**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
def can_transform(s, t):
    from functools import lru_cache

    @lru_cache(None)
    def dp(i, j):
        # Base case: if both pointers reach the end of their respective strings
        if i == len(s) and j == len(t):
            return True

        # If s is exhausted but t is not
        if i == len(s):
            return False

        # If t is exhausted but s is not
        if j == len(t):
            # Check if the remaining part of s can be matched with '*'
            return all(c == '*' for c in s[i:])

        # If s[i] is a lowercase letter
        if s[i].isalpha():
            if s[i] != t[j]:
                return False
            return dp(i + 1, j + 1)

        # If s[i] is '-'
        if s[i] == '-':
            return dp(i + 1, j + 1)

        # If s[i] is '*'
        # Option 1: Skip '*'
        if dp(i + 1, j):
            return True

        # Option 2: Match '*' with one or more characters in t
        for k in range(j, len(t)):
            if dp(i + 1, k + 1):
                return True

        return False

    return "Yes" if dp(0, 0) else "No"

# Read input
n, m = map(int, input().split())
s = input().strip()
t = input().strip()

# Output result
print(can_transform(s, t))

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

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

Think step by step.

Response:
```plaintext
2 3
a*b
abc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
def can_transform(s, t):
    # Remove all '*' from both strings
    s = [char for char in s if char != '*']
    t = [char for char in t if char != '*']

    i, j = 0, 0
    len_s, len_t = len(s), len(t)

    while i < len_s and j < len_t:
        if s[i] == t[j]:
            i += 1
            j += 1
        elif s[i] == '-':
            i += 1
        elif t[j] == '-':
            j += 1
        else:
            return "No"

    # If there are remaining unmatched '-' in either string, continue checking
    while i < len_s and s[i] == '-':
        i += 1
    while j < len_t and t[j] == '-':
        j += 1

    # Both pointers should reach the end of their respective strings
    if i == len_s and j == len_t:
        return "Yes"
    else:
        return "No"

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

n = int(data[0])
m = int(data[1])
s = data[2]
t = data[3]

print(can_transform(s, t))

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

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

Think step by step.

Response:
```plaintext
5 5
a*b*c
a-b-c
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        sc.nextLine();  // Consume the newline
        String s = sc.nextLine();
        String t = sc.nextLine();

        if (canTransform(s, t, n, m)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }

    public static boolean canTransform(String s, String t, int n, int m) {
        // Locate the positions of the first and last '*' in s
        int firstStar = s.indexOf('*');
        int lastStar = s.lastIndexOf('*');

        // No '*' present, exact match required
        if (firstStar == -1) {
            return n == m && matchesDirectly(s, t);
        }

        // Match prefix and suffix
        if (!matchPrefix(s, t, firstStar) || !matchSuffix(s, t, lastStar, n, m)) {
            return false;
        }

        // If prefix and suffix match, the middle '*' can match any substring
        return true;
    }

    public static boolean matchesDirectly(String s, String t) {
        int len = s.length();
        for (int i = 0; i < len; i++) {
            if (s.charAt(i) != '-' && s.charAt(i) != t.charAt(i)) {
                return false;
            }
        }
        return true;
    }

    public static boolean matchPrefix(String s, String t, int upTo) {
        for (int i = 0; i < upTo; i++) {
            if (i >= t.length() || (s.charAt(i) != '-' && s.charAt(i) != t.charAt(i))) {
                return false;
            }
        }
        return true;
    }

    public static boolean matchSuffix(String s, String t, int fromStar, int n, int m) {
        int suffixLength = n - fromStar - 1;
        if (suffixLength > m - fromStar) {
            return false;
        }
        for (int i = 0; i < suffixLength; i++) {
            if (s.charAt(n - 1 - i) != '-' && s.charAt(n - 1 - i) != t.charAt(m - 1 - i)) {
                return false;
            }
        }
        return true;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the code might fail to correctly match the strings. The code assumes that the presence of a single `*` can match any substring, but it does not handle multiple `*` characters or more complex scenarios correctly.

Let's consider a test case where the string `s` contains multiple `*` characters, and the string `t` has a specific structure that should be matched but is not handled correctly by the code.

### Test Case:
- **Input:**
  - `n = 7`
  - `m = 7`
  - `s = "*a*b*c*"`
  - `t = "aabcba"`

### Explanation:
- The string `s` contains multiple `*` characters, which can be replaced by any substring.
- The string `t` is "aabcba".
- The correct transformation would be:
  - `*` -> "a" (first `*`)
  - `*` -> "ab" (second `*`)
  - `*` -> "c" (third `*`)
  - `*` -> "b" (fourth `*`)
- After transformation, `s` becomes "aabcba", which matches `t`.

However, the given code will fail to handle this case correctly because it only considers the first and last `*` and assumes that the middle `*` can match any substring, but it does not handle multiple `*` characters correctly.

### Test Input:
```plaintext
7 7
*a*b*c*
aabcba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
import java.util.Scanner;
import java.util.Arrays;

public class stringCheck {

    public static boolean isEqual(char c1, char c2) {
        if (c1 == '-' || c2 == '-' || c1 == c2)
            return true;
        return false;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str1 = sc.nextLine();
        String str2 = sc.nextLine();
        String[] arr1 = str1.split("\\*");
        String[] arr2 = str2.split("\\*");
        String subStrF1 = arr1[0];
        String subStrF2 = arr2[0];
        // System.out.println(Arrays.toString(arr1));
        // System.out.println(Arrays.toString(arr2));
        if (subStrF1.length() > subStrF2.length()) {
            for (int i = 0; i < subStrF2.length(); i++) {
                if (!isEqual(subStrF2.charAt(i), subStrF1.charAt(i))) {
                    System.out.println("No");
                    return;
                }
            }
        } else {
            for (int i = 0; i < subStrF1.length(); i++) {
                if (!isEqual(subStrF2.charAt(i), subStrF1.charAt(i))) {
                    System.out.println("No");
                    return;
                }
            }
        }

        String subStrB1 = arr1[arr1.length - 1];
        String subStrB2 = arr2[arr2.length - 1];
        // System.out.println(subStrB1);
        // System.out.println(subStrB2);
        if (subStrB1.length() > subStrB2.length()) {
            for (int i = 0; i < subStrB2.length(); i++) {
                // System.out.println(
                // subStrB2.charAt(subStrB2.length() - 1 - i) +
                // subStrB1.charAt(subStrB2.length() - 1 - i));
                if (!isEqual(subStrB2.charAt(subStrB2.length() - 1 - i), subStrB1.charAt(subStrB1.length() - 1 - i))) {
                    System.out.println("No");
                    return;
                }
            }
        } else {
            for (int i = 0; i < subStrB1.length(); i++) {
                // System.out.println(
                // subStrB2.charAt(subStrB1.length() - 1 - i) +
                // subStrB1.charAt(subStrB1.length() - 1 - i));
                if (!isEqual(subStrB2.charAt(subStrB1.length() - 1 - i), subStrB1.charAt(subStrB1.length() - 1 - i))) {
                    System.out.println("No");
                    return;
                }
            }
        }

        System.out.println("Yes");

    }
}

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

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

Think step by step.

Response:
```plaintext
5 5
a*b*c
a*d*c
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

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

int main() {
    int n, m;
    cin >> n >> m;
    string s, t;
    cin >> s >> t;

    if (n > m) {
        cout << "No" << endl;
        return 0;
    }

    int starCountS = 0, dashCountS = 0;
    int starCountT = 0, dashCountT = 0;

    // Count the number of '*' and '-' in both strings
    for (char c : s) {
        if (c == '*') starCountS++;
        else if (c == '-') dashCountS++;
    }
    for (char c : t) {
        if (c == '*') starCountT++;
        else if (c == '-') dashCountT++;
    }

    // Check if the number of '*' and '-' in s is less than or equal to t
    if (starCountS > starCountT || dashCountS > dashCountT) {
        cout << "No" << endl;
        return 0;
    }

    // Use two pointers to traverse both strings
    int i = 0, j = 0;
    while (i < n && j < m) {
        if (s[i] != '-' && s[i] != '*') {
            if (s[i] != t[j]) {
                cout << "No" << endl;
                return 0;
            }
            i++;
            j++;
        } else {
            if (s[i] == '-') {
                i++;
            } else {
                j++;
            }
        }
    }

    // Check if all characters in t have been processed
    if (j != m) {
        cout << "No" << endl;
        return 0;
    }

    cout << "Yes" << 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
5 5
a*b*c
ab*c*
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

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

int main() {
    int n, m;
    cin >> n >> m;

    string s, t;
    cin >> s >> t;

    // Check if lengths are equal
    if (n != m) {
        cout << "No" << endl;
        return 0;
    }

    // Count the number of '-' and '*' in both strings
    int count_s_minus = 0, count_s_asterisk = 0;
    int count_t_minus = 0, count_t_asterisk = 0;

    for (int i = 0; i < n; ++i) {
        if (s[i] == '-') count_s_minus++;
        else if (s[i] == '*') count_s_asterisk++;

        if (t[i] == '-') count_t_minus++;
        else if (t[i] == '*') count_t_asterisk++;
    }

    // Check if the counts of '-' and '*' match
    if (count_s_minus != count_t_minus || count_s_asterisk != count_t_asterisk) {
        cout << "No" << endl;
        return 0;
    }

    // Check if the characters can be matched
    for (int i = 0; i < n; ++i) {
        if (s[i] != '-' && s[i] != '*' && s[i] != t[i]) {
            cout << "No" << endl;
            return 0;
        }
    }

    cout << "Yes" << 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
5 5
a*b*c
a*c*b
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
def can_transform(s, t):
    n, m = len(s), len(t)

    # List to store mismatches
    mismatches = []

    # Identify mismatches
    for i in range(min(n, m)):
        if s[i] == '-':
            if t[i] != '-':
                mismatches.append(('-', '-', i))
        elif s[i] == '*':
            if t[i] != '*':
                mismatches.append(('*', '*', i))
            else:
                # Check if there's a matching sequence
                j = i
                while j < min(n, m) and (s[j] == '*' or t[j] == '*'):
                    if s[j] == '*':
                        j += 1
                    else:
                        j += 1

    # Try to match sequences greedily
    unmatched_s = set(range(n))
    unmatched_t = set(range(m))

    for _, _, index in mismatches:
        if s[index] == '-':
            # Replace '-' in s with a single character
            unmatched_s.remove(index)
        elif s[index] == '*':
            # Find a matching sequence in t
            found = False
            for j in range(index + 1, min(n, m)):
                if t[j] != '*':
                    break
                if t[index:j+1] == s[index:j+1]:
                    found = True
                    unmatched_t.remove(j)
                    break

            if not found:
                return "No"

    # Check if all characters are matched
    return "Yes" if not unmatched_s and not unmatched_t else "No"

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

n, m = map(int, data[0:2])
s = data[2]
t = data[3]

# Output result
print(can_transform(s, t))

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

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

Think step by step.

Response:
```plaintext
3 5
a*b
a*bb
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
def can_match(s, t):
    n, m = len(s), len(t)
    dp = [[-1]*(m+1) for _ in range(n+1)]

    # Initialize the base cases
    for i in range(n+1):
        dp[i][0] = -1
    for j in range(m+1):
        dp[0][j] = -1

    # Fill the dp table
    for i in range(1, n+1):
        for j in range(1, m+1):
            if s[i-1] == t[j-1]:
                dp[i][j] = dp[i-1][j-1]
            elif s[i-1] == '*':
                dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + 1
            elif t[j-1] == '*':
                dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + 1
            else:
                dp[i][j] = -1

    return dp[n][m] >= 0

# Read input from standard input
n, m = map(int, input().split())
s = input()
t = input()

if can_match(s, t):
    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:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on the following aspects:

1. **Edge Cases**: Consider scenarios where the strings contain a mix of lowercase letters, `-`, and `*`.
2. **Length Mismatch**: Ensure that the lengths of `s` and `t` are such that they can be transformed to match each other, but the current implementation might fail to handle it correctly.
3. **Special Characters**: Test with strings that have multiple `-` and `*` characters in various positions.

Given the constraints and the nature of the problem, a small-scale test input that can expose the bug might look like this:

```plaintext
5 5
a-b*c
a*b-c
```

### Explanation:

- **Input**:
  - `n = 5`, `m = 5`
  - `s = "a-b*c"`
  - `t = "a*b-c"`

- **Expected Output**:
  - `Yes`

### Reasoning:

- **Transformation**:
  - In `s = "a-b*c"`, the `-` can be replaced with any lowercase letter, and the `*` can be replaced with any string of any length.
  - In `t = "a*b-c"`, the `*` can be replaced with any string of any length, and the `-` can be replaced with any lowercase letter.

- **Possible Match**:
  - We can transform `s` to `t` by:
    - Replacing the `-` in `s` with `b`.
    - Replacing the `*` in `s` with `b`.
  - This results in `s' = "abbbc"`.
  - Similarly, we can transform `t` to `s` by:
    - Replacing the `*` in `t` with `b`.
    - Replacing the `-` in `t` with `b`.
  - This results in `t' = "abbbc"`.

- **Bug Exposure**:
  - The buggy code incorrectly handles the `*` character, especially when it can be replaced with an empty string or a string of any length. The `max` function used in the DP table might not correctly account for all possible replacements, leading to a false negative.

This test input is small and valid, and it should effectively expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        String s = scanner.next();
        String t = scanner.next();

        String sModified = s.replace("*", "abc").replace("-", "a");
        String tModified = t.replace("*", "abc").replace("-", "a");

        if (sModified.equals(tModified)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to create a test case that highlights the issue with the way the code handles the replacement of `*` and `-`. Specifically, the code always replaces `*` with "abc" and `-` with "a", which is not a valid approach for all cases. We need a test case where the replacement of `*` and `-` can lead to a different outcome than what the code currently handles.

### Key Points to Consider:
1. **Length Mismatch**: The code does not handle cases where the length of the modified strings can be different due to the replacement of `*`.
2. **Different Replacements for `*` and `-`**: The code uses fixed replacements, which may not be valid for all cases.

### Test Case:
Let's create a test case where the length of the strings after replacing `*` and `-` can be different, and the fixed replacements used by the code do not match.

```plaintext
5 6
a*b*c
ab*c
```

### Explanation:
- **Input**:
  - `n = 5`, `m = 6`
  - `s = "a*b*c"`
  - `t = "ab*c"`

- **Expected Output**:
  - "Yes"

### Reasoning:
- In the given test case, `s = "a*b*c"` can be transformed to "ababc" by replacing the first `*` with "b" and the second `*` with "c".
- Similarly, `t = "ab*c"` can be transformed to "ababc" by replacing the `*` with "abc".
- The code, however, will replace both `*` in `s` with "abc" and both `-` with "a", resulting in `sModified = "abcaabc"` and `tModified = "ababc"`, which are not equal.

This test case will expose the bug in the code because the fixed replacements used by the code do not allow for the flexibility needed to match the strings correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
import java.util.Scanner;

public class ZimphaFanClub {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        scanner.nextLine(); // Consume the newline character
        String s = scanner.nextLine();
        String t = scanner.nextLine();

        System.out.println(canTransform(s, t) ? "Yes" : "No");
    }

    private static boolean canTransform(String s, String t) {
        int n = s.length();
        int m = t.length();

        // Find the first occurrence of '*' in s
        int starIndex = s.indexOf('*');

        // If there is no '*', simple case, s should match t exactly after replacing '-'
        if (starIndex == -1) {
            return matchWithoutStar(s, t);
        }

        // Check the prefix part (before the '*')
        if (!matchWithoutStar(s.substring(0, starIndex), t.substring(0, Math.min(n, m)))) {
            return false;
        }

        // Check the suffix part (after the '*')
        int sSuffixStart = starIndex + 1;
        int tSuffixStart = m - (n - starIndex - 1);

        if (tSuffixStart < 0 || !matchWithoutStar(s.substring(sSuffixStart), t.substring(tSuffixStart))) {
            return false;
        }

        return true;
    }

    private static boolean matchWithoutStar(String s, String t) {
        int len = s.length();
        if (len > t.length()) {
            return false;
        }
        for (int i = 0; i < len; i++) {
            if (s.charAt(i) != '-' && s.charAt(i) != t.charAt(i)) {
                return false;
            }
        }
        return true;
    }
}

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

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

Think step by step.

Response:
```plaintext
3 5
a*b
abcde
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
// LUOGU_RID: 160627568
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define bs basic_string
#define pa pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define c2(x) ((x)*(x))
#define c3(x) ((x)*(x)*(x))
inline int read()
{
	int x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+(ch-'0');ch=getchar();}
	return x*f;
}
inline ll lread()
{
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+(ch-'0');ch=getchar();}
	return x*f;
}
const int N=2e6+10;
int n,m;
char s[N],t[N];
namespace INIT
{
	int n1,m1;
	char s1[N],t1[N];
	void rd()
	{
		n1=read(),m1=read();
		scanf("%s",s1+1);
		scanf("%s",t1+1);
		bool tgs=0,tgt=0;
		for(int i=1;i<=n1;i++)
			if(s1[i]=='*')
			{
				tgs=1;
				break;
			}
		for(int i=1;i<=m1;i++)
			if(t1[i]=='*')
			{
				tgt=1;
				break;
			}
		if(tgs==0&&tgt==0)
		{
			if(n1!=m1)puts("No");
			else
			{
				bool flag=0;
				for(int i=1;i<=n1;i++)
					if(s1[i]!='-'&&t1[i]!='-'&&s1[i]!=t1[i])
					{
						flag=1;
						break;
					}
				if(flag)puts("No");
				else puts("Yes");
			}
			exit(0);
		}
		int l1=0,l2=0;
		while(s1[1+l1]!='*'&&t1[1+l1]!='*')
		{
			l1++;
			if(s1[l1]!='-'&&t1[l1]!='-'&&s1[l1]!=t1[l1])
			{
				puts("No");
				exit(0);
			}
		}
		while(s1[n1-l2]!='*'&&t1[m1-l2]!='*')
		{
			l2++;
			if(s1[n1-l2+1]!='-'&&t1[m1-l2+1]!='-'&&s1[n1-l2+1]!=t1[m1-l2+1])
			{
				puts("No");
				exit(0);
			}
		}
		if(tgs&&tgt)
		{
			puts("Yes");
			exit(0);
		}
		if(tgs)swap(n1,m1),swap(s1,t1);
		n=n1-l1-l2;
		m=m1-l1-l2;
		for(int i=l1+1;i<=n1-l2;i++)s[i-l1]=s1[i];
		for(int i=l1+1;i<=m1-l2;i++)t[i-l1]=t1[i];
	}
}
namespace yzy
{
	const int p=998244353,G=3,invG=332748118;
	const int maxn=8e6+10;
	int M,N;
	int A[maxn],B[maxn];
	int rev[maxn];
	int a[maxn],b[maxn],c[maxn];
	int val[maxn];
	void clear(int n)
	{
		M=0,N=0;
		memset(A,0,sizeof(int)*(n+5));
		memset(B,0,sizeof(int)*(n+5));
		memset(rev,0,sizeof(int)*(n+5));
		memset(a,0,sizeof(int)*(n+5));
		memset(b,0,sizeof(int)*(n+5));
		memset(c,0,sizeof(int)*(n+5));
		memset(val,0,sizeof(int)*(n+5));
	}
	void add(int &x,int y){x=(x+y)%p;}
	int pw(int x,int y)
	{
		int ans=1;
		while(y)
		{
			if(y&1)ans=(1ll*ans*x)%p;
			x=(1ll*x*x)%p;
			y>>=1;
		}
		return ans;
	}
	int rd(char ch)
	{
		if(islower(ch))return ch-'a'+1;
		else return 0;
	}
	int inv(int x){return pw(x,p-2);}
	void init(int n)
	{
		rev[0]=0;
		for(int i=1;i<n;i++)rev[i]=((rev[i>>1]>>1)|((i&1)?(n>>1):0));
	}
	void NTT(int n,int a[],bool dft)
	{
		for(int i=0;i<n;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
		for(int len=2;len<=n;len<<=1)
		{
			int s=pw(dft?G:invG,(p-1)/len);
			for(int l=0;l<n;l+=len)
			{
				int now=1;
				for(int i=l;i<l+len/2;i++)
				{
					int v1=a[i],v2=a[i+len/2];
					a[i]=(v1+1ll*now*v2)%p;
					a[i+len/2]=(v1-1ll*now*v2)%p;
					now=(1ll*now*s)%p;
				}
			}
		}
		for(int i=0;i<n;i++)add(a[i],p);
		if(dft)
		{
			int invn=inv(n);
			for(int i=0;i<n;i++)a[i]=(1ll*a[i]*invn)%p;
		}
	}
	int get(const string &S,const string &T)
	{
		M=S.size(),N=T.size();
		for(int i=1;i<=M;i++)A[i]=rd(S[i-1]);
		for(int i=1;i<=N;i++)B[i]=rd(T[i-1]);
		int n;
		for(n=1;!(N+M<n);n<<=1);
		init(n);
		memset(a,0,sizeof(int)*(n+5));
		for(int i=1;i<=M;i++)a[i]=c3(A[M-i+1]);
		for(int i=1;i<=N;i++)b[i]=B[i];
		NTT(n,a,0);
		NTT(n,b,0);
		for(int i=0;i<n;i++)add(c[i],1ll*a[i]*b[i]%p);
		memset(a,0,sizeof(int)*(n+5));
		for(int i=1;i<=M;i++)a[i]=A[M-i+1];
		for(int i=1;i<=N;i++)b[i]=c3(B[i]);
		NTT(n,a,0);
		NTT(n,b,0);
		for(int i=0;i<n;i++)add(c[i],1ll*a[i]*b[i]%p);
		memset(a,0,sizeof(int)*(n+5));
		for(int i=1;i<=M;i++)a[i]=-2*c2(A[M-i+1]);
		for(int i=1;i<=N;i++)b[i]=c2(B[i]);
		NTT(n,a,0);
		NTT(n,b,0);
		for(int i=0;i<n;i++)add(c[i],1ll*a[i]*b[i]%p);
		NTT(n,c,1);
		for(int i=M;i<=N;i++)val[i-M+1]=c[i+1];
		int ans=-1;
		for(int i=1;i+M-1<=N;i++)
			if(val[i]==0)
			{
				ans=i;
				break;
			}
		clear(n);
		return ans;
	}
}
signed main()
{
	INIT::rd();
	int p=0;
	int zyz=0;
	for(int l=1,r;l<m;l=r+1)
	{
		if(n>20)
		{
			zyz++;
			if(zyz%100==0)printf("test : %d\n",zyz);
		}
		r=l+1;
		while(t[r]!='*')r++;
		int len=r-l-1;
		string S="";
		for(int i=l+1;i<=r-1;i++)S+=t[i];
		while(1)
		{
			int pl=p+1,pr=min(n,pl+2*len);
			if(pr-pl+1<len)
			{
				puts("No");
				exit(0);
			}
			string T="";
			for(int i=pl;i<=pr;i++)T+=s[i];
			int x=yzy::get(S,T);
			if(x==-1)
			{
				p=pr-len;
				continue;
			}
			else
			{
				p=x+len-1;
				break;
			}
		}
	}
	puts("Yes");
	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
10 10
a*b*c*d*e
f*g*h*i*j
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define all(a) (a).begin(), (a).end()
#define sz(a) (a).size()

using namespace std;

template<class T>
using min_prority_queue = priority_queue<T, vector<T>, greater<T>>;

const int MOD = 1e9 + 7;
const int N = 2e6 + 11;

#define rep(i, j, k) for (int i = j; i < k; i++)
typedef vector<int> vi;
typedef complex<long double> C;
typedef vector<double> vd;
void fft(vector<C>& a) {
	int n = sz(a), L = 31 - __builtin_clz(n);
	static vector<complex<long double>> R(2, 1);
	static vector<C> rt(2, 1);  // (^ 10% faster if double)
	for (static int k = 2; k < n; k *= 2) {
		R.resize(n); rt.resize(n);
		auto x = polar(1.0L, acos(-1.0L) / k);
		rep(i,k,2*k) rt[i] = R[i] = i&1 ? R[i/2] * x : R[i/2];
	}
	vi rev(n);
	rep(i,0,n) rev[i] = (rev[i / 2] | (i & 1) << L) / 2;
	rep(i,0,n) if (i < rev[i]) swap(a[i], a[rev[i]]);
	for (int k = 1; k < n; k *= 2)
		for (int i = 0; i < n; i += 2 * k) rep(j,0,k) {
			// C z = rt[j+k] * a[i+j+k]; // (25% faster if hand-rolled)  /// include-line
			auto x = (long double *)&rt[j+k], y = (long double *)&a[i+j+k];        /// exclude-line
			C z(x[0]*y[0] - x[1]*y[1], x[0]*y[1] + x[1]*y[0]);           /// exclude-line
			a[i + j + k] = a[i + j] - z;
			a[i + j] += z;
		}
}
vector<C> conv(const vector<C>& a, const vector<C>& b) {
	if (a.empty() || b.empty()) return {};
	vd res(sz(a) + sz(b) - 1);
	int L = 64 - __builtin_clzll(sz(res)), n = 1 << L;
	vector<C> X(n), Y(n);
    rep(i, 0, n) X[i] = i < a.size() ? a[i] : 0;
    rep(i, 0, n) Y[i] = i < b.size() ? b[i] : 0;
    fft(X); fft(Y);
    rep(i, 0, n) X[i] *= Y[i];
    reverse(X.begin() + 1, X.end());
	fft(X);
    rep(i, 0, n) X[i] /= n;
	return X;
}

bool not_match(char c, char d){
    if (c == '-') return false;
    if (d == '-') return false;
    if (c == d) return false;
    return true;
}

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
void solve(){
    int n, m; cin >> n >> m;
    string s, t; cin >> s >> t;
    deque<char> S(all(s)), T(all(t));
    while (S.size() && T.size() && S.front() != '*' && T.front() != '*') {
        if (not_match(S.front(), T.front())) {
            cout << "No" << '\n';
            return;
        }
        S.pop_front(); T.pop_front();
    }
    while (S.size() && T.size() && S.back() != '*' && T.back() != '*') {
        if (not_match(S.back(), T.back())) {
            cout << "No" << '\n';
            return;
        }
        S.pop_back(); T.pop_back();
    }

    if (S.empty() && T.empty()) {
        cout << "Yes" << '\n';
        return;
    }

    if (S.empty() || S.front() != '*') swap(S, T);

    if (T.empty()) {
        for (auto c : S) {
            if (c != '*') {
                cout << "No" << '\n';
                return;
            }
        }
        cout << "Yes" << '\n';
        return;
    }

    assert(S.size() && T.size());
    if (S.front() == '*' && T.back() == '*') {
        cout << "Yes" << '\n';
        return;
    }

    assert(S.back() == '*');
    if (S.size() == 1) {
        cout << "Yes" << '\n';
        return;
    }

    for (auto c : T) {
        if (c == '*') {
            cout << "Yes" << '\n';
            return;
        }
    }

    string _s(S.begin(), S.end());
    vector<string> V;

    {
        string tmp = "";
        for (auto c : S) {
            if (c == '*') {
                if (!tmp.empty()) V.push_back(tmp);
                tmp.clear();
            } else {
                tmp += c;
            }
        }
    }

    // for (auto x : S) cout << x; cout << endl;
    // for (auto x : T) cout << x; cout << endl;

    int pos = 0, k = T.size();
    vector<long double> ratio;
    uniform_real_distribution<long double> dist(0.5, 2);
    for (int i = 0; i < 26; i++) {
        ratio.push_back(dist(rng));
    }
    for (auto s : V) {
        if (k - pos < s.size()) {
            cout << "No" << '\n';
            return;
        }
        vector<C> a(s.size());
        for (int i = 0; i < s.size(); i++)
            a[i] = (s[i] == '-' ? C(0.0, 0.0) : polar(ratio[s[i] - 'a'], acosl(-1) / 26 * (s[i] - 'a')));
        // for (auto x : a) cout << x << ' '; cout << endl;
        int matched = -1;
        while (k - pos >= s.size()) {
            vector<C> b(min((int) s.size() * 2, k - pos));
            for (int i = 0; i < b.size(); i++) {
                b[i] = (T[pos + i] == '-' ? C(0.0, 0.0) : polar(1.0L / ratio[T[pos + i] - 'a'], acosl(-1) / 26 * (26 - (T[pos + i] - 'a'))));
                // cout << 1.0L / ratio[T[pos + i] - 'a'] << ' ' << pos << ' ' << i << ' ' << k << ' ' << T[pos + i] << endl;
                // assert((!isnan(b[i].real())));
            }
            // for (auto x : b) cout << x << ' '; cout << endl;
            int c = a.size(), d = b.size();
            reverse(b.begin(), b.end());
            vector<C> res = conv(a, b);
            // cout << "RES: "; for (auto x : res) cout << x << ' '; cout << endl;
            for (int i = (int) b.size() - 1; i >= (int) a.size() - 1; i--) {
                auto val = res[i];
                if ((fabs(val.imag()) < 1e-8)) {
                    matched = pos + (((int) b.size() - 1) - i);
                    break;
                }
            }
            // cout << matched << endl;
            if (matched != -1) break;
            pos += (int) b.size() - (int) a.size() + 1;
        }
        pos = matched + s.size();
    }

    // for (auto s : V) cout << s << endl;
    // for (auto c : T) cout << c; cout << endl;
    cout << "Yes" << '\n';
}

int32_t main(){
    cin.tie(0); cout.tie(0)->sync_with_stdio(false);
    int t = 1;
#ifdef LOCAL
    cin >> t;
#endif
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
10 10
-a-b-c-d-e
a*b*c*d*e
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
for test in range(1):
    #n = int(input())
    n, m = map(int, input().split())
    #c = list(map(int, input().split()))
    s = str(input())
    t = str(input())
    if '*' in s:
        if '*' in t:
            verd = True
            for i in range(min(n, m)):
                if s[i] == '*' or t[i] == '*':
                    break
                if not (s[i] == t[i] or s[i] == '-' or t[i] == '-'):
                    verd = False
                    break
            if verd:
                verd = True
                for i in range(1, min(n, m) + 1):
                    if s[n - i] == '*' or t[m - i] == '*':
                        break
                    if not (s[n - i] == t[m - i] or s[n - i] == '-' or t[m - i] == '-'):
                        verd = False
                        break
                if verd:
                    print("Yes")
                else:
                    print("No")
            else:
                print("No")
        else:
            sa = list(s.split('*'))
            nuch = 0
            t2 = ''
            for elem in range(len(t) - 1, -1, -1):
                t2 += t[elem]
            t = t2
            #print(t)
            for x in sa:
                if x == '':
                    continue
                x2 = ''
                for elem in range(len(x) - 1, -1, -1):
                    x2 += x[elem]
                x = x2
                s2 = t + ' ' + x
                dlin = len(s2) - 1
                ans = [0]
                for i in range(len(s2) - 1):
                    k = ans[i]
                    while True:
                        if s2[dlin - k] == s2[dlin - i - 1] or s2[dlin - k] == '-' or s2[dlin - i - 1] == '-':
                            ans.append(k + 1)
                            break
                        elif k == 0:
                            ans.append(0)
                            break
                        else:
                            k = ans[k - 1]
                    if ans[-1] == len(x):
                        nuch = len(ans) - 1 - len(x)
                        t = t[:len(t) - nuch]
                        #print(t, x)
                        break
                else:
                    print("No")
                    break
            else:
                print("Yes")
    else:
        if '*' in t:
            s, t = t, s
            sa = list(s.split('*'))
            nuch = 0
            t2 = ''
            for elem in range(len(t) - 1, -1, -1):
                t2 += t[elem]
            t = t2
            #print(t)
            for x in sa:
                if x == '':
                    continue
                x2 = ''
                for elem in range(len(x) - 1, -1, -1):
                    x2 += x[elem]
                x = x2
                s2 = t + ' ' + x
                dlin = len(s2) - 1
                ans = [0]
                for i in range(len(s2) - 1):
                    k = ans[i]
                    while True:
                        if s2[dlin - k] == s2[dlin - i - 1] or s2[dlin - k] == '-' or s2[dlin - i - 1] == '-':
                            ans.append(k + 1)
                            break
                        elif k == 0:
                            ans.append(0)
                            break
                        else:
                            k = ans[k - 1]
                    if ans[-1] == len(x):
                        nuch = len(ans) - 1 - len(x)
                        t = t[:len(t) - nuch]
                        #print(t, x)
                        break
                else:
                    print("No")
                    break
            else:
                print("Yes")
        else:
            if n == m:
                for i in range(n):
                    if not (s[i] == t[i] or s[i] == '-' or t[i] == '-'):
                        print("No")
                        break
                else:
                    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
10 10
a*b*c*d*e
a-b-c-d-e
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
import re

def is_letter(x):
    return ord(x) >= ord('a') and ord(x) <= ord('z')

def solve():
    n, m = map(int, input().split())
    s = input()
    t = input()

    ln, lm = 0, 0
    rn = n
    rm = m

    while ln < rn and lm < rm:
        if is_letter(s[ln]) and is_letter(t[lm]):
            if s[ln] != t[lm]:
                print('No')
                return
            ln += 1
            lm += 1
            continue
        if s[ln] != '*' and t[lm] != '*':
            ln += 1
            lm += 1
            continue
        break
    while ln < rn and lm < rm:
        if is_letter(s[rn-1]) and is_letter(t[rm-1]):
            if s[rn-1] != t[rm-1]:
                print('No')
                return
            rn -= 1
            rm -= 1
            continue
        if s[rn-1] != '*' and t[rm-1] != '*':
            rn -= 1
            rm -= 1
            continue
        break

    s = s[ln: rn]
    t = t[lm: rm]
    if len(t) == 0: s, t = t, s
    if len(s) == 0:
        if t != '*' * len(t):
            print('No')
            return
        print('Yes')
        return
    if (s[0] == '*' or s[-1] == '*') and (t[0] == '*' and t[-1] == '*'):
        print('Yes')
        return
    if s[0] != '*': s, t = t, s
    if s == '*' * len(s):
        print('Yes')
        return
    if '*' in t:
        print('Yes')
        return
    s = ''.join('.*' if i == '*' else '[a-z]' if i == '-' else f'(-|{i})' for i in s)
    if re.match(s, t):
        print('Yes')
    else:
        print('No')

solve()

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

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

Think step by step.

Response:
```plaintext
7 7
a*b*c*
a*b*c*
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
import java.util.*;

public class Main {
    public static boolean f(String s, String t, int idx1, int idx2, int dp[][]) {
        if(idx1 >= s.length() && idx2 >= t.length()) return true;
        if(idx1 >= s.length() || idx2 >= t.length()) {
            for(int i=idx1; i<s.length(); i++) {
                if(s.charAt(i) != '*') return false;
            }

            for(int i=idx2; i<t.length(); i++) {
                if(t.charAt(i) != '*') return false;
            }
            return true;
        }
        if(dp[idx1][idx2] != -1) return dp[idx1][idx2]==1 ? true : false;

        boolean x, y, z;
        x = y = z = false;
        if(s.charAt(idx1) == t.charAt(idx2) ||
            s.charAt(idx1) == '-' || t.charAt(idx2) == '-') {
            x = f(s, t, idx1+1, idx2+1, dp);
        }

        if(s.charAt(idx1) == '*' || t.charAt(idx2) == '*') {
            y = f(s, t, idx1+1, idx2, dp);
            z = f(s, t, idx1, idx2+1, dp);
        }
        dp[idx1][idx2] = x || y || z ? 1 : 0;
        return x || y || z;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        String s = sc.next();
        String t = sc.next();

        int dp[][] = new int[s.length()+5][t.length()+5];
        for(int i=0; i<dp.length; i++)
            Arrays.fill(dp[i], -1);

        if(f(s, t, 0, 0, dp))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

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

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

Think step by step.

Response:
```plaintext
10 10
a*b*c*d*e
abc-de-fg*
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

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

public class Main {
    private static final int N = (1 << 22) + 5, g = 31, p = 2013265921, gi = (int)pow(g, p - 2);
//            g = 3, p = 998244353, gi = 332748118;
    private static final PrintWriter out = new PrintWriter(System.out, true);
    private static final int[] r = new int[N];
    private static final Input input = new Input(System.in);

    private static long pow(long a, long b) {
        long ans = 1L;
        long c = a;
        while (b > 0) {
            if (b % 2 == 1) ans = ans * c % p;
            b >>= 1;
            c = c * c % p;
        }
        return ans;
    }

    private static void ntt(long[] A, int n, int op) {
        for (int i = 0; i < n; ++i) {
            if (i < r[i]) {
                long t = A[i];
                A[i] = A[r[i]];
                A[r[i]] = t;
            }
        }

        for (int i = 2; i <= n; i <<= 1) {
            long g1 = pow(op == 1 ? g : gi, (p - 1) / i);
            for (int j = 0; j < n; j += i) {
                long gk = 1L;
                for (int k = 0; k < i / 2; ++k) {
                    long x = A[j + k];long y = A[j + k + i / 2] * gk % p;
                    A[j + k] = (x + y) % p;A[j + k + i / 2] = (x - y + p) % p;gk = gk * g1 % p;
                }
            }
        }
    }

    public static int indexOf(String s, String t, int s1, int e1, int s2, int e2) {
        int m = e1 - s1;int n = e2 - s2;
        int len = 1;
        while (len < n + m) len <<= 1;
        for (int i = 0; i < len; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
        long[] A = new long[len], B = new long[len];
        for (int i = 0; i < m; ++i) A[m - 1 - i] = s.charAt(s1 + i) == '-' ? 0 : s.charAt(s1 + i);
        for (int i = 0; i < n; ++i) B[i] = t.charAt(s2 + i) == '-' ? 0 : t.charAt(s2 + i);
        long[] P = new long[len], C = new long[len], D = new long[len];
        for (int i = 0; i < len; ++i) {
            C[i] = A[i] * A[i] * A[i];
            D[i] = B[i];
        }
        ntt(C, len, 1);ntt(D, len, 1);
        for (int i = 0; i < len; ++i) P[i] = C[i] * D[i] % p;
        for (int i = 0; i < len; ++i) {
            C[i] = A[i];
            D[i] = B[i] * B[i] * B[i];
        }
        ntt(C, len, 1);ntt(D, len, 1);
        for (int i = 0; i < len; ++i) P[i] = (P[i] + (C[i] * D[i] % p)) % p;
        for (int i = 0; i < len; ++i) {
            C[i] = A[i] * A[i];
            D[i] = B[i] * B[i];
        }
        ntt(C, len, 1);ntt(D, len, 1);
        for (int i = 0; i < len; ++i) P[i] = (P[i] - (2 * (C[i] * D[i] % p) % p) + p) % p;
        ntt(P, len, -1);
        long rev = pow(len, p - 2);
        for (int x = 0; x + m - 1 < n; ++x) {
            if (P[x + m - 1] * rev % p == 0) return s2 + x;
        }
        return -1;
    }

    public static void main(String[] args) {
        final int m = nextInt(), n = nextInt();
        String s = next(), t = next();
        if (m == 500001)
            System.out.print(s.substring(0, 30) + " " + s.substring(s.length() - 30) + " " + t.substring(0, 30) + " " + t.substring(t.length() - 30));
        int l = 0;
        for (; l < t.length() && l < s.length() && s.charAt(l) != '*' && t.charAt(l) != '*'; ++l) {
            if (s.charAt(l) != '-' && t.charAt(l) != '-' && s.charAt(l) != t.charAt(l)) {
                System.out.println("No");
                return;
            }
        }
        int r1 = s.length() - 1;
        int r2 = t.length() - 1;
        while (r1 >= l && r2 >= l && s.charAt(r1) != '*' && t.charAt(r2) != '*') {
            if (s.charAt(r1) != '-' && t.charAt(r2) != '-' && s.charAt(r1) != t.charAt(r2)) {
                System.out.println("No");
                return;
            }
            --r1;--r2;
        }

        boolean f1 = false;boolean f2 = false;
        for (int i = l; i <= r1; ++i) {
            if (s.charAt(i) == '*') {
                f1 = true;
                break;
            }
        }
        for (int i = l; i <= r2; ++i) {
            if (t.charAt(i) == '*') {
                f2 = true;
                break;
            }
        }
        if (f1 && f2) {
            System.out.println("Yes");
            return;
        }

        if (l < r2 && t.charAt(l) == '*') {
            String tmp = s;
            s = t;
            t = tmp;
            int ti = r1;
            r1 = r2;
            r2 = ti;
        }

        int l1 = l, l2 = l;
        while (l1 <= r1 && l2 <= r2) {
            while (l1 <= r1 && s.charAt(l1) == '*') ++l1;
            if (l1 <= r1) {
                int nl1 = l1;
                while (nl1 <= r1 && s.charAt(nl1) != '*') ++nl1;
                int nl2 = Math.min(l2 + (nl1 - l1) * 2, r2 + 1);
                if (nl2 - l2 < nl1 - l1) break;
                int index = indexOf(s, t, l1, nl1, l2, nl2);

                if (index != -1) {
                    l2 = index + nl1 - l1;
                    l1 = nl1;
                } else {
                    l2 = nl2;
                }
            }
        }

        while (l1 <= r1 && s.charAt(l1) == '*') ++l1;
        if (l1 > r1) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
    private static String next() {
        return input.next();
    }
    private static int nextInt() {
        return Integer.parseInt(next());
    }

    private static class Input {
        private final BufferedReader br;
        private StringTokenizer tokenizer;
        public Input(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is), 32768);
        }
        private String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
    }
}

**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
10 10
a*b*c*d*
a*b*c*d*
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

Examples:
input:
5 1
1 2
1 3
3 4
3 5
1
output:
Ron

Note:

Buggy Code:
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<algorithm>
#include<string>

using namespace std;

void DFS(int v, vector<vector<int>> & adj, int dp[], int vis[]){
	vis[v] = 1;
	bool is_win = 0;
	for(auto i:adj[v]){
		if(vis[i]) continue;
		else{
			DFS(i, adj, dp, vis);
			is_win |= (dp[i] == 0);
		}
	}
	dp[v] = is_win;
}

void solve(){
	int n, t;
	cin >> n >> t;
	int dp[n];

	vector<vector<int>> adj(n);

	int x, y;
	for(int i=0; i<n-1; i++){
		cin >> x >> y;
		adj[x-1].push_back(y-1);
		adj[y-1].push_back(x-1);
	}
	cin >> t;
	int vis[n];
	for(int i=0; i<n; i++) vis[i] = 0;
	DFS(0, adj, dp, vis);

	cout << (dp[0] ? "Ron" : "Hermione" )<< endl;

}

int main(){
	for(int i=0; i<1; 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 focus on edge cases and scenarios that can highlight the issue. The buggy code seems to have a problem with the initialization of the `vis` array and the `dp` array, as well as the handling of the tree structure.

Here are the steps to think through the problem:

1. **Tree Structure**: The tree should be such that the DFS traversal can expose the bug. A linear tree (a chain) or a tree with a specific structure can be useful.
2. **Initial Node**: The initial node for the game should be chosen to maximize the exposure of the bug.
3. **Edge Cases**: Consider edge cases where the tree has a single child or a single parent.

Let's create a test input with a simple linear tree where the initial node is not the root. This should expose the bug in the DFS traversal and the `vis` array initialization.

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

### Explanation
- **Tree Structure**: The tree is a simple linear chain: `1 - 2 - 3 - 4`.
- **Initial Node**: The initial node for the game is `2`.

### Expected Behavior
- **DFS Traversal**: The DFS should correctly visit all nodes and determine the winning/losing states.
- **Winning/Losing States**: In a linear tree, the player who starts at an even-indexed node (0-based) will lose, and the player who starts at an odd-indexed node will win.

### Expected Output
- **Output**: The expected output is "Hermione" because starting at node `2` (1-based, which is 1 in 0-based) is an odd-indexed node, and thus Hermione wins.

This test input should expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

Examples:
input:
5 1
1 2
1 3
3 4
3 5
1
output:
Ron

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
const int mod = 1e9 + 7;

#define vi vector<int>
#define pi pair<int, int>
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()

int n, T;
const int N = 2e5 + 1;

int dp[N];

void f(int node, int par, vector<vi> &adj){

    int state = 1;

    for(auto it: adj[node]){
        if(it == par) continue;

        f(it, node, adj);
        state = (state and dp[it]);
    }

    dp[node] = state ? 0 : 1;
}

void solve(){
    cin >> n >> T;
    int node;

    vector<vi> adj(n + 1);

    for(int i=0; i<n-1; i++){
        int x, y;
        cin>> x >> y;

        adj[x].push_back(y);
        adj[y].push_back(x);
    }

    cin >> node;

    f(1, 0, adj);

    if(dp[1]) cout<<"Ron"<<endl;
    else cout<<"Hermione"<<endl;
}

signed main() {

    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(NULL);

    // int t;  cin>>t;
    int t = 1;

    while(t--) solve();

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on the logic of the game and the conditions under which the code might fail. The key points to consider are:

1. **Tree Structure**: The tree structure can significantly affect the outcome. For example, a tree with a long path or a tree with a high branching factor can lead to different results.
2. **Starting Node**: The starting node can influence the game's outcome. If the starting node is a leaf or a node with a specific number of neighbors, it can change the result.
3. **Optimal Play**: Both players play optimally, so the test input should force the code to consider optimal moves.

Let's think through a few scenarios:

### Scenario 1: Long Path Tree
A tree that is a long path can be tricky because the optimal strategy might differ based on the starting node. If the starting node is near the middle of the path, the first player (Ron) might have a better chance of winning.

### Scenario 2: Tree with a High Branching Factor
A tree with a high branching factor can also be challenging. If the starting node has many neighbors, the first player might have more options, making it harder for the second player to win.

### Scenario 3: Specific Node Positions
Starting the game from a specific node that has a particular number of neighbors can also expose bugs in the code.

### Chosen Test Input
Let's choose a tree with a high branching factor and a specific starting node to test the code. We will create a star graph, where one central node is connected to all other nodes. Starting from the central node should make the game interesting.

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

### Explanation
- **Tree Structure**: The tree is a star graph with node 1 as the center.
- **Starting Node**: The starting node is the central node (1).
- **Expected Outcome**: In this configuration, Ron (the first player) can always win by moving to any of the leaf nodes, and then Hermione (the second player) will be forced to move to another leaf node, and so on. Since Ron can always force Hermione to move to a leaf node, Ron will always win.

This test input should expose any bugs in the code's logic for handling such a tree structure and starting node.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

Examples:
input:
5 1
1 2
1 3
3 4
3 5
1
output:
Ron

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
//if meet the leaf node at time =0 Ron win else Hermione
bool dfs(int u,int p,vector<vector<int>>&g,int time) {
	if (g[u].size()==1) return !time;
	if (time==0) {
		//Bob
		for (int v: g[u]) {
			if (v==p) continue;
			if (dfs(v,u,g,1-time)) return true;
		}
		return false;
	} else {
		//Hermione
		for (int v: g[u]) {
			if (v==p) continue;
			if (!dfs(v,u,g,0)) return false;
		}
		return true;
	}
}
int main() {
	int n,t; cin>>n>>t;
	vector<vector<int>> g(n+1);
	int u,v;
	for (int i=1;i<n;i++) {
		cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
	}

	int p; cin>>p;
	if (dfs(p,p,g,0)) cout<<"Ron"; else cout<<"Hermione";
}

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the code is that it does not correctly handle the case where the starting node is a leaf node. The function `dfs` assumes that if the starting node is a leaf node and it's Ron's turn, Ron wins. However, this is not always correct. Ron can only win if there are no other valid moves available, which is not necessarily true for a leaf node.

2. **Edge Cases**:
   - A tree with only two nodes (the smallest tree).
   - A tree with a single leaf node where the starting node is the leaf.
   - A tree with multiple leaf nodes where the starting node is one of the leaf nodes.

3. **Challenging Scenario**:
   - A tree with a structure that forces the game to end in a specific way, depending on the starting node.

Let's create a test case where the starting node is a leaf node, and the tree structure is such that the game should end in a specific way, but the buggy code might not handle it correctly.

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

### Explanation:
- **Tree Structure**:
  - Node 1 is connected to Node 2.
  - Node 2 is connected to Node 3.
  - Node 3 is a leaf node.

- **Starting Node**:
  - The starting node is Node 3.

- **Expected Output**:
  - Hermione should win because Ron has no valid move from Node 3 (it's a leaf node), and Hermione will win by default.

This test case should expose the bug in the buggy code, where the function `dfs` incorrectly assumes that Ron wins if the starting node is a leaf node.