**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import sys

def main():
    sys.setrecursionlimit(1 << 25)
    t = int(sys.stdin.readline())
    for _ in range(t):
        n, m = map(int, sys.stdin.readline().split())
        A = [list(map(int, sys.stdin.readline().split())) for _ in range(n)]
        B = [list(map(int, sys.stdin.readline().split())) for _ in range(n)]
        possible = True
        for k in range(31):
            Ak = [[(A[i][j] >> k) &1 for j in range(m)] for i in range(n)]
            Bk = [[(B[i][j] >> k) &1 for j in range(m)] for i in range(n)]
            total_vars = n + m
            nodes = 2 * total_vars
            graph = [[] for _ in range(nodes)]
            for i in range(n):
                for j in range(m):
                    A_bit = Ak[i][j]
                    B_bit = Bk[i][j]
                    if B_bit ==1 and A_bit ==0:
                        # R_i_k ⇒ ~R_i_k
                        u = 2 * i
                        v = 2 * i +1
                        graph[u].append(v)
                        # ~C_j_k ⇒ C_j_k
                        u = 2 * (n + j) +1
                        v = 2 * (n + j)
                        graph[u].append(v)
                    elif B_bit ==1 and A_bit ==1:
                        # R_i_k ⇒ ~R_i_k
                        u = 2 * i
                        v = 2 * i +1
                        graph[u].append(v)
                    elif B_bit ==0 and A_bit ==1:
                        # ~R_i_k ⇒ R_i_k
                        u = 2 * i +1
                        v = 2 * i
                        graph[u].append(v)
                    elif B_bit ==0 and A_bit ==0:
                        # ~R_i_k ⇒ ~C_j_k
                        u = 2 * i +1
                        v = 2 * (n + j) +1
                        graph[u].append(v)
                        # C_j_k ⇒ R_i_k
                        u = 2 * (n + j)
                        v = 2 * i
                        graph[u].append(v)
            # Tarjan's SCC
            index_counter = 0
            indices = [-1] * nodes
            lowlinks = [0] * nodes
            on_stack = [False] * nodes
            stack = []
            scc_index = 0
            scc = [0] * nodes
            def strongconnect(v):
                nonlocal index_counter, scc_index
                indices[v] = index_counter
                lowlinks[v] = index_counter
                index_counter +=1
                stack.append(v)
                on_stack[v] = True
                for w in graph[v]:
                    if indices[w] == -1:
                        strongconnect(w)
                        lowlinks[v] = min(lowlinks[v], lowlinks[w])
                    elif on_stack[w]:
                        lowlinks[v] = min(lowlinks[v], indices[w])
                if indices[v] == lowlinks[v]:
                    scc_index +=1
                    while True:
                        w = stack.pop()
                        on_stack[w] = False
                        scc[w] = scc_index
                        if w == v:
                            break
            for v in range(nodes):
                if indices[v] == -1:
                    strongconnect(v)
            # Check conflict
            conflict = False
            for x in range(total_vars):
                if scc[2 * x] == scc[2 * x +1]:
                    conflict = True
                    break
            if conflict:
                possible = False
                break
        if possible:
            print("Yes")
        else:
            print("No")

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
2 2
1 1
1 1
0 0
0 0
2 2
1 1
1 1
1 0
0 1
2 2
1 1
1 1
2 2
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
def solve():
    import sys
    sys.setrecursionlimit(10**7)
    inputs = sys.stdin.read().strip().split()
    num_tests = int(inputs[0])
    index = 1

    def detect_cycle(graph, size):
        state = [0] * size

        def dfs(node):
            state[node] = 1
            for neighbor in graph[node]:
                if state[neighbor] == 0:
                    if dfs(neighbor):
                        return True
                elif state[neighbor] == 1:
                    return True
            state[node] = 2
            return False

        for start_node in range(size):
            if state[start_node] == 0:
                if dfs(start_node):
                    return True
        return False

    def check_bit(value, position):
        return ((value >> position) & 1) != 0

    results = []

    for _ in range(num_tests):
        rows = int(inputs[index])
        index += 1
        cols = int(inputs[index])
        index += 1

        matrix_a = []
        for _ in range(rows):
            matrix_a.append(list(map(int, inputs[index:index + cols])))
            index += cols

        matrix_b = []
        for _ in range(rows):
            matrix_b.append(list(map(int, inputs[index:index + cols])))
            index += cols

        is_possible = True

        for bit in range(31):
            connections = []
            for i in range(rows):
                for j in range(cols):
                    val_a = check_bit(matrix_a[i][j], bit)
                    val_b = check_bit(matrix_b[i][j], bit)
                    row_node = i
                    col_node = rows + j

                    if val_a and not val_b:
                        connections.append((col_node, row_node))
                    elif not val_a and val_b:
                        connections.append((row_node, col_node))

            total_nodes = rows + cols
            adjacency_list = [[] for _ in range(total_nodes)]
            for u, v in connections:
                adjacency_list[u].append(v)

            if detect_cycle(adjacency_list, total_nodes):
                is_possible = False
                break

        results.append("Yes" if is_possible else "No")

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

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
4
1 1
1
0
2 2
1 1
1 1
0 0
0 0
2 2
1 0
0 1
0 1
1 0
2 2
1 2
3 4
3 4
5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
def main():
    import sys
    input_data = sys.stdin.read().strip().split()
    t = int(input_data[0])
    idx = 1

    answers = []

    for _ in range(t):
        n = int(input_data[idx]); idx += 1
        m = int(input_data[idx]); idx += 1

        # Read matrix A (not actually used in the logic, but we must consume it)
        A = []
        for _row in range(n):
            _ = input_data[idx : idx + m]  # not used
            idx += m

        # Read matrix B
        B = []
        for _row in range(n):
            row = list(map(int, input_data[idx : idx + m]))
            idx += m
            B.append(row)

        possible = True

        # We only need to check bits up to ~30 (since 0 <= B[i][j] <= 10^9)
        for bit in range(31):
            mask = 1 << bit

            # 1) Figure out which rows are entirely zero in this bit
            #    i.e. row i is in R_off if for all j, (B[i][j] & mask) == 0
            row_off = [True]*n  # row_off[i] = True if row i is all 0 in bit 'bit'
            for i in range(n):
                # if we find any B[i][j] with bit set => row i NOT fully off
                for j in range(m):
                    if (B[i][j] & mask) != 0:
                        row_off[i] = False
                        break

            # 2) Figure out which columns must be "on" for this bit
            #    A column j must be "on" if it has at least one '1' in a row
            #    that is not in row_off.
            col_on = [False]*m
            for j in range(m):
                # check if there's some row i not in row_off,
                # for which B[i][j]'s bit is 1 => column j must be on
                need_on = False
                for i in range(n):
                    if not row_off[i]:  # row i is not forcing the bit 0
                        if (B[i][j] & mask) != 0:
                            need_on = True
                            break
                col_on[j] = need_on

            # 3) Check consistency
            #    If row_off[i], then for all j, B[i][j]'s bit must be 0
            #    If row i is not off, but col_on[j] is True, B[i][j]'s bit must be 1
            #    If row i is not off, but col_on[j] is False, B[i][j]'s bit must be 0
            for i in range(n):
                for j in range(m):
                    has_bit = ((B[i][j] & mask) != 0)
                    if row_off[i]:
                        # then this bit must be 0 in B[i][j]
                        if has_bit:
                            possible = False
                            break
                    else:
                        # row i is not forced off
                        if col_on[j]:
                            # then bit must be 1
                            if not has_bit:
                                possible = False
                                break
                        else:
                            # then bit must be 0
                            if has_bit:
                                possible = False
                                break
                if not possible:
                    break
            if not possible:
                break

        answers.append("Yes" if possible else "No")

    print("\n".join(answers))

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
4
1 1
1
0
2 2
1 1
1 1
0 0
0 0
2 2
1 0
0 1
0 1
1 0
2 2
1 0
0 1
0 0
0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int test = 0; test < t; test++) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            int[][] A = new int[n][m];
            int[][] B = new int[n][m];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    A[i][j] = scanner.nextInt();
                }
            }
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    B[i][j] = scanner.nextInt();
                }
            }
            if (canTransform(A, B, n, m)) {
                System.out.println("Yes");
            } else {
                System.out.println("No");
            }
        }
    }

    private static boolean canTransform(int[][] A, int[][] B, int n, int m) {
        for (int k = 0; k < 31; k++) {
            int[] rowOps = new int[n]; // 0: must set to 0, 1: must set to 1, -1: undefined
            for (int i = 0; i < n; i++) {
                rowOps[i] = -1;
                for (int j = 0; j < m; j++) {
                    boolean aHasK = (A[i][j] & (1 << k)) != 0;
                    boolean bHasK = (B[i][j] & (1 << k)) != 0;
                    if (aHasK && !bHasK) {
                        if (rowOps[i] == 1) {
                            return false;
                        }
                        rowOps[i] = 0;
                    } else if (!aHasK && bHasK) {
                        if (rowOps[i] == 0) {
                            return false;
                        }
                        rowOps[i] = 1;
                    }
                }
                if (rowOps[i] == 0) {
                    for (int j = 0; j < m; j++) {
                        if ((B[i][j] & (1 << k)) != 0) {
                            return false;
                        }
                    }
                }
            }
            boolean[] colNeedsK = new boolean[m];
            for (int i = 0; i < n; i++) {
                if (rowOps[i] == 1) {
                    for (int j = 0; j < m; j++) {
                        boolean aHasK = (A[i][j] & (1 << k)) != 0;
                        boolean bHasK = (B[i][j] & (1 << k)) != 0;
                        if (!aHasK && bHasK) {
                            colNeedsK[j] = true;
                        } else if (aHasK && !bHasK) {
                            return false;
                        }
                    }
                }
            }
            for (int i = 0; i < n; i++) {
                if (rowOps[i] == 1) {
                    for (int j = 0; j < m; j++) {
                        boolean aHasK = (A[i][j] & (1 << k)) != 0;
                        boolean bHasK = (B[i][j] & (1 << k)) != 0;
                        if (!aHasK && bHasK != colNeedsK[j]) {
                            return false;
                        }
                    }
                }
            }
        }
        return true;
    }
}

// Q073mJEwqFaikYtdlsz

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

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

Think step by step.

Response:
```plaintext
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import java.util.Scanner;

//TIP To <b>Run</b> code, press <shortcut actionId="Run"/> or
// click the <icon src="AllIcons.Actions.Execute"/> icon in the gutter.
public class Main {

    private static int helper_row(int row,int[][] graph, int[] not_cycle, int[] cycle, int[] visited, int n, int m) {//return value 0 means cycle
        if(visited[row]==1) {
            cycle[row]=1;
            return 0;
        }
        if(cycle[row]==1) {
            return 0;
        }
        if(not_cycle[row]==1) {
            return 1;
        }
        visited[row]=1;
        int temp = 1;
        for (int i = 0; i < m; ++i) {
            if(graph[row][n+i]==1) {
                temp&=helper_col(i, graph, not_cycle, cycle, visited, n, m);
            }
        }
        visited[row]=0;
        if(temp==0) {
            cycle[row]=1;
            return 0;
        }
        not_cycle[row]=1;
        return 1;
    }

    private static int helper_col(int col, int[][] graph, int[] not_cycle, int[] cycle, int[] visited, int n, int m) {
        if(visited[n+col]==1) {
            cycle[n+col]=1;
            return 0;
        }
        if(cycle[n+col]==1) {
            return 0;
        }
        if(not_cycle[n+col]==1) {
            return 1;
        }
        visited[n+col]=1;
        int temp = 1;
        for (int i = 0; i < n; ++i) {
            if(graph[col+n][i]==1) {
                temp&=helper_row(i, graph, not_cycle, cycle, visited, n, m);
            }
        }
        visited[n+col]=0;
        if(temp==0) {
            cycle[n+col]=1;
            return 0;
        }
        not_cycle[n+col]=1;
        return 1;
    }

    static void solve(Scanner in) {
        int n = in.nextInt();
        int m = in.nextInt();
        long[][] A = new long[n][m];
        long[][] B = new long[n][m];
        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                A[i][j] = in.nextLong();
            }
        }
        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                B[i][j] = in.nextLong();
            }
        }
        int valid = 1;
        while(valid==1) {
            int fin = 1;
            for(int i=0;i<n;i++) {
                for(int j=0;j<m;j++) {
                    if (A[i][j] != B[i][j]) {
                        fin = 0;
                        break;
                    }
                }
            }
            if(fin==1)break;

            int[][] graph = new int[n+m][n+m];
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    if((B[i][j]%2)==1) {
                        graph[i][n+j] = 1;
                    }else {
                        graph[i][n+j] = 0;
                    }
                }
            }

            for (int j = 0; j < m; ++j) {
                for (int i = 0; i < n; ++i) {
                    if((B[i][j]%2)==0) {
                        graph[n+j][i] = 1;
                    }else {
                        graph[n+j][i] = 0;
                    }
                }
            }
            // for (int i = 0; i < n+m; ++i) {
            //     for (int j = 0; j < n+m; ++j) {
            //         cout << graph[i][j] << " ";
            //     }
            //     cout << endl;
            // }
            // return;
            int[] not_cycle = new int[n+m];
            for(int i=0;i<n+m;i++)not_cycle[i]=0;
            int[] cycle = new int[n+m];
            for(int i=0;i<n;i++)cycle[i]=0;

            for (int i = 0; i < n; ++i) {
                int[] visited = new int[n+m];
                for (int j = 0; j < n+m; ++j) {
                    visited[j]=0;
                }
                helper_row(i, graph, not_cycle, cycle, visited, n, m);
            }

            for (int i = 0; i < m; ++i) {
                int[] visited = new int[n+m];
                for (int j = 0; j < n+m; ++j) {
                    visited[j]=0;
                }
                helper_col(i, graph, not_cycle, cycle, visited, n, m);
            }
            // cout << "cycle" << endl;
            // for (int i = 0; i < n+m; ++i) {
            //     cout << cycle[i] << " ";
            // }
            // cout << endl;
            // cout << "not_cycle" << endl;
            // for (int i = 0; i < n+m; ++i) {
            //     cout << not_cycle[i] << " ";
            // }
//            cout << endl << endl;
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    if((A[i][j]%2)!=(B[i][j]%2)) {
                        if((B[i][j]%2)==1) {//col_op_j
                            if(cycle[n+j]==1) {
                                valid = 0;
                                break;
                            }
                        }else {//row_op_i
                            if(cycle[i]==1) {
                                valid = 0;
                                break;
                            }
                        }
                    }
                }
                if(valid==0)break;
            }
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    A[i][j]>>=1;
                    B[i][j]>>=1;
                }
            }
        }
        if(valid==1) {
            System.out.println("YES");
        }else {
            System.out.println("NO");
        }
    }

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

**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 provided buggy code, let's consider the following aspects:

1. **Edge Cases**:
   - Matrices with dimensions that are at the boundary of the constraints (e.g., 1x1, 1x1000, 1000x1).
   - Matrices with all elements being the same or very similar values.
   - Matrices where the transformation is possible but requires a specific sequence of operations.

2. **Specific Scenarios**:
   - Matrices where the transformation is impossible due to the bitwise operations.
   - Matrices where the transformation is possible but the code fails to detect it.

Given the constraints and the nature of the problem, a small-scale test input that is complete and valid can be:

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

### Explanation:

1. **Test Case 1**:
   - Matrices `A` and `B` are identical.
   - Expected Output: `Yes`
   - This tests the basic case where no operations are needed.

2. **Test Case 2**:
   - Matrix `A` is:
     ```
     1 2
     3 4
     ```
   - Matrix `B` is:
     ```
     1 3
     3 4
     ```
   - Expected Output: `No`
   - This tests a case where the transformation is impossible. The element `A[0][1]` (2) cannot be transformed to `B[0][1]` (3) using the given operations.

3. **Test Case 3**:
   - Matrix `A` is:
     ```
     1 2
     3 4
     ```
   - Matrix `B` is:
     ```
     1 3
     3 5
     ```
   - Expected Output: `No`
   - This tests a more complex case where the transformation is impossible. The element `A[0][1]` (2) cannot be transformed to `B[0][1]` (3) and the element `A[1][1]` (4) cannot be transformed to `B[1][1]` (5) using the given operations.

This test input should help expose any issues in the buggy code, particularly in handling cases where the transformation is impossible.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
// Source: https://usaco.guide/general/io

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

vector<vector<int>> nodes;
vector<int> vis;

int leastPow(int n){
	int x = 1;
	while(x <= n){
		x*=2;
	}
	return x/2;
}

bool dfs(int node, bool f){
	for(int neighbour: nodes[node]){
		if(vis[neighbour] == node && !f){
			return false;
		}
		else{
			if(vis[neighbour] != -1){
				f = true;
			}
			vis[neighbour] = node;
			if(!dfs(neighbour, f)){
				return false;
			}
		}
	}
	return true;
}

int main() {
	int tt;
	cin >> tt;
	while(tt--){
		int n, m;
		int pow2 = 0;
		cin >> n >> m;
		vector<vector<int>> start(n, vector<int> (m));
		vector<vector<int>> end(n, vector<int> (m));
		for(int i = 0; i < n; i++){
			for(int j = 0; j < m; j++){
				cin >> start[i][j];
				pow2 = max(pow2, start[i][j]);
			}
		}

		for(int i = 0; i < n; i++){
			for(int j = 0; j < m; j++){
				cin >> end[i][j];
				pow2 = max(pow2, end[i][j]);
			}
		}
		pow2 = leastPow(pow2);

		bool f = true;

		nodes.resize(n+m);
		vis.resize(n+m);

		while(pow2 > 0 && f){

			//clear everything first
			for(int i = 0; i < n+m; i++){
				nodes[i].clear();
				vis[i] = -1;
			}

			//load binary array
			vector<vector<int>> startBin(n, vector<int> (m, 0));
			vector<vector<int>> endBin(n, vector<int> (m, 0));
			for(int i = 0; i < n; i++){
				for(int j = 0; j < m; j++){
					if(start[i][j] >= pow2){
						start[i][j] -= pow2;
						startBin[i][j] = 1;
					}
					if(end[i][j] >= pow2){
						end[i][j] -= pow2;
						endBin[i][j] = 1;

					}
				}
			}

			//construct tree
			for(int i = 0; i < n; i++){
				for(int j = 0; j < m; j++){
					if(endBin[i][j] == 1){
						nodes[i].push_back(n+j);

					}
					else{
						nodes[n+j].push_back(i);
					}
				}
			}

			//run dfs for every tree required
			//maybe memoize a few cases? **
			for(int i = 0; i < n; i++){
				for(int j = 0; j < m; j++){
					if(startBin[i][j] != endBin[i][j]){
						for(int k = 0; k < n+m; k++){
							vis[k] = -1;
						}
						if(endBin[i][j] == 1){
							if(!dfs(n+j, false)){
								f = false;
								break;
							}
						}
						else{
							if(!dfs(i, false)){
								f = false;
								break;
							}
						}
					}
				}
				if(!f){
					break;
				}
			}

			pow2/=2;

		}

		if(f){
			cout << "YES" << endl;
		}
		else{
			cout << "NO" << endl;
		}

	}
}
/* implement array/elements
we will need to extract several binary arrays
from there , we can form necessary trees --> nodes will be single digit
*/

**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
2 2
1 0
0 1
0 1
1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
#line 1 "main.cpp"
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rrep(i, n) for (int i = (int)(n - 1); i >= 0; i--)
#define all(x) (x).begin(), (x).end()
#define sz(x) int(x.size())
using namespace std;
using ll = long long;
constexpr int INF = 1e9;
constexpr ll LINF = 1e18;
string YesNo(bool cond) {
    return cond ? "Yes" : "No";
}
string YESNO(bool cond) {
    return cond ? "YES" : "NO";
}
template <class T>
bool chmax(T& a, const T& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <class T>
bool chmin(T& a, const T& b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}
template <typename T, class F>
T bisect(T ok, T ng, const F& f) {
    while (abs(ok - ng) > 1) {
        T mid = min(ok, ng) + (abs(ok - ng) >> 1);
        (f(mid) ? ok : ng) = mid;
    }
    return ok;
}
template <typename T>
T floor_div(const T a, const T b) {
    assert(b != 0);
    return a / b - static_cast<T>((a ^ b) < 0 && a % b != 0);
}
template <typename T>
T ceil_div(const T a, const T b) {
    assert(b != 0);
    return a / b + static_cast<T>((a ^ b) >= 0 && a % b != 0);
}
template <typename T>
bool a_mul_b_eq_c(const T a, const T b, const T c) {
    return c % a == 0 && b == c / a;
}
template <typename T>
bool a_mul_b_le_c(const T a, const T b, const T c) {
    return b <= floor_div(c, a);
}
template <typename T>
bool a_mul_b_lt_c(const T a, const T b, const T c) {
    return a_mul_b_le_c(a, b, c) && !a_mul_b_eq_c(a, b, c);
}
template <typename T>
bool a_mul_b_ge_c(const T a, const T b, const T c) {
    return !a_mul_b_lt_c(a, b, c);
}
template <typename T>
bool a_mul_b_gt_c(const T a, const T b, const T c) {
    return !a_mul_b_le_c(a, b, c);
}
template <class T>
vector<T> make_vec(size_t a) {
    return vector<T>(a);
}
template <class T, class... Ts>
auto make_vec(size_t a, Ts... ts) {
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
    for (int i = 0; i < int(v.size()); i++) {
        is >> v[i];
    }
    return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
    for (int i = 0; i < int(v.size()); i++) {
        os << v[i];
        if (i < sz(v) - 1) os << ' ';
    }
    return os;
}
#line 1 "/Users/gyouzasushi/kyopro/library/atcoder/scc.hpp"

#line 7 "/Users/gyouzasushi/kyopro/library/atcoder/scc.hpp"

#line 1 "/Users/gyouzasushi/kyopro/library/atcoder/internal_scc.hpp"

#line 7 "/Users/gyouzasushi/kyopro/library/atcoder/internal_scc.hpp"

#line 1 "/Users/gyouzasushi/kyopro/library/atcoder/internal_csr.hpp"

#line 7 "/Users/gyouzasushi/kyopro/library/atcoder/internal_csr.hpp"

namespace atcoder {
namespace internal {

template <class E> struct csr {
    std::vector<int> start;
    std::vector<E> elist;
    explicit csr(int n, const std::vector<std::pair<int, E>>& edges)
        : start(n + 1), elist(edges.size()) {
        for (auto e : edges) {
            start[e.first + 1]++;
        }
        for (int i = 1; i <= n; i++) {
            start[i] += start[i - 1];
        }
        auto counter = start;
        for (auto e : edges) {
            elist[counter[e.first]++] = e.second;
        }
    }
};

}  // namespace internal

}  // namespace atcoder

#line 9 "/Users/gyouzasushi/kyopro/library/atcoder/internal_scc.hpp"

namespace atcoder {
namespace internal {

// Reference:
// R. Tarjan,
// Depth-First Search and Linear Graph Algorithms
struct scc_graph {
  public:
    explicit scc_graph(int n) : _n(n) {}

    int num_vertices() { return _n; }

    void add_edge(int from, int to) { edges.push_back({from, {to}}); }

    // @return pair of (# of scc, scc id)
    std::pair<int, std::vector<int>> scc_ids() {
        auto g = csr<edge>(_n, edges);
        int now_ord = 0, group_num = 0;
        std::vector<int> visited, low(_n), ord(_n, -1), ids(_n);
        visited.reserve(_n);
        auto dfs = [&](auto self, int v) -> void {
            low[v] = ord[v] = now_ord++;
            visited.push_back(v);
            for (int i = g.start[v]; i < g.start[v + 1]; i++) {
                auto to = g.elist[i].to;
                if (ord[to] == -1) {
                    self(self, to);
                    low[v] = std::min(low[v], low[to]);
                } else {
                    low[v] = std::min(low[v], ord[to]);
                }
            }
            if (low[v] == ord[v]) {
                while (true) {
                    int u = visited.back();
                    visited.pop_back();
                    ord[u] = _n;
                    ids[u] = group_num;
                    if (u == v) break;
                }
                group_num++;
            }
        };
        for (int i = 0; i < _n; i++) {
            if (ord[i] == -1) dfs(dfs, i);
        }
        for (auto& x : ids) {
            x = group_num - 1 - x;
        }
        return {group_num, ids};
    }

    std::vector<std::vector<int>> scc() {
        auto ids = scc_ids();
        int group_num = ids.first;
        std::vector<int> counts(group_num);
        for (auto x : ids.second) counts[x]++;
        std::vector<std::vector<int>> groups(ids.first);
        for (int i = 0; i < group_num; i++) {
            groups[i].reserve(counts[i]);
        }
        for (int i = 0; i < _n; i++) {
            groups[ids.second[i]].push_back(i);
        }
        return groups;
    }

  private:
    int _n;
    struct edge {
        int to;
    };
    std::vector<std::pair<int, edge>> edges;
};

}  // namespace internal

}  // namespace atcoder

#line 9 "/Users/gyouzasushi/kyopro/library/atcoder/scc.hpp"

namespace atcoder {

struct scc_graph {
  public:
    scc_graph() : internal(0) {}
    explicit scc_graph(int n) : internal(n) {}

    void add_edge(int from, int to) {
        int n = internal.num_vertices();
        assert(0 <= from && from < n);
        assert(0 <= to && to < n);
        internal.add_edge(from, to);
    }

    std::vector<std::vector<int>> scc() { return internal.scc(); }

  private:
    internal::scc_graph internal;
};

}  // namespace atcoder

#line 94 "main.cpp"
int solve() {
    int h, w;
    cin >> h >> w;
    vector aa(h, vector<int>(w));
    vector bb(h, vector<int>(w));
    cin >> aa >> bb;

    vector a(h, vector<int>(w));
    vector b(h, vector<int>(w));
    bool ok = true;
    rep(d, 31) {
        atcoder::scc_graph g(h + w);
        vector<int> used(h + w, false);
        rep(i, h) rep(j, w) {
            a[i][j] = aa[i][j] >> d & 1;
            b[i][j] = bb[i][j] >> d & 1;
        }
        rep(i, h) rep(j, w) {
            if (a[i][j] == 0 && b[i][j] == 1) {
                used[j + h] = true;
            }
            if (a[i][j] == 1 && b[i][j] == 0) {
                used[i] = true;
            }
        }
        rep(i, h) rep(j, w) if (used[i] && used[j + h]) {
            if (b[i][j] == 0) {
                g.add_edge(j + h, i);
            } else {
                g.add_edge(i, j + h);
            }
        }
        ok &= sz(g.scc()) == h + w;
        if (!ok) break;
        for (vector<int> c : g.scc()) {
            assert(sz(c) == 1);
            if (!used[c[0]]) continue;
            if (c[0] < h) {
                int i = c[0];
                int mask = ((1 << 30) - 1) ^ (1 << d);
                rep(j, w) aa[i][j] &= mask;
            } else {
                int j = c[0] - h;
                int mask = 1 << d;
                rep(i, h) aa[i][j] |= mask;
            }
        }
    }
    if (ok) {
        rep(i, h) cerr << aa[i] << '\n';
        assert(aa == bb);
    }
    cout << YesNo(ok) << '\n';
    return 0;
}
int main() {
    int tt = 1;
    cin >> tt;
    while (tt--) solve();
}

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

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

Think step by step.

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

1. **Bitwise Operations**: The operations involve bitwise AND and OR, which can lead to subtle issues if not handled correctly.
2. **Matrix Dimensions**: Small matrices can be easier to debug but still expose issues.
3. **Specific Bit Patterns**: Matrices with specific bit patterns can highlight issues in the bitwise operations.

Let's construct a test case that is small but challenging:

### Test Case Construction
- **Matrix Size**: 2x2
- **Bit Patterns**: Use specific bit patterns that can highlight issues in the bitwise operations.

### Test Case
```plaintext
1
2 2
3 1
2 0
2 2
1 0
1 0
```

### Explanation
- **Matrix A**:
  ```
  3 1
  2 0
  ```
- **Matrix B**:
  ```
  1 0
  1 0
  ```

### Why This Test Case?
- **Initial State**: Matrix A has values that are not directly transformable to Matrix B using simple operations.
- **Bitwise AND and OR**: The values in Matrix A and Matrix B are such that they require careful application of bitwise AND and OR operations to transform A into B.
- **Edge Case**: The values in Matrix A and Matrix B are chosen to ensure that the transformation is not straightforward, which can expose bugs in the algorithm.

This test case should be challenging enough to expose any issues in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import sys
input = sys.stdin.readline
for _ in range(int(input())):
    n,m = map(int,input().split())
    A = [list(map(int,input().split())) for _ in range(n)]
    B = [list(map(int,input().split())) for _ in range(n)]
    for _ in range(60):
        for i in range(n):
            BIT = (1 << 34) - 1
            GONE = 0
            for j in range(m):
                a = A[i][j]
                b = B[i][j]
                c = 0
                while a or b:
                    if a & 1 > b & 1:
                        GONE |= 1 << c
                    c  += 1
                    a >>= 1
                    b >>= 1
            BIT ^= GONE
            for j in range(m):
                A[i][j] &= BIT
            for j in range(m):
                ADD = 0
                for i in range(n):
                    a = A[i][j]
                    b = B[i][j]
                    c = 0
                    while a or b:
                        if a & 1 < b & 1:
                            ADD |= 1 << c
                        c += 1
                        a >>= 1
                        b >>= 1
                for i in range(n):
                    A[i][j] |= ADD
    for x in range(n):
        if A[x] != B[x]:print("NO");break
    else:print("YES")

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

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

Think step by step.

Response:
```plaintext
1
2 2
3 3
2 2
1 1
0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import sys

input = lambda: sys.stdin.readline().strip()
maput = lambda: map(int, input().split())
printf = sys.stdout.write

t = int(input())

for _ in range(t):
    n,m = maput()
    a = [list(maput()) for i in range(n)]
    b = [list(maput()) for i in range(n)]

    for _ in range(min(n, m)):
        for i in range(n):
            ands = 0
            for j in range(m):
                c = a[i][j] ^ b[i][j]
                ands |= c & a[i][j]

            for k in range(m):
                a[i][k] &= ~ands

        for j in range(m):
            ors = 0
            for i in range(n):
                c = a[i][j] ^ b[i][j]
                ors |= c & b[i][j]

            for k in range(n):
                a[k][j] |= ors

        if a == b:
            break
    else:
        # print(a, b)
        printf("No\n")
        continue
    printf('Yes\n')

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
1 2 3
4 5 6
7 8 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import java.util.*;

public class Main {
    static Scanner in = new Scanner(System.in);

    static int t;
    static int n, m;
    static long[][] A, B;

    public static void main(String[] args) {
        t = in.nextInt();
        while (t-- > 0) {
            n = in.nextInt();
            m = in.nextInt();
            A = new long[n][m];
            B = new long[n][m];

            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    A[i][j] = in.nextLong();
                }
            }
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    B[i][j] = in.nextLong();
                }
            }

            if (solveOneTest()) {
                System.out.println("Yes");
            } else {
                System.out.println("No");
            }
        }
    }
    static boolean solveOneTest() {
        for (int bit = 0; bit < 31; bit++) {
            if (!checkBit(bit)) {
                return false;
            }
        }
        return true;
    }
    static boolean checkBit(int bit) {
        int totalNodes = n + m;
        List<List<Integer>> adj = new ArrayList<>(totalNodes);
        for (int i = 0; i < totalNodes; i++) {
            adj.add(new ArrayList<>());
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                long aVal = A[i][j];
                long bVal = B[i][j];
                int aBit = (int)((aVal >> bit) & 1L);
                int bBit = (int)((bVal >> bit) & 1L);

                if (aBit == 1 && bBit == 0) {
                    int colNode = n + j;
                    adj.get(colNode).add(i);
                } else if (aBit == 0 && bBit == 1) {
                    int colNode = n + j;
                    adj.get(i).add(colNode);
                }
            }
        }

        return !hasCycle(adj, totalNodes);
    }

    static boolean hasCycle(List<List<Integer>> adj, int totalNodes) {

        int[] state = new int[totalNodes];

        for (int u = 0; u < totalNodes; u++) {
            if (state[u] == 0) {
                if (dfsCycle(u, adj, state)) {
                    return true;
                }
            }
        }
        return false;
    }

    static boolean dfsCycle(int u, List<List<Integer>> adj, int[] state) {
        state[u] = 1;
        for (int v : adj.get(u)) {
            if (state[v] == 0) {
                if (dfsCycle(v, adj, state)) {
                    return true;
                }
            } else if (state[v] == 1) {

                return true;
            }
        }
        state[u] = 2;
        return false;
    }
}

**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
2 2
3 3
3 3
2 2
3 3
2 2
```

### Explanation:
This test case is designed to expose a potential issue with the logic used in the `checkBit` function, particularly in how it handles the transformation of the matrix.

1. **Initial Matrices:**
   - Matrix `A`:
     ```
     3 3
     3 3
     ```
   - Matrix `B`:
     ```
     2 2
     2 2
     ```

2. **Bitwise Analysis:**
   - For the bit `0` (least significant bit):
     - `A[0][0]` (3) & `B[0][0]` (2): `1 & 0 = 0`
     - `A[0][1]` (3) & `B[0][1]` (2): `1 & 0 = 0`
     - `A[1][0]` (3) & `B[1][0]` (2): `1 & 0 = 0`
     - `A[1][1]` (3) & `B[1][1]` (2): `1 & 0 = 0`
   - For the bit `1`:
     - `A[0][0]` (3) & `B[0][0]` (2): `1 & 1 = 1`
     - `A[0][1]` (3) & `B[0][1]` (2): `1 & 1 = 1`
     - `A[1][0]` (3) & `B[1][0]` (2): `1 & 1 = 1`
     - `A[1][1]` (3) & `B[1][1]` (2): `1 & 1 = 1`
   - For the bit `2`:
     - `A[0][0]` (3) & `B[0][0]` (2): `0 & 1 = 0`
     - `A[0][1]` (3) & `B[0][1]` (2): `0 & 1 = 0`
     - `A[1][0]` (3) & `B[1][0]` (2): `0 & 1 = 0`
     - `A[1][1]` (3) & `B[1][1]` (2): `0 & 1 = 0`

3. **Transformation Feasibility:**
   - The transformation from `3` to `2` requires setting the least significant bit to `0` and the second bit to `1`.
   - The `&` operation can only set bits to `0`, and the `|` operation can only set bits to `1`.
   - Since all elements in `A` are `3` and all elements in `B` are `2`, the transformation is possible by performing the `&` operation on the least significant bit and the `|` operation on the second bit.

4. **Expected Output:**
   - The expected output for this test case is `Yes`.

This test case is small and valid, and it should expose any issues in the logic of the `checkBit` function, especially in how it handles the transformation of bits.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
// package codeforces.C2043_E_MatrixTransformation;

import java.io.*;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.Queue;

/*
https://codeforces.com/contest/1679/submission/157139001
https://codeforces.com/blog/entry/116603
 */
public class Naive1 {

    InputStream is;
    FastWriter out;
    String INPUT = "4\n" +
            "1 1\n" +
            "12\n" +
            "13\n" +
            "2 2\n" +
            "10 10\n" +
            "42 42\n" +
            "21 21\n" +
            "21 21\n" +
            "2 2\n" +
            "74 10\n" +
            "42 106\n" +
            "21 85\n" +
            "85 21\n" +
            "2 4\n" +
            "1 2 3 4\n" +
            "5 6 7 8\n" +
            "3 2 3 4\n" +
            "1 0 1 0\n";

    public static void main(String[] args) throws Exception {
        new Naive1().run();
    }

    void run() throws Exception {
        is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
        out = new FastWriter(System.out);

        long s = System.currentTimeMillis();
        solve();
        out.flush();
        objTr(System.currentTimeMillis() - s + "ms");
    }

    void solve() {
        int t = ni();
        for (int i = 0; i < t; i++) {
            int r = ni();
            int c = ni();
            int[][] mat1 = nmi(r, c);
            int[][] mat2 = nmi(r, c);

            if (i == 8) {
                out.println(r + " " + c);
                out.println(Arrays.deepToString(mat1));
                out.println(Arrays.deepToString(mat2));
            }

            transform(r, c, mat1, mat2);
        }
    }

    private void transform(int r, int c, int[][] mat1, int[][] mat2) {
        int[] and = new int[r];
        for (int i = 0; i < r; i++) {
            int mask = Integer.MAX_VALUE;
            for (int j = 0; j < c; j++) {
                int m = 1;
                while (m < mat1[i][j] || m < mat2[i][j]) {
                    if ((mat1[i][j] & m) != 0 && (mat2[i][j] & m) == 0) {
                        if ((mask & m) != 0) {
                            mask -= m;
                        }
                    }
                    m <<= 1;
                }
            }
            and[i] = mask;
        }

        int[] or = new int[c];
        for (int i = 0; i < c; i++) {
            int mask = 0;
            for (int j = 0; j < r; j++) {
                int m = 1;
                while (m < mat1[j][i] || m < mat2[j][i]) {
                    if ((mat1[j][i] & m) == 0 && (mat2[j][i] & m) != 0) {
                        mask |= m;
                    }
                    m <<= 1;
                }
            }
            or[i] = mask;
        }

        int[][][] order = new int[r][c][32];
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                int ind = 0;
                int m = 1;
                while (m < mat1[i][j] || m < mat2[i][j]) {
                    int a = mat1[i][j] & m;
                    int b = mat2[i][j] & m;

                    if (a == 0 && b == 0) {
                        if ((and[i] & m) != 0) {
                            and[i] -= m;
                        }
                        if ((or[j] & m) != 0) {
                            order[i][j][ind] = -1;
                        }
                    } else if (a != 0 && b != 0) {
                        if ((or[j] & m) == 0) {
                            or[j] |= m;
                        }
                        if ((and[i] & m) == 0) {
                            order[i][j][ind] = 1;
                        }
                    } else if (a != 0 && b == 0) {
                        if ((and[i] & m) != 0) {
                            and[i] -= m;
                        }
                        if ((or[j] & m) != 0) {
                            order[i][j][ind] = -1;
                        }
                    } else {
                        if ((or[j] & m) == 0) {
                            or[j] |= m;
                        }
                        if ((and[i] & m) == 0) {
                            order[i][j][ind] = 1;
                        }
                    }

                    m <<= 1;
                    ind++;
                }
            }
        }

        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                int m = 1;
                int ind = 0;
                while (m < mat1[i][j] || m < mat2[i][j]) {
                    if (order[i][j][ind] == 1) {
                        for (int t = j + 1; t < c; t++) {
                            if (order[i][t][ind] == -1) {
                                for (int k = i + 1; k < r; k++) {
                                    if (order[k][j][ind] == -1 && order[k][t][ind] == 1) {
                                        out.println("no");
                                        return;
                                    }
                                }
                            }
                        }
                    } else if (order[i][j][ind] == -1) {
                        for (int t = j + 1; t < c; t++) {
                            if (order[i][t][ind] == 1) {
                                for (int k = i + 1; k < r; k++) {
                                    if (order[k][j][ind] == 1 && order[k][t][ind] == -1) {
                                        out.println("no");
                                        return;
                                    }
                                }
                            }
                        }
                    }

                    m <<= 1;
                    ind++;
                }
            }
        }

        out.println("yes");
    }

    private final byte[] inBuf = new byte[1024];
    public int lenBuf = 0, ptrBuf = 0;

    private int readByte() {
        if (lenBuf == -1) {
            throw new InputMismatchException();
        }
        if (ptrBuf >= lenBuf) {
            ptrBuf = 0;
            try {
                lenBuf = is.read(inBuf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (lenBuf <= 0) {
                return -1;
            }
        }
        return inBuf[ptrBuf++];
    }

    private boolean isSpaceChar(int c) {
        return !(c >= 33 && c <= 126);
    }

    private int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b)) ;
        return b;
    }

    private double nd() {
        return Double.parseDouble(ns());
    }

    private char nc() {
        return (char) skip();
    }

    private String ns() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    private char[] ns(int n) {
        char[] buf = new char[n];
        int b = skip(), p = 0;
        while (p < n && !(isSpaceChar(b))) {
            buf[p++] = (char) b;
            b = readByte();
        }
        return n == p ? buf : Arrays.copyOf(buf, p);
    }

    private int[] na(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = ni();
        }
        return a;
    }

    private long[] nal(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nl();
        }
        return a;
    }

    private char[][] nm(int n, int m) {
        char[][] map = new char[n][];
        for (int i = 0; i < n; i++) {
            map[i] = ns(m);
        }
        return map;
    }

    private int[][] nmi(int n, int m) {
        int[][] map = new int[n][];
        for (int i = 0; i < n; i++) {
            map[i] = na(m);
        }
        return map;
    }

    private int ni() {
        return (int) nl();
    }

    private long nl() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
        if (b == '-') {
            minus = true;
            b = readByte();
        }

        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    public static class FastWriter {
        private static final int BUF_SIZE = 1 << 13;
        private final byte[] buf = new byte[BUF_SIZE];
        private final OutputStream out;
        private int ptr = 0;

        private FastWriter() {
            out = null;
        }

        public FastWriter(OutputStream os) {
            this.out = os;
        }

        public FastWriter(String path) {
            try {
                this.out = new FileOutputStream(path);
            } catch (FileNotFoundException e) {
                throw new RuntimeException("FastWriter");
            }
        }

        public FastWriter write(byte b) {
            buf[ptr++] = b;
            if (ptr == BUF_SIZE) {
                innerFlush();
            }
            return this;
        }

        public FastWriter write(char c) {
            return write((byte) c);
        }

        public FastWriter write(char[] s) {
            for (char c : s) {
                buf[ptr++] = (byte) c;
                if (ptr == BUF_SIZE) {
                    innerFlush();
                }
            }
            return this;
        }

        public FastWriter write(String s) {
            s.chars().forEach(c -> {
                buf[ptr++] = (byte) c;
                if (ptr == BUF_SIZE) {
                    innerFlush();
                }
            });
            return this;
        }

        private static int countDigits(int l) {
            if (l >= 1000000000) return 10;
            if (l >= 100000000) return 9;
            if (l >= 10000000) return 8;
            if (l >= 1000000) return 7;
            if (l >= 100000) return 6;
            if (l >= 10000) return 5;
            if (l >= 1000) return 4;
            if (l >= 100) return 3;
            if (l >= 10) return 2;
            return 1;
        }

        public FastWriter write(int x) {
            if (x == Integer.MIN_VALUE) {
                return write((long) x);
            }
            if (ptr + 12 >= BUF_SIZE) {
                innerFlush();
            }
            if (x < 0) {
                write((byte) '-');
                x = -x;
            }
            int d = countDigits(x);
            for (int i = ptr + d - 1; i >= ptr; i--) {
                buf[i] = (byte) ('0' + x % 10);
                x /= 10;
            }
            ptr += d;
            return this;
        }

        private static int countDigits(long l) {
            if (l >= 1000000000000000000L) return 19;
            if (l >= 100000000000000000L) return 18;
            if (l >= 10000000000000000L) return 17;
            if (l >= 1000000000000000L) return 16;
            if (l >= 100000000000000L) return 15;
            if (l >= 10000000000000L) return 14;
            if (l >= 1000000000000L) return 13;
            if (l >= 100000000000L) return 12;
            if (l >= 10000000000L) return 11;
            if (l >= 1000000000L) return 10;
            if (l >= 100000000L) return 9;
            if (l >= 10000000L) return 8;
            if (l >= 1000000L) return 7;
            if (l >= 100000L) return 6;
            if (l >= 10000L) return 5;
            if (l >= 1000L) return 4;
            if (l >= 100L) return 3;
            if (l >= 10L) return 2;
            return 1;
        }

        public FastWriter write(long x) {
            if (x == Long.MIN_VALUE) {
                return write("" + x);
            }
            if (ptr + 21 >= BUF_SIZE) {
                innerFlush();
            }
            if (x < 0) {
                write((byte) '-');
                x = -x;
            }
            int d = countDigits(x);
            for (int i = ptr + d - 1; i >= ptr; i--) {
                buf[i] = (byte) ('0' + x % 10);
                x /= 10;
            }
            ptr += d;
            return this;
        }

        public FastWriter write(double x, int precision) {
            if (x < 0) {
                write('-');
                x = -x;
            }
            x += Math.pow(10, -precision) / 2;
            //		if(x < 0){ x = 0; }
            write((long) x).write(".");
            x -= (long) x;
            for (int i = 0; i < precision; i++) {
                x *= 10;
                write((char) ('0' + (int) x));
                x -= (int) x;
            }
            return this;
        }

        public FastWriter writeln(char c) {
            return write(c).writeln();
        }

        public FastWriter writeln(int x) {
            return write(x).writeln();
        }

        public FastWriter writeln(long x) {
            return write(x).writeln();
        }

        public FastWriter writeln(double x, int precision) {
            return write(x, precision).writeln();
        }

        public FastWriter write(int... xs) {
            boolean first = true;
            for (int x : xs) {
                if (!first) {
                    write(' ');
                }
                first = false;
                write(x);
            }
            return this;
        }

        public FastWriter write(long... xs) {
            boolean first = true;
            for (long x : xs) {
                if (!first) {
                    write(' ');
                }
                first = false;
                write(x);
            }
            return this;
        }

        public FastWriter writeln() {
            return write((byte) '\n');
        }

        public FastWriter writeln(int... xs) {
            return write(xs).writeln();
        }

        public FastWriter writeln(long... xs) {
            return write(xs).writeln();
        }

        public FastWriter writeln(char[] line) {
            return write(line).writeln();
        }

        public FastWriter writeln(char[]... map) {
            for (char[] line : map) {
                write(line).writeln();
            }
            return this;
        }

        public FastWriter writeln(String s) {
            return write(s).writeln();
        }

        private void innerFlush() {
            try {
                out.write(buf, 0, ptr);
                ptr = 0;
            } catch (IOException e) {
                throw new RuntimeException("innerFlush");
            }
        }

        public void flush() {
            innerFlush();
            try {
                out.flush();
            } catch (IOException e) {
                throw new RuntimeException("flush");
            }
        }

        public FastWriter print(byte b) {
            return write(b);
        }

        public FastWriter print(char c) {
            return write(c);
        }

        public FastWriter print(char[] s) {
            return write(s);
        }

        public FastWriter print(String s) {
            return write(s);
        }

        public FastWriter print(int x) {
            return write(x);
        }

        public FastWriter print(long x) {
            return write(x);
        }

        public FastWriter print(double x, int precision) {
            return write(x, precision);
        }

        public FastWriter println(char c) {
            return writeln(c);
        }

        public FastWriter println(int x) {
            return writeln(x);
        }

        public FastWriter println(long x) {
            return writeln(x);
        }

        public FastWriter println(double x, int precision) {
            return writeln(x, precision);
        }

        public FastWriter print(int... xs) {
            return write(xs);
        }

        public FastWriter print(long... xs) {
            return write(xs);
        }

        public FastWriter println(int... xs) {
            return writeln(xs);
        }

        public FastWriter println(long... xs) {
            return writeln(xs);
        }

        public FastWriter println(char[] line) {
            return writeln(line);
        }

        public FastWriter println(char[]... map) {
            return writeln(map);
        }

        public FastWriter println(String s) {
            return writeln(s);
        }

        public FastWriter println() {
            return writeln();
        }
    }

    public void trnz(int... o) {
        for (int i = 0; i < o.length; i++) {
            if (o[i] != 0) {
                System.out.print(i + ":" + o[i] + " ");
            }
        }
        System.out.println();
    }

    // print ids which are 1
    public void trt(long... o) {
        Queue<Integer> stands = new ArrayDeque<>();
        for (int i = 0; i < o.length; i++) {
            for (long x = o[i]; x != 0; x &= x - 1) {
                stands.add(i << 6 | Long.numberOfTrailingZeros(x));
            }
        }
        System.out.println(stands);
    }

    public void tf(boolean... r) {
        for (boolean x : r) {
            System.out.print(x ? '#' : '.');
        }
        System.out.println();
    }

    public void tf(boolean[]... b) {
        for (boolean[] r : b) {
            for (boolean x : r) {
                System.out.print(x ? '#' : '.');
            }
            System.out.println();
        }
        System.out.println();
    }

    public void tf(long[]... b) {
        if (INPUT.length() != 0) {
            for (long[] r : b) {
                for (long x : r) {
                    for (int i = 0; i < 64; i++) {
                        System.out.print(x << ~i < 0 ? '#' : '.');
                    }
                }
                System.out.println();
            }
            System.out.println();
        }
    }

    public void tf(long... b) {
        if (INPUT.length() != 0) {
            for (long x : b) {
                for (int i = 0; i < 64; i++) {
                    System.out.print(x << ~i < 0 ? '#' : '.');
                }
            }
            System.out.println();
        }
    }

    private final boolean oj = System.getProperty("ONLINE_JUDGE") != null;

    private void objTr(Object... o) {
        if (!oj) {
            System.out.println(Arrays.deepToString(o));
        }
    }

}

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

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

Think step by step.

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

1. **Bitwise Operations**: The operations involve bitwise AND and OR, so we need to ensure that the test case includes numbers that have different bit patterns, especially where the AND and OR operations can lead to different results.
2. **Matrix Dimensions**: Use small matrices (e.g., 2x2 or 3x3) to keep the input manageable and easy to verify.
3. **Edge Cases**: Consider matrices where some elements are 0 or the maximum value (10^9), as these can cause interesting behavior in bitwise operations.
4. **Infeasibility**: Ensure that the test case is designed in such a way that it is impossible to transform matrix A into matrix B, to test the algorithm's ability to detect infeasibility.

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

```plaintext
4
2 2
1 2
3 4
0 0
0 0
2 2
1 2
3 4
5 6
7 8
2 2
1 2
3 4
3 2
1 4
2 2
1 2
3 4
1 3
3 1
```

### Explanation:
1. **Test Case 1**:
   - Matrix A: `[[1, 2], [3, 4]]`
   - Matrix B: `[[0, 0], [0, 0]]`
   - This case should be `Yes` because we can set all elements to 0 using bitwise AND with 0.

2. **Test Case 2**:
   - Matrix A: `[[1, 2], [3, 4]]`
   - Matrix B: `[[5, 6], [7, 8]]`
   - This case should be `No` because it is impossible to transform the elements of A into the elements of B using the given operations.

3. **Test Case 3**:
   - Matrix A: `[[1, 2], [3, 4]]`
   - Matrix B: `[[3, 2], [1, 4]]`
   - This case should be `No` because the elements of A cannot be transformed to match B using the given operations.

4. **Test Case 4**:
   - Matrix A: `[[1, 2], [3, 4]]`
   - Matrix B: `[[1, 3], [3, 1]]`
   - This case should be `No` because the elements of A cannot be transformed to match B using the given operations.

This test input is designed to cover a variety of scenarios, including infeasible transformations and simple transformations, to thoroughly test the algorithm's ability to handle different cases.