try {
                size = in.read(buf);
            } catch (Exception e) {
                return NC;
            }
            if (size == -1) return NC;
            bId = 0;
        }
        return (char) buf[bId++];
    }

    public int nextInt() {
        return (int) nextLong();
    }

    public int[] nextInts(int N) {
        int[] res = new int[N];
        for (int i = 0; i < N; i++) {
            res[i] = (int) nextLong();
        }
        return res;
    }

    public long[] nextLongs(int N) {
        long[] res = new long[N];
        for (int i = 0; i < N; i++) {
            res[i] = nextLong();
        }
        return res;
    }

    public long nextLong() {
        cnt = 1;
        boolean neg = false;
        if (c == NC) c = getChar();
        for (; (c < '0' || c > '9'); c = getChar()) {
            if (c == '-') neg = true;
        }
        long res = 0;
        for (; c >= '0' && c <= '9'; c = getChar()) {
            res = (res << 3) + (res << 1) + c - '0';
            cnt *= 10;
        }
        return neg ? -res : res;
    }

    public double nextDouble() {
        double cur = nextLong();
        return c != '.' ? cur : cur + nextLong() / cnt;
    }

    public double[] nextDoubles(int N) {
        double[] res = new double[N];
        for (int i = 0; i < N; i++) {
            res[i] = nextDouble();
        }
        return res;
    }

    public String next() {
        StringBuilder res = new StringBuilder();
        while (c <= 32) c = getChar();
        while (c > 32) {
            res.append(c);
            c = getChar();
        }
        return res.toString();
    }

    public String nextLine() {
        StringBuilder res = new StringBuilder();
        while (c <= 32) c = getChar();
        while (c != '\n') {
            res.append(c);
            c = getChar();
        }
        return res.toString();
    }

    public boolean hasNext() {
        if (c > 32) return true;
        while (true) {
            c = getChar();
            if (c == NC) return false;
            else if (c > 32) 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 expose the bug in the given code, we need to focus on the logic that constructs the tree and the labeling process. The buggy code seems to have issues with how it handles the splitting and labeling of the arrays. Specifically, it might not correctly handle cases where the same element is split multiple times.

Let's consider a small-scale test input that can highlight these issues:

1. **Initial Array and Operations:**
   - Initial array: `[1, 2, 3, 4]`
   - Operations:
     - Split `[1, 2, 3, 4]` into `[1, 2]` and `[3, 4]`. Write `l_1 = 2` and `r_1 = 4`.
     - Split `[1, 2]` into `[1]` and `[2]`. Write `l_2 = 1` and `r_2 = 2`.
     - Split `[3, 4]` into `[3]` and `[4]`. Write `l_3 = 3` and `r_3 = 4`.

2. **Input:**
   - `n = 4`
   - `q = 3`
   - `l = [2, 1, 3]`
   - `r = [4, 2, 4]`

This input will test the following:
- How the code handles multiple splits of the same element.
- How the code constructs the tree and labels the nodes.

Here is the test input:

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