**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
/*Rahul Bhuiya 2312042*/

#include<bits/stdc++.h>
using namespace std;
/*<----------------------------template starts--------------------------------->*/
#define fast_io     ios_base::sync_with_stdio(false);cin.tie(NULL)
#define fr(s,n)     for(int i=s;i<n;i++)
#define frj(j,s,n)  for(int j=s;j<n;j++)
#define ll          long long
#define ld          long double
#define sort(v)     sort(v.begin(),v.end())
// #define rsort(v)    sort(v.rbegin(),v.rend())
/*<----------------------------template ends--------------------------------->*/

int main(){
    int t;
    cin>>t;
    while(t--){
        int n,m,k;
        cin>>n>>m>>k;
            if(n/m*(m-1)>k)
            cout<<"YES";
            else
            cout<<"NO";
        cout<<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:
To expose the bug in the given code, we need to focus on scenarios where the logic of the condition `if(n/m*(m-1)>k)` fails. This condition is intended to check if it's possible for Alice to paint the ribbon in such a way that Bob cannot repaint at most `k` parts to make all parts the same color. However, the condition is not correctly formulated to handle all edge cases.

Let's break down the problem and identify a test case that can expose the bug:

1. **Condition Analysis:**
   - The condition `n/m*(m-1)>k` is supposed to check if there are enough parts with different colors such that Bob cannot repaint them all to a single color.
   - If `n/m` is the number of parts per color and `m-1` is the number of different colors used, then `n/m*(m-1)` should be the number of parts that are not of the same color.
   - This condition is not correct because it does not account for the distribution of colors and the actual number of parts that can be painted differently.

2. **Edge Cases:**
   - **Case 1:** When `n` is exactly divisible by `m` and `k` is just one less than `n/m*(m-1)`.
   - **Case 2:** When `n` is small and `k` is large, but Alice can still paint the ribbon in such a way that Bob cannot repaint it.

3. **Test Case Selection:**
   - A good test case to expose the bug is when `n` is small, `m` is large, and `k` is just one less than the number of parts that can be painted differently.

Let's choose a test case where `n = 6`, `m = 3`, and `k = 2`:

- Alice can paint the ribbon as `[1, 2, 3, 1, 2, 3]`.
- Bob can only repaint 2 parts, but it's impossible for him to make all parts the same color.

Here is the test input:

```plaintext
1
6 3 2
```

This test case should output `YES` because Alice can paint the ribbon in such a way that Bob cannot make all parts the same color, but the buggy code might incorrectly output `NO`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
t = int(input())
lst = []
for i in range(t):
    n, m , k = map(int, input().split())
    if m > k:
        lst.append("YES")
    else:
        lst.append("NO")
for j in lst:
    print(j)

**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 where the conditions \( m > k \) and \( m \le k \) do not correctly determine whether Alice can paint the ribbon such that Bob cannot make all parts have the same color.

### Key Observations:
1. **Case 1: \( n = 1 \)**
   - If the ribbon has only one part, Bob can always make all parts have the same color regardless of \( m \) and \( k \).
2. **Case 2: \( m = 1 \)**
   - If there is only one color available, Bob can always make all parts have the same color.
3. **Case 3: \( k \ge n \)**
   - If Bob can repaint all parts, he can always make all parts have the same color.
4. **Case 4: \( k < n \) and \( m > k \)**
   - Alice can paint the ribbon in such a way that Bob cannot make all parts have the same color.
5. **Case 5: \( k < n \) and \( m \le k \)**
   - Bob can always make all parts have the same color if \( k \ge n \) or if \( m \le k \).

### Challenging Test Input:
Let's create a test input that includes all these edge cases to ensure the buggy code fails:

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

### Explanation:
1. **Test Case 1: \( n = 1, m = 1, k = 1 \)**
   - Expected output: NO (Bob can always make all parts have the same color)
2. **Test Case 2: \( n = 2, m = 1, k = 1 \)**
   - Expected output: NO (Bob can always make all parts have the same color)
3. **Test Case 3: \( n = 3, m = 2, k = 1 \)**
   - Expected output: YES (Alice can paint the ribbon as \([1, 2, 1]\) and Bob cannot make all parts have the same color)
4. **Test Case 4: \( n = 4, m = 2, k = 2 \)**
   - Expected output: NO (Bob can always make all parts have the same color)
5. **Test Case 5: \( n = 5, m = 3, k = 2 \)**
   - Expected output: YES (Alice can paint the ribbon as \([1, 2, 3, 1, 2]\) and Bob cannot make all parts have the same color)

This test input should expose the bug in the given buggy code, as it covers various edge cases where the conditions \( m > k \) and \( m \le k \) do not correctly determine the outcome.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
t=int(input())
for i in range(t):
    n,m,k=map(int, input().strip().split())
    if m>k:
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
import java.util.*;

public class osumania {
    public static void main(String[] args) {
        Scanner user = new Scanner(System.in);
        int t = user.nextInt();
        while (t-- > 0) {
            solve(user);
        }
        user.close();
    }
    public static void solve(Scanner user) {
        int n = user.nextInt();
        int m = user.nextInt();
        int k = user.nextInt();
        if (n == 1 || m == 1) {
            System.out.println("NO");
            return;
        }
        int parts = (int)Math.floor(n / m);
        if((m - 1) * parts > k){
            System.out.println("YES");
        }
        else System.out.println("NO");
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
/*package whatever //do not write package name here */

import java.io.*;

public class GFG {
	public static void main (String[] args) throws IOException
	 {
BufferedReader br =
            new BufferedReader(new InputStreamReader(System.in));

        int t = Integer.parseInt(br.readLine());

        while(t-->0){
            String st=br.readLine();
            String[] in=st.split("\\s+");

	    int n=Integer.parseInt(in[0]);
            int m=Integer.parseInt(in[1]);
            int k=Integer.parseInt(in[2]);

            if(m==k)
            System.out.println("NO");
            else
            System.out.println("YES");

        }}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the logic of the algorithm and identify scenarios where the code might fail. The buggy code checks if `m == k` and prints "NO" if true, otherwise it prints "YES". This logic is incorrect because it doesn't consider the actual constraints and the problem requirements.

### Step-by-Step Analysis:
1. **Understanding the Bug**:
   - The bug in the code is that it only checks if `m == k` and prints "NO" if true. This is incorrect because the condition for Bob to be able to make all parts the same color depends on the number of parts `n`, the number of colors `m`, and the number of parts Bob can repaint `k`.

2. **Correct Condition**:
   - Bob can make all parts the same color if he can repaint at least `n - m + 1` parts. This is because if there are `m` colors, the worst-case scenario for Alice is that she uses all `m` colors, and Bob needs to repaint at least `n - m + 1` parts to ensure all parts are the same color.

3. **Edge Cases**:
   - **Case 1**: `m == k` and `n > m` (Bob can still make all parts the same color).
   - **Case 2**: `m > k` and `n > m` (Bob cannot make all parts the same color).
   - **Case 3**: `m == k` and `n == m` (Bob can make all parts the same color).
   - **Case 4**: `m > k` and `n == m` (Bob cannot make all parts the same color).

### Challenging Test Input:
```plaintext
4
3 2 2
5 3 3
4 4 4
6 2 1
```

### Explanation:
1. **Test Case 1: `3 2 2`**
   - `n = 3`, `m = 2`, `k = 2`
   - Alice can paint the ribbon as `[1, 2, 1]`. Bob can repaint at most 2 parts, but he cannot make all parts the same color.
   - Expected output: `YES`

2. **Test Case 2: `5 3 3`**
   - `n = 5`, `m = 3`, `k = 3`
   - Alice can paint the ribbon as `[1, 2, 3, 1, 2]`. Bob can repaint at most 3 parts, but he cannot make all parts the same color.
   - Expected output: `YES`

3. **Test Case 3: `4 4 4`**
   - `n = 4`, `m = 4`, `k = 4`
   - Alice can paint the ribbon as `[1, 2, 3, 4]`. Bob can repaint all 4 parts, so he can make all parts the same color.
   - Expected output: `NO`

4. **Test Case 4: `6 2 1`**
   - `n = 6`, `m = 2`, `k = 1`
   - Alice can paint the ribbon as `[1, 2, 1, 2, 1, 2]`. Bob can repaint at most 1 part, but he cannot make all parts the same color.
   - Expected output: `YES`

This test input will expose the bug in the buggy code by providing cases where the incorrect logic of `m == k` will fail to give the correct output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
#include <iostream>
#include <vector>

using namespace std;

const int MOD = 998244353;

long long count_possible_arrays(int n) {
    if (n == 2) {
        return 1;
    }
    // According to the sample output in the problem statement, we have:
    // n=4 -> 5
    // n=7 -> 682
    // n=69 -> 325188814
    // To find a pattern, we need additional information or constraints which is missing.
    // Based on the problem's constraints, a better approach is needed.
    // However, considering the problem's specific nature, it seems it requires a specific combinatorial approach or precomputed values.
    // Since a general solution isn't immediately obvious from the problem description, we assume that there's a pattern that can be programmed for smaller values.

    // For now, based on the sample input/output, we return the values.
    if (n == 4) {
        return 5;
    } else if (n == 7) {
        return 682;
    } else if (n == 69) {
        return 325188814;
    }
    return 0;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        cout << count_possible_arrays(n) << endl;
    }
    return 0;
}

// OKwPFZJUqbegxl8VmoQ

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code currently handles only specific values of `n` (4, 7, and 69) and returns hard-coded results. For any other value, it returns 0, which is incorrect.

### Step-by-Step Analysis:
1. **Edge Cases**:
   - The smallest possible value for `n` is 2.
   - The largest possible value for `n` is 1,000,000.
   - Values close to the boundaries (2 and 1,000,000) are good candidates for testing.

2. **Intermediate Values**:
   - Values that are not explicitly handled by the hard-coded results (4, 7, 69) should be tested to ensure the code fails.

3. **Multiple Test Cases**:
   - We can include multiple test cases to cover a range of values, including edge cases and intermediate values.

### Chosen Test Input:
- Test case 1: `n = 2` (smallest value)
- Test case 2: `n = 3` (a value not handled by the hard-coded results)
- Test case 3: `n = 10` (an intermediate value)
- Test case 4: `n = 1000000` (largest value)

### Output:
```plaintext
4
2
3
10
1000000
```

This test input will expose the bug in the buggy code, as it will return 0 for `n = 3`, `n = 10`, and `n = 1000000`, which are incorrect results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<queue>
#include<bitset>
#include<cstdio>
#include<vector>
#include<random>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
#define I ll
#define her1 20081214
#define IV void
#define cht 998244353
#define ld long double
#define Aestas16 392699
#define ull unsigned long long
#define cp(x,y)memcpy(x,y,sizeof y)
#define mem(x,val)memset(x,val,sizeof x)
#define D(i,j,n)for(register int i=j;i>=n;i--)
#define E(i,now)for(register int i=first[now];i;i=e[i].nxt)
#define F(i,j,n)for(register int i=j;i<=n;i++)
#define DL(i,j,n)for(register i64 i=j;i>=n;i--)
#define EL(i,now)for(register i64 i=first[now];i;i=e[i].nxt)
#define FL(i,j,n)for(register i64 i=j;i<=n;i++)
//#define D(i,j,n)for(int i=j;i>=n;i--)
//#define E(i,now)for(int i=first[now];i;i=e[i].nxt)
//#define F(i,j,n)for(int i=j;i<=n;i++)
//#define DL(i,j,n)for(register ll i=j;i>=n;i--)
//#define EL(i,now)for(register ll i=first[now];i;i=e[i].nxt)
//#define FL(i,j,n)for(register ll i=j;i<=n;i++)
ll read(){
	ll ans=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')ans=ans*10+c-'0',c=getchar();
	return ans*f;
}
#undef ll
#include "assert.h"
mt19937_64 rnd(her1);
#include "functional"
using i64 = long long;
const int MAX = 1e3;
const int maxn = 1e6+5;
IV cadd(i64&x,i64 val){x=(x+val)%cht;}
i64 n,f[maxn],g[maxn];
i64 s2[maxn];
IV init(){
	i64 sum=0;
	F(i,1,MAX){
		cadd(sum,i+1);
		cadd(f[i+3],sum);
		// F(j,0,MAX)cadd(f[i+j+3],i+1);
	}
	i64 s1=0,s2=0;
	F(i,1,MAX){
		cadd(s1,f[i]*i);
		cadd(f[i+1],s1);
		cadd(s2,f[i]);cadd(g[i],s2);
	}
	F(i,1,MAX)g[i]++;
	F(i,2,MAX)cadd(g[i],i-2);
}
IV solve(){
	n=read();
	printf("%lld\n",g[n]);
}
int main(){
	// freopen("1.in","r",stdin);
	// freopen("1.out","w",stdout);
	i64 T=read();init();while(T--)solve();return 0;
}

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

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

Think step by step.

Response:
To create 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 aspects to consider are:

1. **Small Test Cases**: Start with small values of \( n \) to ensure the input is small-scale.
2. **Edge Cases**: Test the boundaries of the input constraints, such as the minimum and maximum values of \( n \).
3. **Special Cases**: Test specific values of \( n \) that might reveal logical errors in the code.

Let's consider the following test cases:

1. **Minimum Value of \( n \)**: \( n = 2 \)
2. **Small Value of \( n \)**: \( n = 3 \)
3. **Intermediate Value of \( n \)**: \( n = 10 \)
4. **Large Value of \( n \)**: \( n = 1000000 \)

These test cases will help us cover a range of scenarios and potentially expose any issues in the code.

Here is the test input:

```plaintext
4
2
3
10
1000000
```

This input includes a variety of values for \( n \) to thoroughly test the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

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

#define int long long
const int mod = 998244353;

int t, n, f[(int) 1e6 + 5], s[(int) 1e6 + 5];

void solve()
{
	cin >> n;
	f[3] = 0;
	s[1] = 1, s[2] = 2, s[3] = 5;
	for(int i = 4; i <= 100; i++)
		f[i] = (f[i - 1] * i % mod + i - 2) % mod;
	// cout << '\n';
	for(int i = 2; i <= n; i++)
		s[i] = (s[i - 1] + f[i + 1] + 1) % mod;
	cout << s[n - 1] << '\n';
	return;
}

signed main()
{
	ios :: sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cin >> t;
	while(t--) solve();
	return 0;
}

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

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

Think step by step.

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

1. **Initialization of `f` and `s` Arrays:**
   - The array `f` is initialized only up to `100`, which is a limitation for larger values of `n`.
   - The array `s` is initialized with some specific values, but the logic for updating `s` might be incorrect.

2. **Logic for Updating `f` and `s` Arrays:**
   - The update logic for `f` and `s` arrays seems to have some issues, particularly in the way `f` is calculated and used to update `s`.

3. **Edge Cases:**
   - Small values of `n` (e.g., 2, 3, 4) to check if the initialization and logic work correctly.
   - Large values of `n` to check if the array size limitation and logic updates work correctly.

Given these points, a test input that could expose the bug would be a combination of small and large values of `n` to cover both edge cases and large input scenarios.

Here is a small-scale test input that should be effective:

```plaintext
4
2
3
101
1000
```

This test input includes:
- `2` and `3` to check the initialization and logic for small values.
- `101` to check the limitation of the `f` array size.
- `1000` to check the overall logic and performance for a larger value.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
MOD = 998244353

def count_distinct_arrays(t, n_values):
    results = []
    for n in n_values:
        if n == 2:
            results.append(1)
        elif n == 4:
            results.append(5)
        elif n == 7:
            results.append(682)
        elif n == 69:
            results.append(325188814)
        else:
            # Assuming generalization for other values
            results.append(0)  # Adjust based on detailed pattern
    return results

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

t = int(data[0])
n_values = list(map(int, data[1:t+1]))

# Get and print results
results = count_distinct_arrays(t, n_values)
for res in results:
    print(res)

# kuo2CdxnHRQzUiBh743

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

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

Think step by step.

Response:
```plaintext
3
2
3
1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
M=998244353
for _ in range(int(input())):
    n=int(input())
    F=[0]*(n+1)
    F[n]=1
    Fs=F[n]
    for i in range(n-1,2,-1):
        F[i]=i*Fs+1
        Fs+=F[i]
    print((n-1+sum(F[i]*(((i-2)*(i-1))//2-1) for i in range(3,n+1)))%M)

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code seems to have issues with the way it calculates the number of possible arrays. Specifically, the logic for updating the array `F` and the final summation might be incorrect.

2. **Edge Cases**:
   - **Small `n`**: Test with the smallest possible value of `n` to see if the base cases are handled correctly.
   - **Medium `n`**: Test with a medium value of `n` to check if the intermediate steps are correct.
   - **Large `n`**: Test with a large value of `n` to ensure the algorithm handles large inputs efficiently and correctly.

3. **Challenging Scenarios**:
   - **Boundary Values**: Values of `n` that are just above the smallest and just below the largest possible values.
   - **Special Values**: Values of `n` that might trigger specific conditions in the algorithm.

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

```plaintext
4
3
4
10
100
```

This input includes:
- `n = 3`: A small value to test the base case.
- `n = 4`: A value that is just above the smallest possible value to check the intermediate steps.
- `n = 10`: A medium value to test more complex scenarios.
- `n = 100`: A large value to ensure the algorithm handles large inputs correctly.

This test input should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
# standard inputs
import itertools
from collections import *
from bisect import *
import heapq
from functools import *
import math
import sys
import random
from copy import deepcopy
# templates import *; fin=open('i', 'r'); input = lambda: fin.readline().strip(); from icecream import ic; from sortedcontainers import SortedList

R = lambda: [*map(int, input().split())]; # read one line(turn to int array)
lines = lambda x: [R() for i in range(x)]; # read multiple input lines

M = int(998244353)

def solve():
  n = R()[0]
  SUFF = 0

  @cache
  def dp(i, flag):
    nonlocal SUFF

    ans = 1 + (i-flag)*SUFF
    return ans

  ans = 1
  for i in range(n, 3-1, -1):
    ans += dp(i, 1)
    SUFF += dp(i, 0)
    ans %= M

  return ans

for i in range(int(input())):
  print(solve())

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

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

Think step by step.

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

1. **Small Input Size**: Since the input size can be large (up to \(10^6\)), a small-scale test input is required to ensure the output is manageable and can be verified easily.
2. **Edge Cases**: Test with values that are at the boundaries of the input constraints, such as the minimum and maximum values.
3. **Special Cases**: Test with values that might cause the algorithm to behave unexpectedly, such as values that are powers of 2 or other special numbers.

Given the constraints:
- \(1 \le t \le 10^4\)
- \(2 \le n \le 10^6\)

Let's consider the following test cases:
- \(n = 2\): The minimum value for \(n\).
- \(n = 3\): A small value that is not the minimum.
- \(n = 4\): A value that is used in the provided example.
- \(n = 5\): A slightly larger value to see if the pattern continues.
- \(n = 10\): A value that is larger but still manageable.

Here is the test input:

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

This test input includes a variety of small values that can help identify any issues in the algorithm, especially in the recursive and dynamic programming parts.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    sl(a[nextInt() - 1]);
//    List<Integer> a = new ArrayList<>();
//    a.add(0);
//    a.add(1);
//    Set<List<Integer>> s = new HashSet<>();
//    s.add(a);
//    Set<List<Integer>> s2 = new HashSet<>();
//    int[] f = new int[100];
//    f[1] = 1;
//    for (int ii = 2; ii < 10; ii++) {
//        s2 = new HashSet<>();
//        for (List<Integer> t : s) {
//            int inv = 0;
//            for (int i = 0; i < t.size(); i++) {
//                for (int j = i + 1; j < t.size(); j++) {
//                    if (t.get(i) > t.get(j)) inv++;
//                }
//            }
//            for (int i = 0; i <= t.size(); i++) {
//                List<Integer> t2 = new ArrayList<>();
//                for (int j = 0; j < t.size(); j++) {
//                    if (j == i) t2.add(inv);
//                    t2.add(t.get(j));
//                }
//                if (i == t.size()) t2.add(inv);
//                s2.add(t2);
//            }
//        }
//        s = s2;
////        H.out(ii, s.size());
//        f[ii] = s.size();
//    }
//    H.out("F---");
//    H.out(f, 20);
//    // a(n+2) - 2*a(n+1) + a(n)
//    // 2,11,69,497,4059,37111,375749
//    // (0, 2, 13, 82, 579, 4638, 41749, 417498)
//    H.out("F---a");
//    for (int i = 1; i < 10; i++) {
//        so(f[i + 2] - 2 * f[i + 1] + f[i] + ",");
//    }
//
//    //a(3)=0, a(n) = n * a(n-1) + n - 2 for n>=4
//    int[] g = new int[20];
//    for (int i = 4; i < 20; i++) {
//        g[i] = i * g[i - 1] + i - 2;
//    }
//    H.out(g);
}

static int N = 1000100;
static long M = 998244353;
// 0 0 0 2 13 82 579 4638 41749
static long[] g;
// 0,2,11,69,497,4059,37111,375749
static long[] d;
static long[] a;

static {
    g = new long[N];
    for (int i = 4; i < N; i++) {
        g[i] = (i * g[i - 1] + i - 2 + M) % M;
    }
    d = new long[N];
    for (int i = 1; i + 2 < N; i++) {
        d[i] = (g[i + 2] - g[i + 1] + M) % M;
    }
//    H.out(g);
//    H.out(d);
    // a(n+2) - 2*a(n+1) + a(n) = d[n]
    // 2,11,69,497,4059,37111,375749
    a = new long[N];
    a[1] = 1;
    a[2] = 2;
    a[3] = 5;
    for (int i = 4; i < N; i++) {
        a[i] = (d[i - 1] - a[i - 2] + 2 * a[i - 1]) % M;
    }
//    H.out(a);
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are the steps to think through this:

1. **Initial Array and Inversions**: The initial array is `[0, 1]` with 0 inversions. This is a simple starting point.
2. **Small Arrays**: Test cases with small values of `n` (e.g., 2, 3, 4) to ensure the base cases are handled correctly.
3. **Intermediate Values**: Test cases with intermediate values of `n` (e.g., 7, 10) to check the correctness of the recurrence relations.
4. **Large Values**: Test cases with large values of `n` (e.g., 1000, 1000000) to check the performance and correctness of the algorithm for larger inputs.
5. **Modulo Operation**: Ensure that the modulo operation is correctly applied to handle large numbers and avoid overflow.

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

```plaintext
5
2
3
4
7
10
```

This test input includes:
- The smallest possible array length (2) to check the base case.
- A slightly larger array length (3) to check the next base case.
- An array length (4) that requires the first insertion operation.
- An intermediate array length (7) to check the correctness of the recurrence relations.
- A larger array length (10) to check the performance and correctness for slightly larger inputs.

This should help identify any issues in the algorithm's implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

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

#define MOD 998244353

typedef long long ll;

int t;
int n;

ll ans;

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf(" %d",&n);
        ans=0;
        ll add=0;

        for(int i=n-3;i>0;i--)
        {
            ans+=add;
            ans%=MOD;
            add=add*(n-i+1);
            add%=MOD;
            add+=n-i-1;
            add%=MOD;
        }

        printf("%lld\n",ans+add+n-1);
    }

}

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

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

Think step by step.

Response:
```plaintext
1
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

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

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define per(i, a, b) for(int i = a; i > (b); --i)
#define ar array
#define sz(x) (int) (x).size()
#define pii pair<int,int>
#define fi first
#define se second
typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;
typedef pair<double,int> pdi;
typedef vector<int> vi;
#define all(x) (x).begin(), (x).end()

template<typename T>
void min_self(T& A, T B) {
    A = min(A,B);
}
template<typename T>
void max_self(T& A, T B) {
    A = max(A,B);
}

const int mxn=1e6;
const ll mod=998244353;
int n,t;

void solve() {
    cin >>n;
    ll ans = 0;
    if(n==2) {
        ans +=1;
    } else {
        ans += 2;
    }
    if(n>3) {
        ll tmp = (n-1)*(n)/2 - 3;
        ans = (ans + tmp)%mod;
    }
    vector<ll> to(n,0);
    to[n-1] = n;
    ll suf = n;
    per(i,n-2,3) {
        to[i] = suf*i%mod;
        to[i] = (to[i] + (i+1))%mod;
        suf = (suf + to[i])%mod;
    }
    rep(x,4,n) {
        ll tmp = 1ll*(x-2)*(x-1)/2 - 1;
        tmp = tmp*to[x]%mod;
        ans = (ans + tmp)%mod;
    }
    cout <<ans <<"\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >>t;
    while(t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following points:

1. **Edge Cases**: The code might have issues with small values of `n` or specific values that trigger incorrect calculations.
2. **Intermediate Values**: The code involves calculations that might overflow or produce incorrect results if not handled properly.
3. **Boundary Conditions**: Values of `n` that are just above the edge cases (e.g., 3, 4, 5) might reveal bugs.

Given these points, let's focus on a small-scale test input that includes values of `n` that are known to be problematic or edge cases.

### Test Input
```plaintext
5
2
3
4
5
10
```

### Explanation
- **n = 2**: This is the smallest possible value of `n` and should always return 1.
- **n = 3**: This is the next smallest value, and the code should handle it correctly.
- **n = 4**: This is a small value where the code might start to show issues with the initial logic.
- **n = 5**: This is a slightly larger value, which might expose bugs in the loop and intermediate calculations.
- **n = 10**: This is a value that is large enough to test the full logic of the code, but small enough to be manageable.

By including these values, we can test the code's ability to handle small, intermediate, and slightly larger values, which should help in identifying any bugs in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn=1e6+5;
const int tt=998244353;
int T,n;
LL dp[maxn];
int main(){
	int T; scanf("%d",&T);
	while (T--){
		scanf("%d",&n);
		LL sum=0;
		for (int i=n; i; i--){
			dp[i]=(sum*i+1)%tt;
			sum=(sum+dp[i])%tt;
		}
		LL ans=n-1;
		for (int i=3; i<=n; i++){
			ans=(ans+dp[i]*((i-2)*(i-1)/2-1)%tt)%tt;
		}
		printf("%lld\n",ans);
	}
	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases or scenarios that might cause the code to fail. The key points to consider are:

1. **Initialization and Base Cases**: Ensure the base cases are handled correctly.
2. **Loop Invariants**: Check if the loop invariants are maintained correctly.
3. **Modulo Operations**: Ensure that modulo operations are handled correctly to avoid overflow.
4. **Array Lengths**: Test with small and large array lengths to see if the code handles them correctly.

Given the nature of the problem, a good test input to expose the bug would be:

- **Small Test Case**: To check basic functionality.
- **Edge Case**: To check if the code handles the minimum and maximum limits correctly.
- **Boundary Case**: To check if the code handles the transition from small to large values correctly.

Here is a small-scale test input that is complete and valid:

```plaintext
4
2
3
4
10
```

This test input includes:
- The smallest possible value (2) to check the base case.
- A small value (3) to check the transition from base case to a slightly larger value.
- A medium value (4) to check the general case.
- A larger value (10) to check the handling of more complex scenarios.

By using this test input, we can observe if the buggy code fails to produce the correct output for any of these cases, thereby exposing the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define st first
#define nd second
typedef long long ll;
#define all(x) begin(x),end(x)
int main()
{
   ios_base::sync_with_stdio(false);cin.tie(0); cout.tie(0);
   ll z;
   cin>>z;
   while(z--)
   {
      ll n;
      ll mod=998244353;
      cin>>n;
      vector<ll>a(n+1,1);
      ll sum=2;
      for(int i=4;i<n+1;i++)
      {
            a[i]=sum-i+5;
            a[i]%=mod;
            sum+=(a[i]*i)%mod;
            sum%=mod;
      }
      ll wyn=0;
      for(int i=2;i<n+1;i++)
      {
            wyn=(wyn+a[i])%mod;
      }
      cout<<wyn<<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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has a few potential issues:

1. **Incorrect Initialization and Calculation**: The code initializes `a` with all elements set to 1, which is incorrect for the problem's requirements. The correct initialization should be `a[2] = 1` and `a[3] = 2`.

2. **Incorrect Loop Boundaries and Logic**: The loop starts from `i = 4` and calculates `a[i]` based on a formula that might not be correct for the problem's requirements.

3. **Sum Calculation**: The sum calculation inside the loop might be incorrect, leading to wrong values for `a[i]`.

Given these issues, we can focus on the following scenarios:

- **Small `n` values**: To ensure the initialization and early calculations are correct.
- **Large `n` values**: To ensure the loop and sum calculations are correct and handle large values correctly.
- **Boundary cases**: Values like `n = 2`, `n = 3`, and `n = 4` to check the correctness of the initial conditions and early calculations.

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

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

This test input covers:
- The smallest possible value of `n` (2).
- The next few values (3, 4, 5, 6) to check the correctness of the loop and sum calculations.

This should help in identifying the bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

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

using namespace std;

typedef long long LL;
#define FORU(loopvar, init, end) for (int loopvar = (init); loopvar < (end); loopvar++)
#define FOR(loopvar, init, end) for (int loopvar = (init); loopvar <= (end); loopvar++)
template<typename T>
vector<T> readvec(int n) {
  vector<T> rtn(n + 1);
  FOR (i, 1, n) { cin >> rtn[i]; }
  return rtn;
}

#define CASES                                                                                \
  int _cs;                                                                                \
  cin >> _cs;                                                                                \
  FOR (__cs, 1, _cs)

template<class Int>
Int power(Int x, int k) {
  Int ans = 1;
  for (LL i = (1LL << 30); i; i >>= 1) {
    ans = ans * ans;
    if (k & i) { ans = ans * x; }
  }
  return ans;
}

template<int MOD>
struct modint_ {
  modint_() {}

  template<class Int>
  modint_(Int x_) : x(((LL)x_ + MOD) % MOD) {}

  modint_ operator+(const modint_& rhs) const { return ((LL)x + rhs.x) % MOD; }
  modint_ operator-(const modint_& rhs) const { return ((LL)x - rhs.x + MOD) % MOD; }
  modint_ operator*(const modint_& rhs) const { return ((LL)x * rhs.x) % MOD; }
  modint_ operator/(const modint_& rhs) const { return ((LL)x * power(rhs, MOD - 2).x) % MOD; }

  modint_& operator+=(const modint_& rhs) {
    *this = *this + rhs;
    return *this;
  }
  modint_& operator-=(const modint_& rhs) {
    *this = *this - rhs;
    return *this;
  }
  modint_& operator*=(const modint_& rhs) {
    *this = *this * rhs;
    return *this;
  }
  modint_& operator/=(const modint_& rhs) {
    *this = *this / rhs;
    return *this;
  }

  int x;
};

template<int MOD>
ostream& operator<<(ostream& os, const modint_<MOD>& x) {
  return (os << x.x);
}

using modint = modint_<998244353>;

modint solve(int n) {
  vector<vector<modint>> g(n + 1, vector<modint>(3, 0));
  FOR (i, 2, n) { g[i][0] = 1; }
  FOR (i, 3, n) { g[i][1] = i - 2; }
  FOR (i, 4, n) { g[i][2] = i * (i - 3) / 2; }
  if (n <= 4) { return g[n][0] + g[n][1] + g[n][2]; }
  vector<vector<modint>> f(n + 1, vector<modint>(2, 0));
  FOR (i, 1, 4) { f[i][0] = g[i][2]; }
  modint sum_f = 0;
  FOR (i, 5, n) {
    sum_f += f[i - 1][0] * (i - 1);
    f[i][0] = g[i][2] + sum_f;
    f[i][1] = f[i - 1][0] + f[i - 1][1];
  }
  return f[n][0] + f[n][1] + g[n][0] + g[n][1];
}

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

  CASES {
    int n;
    cin >> n;
    cout << solve(n) << endl;
  }

  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 buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The key aspects to consider are:

1. **Small-scale Input**: The input should be small enough to be valid but complex enough to expose the bug.
2. **Edge Cases**: Consider the smallest and largest possible values for `n` within the constraints.
3. **Boundary Conditions**: Test the boundary values for `n` where the behavior of the algorithm might change.

Given the constraints:
- \(1 \le t \le 10^4\)
- \(2 \le n \le 10^6\)
- The sum of \(n\) over all test cases does not exceed \(10^6\)

Let's create a test input that includes:
- The smallest possible value for \(n\) (2).
- A value just above the smallest (3).
- A value just below the largest possible value (10^6 - 1).
- The largest possible value for \(n\) (10^6).

Here is the test input:

```plaintext
4
2
3
999999
1000000
```

This input includes:
- \(n = 2\): The smallest possible value.
- \(n = 3\): Just above the smallest value.
- \(n = 999999\): Just below the largest possible value.
- \(n = 1000000\): The largest possible value.

These values will help in identifying any issues in the algorithm, especially at the boundaries and with large inputs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
#ifdef _DEBUG
#define _CRT_SECURE_NO_WARNINGS
#endif
#include <iostream>
#include <fstream>
#include <sstream>
#include <climits>
#include <iomanip>
#include <cstdio>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <chrono>
#include <random>
#include <complex>
#include <numeric>
#include <assert.h>

using namespace std;

using ll = long long;
using ull = unsigned long long;
using ui = unsigned int;
using ld = long double;
using pii = pair<int, int>;
using pil = pair<int, long long>;
using pli = pair<long long, int>;
using pll = pair<long long, long long>;
template <typename T>
using uset = unordered_set<T>;
template <typename T1, typename T2>
using umap = unordered_map<T1, T2>;

#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

template <uint32_t mod>
class Montgomery {
    using mint = Montgomery<mod>;

private:
    uint32_t value;

    static constexpr uint32_t inv_neg_mod = (mod == 998'244'353) ? 998'244'351 : ((mod == 1'000'000'007) ? 2226617417 : 0);

    static constexpr uint32_t neg_mod = (0 - uint64_t(mod)) % mod; // since R = 2^32, R^2 = 2^64 and 2^64 mod MOD = (2^64 - MOD) mod MOD

    static uint32_t reduce(const uint64_t& value) {
        return (value + uint64_t(uint32_t(value) * inv_neg_mod) * mod) >> 32;
    }

public:
    Montgomery() : value(0) {}

    Montgomery(const mint& other) : value(other.value) {}

    template<typename T>
    Montgomery(const T& x) : value(!x ? 0 : reduce(uint64_t(x % int32_t(mod) + int32_t(mod)) * neg_mod)) {}

    uint32_t get() const {
        uint32_t real = reduce(value);
        return real < mod ? real : real - mod;
    }

    mint& operator=(const mint& other) {
        value = other.value;
        return *this;
    }

    mint& operator+=(const mint& other) { // operations are made in range [0, 2 * mod)
        if (int32_t(value += other.value - (mod << 1)) < 0) {
            value += mod << 1;
        }
        return *this;
    }

    mint& operator-=(const mint& other) {
        if (int32_t(value -= other.value) < 0) {
            value += mod << 1;
        }
        return *this;
    }

    mint& operator*=(const mint& other) {
        value = reduce(uint64_t(value) * other.value);
        return *this;
    }

    mint& operator/=(const mint& other) {
        return *this *= other.inv();
    }

    friend mint operator+(const mint& a, const mint& b) {
        return mint(a) += b;
    }

    friend mint operator-(const mint& a, const mint& b) {
        return mint(a) -= b;
    }

    friend mint operator*(const mint& a, const mint& b) {
        return mint(a) *= b;
    }

    friend mint operator/(const mint& a, const mint& b) {
        return mint(a) /= b;
    }

    mint operator-() const {
        return mint(0) - *this;
    }

    bool operator==(const mint& other) const {
        return get() == other.get();
    }

    bool operator!=(const mint& other) const {
        return get() != other.get();
    }

    bool operator<(const mint& other) const {
        return get() < other.get();
    }

    friend std::istream& operator>>(std::istream& in, mint& other) {
        std::string s;
        in >> s;
        other = 0;
        bool neg = s[0] == '-';
        for (const auto c : s)
            if (c != '-')
                other = other * 10 + (c - '0');

        if (neg)
            other *= -1;

        return in;
    }

    friend std::ostream& operator<<(std::ostream& out, const mint& other) {
        return out << other.get();
    }

    template <typename T>
    mint power(T pw) const {
        mint b = 1, a = *this;
        for (; pw; pw >>= 1, a *= a)
            if (pw & 1)
                b *= a;
        return b;
    }

    mint inv() const {
        return power(mod - 2);
    }

    static int32_t primitive_root() {
        if constexpr (mod == 1'000'000'007)
            return 5;
        if constexpr (mod == 998'244'353)
            return 3;
        if constexpr (mod == 786433)
            return 10;
        return 0;
    }
};

const int mod = 998244353;
using mint = Montgomery<mod>;

set<vector<int>> st;

int n;

void gen(vector<int> arr) {
    if (arr.size() == n) {
        int invs = 0;
        for (int i = 0; i < arr.size(); i++)
            for (int j = 0; j < i; j++)
                if (arr[j] > arr[i])
                    invs++;
        st.insert(arr);
        return;
    }
    int invs = 0;
    for (int i = 0; i < arr.size(); i++)
        for (int j = 0; j < i; j++)
            if (arr[j] > arr[i])
                invs++;
    for (int i = 0; i <= arr.size(); i++) {
        auto b = arr;
        b.insert(b.begin() + i, invs);
        gen(b);
    }
}

const mint i2 = mint(2).inv();

inline void solve() {
    cin >> n;
    vector<mint> dp(n + 1);
    mint sum = 0;
    for (int len = 3; len <= n; len++) {
        if (len < n)
            dp[len + 1] += (len - 3) * (len - 2) * i2 + 2 * (len - 2);
        dp[len] += sum;
        sum += dp[len] * len;
    }
    mint ans = 0;
    for (int i = 1; i <= n; i++)
        ans += dp[i];
    // 0 invs
    ans += 1;
    // 1 inv
    ans += n - 2;
    cout << ans << '\n';
}

signed main() {
#ifdef _DEBUG
    freopen("01.dat", "r", stdin);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int tt = 1;
    cin >> tt;
    while (tt--) {
#ifdef _DEBUG
        cout << "Testcase#\n";
#endif
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Understanding the Buggy Code**:
   - The code uses dynamic programming to count the number of distinct arrays of length `n` that can be obtained.
   - It initializes a DP array and iterates through lengths to update the count.
   - The code has a specific formula for updating the DP values, which might have logical errors.

2. **Identifying Edge Cases**:
   - **Small `n`**: Test cases with small values of `n` (e.g., `n = 2`, `n = 3`) can help verify the base cases.
   - **Large `n`**: Test cases with large values of `n` (e.g., `n = 10^6`) can stress the algorithm and expose potential overflow or logical errors.
   - **Specific Values**: Test cases with specific values that might cause the algorithm to behave unexpectedly (e.g., `n = 100`, `n = 1000`).

3. **Generating the Test Input**:
   - We will create a test input with a mix of small and large values to cover both base cases and stress tests.
   - The number of test cases should be manageable to ensure the test input is small-scale.

Here is a small-scale test input that includes a mix of edge cases and specific values:

```plaintext
5
2
3
100
1000
1000000
```

This test input includes:
- `n = 2` to test the base case.
- `n = 3` to test a slightly larger base case.
- `n = 100` to test a mid-range value.
- `n = 1000` to test a larger value.
- `n = 1000000` to test a very large value, which can stress the algorithm and expose potential issues.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
import java.io.*;
import java.util.*;
public class P6 {
    public static int MOD = 998244353;
    public static void main(String[] args) throws IOException {
        BufferedReader scan = new BufferedReader(new InputStreamReader(System.in)); PrintWriter out = new PrintWriter(System.out);
        StringTokenizer l1 = new StringTokenizer(scan.readLine());
        int t = Integer.parseInt(l1.nextToken());
        for (int q = 0;q<t;q++) {
            StringTokenizer st = new StringTokenizer(scan.readLine());
            int num = Integer.parseInt(st.nextToken());

            long[] dp = new long[num]; dp[num-1] = 1; //Number of final arrays we can make with a current array of length i+1 and having # inversion>max(a_i)
            long[] ss = new long[num]; ss[num-1] = 1; //Suffix Sum
            for (int i = num-2;i>=0;i--) {
                dp[i] = ((i+1)*ss[i+1]+1)%MOD;
                ss[i] = (ss[i+1]+dp[i])%MOD;
            }

            long ans = 0;
            for (int i = 2;i<num;i++) {
                int l = i+1;
                ans = (ans+((long) (l-2)*(l-1)/2-1)*dp[i])%MOD;
                //If the 0,1,0 is at the front, then 2 places. If it's at the back, then total length is l-1 (-1 since we've yet to place the 1) so l-2 total places to put it.
            }
            ans += num-1; //If the final number of inversions is <= 1.
            out.println(ans);
        }
        out.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

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

// #pragma GCC optimize ("Ofast")
// #pragma GCC optimize ("unroll-loops")
// #pragma GCC target ("avx,avx2,tune=native")

#ifdef LOCAL
  #include "algo/debug.h"
#else
  #define debug(...) 42
#endif

using ll = long long;
using db = double;  // or db, if TL is tight
#define int ll

// pairs
#define mp make_pair
#define f first
#define s second

// vectors
#define sz(v) (int)((v).size())
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

const int N = 1e6, MOD = 998244353;

int a[N + 1], d[N + 1], f[N + 1];

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

  for (int i = 4; i <= N; i++)
    a[i] = (i * a[i - 1] % MOD + i - 2 + MOD) % MOD;

  for (int i = 1; i <= N - 2; i++)
    d[i] = (a[i] - a[i - 1] + MOD) % MOD;

  f[2] = 1, f[3] = 2;
  for (int i = 4; i <= N; i++)
    f[i] = (f[i - 1] * 2 - f[i - 2] + d[i] + MOD) % MOD;

  int tc; cin >> tc;

  while (tc--) {
    int n; cin >> n;

    cout << f[n] << "\n";

    // set<vector<int>> S;
    // auto solve = [&](auto &&self, vector<int> a) -> void {
    //   if (sz(a) > n) return;
    //   if (sz(a) == n) {
    //     S.insert(a);
    //     return;
    //   }

    //   int k = 0;
    //   for (int i = 0; i < sz(a); i++) {
    //     for (int j = i + 1; j < sz(a); j++) {
    //       if (a[i] > a[j]) k++;
    //     }
    //   }

    //   for (int i = 0; i <= sz(a); i++) {
    //     vector<int> newA;
    //     for (int j = 0; j < i; j++)
    //       newA.push_back(a[j]);

    //     newA.push_back(k);

    //     for (int j = i; j < sz(a); j++)
    //       newA.push_back(a[j]);

    //     self(self, newA);
    //   }
    // };

    // solve(solve, vector<int>{0, 1});

    // cout << sz(S) << "\n";
  }
}

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

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

Think step by step.

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

1. **Understanding the Bug**:
   - The code calculates the number of distinct arrays of length `n` that can be obtained by repeatedly inserting the number of inversions.
   - The key part of the code is the dynamic programming array `f` which is used to store the number of possible arrays of each length.
   - The bug might be in the transition formula for `f[i]`.

2. **Edge Cases**:
   - Small values of `n` (e.g., `2`, `3`, `4`) are already handled in the example and might not expose the bug.
   - Larger values of `n` (e.g., close to the upper limit `10^6`) might be more likely to expose issues in the transition formula.

3. **Maximizing Failure Probability**:
   - Test cases with values of `n` that are just above the initial small values (e.g., `5`, `6`, `7`) to see if the transition formula starts to fail.
   - Test cases with values of `n` that are close to the upper limit to see if there are any issues with large inputs.

4. **Generating the Test Input**:
   - We will create a small-scale test input with a few test cases that include both small and large values of `n`.

Here is the test input:

```plaintext
4
5
7
10
999999
```

This test input includes:
- A small value (`5`) to check if the transition formula works correctly.
- A slightly larger value (`7`) to further test the transition formula.
- A medium value (`10`) to see if there are any issues with moderate inputs.
- A very large value (`999999`) to check if the code handles large inputs correctly.

This should help in exposing any bugs in the code.