**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

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

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
import bisect
import math
from collections import deque
from decimal import Decimal, getcontext
import sys
sys.setrecursionlimit(100000)
getcontext().prec = 100

for _ in range(int(input())):
    n=int(input())
    a=[n+10]+list(map(int,input().split()))+[n+10]
    b=list(a)
    dk=1
    while(dk):
        dk=0
        for i in range(1,n//2):
            t1=(a[i]==a[i+1]) + (a[i]==a[i-1]) + (a[n-i+1]==a[n-i]) + (a[n-i+1]==a[n-i+2])
            t2=(a[n-i+1]==a[i+1]) + (a[n-i+1]==a[i-1]) + (a[i]==a[n-i]) + (a[i]==a[n-i+2])
            if t1>t2:
                dk=1
                a[i],a[n-i+1]=a[n-i+1],a[i]
        if (a[n//2]==a[n//2-1])+ (a[n-n//2+1]==a[n-n//2+2])>(a[n-n//2+1]==a[n//2-1])+ (a[n//2]==a[n-n//2+2]):
            dk=1
            a[n//2],a[n-n//2+1]=a[n-n//2+1],a[n//2]

    d1=d2=0
    for i in range(1,n+1):
        if a[i]==a[i+1]:
            d1+=1

    print(d1)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

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

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
for _ in range(int(input())):
    n=int(input())
    s1=input()
    a1=s1.split()
    a1=[int(x) for x in a1]
    t=0
    for k in range(2):
        for i in range((n//2)-1):
            '''if i==0 and ((a1[i]==a1[i+1] and a1[i]!=a1[n-i-2]) or (a1[n-i-1]==a1[n-i-2] and a1[n-i-1]!=a1[i+1])):
                a1[i],a1[n-i-1]=a1[n-i-1],a1[i]
           '''
            if(a1[i]==a1[i+1] and a1[i]!=a1[n-i-2] and a1[n-i-1]!=a1[i+1]) or (a1[n-i-1]==a1[n-i-2] and a1[n-i-1]!=a1[i+1] and a1[i]!=a1[n-i-2]):
                a1[i],a1[n-i-1]=a1[n-i-1],a1[i]

    for i in range(n-1):
        if a1[i]==a1[i+1]:
            t+=1

    print(t)

**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
6
1 1 2 2 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

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

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
import java.util.Scanner;

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

        int t = sc.nextInt(); // Number of test cases
        while (t-- > 0) {
            int n = sc.nextInt(); // Number of students
            int[] a = new int[n];

            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt(); // Topic of interest for student i
            }

            // Calculate initial disturbance
            int minDisturbance = calculateDisturbance(a, n);

            // Try swapping elements
            for (int i = 0; i < n / 2; i++) {
                // Swap a[i] with a[n - i - 1]
                swap(a, i, n - i - 1);
                // Calculate disturbance after swap
                int currentDisturbance = calculateDisturbance(a, n);
                // Update minimum disturbance if current is lower
                if (currentDisturbance < minDisturbance) {
                    minDisturbance = currentDisturbance;
                }
                // Restore the original array
                swap(a, i, n - i - 1);
            }

            System.out.println(minDisturbance);
        }

        sc.close();
    }

    // Method to calculate disturbance
    private static int calculateDisturbance(int[] a, int n) {
        int disturbance = 0;
        for (int i = 0; i < n - 1; i++) {
            if (a[i] == a[i + 1]) {
                disturbance++;
            }
        }
        return disturbance;
    }

    // Method to swap elements in the array
    private static void swap(int[] a, int i, int j) {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }
}

**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
1 1
3
1 2 1
4
1 2 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

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

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.LinkedList;

public class Main {
    private static final int BUFFER_SIZE = 1 << 16;
    private static byte[] buffer = new byte[BUFFER_SIZE];
    private static int idx = 0;
    private static int bytesRead = 0;
    private static InputStream is = System.in;

    private static void fillBuffer() throws IOException {
        if (bytesRead == -1)
            return; // EOF reached
        bytesRead = is.read(buffer, 0, BUFFER_SIZE);
        idx = 0;
    }

    public static int nextInt() throws IOException {
        int num = 0;
        byte c = read();
        while (c <= ' ')
            c = read(); // Skip spaces/newlines
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do {
            num = num * 10 + (c - '0');
            c = read();
        } while (c >= '0' && c <= '9');
        return neg ? -num : num;
    }

    public static String nextLine() throws IOException {
        StringBuilder sb = new StringBuilder();
        byte c = read();
        while (c != '\n' && c != '\r') {
            sb.append((char) c);
            c = read();
        }
        if (c == '\r')
            read(); // handle CRLF
        return sb.toString();
    }

    private static byte read() throws IOException {
        if (idx == bytesRead) {
            fillBuffer();
            if (bytesRead == -1)
                return -1; // EOF
        }
        return buffer[idx++];
    }

    public static void main(String[] args) throws IOException {
        int t = nextInt();

        LinkedList<Integer> result = new LinkedList<>();

        while (t-- > 0) {
            int initialDisturbance = 0;
            int n = nextInt();
            int[] arr = new int[n];
            int pre = 0;
            for (int j = 0; j < n; j++) {
                arr[j] = nextInt();
                if (pre == arr[j])
                    initialDisturbance++;
                pre = arr[j];
            }

            for (int j = 0; j < 2; j++) {
                for (int i = 0; i < n / 2; i++) {
                    swap(arr, i, n - i - 1);
                    int newDisturbance = countDisturbance(arr);

                    if (newDisturbance > initialDisturbance) {
                        swap(arr, i, n - i - 1);
                    } else {
                        initialDisturbance = newDisturbance;
                    }
                }

            }

            result.add(initialDisturbance);
        }

        for (int res : result) {
            System.out.println(res);
        }
    }

    static int countDisturbance(int[] arr) {
        int disturbance = 0;
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] == arr[i - 1]) {
                disturbance++;
            }
        }
        return disturbance;
    }

    static void swap(int[] arr, int i, int j) {
        arr[i] ^= arr[j];
        arr[j] ^= arr[i];
        arr[i] ^= arr[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 test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code provided has a few potential issues:

1. **Swapping Logic**: The code swaps elements multiple times and checks if the disturbance increases. If it does, it swaps back. This approach might not always find the optimal solution, especially if the initial disturbance is already low.
2. **Multiple Swaps**: The code performs two rounds of swaps, which might not be necessary and could lead to suboptimal results.
3. **Edge Cases**: The code might not handle edge cases well, such as when the array is already optimal or when the array has very few unique elements.

Let's consider a test input that can expose these issues:

- **Array with repeated elements**: An array where many elements are the same, but not all, to see if the code can minimize the disturbance effectively.
- **Array with a single unique element in the middle**: An array where the unique element is in the middle to check if the code can handle this edge case.
- **Array with a single unique element at the ends**: An array where the unique element is at the ends to see if the code can handle this edge case.

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

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

Explanation:
1. **Test Case 1**: `1 1 2 1 1`
   - Initial disturbance: 3 (1-1, 1-1, 1-1)
   - Optimal disturbance: 1 (after swapping 2 with 1 at index 2, the array becomes `1 1 1 1 2`)

2. **Test Case 2**: `1 1 1 2 2 2`
   - Initial disturbance: 4 (1-1, 1-1, 2-2, 2-2)
   - Optimal disturbance: 2 (after swapping 1 at index 1 with 2 at index 4, the array becomes `1 2 1 2 1 2`)

3. **Test Case 3**: `1 1 1 3 1 1 1`
   - Initial disturbance: 5 (1-1, 1-1, 1-1, 1-1, 1-1)
   - Optimal disturbance: 3 (after swapping 3 with 1 at index 3, the array becomes `1 1 1 1 1 1 3`)

This test input should help expose the issues in the code, particularly in handling repeated elements and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

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

struct str{

};

pair<int, int> find_anc(int x, int y, vector<vector<pair<int, int>>>& parent) {

    if ((x == parent[x][y].first) && (y == parent[x][y].second)) {

        return {x, y};
    }
    return parent[x][y] = find_anc(parent[x][y].first, parent[x][y].second, parent);
}

void union_by_size(int x1, int y1, int x2, int y2, vector<vector<pair<int, int>>>& parent, vector<vector<int>>& size) {

    pair<int, int> p1 = find_anc(x1, y1, parent);
    pair<int, int> p2 = find_anc(x2, y2, parent);

    if (p1 == p2) return ;

    x1 = p1.first;
    y1 = p1.second;

    x2 = p2.first;
    y2 = p2.second;

    if (size[x1][y1] < size[x2][y2]) {

        parent[x1][y1] = {x2, y2};
        size[x2][y2] += size[x1][y1];
    }
    else {

        parent[x2][y2] = {x1, y1};
        size[x1][y1] += size[x2][y2];
    }
}

bool isValid(int x, int y, int n, int m) {

    if ((x < 0) || (y < 0) || (x == n) || (y == m)) return false;
    return true;
}

int main(){
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;

    while (t --) {

        int n, m;
        cin >> n >> m;

        vector<vector<pair<int, int>>> parent(n, vector<pair<int, int>>(m));
        vector<vector<int>> size(n, vector<int>(m, 1));

        for (int i = 0 ; i < n ; i ++) {

            for (int j = 0 ; j < m ; j ++) {

                parent[i][j] = {i, j};
            }
        }

        string a[n];
        for (int i = 0 ; i < n ; i ++) {

            cin >> a[i];
        }

        for (int i = 0 ; i < n ; i ++) {

            for (int j = 0 ; j < m ; j ++) {

                if (a[i][j] == '#') {

                    if (isValid(i + 1, j, n, m) && (a[i + 1][j] == '#')) {

                        union_by_size(i, j, i + 1, j, parent, size);
                    }

                    if (isValid(i, j + 1, n, m) && (a[i][j + 1] == '#')) {

                        union_by_size(i, j, i, j + 1, parent, size);
                    }

                    if (isValid(i - 1, j, n, m) && (a[i - 1][j] == '#')) {

                        union_by_size(i, j, i - 1, j, parent, size);
                    }

                    if (isValid(i, j - 1, n, m) && (a[i][j - 1] == '#')) {

                        union_by_size(i, j, i, j - 1, parent, size);
                    }
                }
            }
        }

        vector<set<pair<int, int>>> row(n);
        vector<int> dcrow(n, 0);
        vector<set<pair<int, int>>> col(m);
        vector<int> dccol(m, 0);

        for (int i = 0 ; i < n ; i ++) {

            for (int j = 0 ; j < m ; j ++) {

                if (a[i][j] == '.') {

                    dcrow[i] ++;
                    dccol[j] ++;
                }
            }
        }

        int ans = 0;

        for (int i = 0 ; i < n ; i ++) {

            set<pair<int, int>> st;

            for (int j = 0 ; j < m ; j ++) {

                if (a[i][j] == '#') {

                    st.insert(find_anc(i, j, parent));
                }
                else {

                    if (isValid(i + 1, j, n, m) && (a[i + 1][j] == '#')) {

                        st.insert(find_anc(i + 1, j, parent));
                    }

                    if (isValid(i, j + 1, n, m) && (a[i][j + 1] == '#')) {

                        st.insert(find_anc(i, j + 1, parent));
                    }

                    if (isValid(i - 1, j, n, m) && (a[i - 1][j] == '#')) {

                        st.insert(find_anc(i - 1, j, parent));
                    }

                    if (isValid(i, j - 1, n, m) && (a[i][j - 1] == '#')) {

                        st.insert(find_anc(i, j - 1, parent));
                    }
                }
            }

            row[i] = st;

        }

        for (int j = 0 ; j < m ; j ++) {

            set<pair<int, int>> st;

            for (int i = 0 ; i < n ; i ++) {

                if (a[i][j] == '#') {

                    st.insert(find_anc(i, j, parent));
                }
                else {

                    if (isValid(i + 1, j, n, m) && (a[i + 1][j] == '#')) {

                        st.insert(find_anc(i + 1, j, parent));
                    }

                    if (isValid(i, j + 1, n, m) && (a[i][j + 1] == '#')) {

                        st.insert(find_anc(i, j + 1, parent));
                    }

                    if (isValid(i - 1, j, n, m) && (a[i - 1][j] == '#')) {

                        st.insert(find_anc(i - 1, j, parent));
                    }

                    if (isValid(i, j - 1, n, m) && (a[i][j - 1] == '#')) {

                        st.insert(find_anc(i, j - 1, parent));
                    }
                }
            }

            col[j] = st;

            for (int i = 0 ; i < n ; i ++) {

                for (int j = 0 ; j < m ; j ++) {

                    int count = dcrow[i] + dccol[j];

                    if (a[i][j] == '.') count --;

                    set<pair<int, int>> st;

                    for (auto itr = row[i].begin() ; itr != row[i].end() ; itr ++) {

                        st.insert(*itr);
                    }

                    for (auto itr = col[j].begin() ; itr != col[j].end() ; itr ++) {

                        st.insert(*itr);
                    }

                    for (auto itr = st.begin() ; itr != st.end() ; itr ++) {

                        count += size[(*itr).first][(*itr).second];
                    }

                    ans = max(ans, count);

                }
            }

        }

        cout << ans << "\n";
    }

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
#..
.#.
..#
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
mt19937 rnd(time(0));
#define int long long
typedef tuple<int,int,int> tp;
#define x first
#define y second
typedef pair<int,int> pii;
constexpr int N=5000010,mod=998244353,inf=1e18;
constexpr double pi=3.1415926535897932384626;
const ll  P=rnd()%mod;
#define all(a) a.begin(),a.end()
#define get_count(x) __builtin_popcount(x)
int dx[]={1,-1,0,0,1,1,-1,-1,0};
int dy[]={0,0,1,-1,1,-1,-1,1,0};
int p[N],sz[N],lx[N],rx[N],ly[N],ry[N];
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    int __=1;
    cin >> __;
    while(__--){
        int n,m;
        cin >> n >> m;
        for(int i=0;i<n*m;i++) p[i]=i,sz[i]=1,lx[i]=ly[i]=inf,rx[i]=ry[i]=-inf;
        vector<vector<char>> g(n+1,vector<char>(m+1));
        vector<vector<int>> a(n+1,vector<int>(m+1));
        auto get_id=[&](int i,int j)->int{
            return i*m+j;
        };
        function<int(int)> find=[&](int x){
            if(x!=p[x]) p[x]=find(p[x]);
            return p[x];
        };
        auto merge=[&](int x,int y)->void{
            x=find(x),y=find(y);
            if(x!=y){
                p[x]=y;
                sz[y]+=sz[x];
            }
        };
        auto insert=[&](int x1,int y1,int x2,int y2,int d)->void{
            a[x1][y1]+=d;
            if(y2+1<m) a[x1][y2+1]-=d;
            if(x2+1<n) a[x2+1][y1]-=d;
            if(x2+1<n&&y1+m<m) a[x2+1][y2+1]+=d;
        };
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin >> g[i][j];
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(g[i][j]=='.') continue;
                for(int k=0;k<4;k++){
                    int x=i+dx[k],y=j+dy[k];
                    if(x<0||x>=n||y<0||y>=m) continue;
                    if(g[x][y]=='.') continue;
                    merge(get_id(i,j),get_id(x,y));
                }
            }
        }
        vector<int> c(n),d(m);
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(g[i][j]=='.'){
                    c[i]++;
                    d[j]++;
                }
                else{
                    int u=get_id(i,j);
                    lx[u]=min(lx[u],i);
                    rx[u]=max(rx[u],i);
                    ly[u]=min(ly[u],j);
                    ry[u]=max(ry[u],j);
                }

            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(g[i][j]=='.') continue;
                int u=get_id(i,j);
                if(find(u)==u){
                    int x1=max(lx[u]-1,0ll),x2=rx[u]+1,y1=max(ly[u]-1,0ll),y2=ry[u]+1;
                    insert(x1,0,x2,m-1,sz[u]);
                    insert(0,y1,n-1,y2,sz[u]);
                    insert(x1,y1,x2,y2,-sz[u]);
                }
            }
        }
        int ans=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(i>0) a[i][j]+=a[i-1][j];
                if(j>0) a[i][j]+=a[i][j-1];
                if(i>0&&j>0) a[i][j]-=a[i-1][j-1];
                int x=c[i]+d[j]+a[i][j]-(g[i][j]=='.');
                ans=max(ans,x);
            }
        }
        cout << ans << '\n';
    }
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - Single cell grid.
   - Grid with all cells as '#'.
   - Grid with all cells as '.'.
   - Grid with a single '.' in the middle.

2. **Complex Scenarios:**
   - Grid with a large connected component and a small isolated component.
   - Grid with multiple small connected components.
   - Grid with a single row or column that is entirely '#'.

3. **Operation Impact:**
   - Ensure the operation (setting all cells in a row and a column to '#') significantly changes the connected component.

Let's create a small-scale test input that fits these criteria:

```plaintext
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
```

This input includes:
- A single cell grid.
- A 4x2 grid with a pattern of '#' and '.'.
- A 3x5 grid with a pattern of '#' and '.'.
- A 5x5 grid with a large connected component and a small isolated component.
- A 6x6 grid with multiple small connected components.
- A 6x8 grid with a complex pattern of '#' and '.'.

Each of these test cases will help in identifying potential issues in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll int
// void dfs(ll num,ll row,ll col,vector<vector<ll>> &vis,vector<vector<char>> &a,map<ll,ll> &mpp){
//     vis[row][col]=num;
//     mpp[num]++;
//     if(row-1>=0&&a[row-1][col]=='#'&&vis[row-1][col]==0){
//         dfs(num,row-1,col,vis,a,mpp);
//     }
//     if(row+1<a.size()&&a[row+1][col]=='#'&&vis[row+1][col]==0){
//         dfs(num,row+1,col,vis,a,mpp);
//     }
//     if(col-1>=0&&a[row][col-1]=='#'&&vis[row][col-1]==0){
//         dfs(num,row,col-1,vis,a,mpp);
//     }
//     if(col+1<a[0].size()&&a[row][col+1]=='#'&&vis[row][col+1]==0){
//         dfs(num,row,col+1,vis,a,mpp);
//     }
// }
void bfs(ll num, ll startRow, ll startCol, vector<vector<ll>> &vis, vector<vector<char>> &a, map<ll, ll> &mpp) {
    queue<pair<ll, ll>> q;
    q.push({startRow, startCol});
    vis[startRow][startCol] = num;
    mpp[num]++;

    ll n = a.size(), m = a[0].size();
    vector<pair<ll, ll>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    while (!q.empty()) {
        auto [row, col] = q.front();
        q.pop();
        for (auto &d : directions) {
            ll newRow = row + d.first;
            ll newCol = col + d.second;
            if (newRow >= 0 && newRow < n && newCol >= 0 && newCol < m && a[newRow][newCol] == '#' && vis[newRow][newCol] == 0) {
                vis[newRow][newCol] = num;
                mpp[num]++;
                q.push({newRow, newCol});
            }
        }
    }
}
void solve(){
    ll n,m;
    cin>>n>>m;
    map<ll,ll> mpp;
    vector<vector<char>> a(n,vector<char>(m));
    for(ll i=0;i<n;i++){
        for(ll j=0;j<m;j++){
            cin>>a[i][j];
        }
    }
    vector<vector<ll>> vis(n,vector<ll>(m,0));
    vector<ll> cols(m,0);
    vector<ll> rows(n,0);
    ll num=1;
    for(ll i=0;i<n;i++){
        for(ll j=0;j<m;j++){
            if(a[i][j]=='#'&&vis[i][j]==0){
                bfs(num, i, j, vis, a, mpp);
                num++;
            }
            if(a[i][j]=='.'){
                cols[j]++;
                rows[i]++;
            }
        }
    }
    set<ll> row[n];
    set<ll> col[m];
    for(ll i=0;i<n;i++){
        for(ll j=0;j<m;j++){
            row[i].insert(vis[i][j]);
            col[j].insert(vis[i][j]);
           if(i-1>=0&&vis[i-1][j]!=0){
              row[i].insert(vis[i-1][j]);
          }
            if(i+1<n&&vis[i+1][j]!=0){
              row[i].insert(vis[i+1][j]);
          }
            if(j-1>=0&&vis[i][j-1]!=0){
                col[j].insert(vis[i][j-1]);
            }
            if(j+1<m&&vis[i][j+1]!=0){
                col[j].insert(vis[i][j+1]);
            }
        }
    }
    ll ans1=0;
    // for(ll i=0;i<n;i++){
    //     ll ans=0;
    //     set<ll> st;
    //     ll cnt=0;
    //     for(ll j=0;j<m;j++){
    //       st.insert(vis[i][j]);
    //       if(i-1>=0&&vis[i-1][j]!=0){
    //           st.insert(vis[i-1][j]);
    //       }
    //       if(i+1<n&&vis[i+1][j]!=0){
    //           st.insert(vis[i+1][j]);
    //       }
    //       if(a[i][j]=='.'){
    //         cnt++;
    //       }
    //     }
    //     row[i]=st;
    //     for(auto it:st){
    //         ans+=mpp[it];
    //     }
    //     ans+=cnt;
    //     ans1=max(ans1,ans);
    //     rows[i]=cnt;
    // }
    // for(ll i=0;i<m;i++){
    //     ll ans=0;
    //     set<ll> st;
    //     ll cnt=0;
    //     for(ll j=0;j<n;j++){
    //       st.insert(vis[j][i]);
    //       if(i-1>=0&&vis[j][i-1]!=0){
    //           st.insert(vis[j][i-1]);
    //       }
    //       if(i+1<m&&vis[j][i+1]!=0){
    //           st.insert(vis[j][i+1]);
    //       }
    //       if(a[j][i]=='.'){
    //         cnt++;
    //       }
    //     }
    //     col[i]=st;
    //     for(auto it:st){
    //         ans+=mpp[it];
    //     }
    //     ans+=cnt;
    //     ans1=max(ans1,ans);
    //     cols[i]=cnt;
    // }
    ll ans2=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            ll ans=0;
            set<ll> st;
            for(auto it:row[i]){
                st.insert(it);
            }
            for(auto it:col[j]){
                st.insert(it);
            }
            for(auto it:st){
                ans+=mpp[it];
            }
            ans+=rows[i]+cols[j];
            if(a[i][j]=='.'){
                ans--;
            }
            ans2=max(ans2,ans);
        }

    }
    cout<<ans2<<endl;
}
int main(){
ios_base::sync_with_stdio(0);
cout.tie(0);
cin.tie(0);
int t;
cin>>t;
while(t--)
solve();
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
#..
...
..#
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
import sys
from collections import deque
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
move = [(0, 1), (0, -1), (-1, 0), (1, 0)]
mrsh = [(1, 1), (1, -1), (-1, 1), (1, 1)]
for __ in range(fg()):
    n, m = fgh()
    a = [[] for i in range(n)]
    r1 = [0] * n
    c1 = [0] * m
    for i in range(n):
        s = input()[:-1]
        s = list(s)
        a[i] = s[:]
        r1[i] = m - s.count('#')
    r0 = [0] * n
    c0 = [0] * m
    for j in range(m):
        cnt = 0
        for i in range(n):
            if a[i][j] == '#':
                cnt += 1
        c1[j] = n - cnt
    vis = [[0] * m for i in range(n)]
    yy = [[0] * m for i in range(n)]
    for i in range(n):
        for j in range(m):
            if a[i][j] == '#' and vis[i][j] == 0:
                vis[i][j] = 1
                vis1 = {}
                sz = 1
                z = [(i, j)]
                dd = {}
                ee = {}
                q = deque()
                q.append((i, j))
                while q:
                    r, c = q.popleft()
                    for u in move:
                        x, y = r + u[0], c + u[1]
                        if 0 <= r + u[0] < n and 0 <= c + u[1] < m and vis[x][y] == 0:
                            if a[x][y] == '.' and (x, y) not in vis1:
                                z.append((x, y))
                                dd[x] = 1
                                ee[y] = 1
                                vis1[(x, y)] = 1
                            elif a[x][y] == '#' and vis[x][y] == 0:
                                z.append((x, y))
                                dd[x] = 1
                                ee[y] = 1
                                q.append((x, y))
                                vis[x][y] = 1
                                vis1[(x, y)] = 1
                                sz += 1
                    for u in mrsh:
                        x, y = r + u[0], c + u[1]
                        if 0 <= x < n and 0 <= y < m and (x, y) not in vis1:
                            z.append((x, y))
                            vis1[(x, y)] = 1
                for ii in dd:
                    for jj in ee:
                        yy[ii][jj] += sz
                for ii in dd:
                    r0[ii] += sz
                for ii in ee:
                    c0[ii] += sz
    ans = 0
    for i in range(n):
        for j in range(m):
            if a[i][j] == '#':
                ans = max(ans, r1[i] + c1[j] + r0[i] + c0[j] - yy[i][j])
            else:
                ans = max(ans, r1[i] + c1[j] - 1 + r0[i] + c0[j] - yy[i][j])
    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
4 4
....
....
....
....
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
def dfs(grid, i, j, visited):
    if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or visited[i][j] or grid[i][j] == '.':
        return 0
    visited[i][j] = True
    size = 1
    directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]
    for di, dj in directions:
        ni, nj = i + di, j + dj
        size += dfs(grid, ni, nj, visited)
    return size

def max_connected_component_size(n, m, grid):
    max_size = 0
    for r in range(n):
        for c in range(m):
            # Create a copy of the grid to simulate turning row r and col c to '#'
            new_grid = [row[:] for row in grid]
            for i in range(n):
                new_grid[i][c] = '#'
            for j in range(m):
                new_grid[r][j] = '#'

            # Initialize visited array
            visited = [[False]*m for _ in range(n)]

            # Perform DFS from each '#' cell to calculate component size
            for i in range(n):
                for j in range(m):
                    if new_grid[i][j] == '#' and not visited[i][j]:
                        component_size = dfs(new_grid, i, j, visited)
                        max_size = max(max_size, component_size)

    return max_size

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    index = 0
    T = int(data[index])
    index += 1
    results = []

    while T > 0:
        n, m = int(data[index]), int(data[index+1])
        index += 2
        grid = []
        for i in range(n):
            row = list(data[index+i])
            grid.append(row)
        index += n
        results.append(str(max_connected_component_size(n, m, grid)))
        T -= 1

    print('\n'.join(results))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
3
1 1
.
2 2
..
..
3 3
.#.
...
.#.
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
from collections import *

from sys import *
input=lambda:stdin.readline().rstrip()

def valid(i,j):
    if 0<=i<n and 0<=j<m:
        return True
    else:
        return False

def bfs(a,b):
    global min_row,max_row,min_col,max_col,cnt
    que=deque()
    min_row=float('inf')
    min_col=float('inf')
    max_row=float('-inf')
    max_col=float('-inf')
    cnt=0
    que.appendleft((a,b))
    v_a[a][b]=1
    while len(que)!=0:
        x,y=que.pop()
        cnt+=1
        min_row=min(x,min_row)
        min_col=min(y,min_col)
        max_row=max(x,max_row)
        max_col=max(y,max_col)
        for i,j in enumerate(steps):
            if valid(x+j[0],y+j[1]) and v_a[x+j[0]][y+j[1]]==0 and l[x+j[0]][y+j[1]]=='#':
                que.appendleft((x+j[0],y+j[1]))
                v_a[x+j[0]][y+j[1]]=1

def solve():
    global v_a,steps,n,m,l
    steps=[[0,-1],[-1,0],[0,1],[1,0]]
    n,m=map(int,input().split())
    v_a=[[0]*m for i in range(n)]
    l=[input() for i in range(n)]
    if m==1:
        print(n)
        return
    elif n==1:
        print(m)
        return
    row=[0]*n
    col=[0]*m
    dot_cnt_row=[0]*n
    dot_cnt_col=[0]*m
    pre=[[0]*m for i in range(n)]

    for i in range(n):
        for j in range(m):
            if l[i][j]=='.':
                dot_cnt_row[i]+=1
                dot_cnt_col[j]+=1
            if v_a[i][j]==0 and l[i][j]=='#':
                bfs(i,j)
                row[max(min_row-1,0)]+=cnt
                col[max(min_col-1,0)]+=cnt
                pre[max(min_row-1,0)][max(min_col-1,0)]+=cnt
                if max_row+2<n:
                    row[max_row+2]-=cnt
                    pre[max_row+2][max(min_col-1,0)]-=cnt
                if max_col+2<m:
                    col[max_col+2]-=cnt
                    pre[max(min_row-1,0)][max_col+2]-=cnt
                if max_row+2<n and max_col+2<m:
                    pre[max_row+2][max_col+2]+=cnt

    ans=float('-inf')
    for i in range(n):
        row[i]+=((i>0)*row[i-1])
        for j in range(m):
            if i==0 and j>0:
                col[j]+=col[j-1]
            pre[i][j]+=((i>0)*pre[i-1][j]+(j>0)*pre[i][j-1]-(i>0 and j>0)*(pre[i-1][j-1]))
            ans=max(ans,row[i]+col[j]-max(pre[i][j],0)+dot_cnt_row[i]+dot_cnt_col[j]-(l[i][j]=='.'))
    print(ans)
    '''print(pre)
    print(row)
    print(col)'''

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

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

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

Think step by step.

Response:
```plaintext
2
3 3
...
...
...
4 4
#...
....
....
....
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
import java.util.*;

public class MaximizeLargestComponent {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t-- > 0) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            char[][] grid = new char[n][m];
            for (int i = 0; i < n; i++) {
                grid[i] = scanner.next().toCharArray();
            }
            System.out.println(maximizeLargestComponent(grid, n, m));
        }
        scanner.close();
    }

    private static int maximizeLargestComponent(char[][] grid, int n, int m) {
        int maxComponentSize = 0;
        for (int r = 0; r < n; r++) {
            for (int c = 0; c < m; c++) {
                char[][] newGrid = copyGrid(grid, n, m);
                setRowAndColumn(newGrid, r, c, n, m);
                maxComponentSize = Math.max(maxComponentSize, largestComponentSize(newGrid, n, m));
            }
        }
        return maxComponentSize;
    }

    private static char[][] copyGrid(char[][] grid, int n, int m) {
        char[][] newGrid = new char[n][m];
        for (int i = 0; i < n; i++) {
            System.arraycopy(grid[i], 0, newGrid[i], 0, m);
        }
        return newGrid;
    }

    private static void setRowAndColumn(char[][] grid, int r, int c, int n, int m) {
        for (int i = 0; i < n; i++) {
            grid[i][c] = '#';
        }
        for (int j = 0; j < m; j++) {
            grid[r][j] = '#';
        }
    }

    private static int largestComponentSize(char[][] grid, int n, int m) {
        boolean[][] visited = new boolean[n][m];
        int maxSize = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == '#' && !visited[i][j]) {
                    maxSize = Math.max(maxSize, dfs(grid, visited, i, j, n, m));
                }
            }
        }
        return maxSize;
    }

    private static int dfs(char[][] grid, boolean[][] visited, int i, int j, int n, int m) {
        if (i < 0 || i >= n || j < 0 || j >= m || grid[i][j] != '#' || visited[i][j]) {
            return 0;
        }
        visited[i][j] = true;
        int size = 1;
        size += dfs(grid, visited, i + 1, j, n, m);
        size += dfs(grid, visited, i - 1, j, n, m);
        size += dfs(grid, visited, i, j + 1, n, m);
        size += dfs(grid, visited, i, j - 1, n, m);
        return size;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
...
...
...
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

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

public class MaximizeLargestComponentHard {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());

        StringBuilder result = new StringBuilder();
        for (int i = 0; i < t; i++) {
            String[] nm = br.readLine().split(" ");
            int n = Integer.parseInt(nm[0]);
            int m = Integer.parseInt(nm[1]);
            char[][] grid = new char[n][m];

            for (int r = 0; r < n; r++) {
                grid[r] = br.readLine().toCharArray();
            }

            result.append(maximizeComponent(n, m, grid)).append("\n");
        }

        System.out.print(result.toString());
    }

    private static int maximizeComponent(int n, int m, char[][] grid) {
        boolean[][] visited = new boolean[n][m];
        List<Set<Point>> components = new ArrayList<>();

        for (int r = 0; r < n; r++) {
            for (int c = 0; c < m; c++) {
                if (grid[r][c] == '#' && !visited[r][c]) {
                    Set<Point> component = new HashSet<>();
                    Queue<Point> queue = new LinkedList<>();
                    queue.add(new Point(r, c));
                    visited[r][c] = true;

                    while (!queue.isEmpty()) {
                        Point p = queue.poll();
                        component.add(p);
                        for (int[] d : directions) {
                            int nr = p.r + d[0], nc = p.c + d[1];
                            if (nr >= 0 && nr < n && nc >= 0 && nc < m && grid[nr][nc] == '#' && !visited[nr][nc]) {
                                queue.add(new Point(nr, nc));
                                visited[nr][nc] = true;
                            }
                        }
                    }

                    components.add(component);
                }
            }
        }

        int maxComponentSize = 0;

        for (int r = 0; r < n; r++) {
            for (int c = 0; c < m; c++) {
                int newSize = calculateNewSize(n, m, grid, components, r, c);
                maxComponentSize = Math.max(maxComponentSize, newSize);
            }
        }

        return maxComponentSize;
    }

    private static int calculateNewSize(int n, int m, char[][] grid, List<Set<Point>> components, int row, int col) {
        boolean[][] newGrid = new boolean[n][m];
        for (Set<Point> comp : components) {
            for (Point p : comp) {
                newGrid[p.r][p.c] = true;
            }
        }

        for (int r = 0; r < n; r++) newGrid[r][col] = true;
        for (int c = 0; c < m; c++) newGrid[row][c] = true;

        boolean[][] visited = new boolean[n][m];
        int newSize = 0;

        for (int r = 0; r < n; r++) {
            for (int c = 0; c < m; c++) {
                if (newGrid[r][c] && !visited[r][c]) {
                    int size = 0;
                    Queue<Point> queue = new LinkedList<>();
                    queue.add(new Point(r, c));
                    visited[r][c] = true;

                    while (!queue.isEmpty()) {
                        Point p = queue.poll();
                        size++;
                        for (int[] d : directions) {
                            int nr = p.r + d[0], nc = p.c + d[1];
                            if (nr >= 0 && nr < n && nc >= 0 && nc < m && newGrid[nr][nc] && !visited[nr][nc]) {
                                queue.add(new Point(nr, nc));
                                visited[nr][nc] = true;
                            }
                        }
                    }

                    newSize = Math.max(newSize, size);
                }
            }
        }

        return newSize;
    }

    static int[][] directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    static class Point {
        int r, c;

        Point(int r, int c) {
            this.r = r;
            this.c = c;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Point point = (Point) o;
            return r == point.r && c == point.c;
        }

        @Override
        public int hashCode() {
            return Objects.hash(r, c);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
2
3 3
...
...
...
4 4
#...
....
....
....
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
vector<pair<int, int>> dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

void dfs(int i, int j, vector<string> &v, vector<vector<int>> &vis, int &ct){
	int n = v.size(), m = v[0].size();
	if(i < 0 || j < 0 || i >= n || j >= m || vis[i][j] || v[i][j] == '.') return;
	vis[i][j] = 1; ct ++;
	for(auto val : dir) dfs(i + val.first, j + val.second, v, vis, ct);
}

void ddfs(int i, int j, vector<string> &v, int &idx, vector<vector<int>> &vis, vector<vector<pair<int, int>>> &init, int &ct){
	int n = v.size(), m = v[0].size();
	if(i < 0 || j < 0 || i >= n || j >= m || vis[i][j] || v[i][j] == '.') return;
	vis[i][j] = 1; init[i][j] = {ct, idx};
	for(auto val : dir) ddfs(i + val.first, j + val.second, v, idx, vis, init, ct);
}

void solve(){
	int n, m, ans = 0, x = 0; cin >> n >> m;
	vector<string> v;
	vector<vector<int>> vis(n, vector<int>(m, 0)), viss(n, vector<int>(m, 0));
	vector<int> dtX(n), dtY(m);
	vector<vector<pair<int, int>>> init(n, vector<pair<int, int>>(m, make_pair(0, 0)));
	vector<set<pair<int, int>>> u(n), w(m);
	for(int i = 0; i < n; i ++){
		string s; cin >> s;
		v.push_back(s);
	}
	for(int i = 0; i < n; i ++){
		for(int j = 0; j < m; j ++){
			if(v[i][j] == '#'){
				int ct = 0;
				if(vis[i][j]) continue;
				dfs(i, j, v, vis, ct);
				ddfs(i, j, v, x, viss, init, ct);
			}
			dtX[i] += v[i][j] == '.';
			dtY[j] += v[i][j] == '.';
			x ++;
		}
	}
	for(int i = 0; i < n; i ++){
		set<pair<int, int>> st;
		int sum = 0, dot = 0;
		for(int j = 0; j < m; j ++) dot += v[i][j] == '.';
		for(int j = 0; j < m; j ++){
			if(init[i][j].first) st.insert(init[i][j]);
			if(i && init[i - 1][j].first) st.insert(init[i - 1][j]);
			if(i < n - 1 && init[i + 1][j].first) st.insert(init[i + 1][j]);
		}
		for(auto val : st) u[i].insert(val);
	}
	for(int j = 0; j < m; j ++){
		set<pair<int, int>> st;
		int sum = 0, dot = 0;
		for(int i = 0; i < n; i ++) dot += v[i][j] == '.';
		for(int i = 0; i < n; i ++){
			if(init[i][j].first) st.insert(init[i][j]);
			if(j && init[i][j - 1].first) st.insert(init[i][j - 1]);
			if(j < m - 1 && init[i][j + 1].first) st.insert(init[i][j + 1]);
		}
		for(auto val : st) w[j].insert(val);
	}
	for(int i = 0; i < n; i ++){
		int sum = 0;
		set<pair<int, int>> st;
		map<int, int> mm;
		for(auto val : u[i]){
			sum += val.first; mm[val.second] ++;
		}
		for(int j = 0; j < m; j ++){
			int s = 0;
			for(auto val : w[j]){
				if(!mm[val.second]) s += val.first;
			}
			ans = max(ans, sum + s + dtX[i] + dtY[j] - (v[i][j] == '.'));
		}
	}
	cout << ans << endl;
}
int main(){
	int t; cin >> t;
	while(t --){
		solve();
	}
}

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

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

Think step by step.

Response:
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 key points to consider are:

1. **Grid with a Single Row or Column**: This can expose issues with handling one-dimensional cases.
2. **Grid with No Initial '#' Cells**: This can test how the code handles an entirely empty grid.
3. **Grid with a Single '#' Cell**: This can test the handling of minimal connected components.
4. **Grid with a Large Number of '.' Cells**: This can test the efficiency and correctness of the code when there are many empty cells.
5. **Grid with a Complex Pattern**: This can test the code's ability to handle more intricate connected components.

Let's create a test input that combines some of these elements:

```plaintext
4
1 1
.
3 3
...
...
...
3 3
#..
...
...
3 3
#..
#..
#..
```

### Explanation:
1. **Test Case 1**: A 1x1 grid with a single '.' cell. This tests the handling of a minimal grid.
2. **Test Case 2**: A 3x3 grid with all '.' cells. This tests the handling of an entirely empty grid.
3. **Test Case 3**: A 3x3 grid with a single '#' cell in the top-left corner. This tests the handling of a minimal connected component.
4. **Test Case 4**: A 3x3 grid with a vertical line of '#' cells. This tests the handling of a more complex connected component and the ability to maximize the size of the component by converting '.' cells to '#'.

This test input should help expose any bugs in the provided code, particularly in how it handles edge cases and complex patterns.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

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

using ll = long long;

class DisjointSet{
private:
	vector<int> parents, componentSize;
public:
	DisjointSet(int cells){
		parents.resize(cells, -1);
		componentSize.resize(cells, 1);
	}

	int findParent(int node){
		if( parents[node] == -1 ){
			return node;
		}
		return parents[node] = findParent(parents[node]);
	}

	void _union(int u, int v){
		int pu = findParent(u);
		int pv = findParent(v);
		if( pu != pv ){
			parents[pv] = pu;
			componentSize[pu] += componentSize[pv];
		}
	}

	int getComponentSize(int node){
		int pnode = findParent(node);
		return componentSize[pnode];
	}
};

class Node{
public:
	unordered_set<int> parents;
	int count, add;
	Node(unordered_set<int> &s, int &count, int &add){
		this->parents = s;
		this->count = count;
		this->add = add;
	}
};

class Solution {
private:
	int dirR[4] = {1,-1,0,0};
	int dirC[4] = {0,0,1,-1};

	int getNode(int m, int i, int j){
		return (m*i)+j;
	}

	bool isCellValid(int &n, int &m, int &r, int &c){
		return !(r<0 || c<0 || r>=n || c>=m);
	}

	Node doRowWise(int &n, int &m, int &i, vector<vector<char>> &grid, DisjointSet &ds){
		unordered_set<int> parents;
		int add = 0;
		int count = 0;
		for(int j=0; j<m; j++){
			add += grid[i][j]=='.';
			for(int k=0; k<4; k++){
				int adjr = i + dirR[k];
				int adjc = j + dirC[k];
				if( isCellValid(n, m, adjr, adjc) && grid[adjr][adjc] == '#' ){
					int node = getNode(m, adjr, adjc);
					int pnode = ds.findParent(node);
					parents.insert(pnode);
				}
			}
		}
		for(auto &p : parents){
			count += ds.getComponentSize(p);
		}
		return Node(parents, count, add);
	}

	Node doColWise(int &n, int &m, int &j, vector<vector<char>> &grid, DisjointSet &ds){
		unordered_set<int> parents;
		int add = 0;
		int count = 0;
		for(int i=0; i<n; i++){
			add += grid[i][j]=='.';
			for(int k=0; k<4; k++){
				int adjr = i + dirR[k];
				int adjc = j + dirC[k];
				if( isCellValid(n, m, adjr, adjc) && grid[adjr][adjc] == '#' ){
					int node = getNode(m, adjr, adjc);
					int pnode = ds.findParent(node);
					parents.insert(pnode);
				}
			}
		}
		for(auto &p : parents){
			count += ds.getComponentSize(p);
		}
		return Node(parents, count, add);
	}

public:

	int solve(int n, int m, vector<vector<char>> &grid){
		int ans = 0;
		int cells = (n*m);
		DisjointSet ds(cells);
		vector<Node> rows, cols;

		// Building the Disjoint Set //
		for(int i=0; i<n; i++){
			for(int j=0; j<m; j++){
				if( grid[i][j] == '#' ){
					int u = getNode(m, i, j);
					for(int k=0; k<4; k++){
						int adjr = i + dirR[k];
						int adjc = j + dirC[k];
						if( isCellValid(n, m, adjr, adjc) && grid[adjr][adjc] == '#' ){
							int v = getNode(m, adjr, adjc);
							ds._union(u, v);
						}
					}
					ans = max(ans, ds.getComponentSize(u));
				}
			}
		}

		// Trying Row Wise //
		for(int r=0; r<n; r++){
			rows.push_back(doRowWise(n, m, r, grid, ds));
		}
		// Trying Column Wise //
		for(int c=0; c<m; c++){
			cols.push_back(doColWise(n, m, c, grid, ds));
		}

		for(int i=0; i<n; i++){
			Node u = rows[i];
			for(int j=0; j<m; j++){
				Node v = cols[j];
				int totalSum = u.count + v.count + u.add + v.add;
				totalSum += (grid[i][j]=='.')? -1 : 0;
				for(auto &p : u.parents){
					if( v.parents.count(p) ){
						totalSum -= ds.getComponentSize(p);
					}
				}
				ans = max(ans, totalSum);
			}
		}
		return ans;
	}

};

int main(){
	#ifndef ONLINE_JUDGE
		// For getting input from input.txt //
		freopen("input.txt", "r", stdin);
		// For writing output to output.txt //
		freopen("output.txt", "w", stdout);
	#endif
	Solution obj;
	int t;
	cin>>t;
	while( t-- > 0 ){
		int n, m;
		cin>>n>>m;
		vector<vector<char>> grid(n, vector<char>(m));
		for(int i=0; i<n; i++){
			for(int j=0; j<m; j++){
				cin>>grid[i][j];
			}
		}
		cout<<obj.solve(n, m, grid);
		cout<<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:
```plaintext
1
4 4
..#.
#...
...#
.##.
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
class UnionFind():
    def __init__(self, n, m):
        mn = m * n
        self.fa = list(range(mn + 1))
        self.sz = [1] * (mn + 1)
        self.rank = [0] * (mn + 1)

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

    def merge(self, x, y):
        rootx, rooty = self.find(x), self.find(y)
        if rootx != rooty:
            if self.rank[rootx] < self.rank[rooty]:
                rootx, rooty = rooty, rootx
            # rank小的合并到rank大的集合中
            self.fa[rooty] = rootx
            if self.rank[rootx] == self.rank[rooty]:
                self.rank[rootx] += 1
            self.sz[rootx] += self.sz[rooty]

        return

    def get_size(self, x):
        return self.sz[self.find(x)]

def solve():
    n, m = [int(x) for x in input().split()]
    g = []
    for _ in range(n):
        g.append(input())
    def f(i, j):
        return i * m + j
    uf = UnionFind(n, m)
    for i in range(n):
        for j in range(m):
            if g[i][j] == '#':
                if i and g[i-1][j] == '#':
                    uf.merge(f(i,j), f(i-1,j))
                if j and g[i][j-1] == '#':
                    uf.merge(f(i,j), f(i,j-1))

    row_st = [set() for _ in range(n)]
    col_st = [set() for _ in range(m)]
    row_dot = [0] * n
    col_dot = [0] * m

    for i in range(n):
        for j in range(m):
            if g[i][j] == '.':
                row_dot[i] += 1
                col_dot[j] += 1
            if g[i][j] == '#':
                row_st[i].add(uf.find(f(i,j)))
                col_st[j].add(uf.find(f(i,j)))

    ans = 0
    for i in range(n):
        for j in range(m):
            sz = 0
            tmp_st = set()
            for u in [i - 1, i, i + 1]:
                if u < 0 or u >= n: continue
                tmp_st |= row_st[u]
            for v in [j - 1, j, j + 1]:
                if v < 0 or v >= m: continue
                tmp_st |= col_st[v]
            for x in tmp_st:
                sz += uf.get_size(x)
            sz += row_dot[i]
            sz += col_dot[j]
            if g[i][j] == '.':
                sz -= 1
            if sz > ans:
                ans = sz
    print(ans)
    return

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

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

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

Think step by step.

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

1. **Edge Cases**: Test with the smallest possible grid and the largest possible grid within the constraints.
2. **Complex Grids**: Test with grids that have complex patterns of '#' and '.' characters to ensure the union-find operations and the logic for calculating the largest connected component are correct.
3. **Special Patterns**: Test with patterns that might cause the union-find to merge incorrectly or miss connections.

Given the constraints and the need for a small-scale test input, let's create a test case that is both valid and challenging.

### Test Input
```plaintext
2
3 3
...
...
...
3 3
#..
...
...
```

### Explanation
1. **Test Case 1**:
   - A 3x3 grid with all '.' characters.
   - This tests the scenario where no '#' characters are present initially, and the operation should create the largest possible connected component.
   - The expected output is 9, as setting any row and column to '#' will create a 3x3 grid of '#'.

2. **Test Case 2**:
   - A 3x3 grid with a single '#' character in the top-left corner.
   - This tests the scenario where there is a single '#' character initially, and the operation should maximize the size of the connected component.
   - The expected output is 5, as setting the first row and first column to '#' will create a 3x3 grid with a single '.' in the bottom-right corner.

This test input is designed to cover both edge cases and simple patterns, which should help expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Largest Component (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: Easy and hard versions are actually different problems, so read statements of both problems completely and carefully. The only difference between the two versions is the operation.

Alex has a grid with $n$ rows and $m$ columns consisting of '.' and '#' characters. A set of '#' cells forms a connected component if from any cell in this set, it is possible to reach any other cell in this set by only moving to another cell in the set that shares a common side. The size of a connected component is the number of cells in the set.

In one operation, Alex selects any row $r$ ($1 \le r \le n$) and any column $c$ ($1 \le c \le m$), then sets every cell in row $r$ and column $c$ to be '#'. Help Alex find the maximum possible size of the largest connected component of '#' cells that he can achieve after performing the operation at most once.

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

The first line of each test case contains two integers $n$ and $m$ ($1 \le n \cdot m \le 10^6$) — the number of rows and columns of the grid.

The next $n$ lines each contain $m$ characters. Each character is either '.' or '#'.

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

Output: For each test case, output a single integer — the maximum possible size of a connected component of '#' cells that Alex can achieve.

Examples:
input:
6
1 1
.
4 2
..
#.
#.
.#
3 5
.#.#.
..#..
.#.#.
5 5
#...#
....#
#...#
.....
...##
6 6
.#..#.
#..#..
.#...#
#.#.#.
.#.##.
###..#
6 8
..#....#
.####.#.
###.#..#
.##.#.##
.#.##.##
#..##.#.
output:
1
7
11
16
22
36

Note: In the fourth test case, it is optimal for Alex to set all cells in row $4$ and column $2$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $16$.

In the fifth test case, it is optimal for Alex to set all cells in row $2$ and column $4$ to be '#'. Doing so will lead to the largest connected component of '#' having a size of $22$.

Buggy Code:
from sys import stdin
input = lambda: stdin.readline()[:-1]

def solve():
    n, m = map(int, input().split())
    a = [[0 for _ in range(m)] for _ in range(n)]
    for i in range(n):
        s = input()
        for j in range(m):
            if s[j] == '#':
                a[i][j] = 1
    direction = [[1, 0], [0, 1], [0, -1], [-1, 0]]
    flag = 10 ** 7
    res = {}
    for i in range(n):
        for j in range(m):
            if a[i][j] == 1:
                cur_pos = [i, j]
                stk = []
                stk.append(cur_pos)
                flag += 1
                a[cur_pos[0]][cur_pos[1]] = flag
                count = 0
                while stk:
                    pos = stk.pop()
                    count += 1
                    for k in range(len(direction)):
                        cur_pos = [pos[0] + direction[k][0], pos[1] + direction[k][1]]
                        if 0 <= cur_pos[0] <= n - 1 and 0 <= cur_pos[1] <= m - 1 and a[cur_pos[0]][cur_pos[1]] == 1:
                            stk.append(cur_pos)
                            a[cur_pos[0]][cur_pos[1]] = flag
                res[flag] = count
    ans = 0
    rows = {}
    rows_comp = {}
    for i in range(n):
        t = {}
        count = 0
        rows_comp[i] = set()
        for j in range(m):
            if a[i][j] == 0:
                count += 1
                if i + 1 < n and a[i + 1][j] != 0:
                    t[a[i + 1][j]] = 1
                if i - 1 >= 0 and a[i - 1][j] != 0:
                    t[a[i - 1][j]] = 1
            else:
                t[a[i][j]] = 1
        for item in t.keys():
            count += res[item]
            rows_comp[i].add(item)
        rows[i] = count

    cols = {}
    cols_comp = {}
    for j in range(m):
        t = {}
        count = 0
        cols_comp[j] = set()
        for i in range(n):
            if a[i][j] == 0:
                count += 1
                if j + 1 < m and a[i][j + 1] != 0:
                    t[a[i][j + 1]] = 1
                if j - 1 >= 0 and a[i][j - 1] != 0:
                    t[a[i][j - 1]] = 1
            else:
                t[a[i][j]] = 1
        for item in t.keys():
            count += res[item]
            cols_comp[j].add(item)
        cols[j] = count

    for i in range(n):
        for j in range(m):
            temp = rows[i] + cols[j]
            stat = {}
            for item in rows_comp[i]:
                if item not in stat:
                    stat[item] = 0
                stat[item] += 1
            for item in cols_comp[j]:
                if item not in stat:
                    stat[item] = 0
                stat[item] += 1
            for item in stat.keys():
                if stat[item] == 2:
                    temp -= res[item]
            if a[i][j] == 0:
                temp -= 1
            ans = max(ans, temp)
    return ans

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

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

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

Think step by step.

Response:
```plaintext
1
5 5
.....
.....
.....
.....
.....
```