**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
def transpose(matrix):
    return list(map(list, zip(*matrix)))

from collections import defaultdict

for _ in range(int(input())):
	m,n = [int(i) for i in input().split()]
	res = 'YES'
	a = []
	b = []
	c = defaultdict(int)

	for i in range(m):
		a.append([i for i in input().split()])
	for i in range(m):
		b.append([i for i in input().split()])

	if m > n:
		a = transpose(a)
		b = transpose(b)

	if m != 1 and n != 1:
		for i in b:
			if a[0][0] in i:
				for j in range(n):
					c[i[j]] = j
				# print(c)
				break
		indx = []

		for i in a[0]:
			indx.append(c[i])
		# print(indx)
		if len(indx) != n:
			res = 'NO'
		else:
			for i in a:
				k = []
				for j in b:
					if i[0] in j:
						for nm in indx:
							k.append(j[nm])
						# print(k,j)
						if k != i:
							res = 'NO'
							break
	print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
import java.io.*;
import java.util.*;
public class E {
  static int HAS_TESTCASES = 1;
  static String INPUT = "1";
  void solve() {
    int n = ni();
    int m = ni();
    int[][] mata = nim(n, m);
    int[][] matb = nim(n, m);
    List<Integer> suma = new ArrayList<>();
    List<Integer> sumb = new ArrayList<>();
    for (int i = 0; i<n; i++) {
      int rowsuma = 0;
      int rowsumb = 0;
      for (int j = 0; j<m; j++) {
        rowsuma+=mata[i][j];
        rowsumb+=matb[i][j];
      }
      suma.add(rowsuma);
      sumb.add(rowsumb);
    }
    List<Integer> cola = new ArrayList<>();
    List<Integer> colb = new ArrayList<>();
    for (int j = 0; j<m; j++) {
      int colsuma = 0;
      int colsumb = 0;
      for (int i = 0; i<n; i++) {
        colsuma+=mata[i][j];
        colsumb+=matb[i][j];
      }
      cola.add(colsuma);
      colb.add(colsumb);
    }
    Collections.sort(suma);
    Collections.sort(sumb);
    Collections.sort(cola);
    Collections.sort(colb);
    for (int i = 0; i<n; i++) {
      if (suma.get(i)!=(int) sumb.get(i)) {
        out.println("no");
        return;
      }
    }
    for (int i = 0; i<m; i++) {
      if (cola.get(i)!=(int) colb.get(i)) {
        out.println("no");
        return;
      }
    }
    out.println("yes");
  }
  static int ni() {
    return Integer.parseInt(ns());
  }
  static int[][] nim(int n, int m) {
    int[][] a = new int[n][m];
    for (int i = 0; i<n; i++) {
      for (int j = 0; j<m; j++) {
        a[i][j]=ni();
      }
    }
    return a;
  }
  static String ns() {
    t();
    return st.nextToken();
  }
  static void t() {
    while (!st.hasMoreTokens()) {
      try {
        st=new StringTokenizer(br.readLine());
      } catch (Exception e) {
        throw new RuntimeException("out of tokens!");
      }
    }
  }
  static BufferedReader br;
  static StringTokenizer st = new StringTokenizer("");
  static PrintWriter out = new PrintWriter(System.out);
  public static void main(String[] a) {
    if (System.getProperty("ONLINE_JUDGE")==null) {
      System.setIn(new ByteArrayInputStream(INPUT.getBytes()));
    }
    br=new BufferedReader(new InputStreamReader(System.in));
    int t = HAS_TESTCASES>0 ? ni() : 1;
    for (int i = 1; i<=t; i++) {
      new E().solve();
    }
    out.flush();
  }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases or scenarios that can highlight the flaw in the logic. The code is checking the sums of rows and columns and sorting them to compare, which might not be sufficient to determine if one matrix can be transformed into another using the allowed operations.

### Key Points to Consider:
1. **Row and Column Sums**: The code only checks the sums of rows and columns and does not consider the actual positions of the elements.
2. **Permutations**: Both matrices are permutations, so the sums of rows and columns will always be the same if the matrices contain the same elements.
3. **Transformations**: Swapping rows and columns can change the positions of elements, but not their values.

### Challenging Test Input:
We need to create a test case where the row and column sums are the same, but the matrices cannot be transformed into each other by swapping rows and columns.

### Example:
Consider the following matrices:
- Matrix `a`:
  ```
  1 2
  3 4
  ```
- Matrix `b`:
  ```
  1 3
  2 4
  ```

In this case:
- The row sums for both matrices are `[3, 7]`.
- The column sums for both matrices are `[4, 6]`.

However, it is impossible to transform matrix `a` into matrix `b` by swapping rows and columns.

### Test Input:
```plaintext
1
2 2
1 2
3 4
1 3
2 4
```

This test input should expose the bug in the code because the row and column sums are the same, but the matrices cannot be transformed into each other using the allowed operations. The code will incorrectly output "YES" for this test case, while the correct output should be "NO".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.*;

// Add your class name here
public class E_Permutation_of_Rows_and_Columns {

  public static void solve() throws IOException {

    int n = in();
    int m = in();
    HashMap<Integer, int[]> hm = new HashMap<>();
    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] = in();
        hm.put(a[i][j], new int[] { i, j });
      }
    }
    boolean isSame = true;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        b[i][j] = in();
        if (a[i][j] != b[i][j]) {
          isSame = false;
        }

      }
    }

    if (isSame || n == 1) {
      out.println("YES");
      return;
    }
    HashSet<Integer> hs = new HashSet<>();
    for (int j = 0; j < m; j++) {
      for (int i = 0; i < n; i++) {
        hs.add(b[i][j]);
      }

      if (!check(a, b[0][j], hm, hs, n)) {
        out.println("NO");
        return;
      }
      hs = new HashSet<>();
    }
    out.println("YES");
  }

  public static boolean check(int[][] a, int val, HashMap<Integer, int[]> hm, HashSet<Integer> hs, int n) {

    int[] index = hm.get(val);
    int row = index[0];
    int col = index[1];

    int x = row;
    while (x >= 0) {
      if (!hs.contains(a[x][col])) {
        return false;
      }

      hs.remove(a[x][col]);
      x--;
    }

    x = row + 1;
    while (x < n) {
      if (!hs.contains(a[x][col])) {
        return false;
      }

      hs.remove(a[x][col]);
      x++;
    }

    return hs.isEmpty();

  }

  static boolean MULTI_CASE = true;

  public static void main(String[] args) throws IOException {
    if (MULTI_CASE) {
      int T = sc.nextInt();
      for (int i = 0; i < T; ++i) {
        solve();
      }
    } else {
      solve();
    }
    out.close();
  }

  static int in() throws IOException {
    return sc.nextInt();
  }

  static long inl() throws IOException {
    return sc.nextLong();
  }

  static void py() {
    out.println("YES");
  }

  static void pn() {
    out.println("NO");
  }

  static InputReader sc = new InputReader();
  static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

  static class InputReader {
    private StringTokenizer st;
    private BufferedReader bf;

    public InputReader() {
      bf = new BufferedReader(new InputStreamReader(System.in));
      st = null;
    }

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

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

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

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

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

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

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

Think step by step.

Response:
```plaintext
2
2 2
1 2
3 4
4 3
2 1
2 3
1 2 3
4 5 6
4 5 6
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
for i in range(int(input())):
    n,m=map(int,input().split())
    a=[]
    b=[]
    for i in range(n):
        a.append(list(map(int,input().split())))
    for i in range(n):
        b.append(list(map(int,input().split())))
    aa=[]
    bb=[]
    for i in zip(*a):
        aa.append(sorted(list(i)))
    for i in zip(*b):
        bb.append(sorted(list(i)))
    p=1
    for i in aa:
        if i not in bb:
            p=0
            break
    print(["NO","YES"][p])

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
import sys
input = sys.stdin.readline

def intput():
    return int(input())

def solve():
    rows, columns = [int(x) for x in input().split()]
    valid = True

    m1 = []
    rowsneeded = []
    rowsset = set()
    for i in range(rows):
        r = tuple(int(x) for x in input().split())
        m1.append(r)
        rowsneeded.append(set(r))
        rowsset.add(r)

    #print(rowsset)

    firstl = tuple(int(x) for x in input().split())

    try:
        idx = rowsneeded.index(set(firstl))
        targetperm = m1[idx]
        #assert set(targetperm) == set(firstl)
    except ValueError:
        valid = False
        for i in range(1, rows):
            newl = tuple(int(x) for x in input().split())
        return False

    perm = []
    for i in firstl:
        perm.append(targetperm.index(i))

    #print(perm)
    #assert set(perm) == set(range(columns))
    for i in range(1, rows):
        newl = tuple(int(x) for x in input().split())
        perml = tuple(newl[idx] for idx in perm)
        #assert set(newl) == set(perml)
        #print(perml)
        if perml not in rowsset:
            valid = False

    if valid or rows == 1 or columns == 1:
        return True
    else:
        return False

test_cases = intput()
for t in range(test_cases):
    if solve():
        print('YES')
    else:
        print('NO')

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

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

public class Main {// Codechef {
    static class pair {
        int val1;
        int val2;

        public pair(int val1, int val2) {
            this.val1 = val1;
            this.val2 = val2;
        }

        public boolean equals(Object o) {
            if (this == o)
                return true;
            if (o == null || getClass() != o.getClass())
                return false;
            pair pair = (pair) o;
            return val1 == pair.val1 && val2 == pair.val2;
        }

        @Override
        public int hashCode() {
            return 31 * Long.hashCode(val1) + Long.hashCode(val2);
        }

        @Override
        public String toString() {
            return "pair{" + "x=" + val1 + ", y=" + val2 + '}';
        }
    }

    public static void main(String[] args) {
        int TTT = 0;
        try {
            FastReader in = new FastReader();
            FastWriter out = new FastWriter();
            int testCases = in.nextInt();
            while (TTT++ != testCases) {
                // write code here
                int n = in.nextInt();
                int m = in.nextInt();
                long arr[][] = new long[n][m];
                for (int i = 0; i < n; i++) {
                    for (int j = 0; j < m; j++) {
                        arr[i][j] = in.nextLong();
                    }
                }
                long brr[][] = new long[n][m];
                for (int i = 0; i < n; i++) {
                    for (int j = 0; j < m; j++) {
                        brr[i][j] = in.nextLong();
                    }
                }
                List<List<Long>> lsta = new ArrayList<>();
                List<List<Long>> lstb = new ArrayList<>();
                for (int i = 0; i < n; i++) {
                    List<Long> tempa = new ArrayList<>();
                    List<Long> tempb = new ArrayList<>();
                    for (int j = 0; j < m; j++) {
                        tempa.add(arr[i][j]);
                        tempb.add(brr[i][j]);
                    }
                    Collections.sort(tempa);
                    Collections.sort(tempb);
                    lsta.add(tempa);
                    lstb.add(tempb);
                }
                Collections.sort(lsta, (a, b) -> Long.compare(a.get(0) , b.get(0)));
                Collections.sort(lstb, (a, b) -> Long.compare(a.get(0) , b.get(0)));
                boolean flag = true;
                for (int i = 0; i < lsta.size(); i++) {
                    for (int j = 0; j < lsta.get(i).size(); j++) {
                        if (lsta.get(i).get(j) != lstb.get(i).get(j)) {
                            flag = false;
                            break;
                        }
                    }
                }
                lsta.clear();
                lstb.clear();
                if (flag == true) {
                    for (int i = 0; i < m; i++) {
                        List<Long> tempa = new ArrayList<>();
                        List<Long> tempb = new ArrayList<>();
                        for (int j = 0; j < n; j++) {
                            tempa.add(arr[j][i]);
                            tempb.add(brr[j][i]);
                        }
                        Collections.sort(tempa);
                        Collections.sort(tempb);
                        lsta.add(tempa);
                        lstb.add(tempb);
                    }
                    Collections.sort(lsta, (a, b) -> Long.compare(a.get(0) , b.get(0)));
                    Collections.sort(lstb, (a, b) -> Long.compare(a.get(0) , b.get(0)));
                    for (int i = 0; i < lsta.size(); i++) {
                        for (int j = 0; j < lsta.get(i).size(); j++) {
                            if (lsta.get(i).get(j) != lstb.get(i).get(j)) {
                                flag = false;
                                break;
                            }
                        }
                    }
                }
                if(flag == false){
                    out.println("NO");
                }
                else{
                    out.println("YES");
                }
            }
            out.close();
        } catch (

        Exception e) {
            System.out.println(e);
            return;
        }
    }

    public static long gcd(long x, long y) {
        if (x > y)
            return gcd(y, x);
        if (y == 0) {
            return x;
        }
        if (x == 0) {
            return y;
        }
        return gcd(y % x, x);
    }

    public static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    public static int lowerbound(long[] arr, long key) {
        // If the key is found multiple times, it returns the index of the x
        // occurrence. If the key is not found, it returns the index of the x
        // element greater than the key.
        int low = 0;
        int high = arr.length - 1;
        int result = -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] >= key) {
                result = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        return result;
    }

    public static int upperbound(long[] arr, long key) {
        // If the key is found multiple times, it returns the index of the last
        // occurrence. If the key is not found, it returns the index of the x
        // element greater than the key.
        int low = 0;
        int high = arr.length - 1;
        int result = -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] <= key) {
                result = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }

        return result;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class FastWriter {

        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to craft a test case that challenges the assumptions and logic of the code. The code checks if the sorted rows and columns of the two matrices are equal, which is not a sufficient condition to determine if one matrix can be transformed into the other using the allowed operations (swapping rows and columns).

### Key Points to Consider:
1. **Permutations and Sorting**: The code sorts the rows and columns of both matrices and checks for equality. This means that the code might incorrectly return "YES" if the sorted rows and columns are the same, even if the original matrices cannot be transformed into each other.

2. **Row and Column Swapping**: The code does not account for the fact that the order of elements within the rows and columns matters. Swapping rows and columns can change the relative positions of elements, and the code needs to ensure that the resulting matrix can be transformed into the target matrix.

### Constructing the Test Case:
To create a test case that exposes this bug, we need a matrix where the sorted rows and columns are the same, but the original matrices cannot be transformed into each other by swapping rows and columns.

### Example Test Case:
Consider the following matrices:
- **Matrix A**:
  ```
  1 2 3
  4 5 6
  7 8 9
  ```

- **Matrix B**:
  ```
  1 4 7
  2 5 8
  3 6 9
  ```

### Explanation:
- **Matrix A** is a simple 3x3 matrix with elements in a natural order.
- **Matrix B** is a matrix where the elements are rearranged such that the columns of **Matrix B** are the rows of **Matrix A**. This means that the sorted rows and columns of both matrices are the same, but **Matrix A** cannot be transformed into **Matrix B** by swapping rows and columns.

### Test Input:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 8
3 6 9
```

### Expected Output:
```plaintext
NO
```

This test case should expose the bug in the code, as the code will incorrectly return "YES" due to the sorted rows and columns being the same, even though the matrices cannot be transformed into each other using the allowed operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
/**
 * LC Contest First Problem
 */

import java.io.*;
import java.util.*;
public class Main1 {
    static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
    static StreamTokenizer st  =new StreamTokenizer(bf);
    static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
    public static void main(String[]args) throws IOException{
        int T = 1;
        T=I();   //根据需要这行可能要注释掉
        while(T-- > 0) {
            solve();
        }
        pw.flush();
    }
    static int I() throws IOException {
        st.nextToken();
        return (int)st.nval;
    }
    static String S() throws IOException{
        String res = bf.readLine();
        while(res.equals(""))
            res=bf.readLine();
        return res;
    }

    static void solve() throws IOException {
          int n = I() , m = I();
          int[][] a = new int[n][m] , b = new int[n][m];
          for (int i = 0 ; i < n ; i++){
              for (int j = 0 ; j < m ; j++){
                  a[i][j] = I();
              }
          }
          for (int i = 0 ; i < n ; i++){
              for (int j = 0 ; j < m ; j++){
                  b[i][j] = I();
              }
          }
          List<Integer>[] aa = new List[n] , bb = new List[n];
          Arrays.setAll(aa , e -> new ArrayList<>());
          Arrays.setAll(bb , e -> new ArrayList<>());
          for (int i = 0 ; i < n ; i++){
              for (int j = 0 ; j < m ; j++){
                  aa[i].add(a[i][j]);
                  bb[i].add(b[i][j]);
              }
              Collections.sort(aa[i]);
              Collections.sort(bb[i]);
          }
          Arrays.sort(aa , (o1 ,o2) -> o1.get(0) - o2.get(0));
          Arrays.sort(bb , (o1 , o2) -> o1.get(0) - o2.get(0));
          for (int i = 0 ; i < n ; i++){
              for (int j = 0 ; j < m ; j++){
                  if (aa[i].get(j) != bb[i].get(j)){
                      pw.println("NO");
                      return;
                  }
              }
          }
        List<Integer>[] aaa = new List[m] , bbb = new List[m];
        Arrays.setAll(aaa , e -> new ArrayList<>());
        Arrays.setAll(bbb , e -> new ArrayList<>());
        for (int j = 0 ; j < m ; j++){
            for (int i = 0 ; i < n ; i++){
                aaa[j].add(a[i][j]);
                bbb[j].add(b[i][j]);
            }
            Collections.sort(aaa[j]);
            Collections.sort(bbb[j]);
        }
        Arrays.sort(aaa , (o1 , o2) -> o1.get(0) - o2.get(0));
        Arrays.sort(bbb , (o1 , o2) -> o1.get(0) - o2.get(0));
        for (int j = 0 ; j < m ; j++){
            for (int i = 0 ; i < n ; i++){
                if (aaa[j].get(i) != bbb[j].get(i)){
                    pw.println("NO");
                    return;
                }
            }
        }
        pw.println("YES");

    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

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

bool cmp(const map<vector<int>, int>& a, const map<vector<int>, int>& b) {
    return a == b;
}

map<vector<int>, int> fMap(const vector<vector<int>>& mat, bool row) {
    map<vector<int>, int> fmap;
    int outer = row ? mat.size() : mat[0].size();
    int inner = row ? mat[0].size() : mat.size();

    for (int i = 0; i < outer; ++i) {
        vector<int> freq(1001, 0);
        for (int j = 0; j < inner; ++j) {
            int elem = row ? mat[i][j] : mat[j][i];
            freq[elem]++;
        }
        fmap[freq]++;
    }

    return fmap;
}

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

    int t;
    cin >> t;

    while (t--) {
        int n, m;
        cin >> n >> m;

        vector<vector<int>> mat1(n, vector<int>(m));
        vector<vector<int>> mat2(n, vector<int>(m));

        for (auto& r : mat1) {
            for (auto& e : r) {
                cin >> e;
            }
        }

        for (auto& r : mat2) {
            for (auto& e : r) {
                cin >> e;
            }
        }

        auto rMap1 = fMap(mat1, true);
        auto rMap2 = fMap(mat2, true);
        auto cMap1 = fMap(mat1, false);
        auto cMap2 = fMap(mat2, false);

        if (cmp(rMap1, rMap2) && cmp(cMap1, cMap2)) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

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

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

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

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

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

#define bug(a) cout<< #a << " : " << a <<endl;
using ll = long long;

void solve( int cs ){
    int n, m;
    cin >> n >> m;

    int a[n + 1][m + 1], b[n + 1][m + 1];
    map<int,pair<int,int>> mp1;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            cin >> a[i][j];
            mp1[a[i][j]] = {i, j};
        }
    }

    map<int,pair<int,int>> mp;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            cin >> b[i][j];
            mp[b[i][j]] = {i, j};
        }
    }

    for(int num = 1; num <= n * m; num++){
        int cur = num;

        int col = mp[cur].second;
        int row = mp[cur].first;

        int i = mp1[cur].first;
        int j = mp1[cur].second;
        if(col != j){
            for(int k = 1; k <= n; k++) {
                swap(a[k][j], a[k][col]);
                swap(mp1[a[k][j]], mp1[a[k][col]]);
            }
        }
        if(row != i){
            for(int k = 1; k <= m; k++) {
                swap(a[i][k], a[row][k]);
                swap(mp1[a[i][k]], mp1[a[row][k]]);
            }
        }

    }

    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            //cout << a[i][j]
            if(a[i][j] != b[i][j]){
                cout << "NO" << '\n';
                return;
            }
        }
    }
    cout << "YES" << '\n';
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);

    int testCase = 1, cs = 0;
    cin >> testCase;

    while(testCase--){
        solve( ++cs );
    }

}

///     < BumbleBee >     ///

**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 2
3 4
4 3
2 1
2 2
1 2
3 4
3 4
1 2
2 1
3 4
4 3
5 6
6 5
3 3
1 2 3
4 5 6
7 8 9
7 4 1
8 5 2
9 6 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#ifdef DEBUG
#include "debug.h"
#else
#define dbg(x...)
#endif
#define fastInOut ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int int64_t
#define endl '\n'
#define mod 1000000007
#define pi 3.14159265358979323846
#define mem(a,b) memset(a, b, sizeof(a) )
typedef double dl;
typedef long long ll;
typedef unsigned long long ull;

void solve()
{
      int n, x, a;
      cin >> n >> x;
      map <int, int> mp;
      set <int> st;
      int dup = -1;
      for (int i = 0; i < n; i ++) {
        cin >> a;
        st.insert (a);
        mp[a] ++;
        if (mp[a] == 2) {
            dup = a;
        }
      }
      int mex = -1;
      int c = 0;
      for (auto u : st) {
        if (c != u) {
           mex = c;
           break;
        }
        c ++;
      }
      dbg (dup, mex)
      if (mex == 0) {
        cout << 0 << endl;
        return;
      }
      if (dup == -1) {
        cout << n << endl;
        return;
      }
      if (mex == -1) {
          int need = n - 1 - dup;
          if (need % x == 0) {
            cout << n << endl;
            return;
          }
          cout << n - 1 << endl;
          return;
      }
      int need = mex - dup;
      if (need % x == 0) {
        cout << n << endl;
        return;
      }
      cout << mex << endl;
}

int32_t main()
{
        fastInOut;
        //int t = 1;
        int t; cin >> t;
        for (int i = 1; i <= t; i ++) {
          solve();
        }
        return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. Let's break down the problem and identify potential issues:

1. **Handling Duplicates**: The code attempts to handle duplicates but might not do it correctly in all cases.
2. **Handling the Initial MEX Calculation**: The code calculates the initial MEX but might not handle it correctly when the array already has a high MEX.
3. **Handling the Need for Operations**: The code calculates the number of operations needed to increase the MEX but might not handle it correctly in all cases, especially when the MEX is already high or when the array has a large number of duplicates.

Let's create a test case that includes these elements:

1. **Array with Duplicates**: To test the handling of duplicates.
2. **Array with High Initial MEX**: To test the handling of high MEX values.
3. **Array with a Mix of Values**: To test the handling of a mix of values and the need for operations.

Here is a small-scale test input that combines these elements:

```plaintext
3
5 2
0 1 2 3 4
5 2
0 1 2 2 4
5 3
0 1 2 2 4
```

Explanation of the test cases:
1. **Test Case 1**:
   - Array: `[0, 1, 2, 3, 4]`
   - MEX: `5` (no operations needed)
2. **Test Case 2**:
   - Array: `[0, 1, 2, 2, 4]`
   - MEX: `3` (one duplicate, operations needed to increase MEX)
3. **Test Case 3**:
   - Array: `[0, 1, 2, 2, 4]`
   - MEX: `3` (one duplicate, different `x` value, operations needed to increase MEX)

This test input should help expose any bugs in the handling of duplicates, initial MEX calculation, and the need for operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define  f(n)  for(int i=0; i<n; i++)
#define in(n) long long n; cin>>n;
#define vi(v) vector<long>v;
#define all(x) (x).begin(), (x).end()
#define ll long long

int main(){
ios_base::sync_with_stdio(0);cin.tie(0);
int t;
cin>>t;
while(t--){
  in(n);in(m);
  ll a[n];
  map<ll,ll>mp;
  set<ll>s;
  f(n)
  {
    cin>>a[i];
    s.insert(a[i]);
    mp[a[i]]++;
  }
  sort(a,a+n);

  int k=0;

  while(s.find(k)!=s.end())
  {
    k++;
  }
  // cout<<k;
  // if(k-1>0) k=k-1;
  //   else k=k;

  // if(ci>0)
  // { cout<<k<<"\n";

  // }

  // else {
  //   k++;
  if(k<*max_element(a,a+n)) cout<<k<<"\n";
  else
    {  int count=0;
    f(n)
    {
      if(mp[a[i]]>0)
      { count++;

      }
    }

    if(count==0) cout<<k<<"\n";
    else
    {int cc=0;
      f(n)
      {
        if(mp[a[i]]>0)
        {

          if((k-a[i])%m==0) {k=k+mp[a[i]]-1;}

        }

      }

        cout<<k<<"\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
3
1 1
0
2 1
1 0
4 2
2 4 6 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

Buggy Code:
/*
         DxKinG
 */
#include <bits/stdc++.h>
using namespace std;
#define ll  long long
#define pub push_back
#define vll vector<long long>
#define vi vector<int>
#define mll map<long long , long long>
#define mii map<int,int>
#define Y cout<<'Y'<<'E'<<'S'<<endl;
#define N cout<<'N'<<'O'<<endl;
#define F(i,s,e) for(int i=s;i<=e;i++)
#define vvll vector<vector<ll>>
#define ip(v) for(auto x: v){cin>>x;}
ll MOD=(1e+9) +7;
ll fact[200001];
void prec_fact( ll mod){
fact[0]=1;
for(int i=1;i<200001;i++){
fact[i]=i*fact[i-1]%mod;
  }
}
long long gcd(long long int a, long long int b){
if (b == 0) return a;
return gcd(b,a%b);}
long long lcm(long long int a,long long int b){
 return (a / gcd(a, b)) * b; }
ll qp(ll a, ll b,ll mod) {
a=a%mod;
if(b==0) return 1;
if(b==1) return a;
ll p=qp(a*a,b/2,mod);
if(b%2) return p*a%mod;
else return p;
}
bool iset(int i,int x){
x=x>>i;return x&1;}
ll MI(ll x,ll mod){
return qp(x,mod-2,mod);}
ll mod_div(ll n,ll r,ll mod){
return (n%mod)*(MI(r%mod,mod)%mod)%mod;}
ll ncr(ll n,ll r,ll mod){
if(n<r) return 0;
return mod_div(fact[n],fact[n-r]*fact[r]%mod,mod);}

int main(){
int t;cin>>t;
F(idc,0,t-1){

    ll n,x;
    cin>>n>>x;
    ll a[n];
     mll mp;
     F(i,0,200001)
     mp[i]=0;
     F(i,0,n-1){
         cin>>a[i];
         if(a[i]<=200001)
         mp[a[i]]++;
     }
     ll m=-1;
     for(int i=0;i<=200001-x;i++){
         if(mp[i]==0){
             m=i;
             break;
         }
         else{

             mp[i+x]+=mp[i]-1;
             mp[i]=1;

         }
     }
     if(m==-1){
         F(i,0,200001){
             if(mp[i]==0){
                 m=i;
                 break;
             }
         }
     }
     cout<<m<<"\n";
}
return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

Buggy Code:
from collections import Counter
for _ in range(int(input())):
    n,x =map(int,input().split())
    arr =list(map(int,input().split()))
    if 0 not in arr:
        print(0)
    else:
        t =Counter(arr)
        arr.sort()
        def bino(a,arr):
            l=0;r=len(arr)-1
            while l<=r:
                m=(l+r)//2
                if arr[m]==a:
                    return True
                elif arr[m]>a:
                    r=m-1
                else:
                    l=m+1
            return False
        op=[]
        a=0;f=True
        while f==True:
            for i in range(a,10**5+1):
                if bino(i,arr)==True:
                    if t[i]>1:
                        op.append(i)
                else:
                    break
            if op==[]:
                print(i)
                break
            else:
                for j in range(len(op)):
                    if (i-op[j])/x ==(i-op[j])//x and (i-op[j])>0:
                        a=i+1
                    else:
                        f=False
                        print(i)
                        break

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

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

Think step by step.

Response:
```plaintext
1
4 1
0 1 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

Buggy Code:
import sys

def intinput():
    return int(sys.stdin.readline().strip())

def string():
    return sys.stdin.readline().strip()

def listinput():
    return list(map(int, sys.stdin.readline().strip().split()))

def digitslist():
    return [int(i) for i in (list(sys.stdin.readline().strip()))]

def listchar():
    return list(sys.stdin.readline().strip())

def solution():
    n,x=listinput()
    values=listinput()
    values.sort()
    check=set(values)
    for i in range(values[-1]):
        if i not in check:
            print(i)
            return
    if values[-1]+x>values[-1]:
        print(values[-1]+2)
    else:
        print(values[-1]+1)

t = 1
t = intinput()
for _ in range(t):
    solution()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios where the code's logic might fail. Let's analyze the code and identify potential issues:

### Issues in the Buggy Code:
1. **Incorrect MEX Calculation**: The code attempts to find the MEX by checking the presence of integers from 0 to the maximum value in the array. However, it does not correctly handle the scenario where the MEX can be greater than the maximum value in the array.
2. **Inefficient MEX Calculation**: The code sorts the array and uses a set to check for the presence of integers, but it does not handle the case where the MEX can be greater than the maximum value in the array plus one.
3. **Incorrect Final MEX Value**: The final MEX value is not correctly calculated, especially when the MEX can be greater than the maximum value in the array plus one.

### Test Input to Expose the Bug:
To expose these issues, we need a test case where the MEX is greater than the maximum value in the array plus one. This will force the code to fail in its current logic.

### Test Case:
Let's create a test case where the MEX is greater than the maximum value in the array plus one:

```plaintext
1
5 1
0 1 2 3 4
```

### Explanation:
- **Input**:
  - `t = 1`: One test case.
  - `n = 5`, `x = 1`: The array has 5 elements, and the increment value is 1.
  - `a = [0, 1, 2, 3, 4]`: The array contains the values 0, 1, 2, 3, and 4.

- **Expected Output**:
  - The MEX of the array `[0, 1, 2, 3, 4]` is 5. Since we can increment any element by 1, the MEX can be increased to 6 by incrementing the last element (4) to 5.

### Why This Test Case Exposes the Bug:
- The code will incorrectly print `6` as the MEX, but the correct MEX is `5` because the MEX is the smallest non-negative integer not in the array.
- The code does not correctly handle the scenario where the MEX can be greater than the maximum value in the array plus one.

### Final Test Input:
```plaintext
1
5 1
0 1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

Buggy Code:
import os
import sys
from io import BytesIO, IOBase
from math import ceil

def main():

    n, k = map(int, input().split())
    arr = list(map(int, input().split()))

    arr.sort()
    extra = []
    curr_mex = 0

    for i in range(n):
        if arr[i] == curr_mex:
            curr_mex += 1
        elif arr[i] > curr_mex:
            break
        else:
            diff = curr_mex - arr[i]
            if diff % k > 0:
                extra.append(arr[i])
            else:
                curr_mex += 1

    able = set()
    for i in range(len(extra)):
        if extra[i] > curr_mex:
            print(curr_mex)
            return
        else:
            if curr_mex not in able:
                can = (((curr_mex - extra[i]) % k) == 0)
                if can == 1:
                    curr_mex += 1
                else:
                    diff = curr_mex - extra[i]
                    need = ceil(diff / k)
                    new = extra[i] + (k * need)
                    able.add(new)
            elif curr_mex in able or extra[i] == curr_mex:
                able.remove(curr_mex)
                curr_mex += 1
    print(curr_mex)

BUFSIZE: int = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

def print(*args, **kwargs):
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

if __name__ == "__main__":
     for _ in range(int(input())):
         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:
To expose the bug in the provided code, let's analyze the potential issues and focus on edge cases or scenarios that could maximize the failure probability.

### Potential Issues in the Code:
1. **Handling of Duplicates**: The code may not handle duplicates correctly, especially when the same value appears multiple times in the array.
2. **Edge Cases for Small Arrays**: Small arrays (e.g., with only one or two elements) might not be handled correctly.
3. **Handling of Large Values**: The code might not handle large values of `x` or large array elements correctly.
4. **Special Cases for MEX**: The code might not correctly handle cases where the MEX is zero or very close to zero.

### Test Case 1: Small Array with Duplicates
```plaintext
1
2 1
0 0
```
- **Explanation**: This test case has a small array with duplicates. The MEX should be 1, but the code might not handle the duplicates correctly.

### Test Case 2: Large Value of x
```plaintext
1
5 1000000000
0 1 2 3 4
```
- **Explanation**: This test case has a large value of `x`. The MEX should be 5, but the code might not handle large values of `x` correctly.

### Test Case 3: Edge Case with MEX Zero
```plaintext
1
3 1
1 2 3
```
- **Explanation**: This test case has an array where the MEX is zero. The code should return 0, but it might not handle this edge case correctly.

### Test Case 4: Large Array with Duplicates
```plaintext
1
10 1
0 0 0 0 0 0 0 0 0 0
```
- **Explanation**: This test case has a large array with all elements being zero. The MEX should be 1, but the code might not handle large arrays with duplicates correctly.

### Test Case 5: Mixed Values and Large x
```plaintext
1
5 10
0 1 2 3 4
```
- **Explanation**: This test case has a mix of small and large values for `x`. The MEX should be 5, but the code might not handle mixed values and large `x` correctly.

### Chosen Test Case
Let's choose a test case that is small and complete, and that is likely to expose the bug in the code. We will use the first test case with a small array and duplicates.

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

This test case should expose issues with handling duplicates and small arrays.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

Buggy Code:
import java.util.*;
public class Main {
    public static int f(ArrayList<Integer>a) {
        int n = a.size();
        for (int i = 0; i < n; i++) {
            if (a.get(i) < 0 || a.get(i) >= n) a.set(i,n+1);
        }
        for(int i=0;i<n;i++){
            int num=Math.abs(a.get(i));
            if(num<n) a.set(num,-Math.abs(a.get(num)));
        }
        for (int i = 0; i < n; i++) {
            if (a.get(i)>0)  return i;
        }
        return n;
    }
    public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        sc.nextLine();
        while(t-->0){
            int n=sc.nextInt();
            int x=sc.nextInt();
            sc.nextLine();
            int arr[]=new int[n];
            boolean flag=true;
            HashMap<Integer,Integer>map=new HashMap<>();
            ArrayList<Integer>arl=new ArrayList<>();
            for(int i=0;i<n;i++) {
                arr[i]=sc.nextInt();
                if(arr[i]==0) flag=false;
                map.put(arr[i],map.getOrDefault(arr[i],0)+1);
                arl.add(arr[i]);
            }

            if(flag)System.out.println(0);
            else{
                boolean flag2=true;
                HashSet<Integer>set=new HashSet<>();
                for(int i:map.keySet()){
                    if(map.get(i)>1) {
                        set.add(i);
                        flag2=false;
                    }
                }
                if(flag2){
                    System.out.println(f(arl));
                }else{
                    for(int i:set){
                        while(map.containsKey(i)){
                            i+=x;
                        }
                        arl.add(i);
                    }
                    System.out.println(f(arl));
                }

            }

        }
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

Buggy Code:
import java.util.*;
import java.util.List;
import java.util.Scanner;

public class MaximizeMex_Codeforces {

    static void solve(int [] ar, int n, int k){
        Map<Integer, Integer> hm1 = new HashMap<>();
        Map<Integer, Integer> hm2 = new HashMap<>();

        for(int i : ar) hm1.put(i, hm1.getOrDefault(i, 0) + 1);
        int c = 0;
        while(c < n){
            if(hm1.containsKey(c) && hm1.get(c) > 0){
                int f = hm1.get(c) - 1;
                hm1.put(c, f);
                hm2.put(c%k, hm1.getOrDefault(c%k, 0) + f);
                c++;
            } else if(hm2.containsKey(c%k) && hm2.get(c%k) > 0){
                hm2.put(c%k, hm2.get(c%k) - 1);
                c++;
            } else {
                break;
            }
        }
        System.out.println(c);
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            int k = sc.nextInt();
            int [] ar = new int [n];
            for(int i=0; i<n; i++)
                ar[i] = sc.nextInt();
            solve(ar, n, k);
        }
    }
}

// 0 - 1
// 1 - 1
// 2 - 2
// 3 - 1
// 5 - 1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize Mex
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ positive integers and an integer $x$. You can do the following two-step operation any (possibly zero) number of times:

  1. Choose an index $i$ ($1 \leq i \leq n$).    2. Increase $a_i$ by $x$, in other words $a_i := a_i + x$.

Find the maximum value of the $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

The $\operatorname{MEX}$ (minimum excluded value) of an array is the smallest non-negative integer that is not in the array. For example:

  * The $\operatorname{MEX}$ of $[2,2,1]$ is $0$ because $0$ is not in the array.    * The $\operatorname{MEX}$ of $[3,1,0,1]$ is $2$ because $0$ and $1$ are in the array but $2$ is not.    * The $\operatorname{MEX}$ of $[0,3,1,2]$ is $4$ because $0$, $1$, $2$ and $3$ are in the array but $4$ is not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5000$). The description of the test cases follows.

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 2 \cdot 10^5$; $1 \le x \le 10^9$) — the length of the array and the integer to be used in the operation.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) — the given array.

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

Output: For each test case, output a single integer: the maximum $\operatorname{MEX}$ of $a$ if you perform the operations optimally.

Examples:
input:
3
6 3
0 3 2 1 5 2
6 2
1 3 4 1 0 2
4 5
2 5 10 3
output:
4
6
0

Note: In the first test case, the $\operatorname{MEX}$ of $a$ is $4$ without performing any operations, which is the maximum.

In the second test case, the $\operatorname{MEX}$ of $a$ is $5$ without performing any operations. If we perform two operations both with $i=1$, we will have the array $a=[5,3,4,1,0,2]$. Then, the $\operatorname{MEX}$ of $a$ will become $6$, which is the maximum.

In the third test case, the $\operatorname{MEX}$ of $a$ is $0$ without performing any operations, which is the maximum.

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

int main() {
	int tt;cin>>tt;
	while(tt--){
	    int n,x;cin>>n>>x;
	    vector<int>a;
	    for(int i=0;i<n;i++){
	        int x;cin>>x;a.push_back(x);
	    }
	    sort(a.begin(),a.end());
	    int j=0;
	    int maxi=n;
	    unordered_map<int,int>mpp,mp;
	    for(int i=0;i<n;i++){
	        while(j<n&&a[j]<i){
	            mp[a[j]]++;
	            if(mp[a[j]]>1)mpp[a[j]%x]++;
	            j++;
	        }
	        if(i==a[j]){
	            mp[a[j]]++;
	            if(mp[a[j]]>1)mpp[a[j]%x]++;
	            j++;
	        }
	        else{
	            if(mpp[i%x]>0)mpp[i%x]--;
	            else{
	                maxi=i;
	                break;
	            }
	            j++;
	        }
	    }
	    cout<<maxi<<endl;
	}

}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the array is very small, contains repeated elements, or has a very large value of `x`.
2. **Optimal Operations**: Test cases where the optimal operations are not straightforward, such as when the array already contains many small values.
3. **Large Test Cases**: Test cases with a large number of elements to ensure the algorithm handles large inputs efficiently.

Given the constraints and the nature of the problem, a small-scale test input that can expose the bug in the provided code is:

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `4 2` and array `[0, 1, 2, 3]`
   - **Expected Output:** `4`
   - **Reason:** The array already contains the values from `0` to `3`. The MEX is `4`, and no operations are needed.

2. **Test Case 2:**
   - **Input:** `5 3` and array `[0, 1, 2, 3, 4]`
   - **Expected Output:** `5`
   - **Reason:** The array already contains the values from `0` to `4`. The MEX is `5`, and no operations are needed.

3. **Test Case 3:**
   - **Input:** `6 1` and array `[0, 1, 2, 3, 4, 5]`
   - **Expected Output:** `6`
   - **Reason:** The array already contains the values from `0` to `5`. The MEX is `6`, and no operations are needed.

These test cases are designed to ensure that the algorithm correctly handles arrays that are already optimal and do not require any operations. The buggy code might fail to correctly handle these scenarios, especially when the array is already in a state where no operations are needed to maximize the MEX.