**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

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

using namespace std;

int main() {
    int t;
    cin>>t;
    while(t--){
        string a,b;
        cin>>a;
        int k=a.size();
        for(int i=0;i<k;i++){
            if(a[k-1-i]=='p'){
                b[i]='q';
            }else if(a[k-1-i]=='q'){
                b[i]='p';
            }else{
                b[i]='w';
            }
        }
        for(int i=0;i<k;i++){
            cout<<b[i];
        }
        cout<<endl;
    }
    return 0;
}

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

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

Think step by step.

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

1. **String Reversal and Character Swapping:**
   - The code attempts to reverse the string and swap 'p' and 'q' characters.
   - The issue is in the way the string `b` is being constructed. The code uses `b[i]` to store the characters, but `b` is not initialized with a size, which can lead to undefined behavior.

2. **Edge Cases:**
   - Strings with only 'p' or only 'q'.
   - Strings with only 'w'.
   - Strings with a mix of 'p', 'q', and 'w'.
   - Empty strings (though the problem guarantees the string length is at least 1).
   - Strings of length 1.

3. **Test Input:**
   - A string that contains a mix of 'p', 'q', and 'w' and is reversed and character-swapped correctly.
   - A string that is already a palindrome to test the character swapping.
   - A string that is very short (length 1) to test edge cases.

Given these considerations, a small-scale test input that can expose the bug is:

```plaintext
3
p
q
w
```

This input includes:
- A single 'p' which should be reversed and swapped to 'q'.
- A single 'q' which should be reversed and swapped to 'p'.
- A single 'w' which should remain 'w' after the transformation.

This test input is small and complete, and it should help expose the issue with the string `b` not being properly initialized.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
  int t;
  cin>>t;
  while(t--){
    string s;
    cin>>s;
    string m;
    for(int i=0;i<s.size();i++){
        m[i]=s[s.size()-1-i];
    }
    for(int i=0;i<s.size();i++){
        if(m[i]=='p') m[i]='q';
        else if(m[i]=='q') m[i]='p';
    }
    for(int i=0;i<s.size();i++){
        cout<<m[i];
    }
    cout<<endl;
  }
}

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

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

Think step by step.

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

1. **Reversing the String**: The code reverses the string by directly assigning characters to the `m` string without initializing it. This can lead to undefined behavior because `m` is not initialized to the correct size.

2. **Character Swapping**: The code swaps 'p' and 'q' after reversing the string. This part of the code seems correct in logic but might fail if the string is not properly reversed.

Given these points, we need to create a test case that will cause the code to fail due to the uninitialized string `m`.

### Test Case
We can create a string that is long enough to ensure that the uninitialized `m` will cause issues, but still within the constraints (1 to 100 characters). A string with a mix of 'p', 'q', and 'w' will be a good candidate.

Let's use the string "pqw" as it is short and contains all possible characters. This string will help us verify if the code correctly handles the reversal and character swapping.

### Test Input
```plaintext
1
pqw
```

### Explanation
- **Input**: The input consists of one test case with the string "pqw".
- **Expected Output**: The expected output should be "wqp" because:
  - The string "pqw" is reversed to "wqp".
  - 'p' and 'q' are swapped, but since there are no 'p' and 'q' in the reversed string, it remains "wqp".

### Why This Test Case is Challenging
- **Reversal**: The string "pqw" will be reversed to "wqp". If the `m` string is not properly initialized, it will cause undefined behavior.
- **Character Swapping**: Since the string contains all possible characters, it will test the character swapping logic.

This test case should expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
def ship_string(a):
    b=a[::-1]
    c=''
    for i in b:
        if i=='p':
            c+='q'
        elif i=='q':
            c+='p'
        elif i=='w':
            c+='w'
    return c

t=int(input())
l=[]
if t>=1 and t<=99:
    for i in range(t):
        n=input()
        if 1<=len(n)<=100:
            l.append(ship_string(n))
for j in l:
    print(j)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has the following issues:

1. **Index Out of Range**: The code checks if `t` is between 1 and 99, but it should check if `t` is between 1 and 100.
2. **String Length Validation**: The code checks if the length of each string `n` is between 1 and 100, but it should handle the case where the string is empty or contains invalid characters.
3. **Character Mapping**: The code correctly maps 'p' to 'q' and 'q' to 'p', but it does not handle the case where the string contains characters other than 'p', 'q', and 'w'.

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

```plaintext
3
p
q
x
```

This test input includes:
- A single character 'p' to check the correct mapping from 'p' to 'q'.
- A single character 'q' to check the correct mapping from 'q' to 'p'.
- An invalid character 'x' to check if the code handles invalid input gracefully.

This input will help identify the following:
- If the code correctly processes valid characters.
- If the code correctly handles invalid characters.
- If the code correctly processes the number of test cases.

Here is the test input:

```plaintext
3
p
q
x
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
t = int(input())

ans = []

for _ in range(t):
    s = list(input())
    for i in range(len(s)):
        s = s[::-1]
        if s[i] == 'q':
            s[i] = 'p'
        elif s[i] == 'p':
            s[i] = 'q'

    ans.append(s)

for line in ans:
    print(*line,sep='')

**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 analyze the issues in the code:

1. **Reversal and Transformation Order**: The code reverses the string and then performs the character transformation. This is done inside the loop, which means the string is reversed and transformed multiple times. This is incorrect because the string should be reversed only once, and then the characters should be transformed.

2. **Character Transformation**: The transformation logic is correct (swapping 'p' and 'q'), but the order of operations (reversal and transformation) is wrong.

To expose this bug, we need to create a test case where the string is long enough to show the effects of multiple reversals and transformations. However, since we are asked to generate a small-scale test input, we can use a string that is short but still demonstrates the issue.

Let's consider the string "pqw". Here's what should happen:
- The string should be reversed once to become "wqp".
- Then, the characters should be transformed: 'p' to 'q' and 'q' to 'p'.
- The final string should be "wqq".

However, in the buggy code:
- The string is reversed and transformed multiple times.
- This will result in incorrect output.

Let's generate the test input:

```plaintext
1
pqw
```

This input will show that the buggy code produces an incorrect result due to the multiple reversals and transformations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
import java.util.Scanner;

public class cpcode {
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        for (int j = 0; j < t; j++) {
            String s = sc.next();

            char[] arr = s.toCharArray();

            for(int i=0;i<=arr.length/2;i++){
                char te;
                if(arr[i]=='p'){
                    te='q';
                }
                else if(arr[i]=='q'){
                    te='p';
                }
                else{
                    te=arr[i];
                }
                if(arr[arr.length-1-i]=='p'){
                    arr[i]='q';
                }
                else if(arr[arr.length-1-i]=='q'){
                    arr[i]='p';
                }
                else{
                    arr[i]=arr[arr.length-1-i];
                }
                arr[arr.length-1-i]=te;
            }

            String string = new String(arr);

            System.out.println(string);
        }

        sc.close();

    }
}

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

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

Think step by step.

Response:
```plaintext
3
p
q
w
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
import java.util.*;

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

        while(t>0){

            String  str = sc.next();
            int n=str.length();
            StringBuilder stri = new StringBuilder(str);
            for(int i=0;i<n;i++){
                if(stri.charAt(i)=='p')
                    stri.setCharAt(i,'q');
               else if(stri.charAt(i)=='q')
                    stri.setCharAt(i,'p');
                stri.reverse();

            }
            System.out.println(stri);

            t--;
        } sc.close();
    }}

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

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

Think step by step.

Response:
```plaintext
3
p
q
w
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
#include <iostream>
#include <string>
using namespace std;
void vol(){
    string s,a;
    cin>>s;
    for(int i=0;i<(int)s.size();i++){
        a[i]=s[(int)s.size()-1-i];
        if(a[i]=='p') a[i]='q';
        else if(a[i]=='q') a[i]='p';
    }
    for(int i=0;i<(int)s.size();i++) cout<<a[i];
    cout<<endl;
}
int main(){
    int t;
    cin>>t;
    while(t--) vol();
    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, let's analyze the buggy code first:

### Buggy Code Analysis:
1. **Reverse and Swap Logic:**
   - The code attempts to reverse the string and swap 'p' with 'q' and vice versa.
   - The issue is in the line `a[i]=s[(int)s.size()-1-i];` where it directly assigns to `a[i]` without ensuring that `a` has been properly initialized to the correct size.

2. **String Initialization:**
   - The string `a` is not initialized before the loop, which leads to undefined behavior when trying to assign values to `a[i]`.

### Challenging Test Input:
To expose this bug, we need a test case where the string `a` is not properly initialized, leading to undefined behavior. A small-scale test case that can trigger this is a single character string, as it will cause the code to access an uninitialized memory location.

### Test Input:
```plaintext
1
p
```

### Explanation:
- **Test Case Count:** `1` (one test case)
- **String `a`:** `p`

When the input is `p`, the code will try to assign `a[0] = s[0]` and then swap `p` to `q`. However, since `a` is not initialized, accessing `a[0]` will result in undefined behavior. This test case is small and valid, but it will likely expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

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

int main() {
    int t;
    cin >> t;

    while (t--) {
        string a;
        cin >> a;
        int c1 = 0, c2 = 0;
        for (char c : a) {
            if (c == 'p') c1++;
            if (c == 'q') c2++;
        }
        if (c1 == c2) {
            cout << a << endl;
        }
        else {
            reverse(a.begin(), a.end());
            if (c1 == 0) {
                for (char &c : a) {
                    if (c == 'q') {
                        c = 'p';
                    }
                }
            }
            else if (c2 == 0) {
                for (char &c : a) {
                    if (c == 'p') {
                        c = 'q';
                    }
                }
            }
            else {
                for (char &c : a) {
                    if (c == 'p') {
                        c = 'q';
                    }
                    else if (c == 'q') {
                        c = 'p';
                    }
                }
            }
            cout << a << endl;
        }
    }}

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

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

Think step by step.

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

### Analysis of the Buggy Code:
1. **Counting 'p' and 'q' characters:**
   - The code counts the number of 'p' and 'q' characters in the string `a`.
2. **Equality Check:**
   - If the counts of 'p' and 'q' are equal, the code simply outputs the string `a` as `b`.
3. **Reversal and Character Replacement:**
   - If the counts are not equal, the string `a` is reversed.
   - If there are no 'p' characters, all 'q' characters are replaced with 'p'.
   - If there are no 'q' characters, all 'p' characters are replaced with 'q'.
   - If there are both 'p' and 'q' characters, all 'p' characters are replaced with 'q' and vice versa.

### Potential Edge Cases:
1. **String with only 'w' characters:**
   - The code does not handle strings with only 'w' characters correctly.
2. **String with equal counts of 'p' and 'q':**
   - The code outputs the original string, which might not be the correct reflection.
3. **String with more 'p' than 'q':**
   - The code reverses the string and swaps 'p' and 'q', but this might not be correct for all cases.
4. **String with more 'q' than 'p':**
   - The code reverses the string and swaps 'p' and 'q', but this might not be correct for all cases.

### Challenging Test Input:
To expose the bug, we can use a string that contains only 'w' characters, as this is a scenario the code does not handle correctly.

```plaintext
1
www
```

### Explanation:
- The input contains one test case with the string "www".
- The string "www" should be reflected as "www" because 'w' does not change when observed from the inside.
- However, the buggy code does not handle this case correctly and might output an incorrect result.

This test input will help expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
import java.util.*;
public class Normal
{
    public static String solution(Scanner sc)
    {
        String s=sc.next();
        ArrayList<Character> res = new ArrayList<>();

        int pcount=0,qcount=0;
        for(int i=0;i<s.length();i++)
        {
            if(s.charAt(i)== 'p') pcount++;
            else if(s.charAt(i)=='q') qcount++;
        }

        if(pcount==qcount) return s;
        else
        {
            for(int i=s.length()-1;i>=0;i--)
            {
                if(s.charAt(i)=='p') res.add('q');
                else if(s.charAt(i)=='q') res.add('p');
                else res.add(s.charAt(i));
            }

            StringBuilder sb = new StringBuilder();
            for (char c : res) {
                sb.append(c);
            }
            return sb.toString();

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

**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 the logic that the code is implementing. The code has a condition that if the counts of 'p' and 'q' are equal, it returns the original string. Otherwise, it reverses the string and swaps 'p' and 'q'.

The bug in the code is that it does not handle the case where the counts of 'p' and 'q' are equal but the string still needs to be reversed and characters swapped. For example, if the string is "pq" or "qp", the counts of 'p' and 'q' are equal, but the output should be the reversed and swapped string, not the original string.

Let's create a test input that exposes this bug:

```plaintext
3
pq
qp
pqp
```

Explanation:
1. For the input "pq", the counts of 'p' and 'q' are equal (1 each). The correct output should be "qp" (reversed and swapped), but the buggy code will return "pq".
2. For the input "qp", the counts of 'p' and 'q' are equal (1 each). The correct output should be "pq" (reversed and swapped), but the buggy code will return "qp".
3. For the input "pqp", the counts of 'p' and 'q' are equal (2 each). The correct output should be "pqp" (reversed and swapped), but the buggy code will return "pqp".

This test input will clearly show that the buggy code is not handling the case where the counts of 'p' and 'q' are equal but the string still needs to be reversed and characters swapped.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
import java.util.*;
public class Normal
{
    public static String solution(Scanner sc)
    {
        String s=sc.nextLine();
        ArrayList<Character> res = new ArrayList<>();

        int pcount=0,qcount=0;
        for(int i=0;i<s.length();i++)
        {
            if(s.charAt(i)== 'p') pcount++;
            else if(s.charAt(i)=='q') qcount++;
        }

        if(pcount==qcount) return s;
        else
        {
            for(int i=s.length()-1;i>=0;i--)
            {
                if(s.charAt(i)=='p') res.add('q');
                else if(s.charAt(i)=='q') res.add('p');
                else res.add(s.charAt(i));
            }

            StringBuilder sb = new StringBuilder();
            for (char c : res) {
                sb.append(c);
            }
            return sb.toString();

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

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider the following:

1. **Edge Cases:**
   - Empty strings or strings with only one type of character.
   - Strings where the counts of 'p' and 'q' are equal.
   - Strings where the counts of 'p' and 'q' are not equal.

2. **Behavior of the Code:**
   - The code checks if the counts of 'p' and 'q' are equal and returns the original string if they are.
   - If the counts are not equal, it reverses the string and swaps 'p' with 'q' and 'q' with 'p'.

3. **Potential Bugs:**
   - The code does not handle cases where the string contains only 'w' characters.
   - The code does not handle cases where the string contains an equal number of 'p' and 'q' but is not symmetric.

Given these points, a small-scale test input that can expose the bug is:

```plaintext
3
ppqq
w
pqp
```

Explanation:
1. **"ppqq"**:
   - The counts of 'p' and 'q' are equal (2 each).
   - The code will return the original string, which is incorrect because the string should be reversed and 'p' and 'q' should be swapped.

2. **"w"**:
   - The string contains only 'w'.
   - The code will return the original string, which is correct.

3. **"pqp"**:
   - The counts of 'p' and 'q' are not equal (2 'p' and 1 'q').
   - The code will reverse the string and swap 'p' and 'q', which is correct.

This test input will help identify the bug in the code where the string is not reversed and swapped correctly when the counts of 'p' and 'q' are equal.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    char a[110][110];
    char b[110][110];
    cin>>n;
    for(int i = 0; i < n; i ++)
    {
        cin>>a[i];
    }
    for(int i = 0; i < n; i ++)
    {
        int temp = strlen(a[i]);
        for(int j = temp-1; j >= 0; j --)
        {
            if(a[i][j] == 'w') b[i][temp - j - 1] = 'w';
            else if(a[i][j] == 'p') b[i][temp - j - 1] = 'q';
            else if(a[i][j] == 'q') b[i][temp - j - 1] = 'p';
        }
    }
    for(int i = 0; i < n; i ++)
    {
        cout<<b[i]<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
p
q
w
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Normal Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: A string consisting of only characters 'p', 'q', and 'w' is painted on a glass window of a store. Ship walks past the store, standing directly in front of the glass window, and observes string $$$a$$$. Ship then heads inside the store, looks directly at the same glass window, and observes string $$$b$$$.

Ship gives you string $$$a$$$. Your job is to find and output $$$b$$$.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases.

The only line of each test case contains a string $$$a$$$ ($$$1 \leq |a| \leq 100$$$) — the string Ship observes from outside the store. It is guaranteed that $$$a$$$ only contains characters 'p', 'q', and 'w'.

Output: For each test case, output string $$$b$$$, the string Ship observes from inside the store, on a new line.

Examples:
input:
5
qwq
ppppp
pppwwwqqq
wqpqwpqwwqp
pqpqpqpq
output:
pwp
qqqqq
pppwwwqqq
qpwwpqwpqpw
pqpqpqpq

Note:

Buggy Code:
import java.util.*;
public class Normal
{
    public static String solution(Scanner sc)
    {
        String s=sc.nextLine();
        ArrayList<Character> res = new ArrayList<>();

        int pcount=0,qcount=0;
        for(int i=0;i<s.length();i++)
        {
            if(s.charAt(i)== 'p') pcount++;
            else if(s.charAt(i)=='q') qcount++;
        }

        if(pcount==qcount) return s;
        else
        {
            for(int i=s.length()-1;i>=0;i--)
            {
                if(s.charAt(i)=='p') res.add('q');
                else if(s.charAt(i)=='q') res.add('p');
                else res.add(s.charAt(i));
            }

            StringBuilder sb = new StringBuilder();
            for (char c : res) {
                sb.append(c);
            }
            return sb.toString();

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

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

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

Think step by step.

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

1. **Bug Analysis:**
   - The code checks if the counts of 'p' and 'q' are equal. If they are, it returns the original string.
   - If the counts are not equal, it reverses the string and swaps 'p' with 'q' and 'q' with 'p', while leaving 'w' unchanged.

2. **Identifying the Bug:**
   - The logic of swapping 'p' and 'q' only when their counts are not equal is incorrect. The correct behavior should be to reverse the string and swap 'p' and 'q' regardless of their counts.

3. **Edge Cases:**
   - Strings with equal counts of 'p' and 'q' that should not be returned as is.
   - Strings with unequal counts of 'p' and 'q' that should be reversed and swapped.

4. **Challenging Test Input:**
   - A string where the counts of 'p' and 'q' are equal but the string should not be returned as is.
   - A string where the counts of 'p' and 'q' are not equal and should be reversed and swapped.

Based on the above analysis, a small-scale test input that exposes the bug could be:

```plaintext
2
pq
qp
```

### Explanation:
- **Test Case 1: "pq"**
  - The counts of 'p' and 'q' are equal (1 each).
  - The correct output should be "qp" (reversed and swapped).
  - The buggy code will return "pq" because it incorrectly returns the original string when the counts are equal.

- **Test Case 2: "qp"**
  - The counts of 'p' and 'q' are equal (1 each).
  - The correct output should be "pq" (reversed and swapped).
  - The buggy code will return "qp" because it incorrectly returns the original string when the counts are equal.

This test input will clearly show that the buggy code is not handling the reversal and swapping correctly when the counts of 'p' and 'q' are equal.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
// semenInRussia 2024
#include <cstdint>
#include <iostream>
#include <vector>
using namespace std;

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,fma,bmi2,tune=native")

#define isz(x) int(x.size())

const int inf = 1e9 + 5;

void solve() {
  string a, b, c;
  cin >> a >> b >> c;

  // f[i][j] - minimum errors that a[0..i) located on c[0..j]
  // answer = f[isz(a)][isz(c)]

  vector<vector<int>> f(isz(a) + 3, vector<int>(isz(c) + 3, 0));

  // f[i][j] - is minimum errors to do
  // that on c[0..j] j of a

  f[0][0] = 0;
  int t = 0;
  for (int j = 0; j < isz(c); j++) {
    bool bad = (t >= isz(b)) || (c[t] != b[j]);
    f[0][j + 1] = f[0][j] + bad;
    if (!bad) {
      t++;
    }
  }

  for (int i = 0; i < isz(a); i++) {
    for (int j = 0; j < isz(c); j++) {
      auto &ans = f[i + 1][j + 1];
      ans = +inf;

      ans = min(ans, f[i][j] + (a[i] != c[j]));

      int k = j - i - 1;
      if (k < 0) {
        continue;
      }

      ans = min(ans, f[i + 1][j] + (b[k] != c[j]));

      // i = 1, a = 2
      // j = 2, c = 3
    }
  }

  cout << f[isz(a)][isz(c)] << "\n";
}

int main() {
  cin.tie(nullptr)->sync_with_stdio(0);
  int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
}

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

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

Think step by step.

Response:
```plaintext
1
abc
def
abcdef
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

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

void solve()
{
   string a, b, c;
    cin >> a >> b >> c;
    vector<vector<int> > dp(a.size()+2, vector<int>(b.size()+2, INT_MAX));
    dp[0][1] = 0;
    dp[1][0] = 0;
    for(int i=1; i<=a.size()+1; i++){
        for(int j=1; j<=b.size()+1; j++){

            if(c[i+j-2]==a[i-1]){
                dp[i][j] = min(dp[i][j],dp[i-1][j]);
            }
            else{
                dp[i][j] = min(dp[i][j],dp[i-1][j]+1);
            }
            if(c[i+j-2]==b[j-2]){
                dp[i][j] = min(dp[i][j],dp[i][j-1]);
            }
            else{
                dp[i][j] = min(dp[i][j],dp[i][j-1]+1);
            }

                        // cout << dp[i][j] << " ";

        }
        // cout << endl;

    }
    cout << dp[a.size()+1][b.size()+1];
    return;
}

int32_t main()
{
    int t;
    cin >> t;
    while (t--){
        solve();
        cout << endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
a
b
ba
ab
cd
acbd
a
bcd
bdc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int T,dp[2000][2000],n,m,x1,x2,ans[5000];
string a,b,c;
int main()
{
	scanf("%d",&T);
	getchar();
	while(T--)
	{
		getline(cin,a);
		getline(cin,b);
		getline(cin,c);
		n=a.length();
		m=b.length();
		dp[0][0]=0;
		for (int i=1;i<=n;i++)
		{
			dp[i][0]=dp[i-1][0];
			if(a[i-1]!=c[i-1])
			dp[i][0]++;
		}
		for (int i=1;i<=m;i++)
		{
			dp[0][i]=dp[0][i-1];
			if(b[i-1]!=c[i-1])
			dp[0][i]++;
		}
		for (int i=1;i<=n;i++)
		{
			for (int j=1;j<=m;j++)
			dp[i][j]=n+m;
		}
		for (int i=1;i<=n+m;i++)
		{
			for (int j=1;j<=min(n,i);j++)
			{
				x1=dp[j-1][i-j];
				if(a[j-1]!=c[i-1])
				x1++;
				x2=dp[j][i-j-1];
				if(b[i-j-1]!=c[i-1])
				x2++;
				dp[j][i-j]=min(x1,x2);
			}
		}
		cout<<dp[n][m]<<endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
abc
def
abcdef
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
def f(a, b, c, i, j, dp):
    if i == 0 and j == 0:
        return 0
    if i == 0:
        return sum(1 for k in range(j) if b[k] != c[k])
    if j == 0:
        return sum(1 for k in range(i) if a[k] != c[k])
    if dp[i][j] != -1:
        return dp[i][j]
    indx = i + j - 1
    ans = float("inf")
    if i > 0 and c[indx] == a[i - 1]:
        ans = min(ans, f(a, b, c, i - 1, j, dp))
    if j > 0 and c[indx] == b[j - 1]:
        ans = min(ans, f(a, b, c, i, j - 1, dp))
    if i > 0 and c[indx] != a[i - 1]:
        ans = min(ans, f(a, b, c, i - 1, j, dp) + 1)
    if j > 0 and c[indx] != b[j - 1]:
        ans = min(ans, f(a, b, c, i, j - 1, dp) + 1)
    dp[i][j] = ans
    return ans
n = int(input())
for _ in range(n):
    a = input()
    b = input()
    c = input()
    i = len(a)
    j = len(b)
    dp = [[-1] * (1000) for _ in range(1000)]
    res = f(a, b, c, i, j, dp)
    print(res)

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

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

Think step by step.

Response:
```plaintext
1
abc
def
abcdef
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
for _ in range(int(input())):
    a = input()
    b = input()
    c = input()
    changes = 1
    while changes:
        changes = 0
        if len(a) and c[0]==a[0]:
            a = a[1:]
            c = c[1:]
            changes += 1
        elif len(b) and c[0]==b[0]:
            b = b[1:]
            c = c[1:]
            changes += 1
        elif len(a) and c[-1]==a[-1]:
            c = c[:-1]
            a = a[:-1]
            changes += 1
        elif len(b) and c[-1]==b[-1]:
            b = b[:-1]
            c = c[:-1]
            changes += 1
    k = min(len(a),len(b))
    ans = k
    for i in range(k,len(c)):
        if len(b) > len(a):
            if c[i]!=b[i-k]:
                ans += 1
        else:
            if c[i]!=a[i-k]:
                ans += 1
    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
abc
def
abcdef
abc
def
defabc
abc
def
abcfde
```

3
abc
def
abcdef
abc
def
defabc
abc
def
abcfde


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
def solve(a,b,c,ind1,ind2,dp):
    if(ind1>=len(a) and ind2>=len(b)):
        return 0
    takea=float("inf")
    takeb=float("inf")
    if(dp[ind1][ind2]!=-1):
        return dp[ind1][ind2]
    if(ind1<len(a)):
        takea=int(c[ind1+ind2]!=a[ind1])+solve(a,b,c,ind1+1,ind2,dp)
    if(ind2<len(b)):
        takeb=int(c[ind1+ind2]!=b[ind2])+solve(a,b,c,ind1,ind2+1,dp)
    dp[ind1][ind2]=min(takea,takeb)
    return min(takea,takeb)
for _ in range(int(input())):
    a=input()
    b=input()
    c=input()
    dp=[[-1 for _ in range(len(b)+1343)] for _ in range(len(a)+1128)]
    ans=solve(a,b,c,0,0,dp)
    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
a
b
ab
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
import java.util.*;

public class three {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine();

        while (t-- > 0) {
            String a = sc.nextLine();
            String b = sc.nextLine();
            String c = sc.nextLine();

            int n = a.length();
            int m = c.length();
            int dp[][] = new int[n + 1][m + 1];
            for(int i=0;i<=n;i++){
               Arrays.fill(dp[i],100000000);
            }
            for (int i = 0; i <= m; i++) {
                dp[0][i] = i;
            }

            for (int i = 1; i <= n; i++) {
                for (int k = 1; k <= m; k++) {
                    int g1 = Integer.MAX_VALUE;
                    int g2 = Integer.MAX_VALUE;

                    if (a.charAt(i - 1) == c.charAt(k - 1)) {
                        g1 = dp[i - 1][k - 1];
                    } else {
                        g1 = 1 + dp[i - 1][k - 1];
                    }

                    int j = k - i;
                    if (j >= 1 && j <= b.length()) {
                        if (b.charAt(j - 1) == c.charAt(k - 1)) {
                            g2 = dp[i][k - 1];
                        } else {
                            g2 = 1 + dp[i][k - 1];
                        }
                    }

                    dp[i][k] = Math.min(dp[i][k],Math.min(g1, g2));
                }
            }

            System.out.println(dp[n][m]);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
ab
ab
baba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;

import java.io.BufferedInputStream;
import java.io.File;
import java.io.PrintWriter;
import java.io.FileInputStream;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Objects;
import java.util.Map;
import java.util.Random;
import java.util.TreeMap;
import java.util.TreeSet;

public class E_Three_Strings {
    static final int MOD = (int) 1e9 + 7;
    static final FastScanner in = new FastScanner();
    static final PrintWriter out = new PrintWriter(System.out);
    static int n;
    static int m;
    static char[] a, b, c;
    static final int INF = (int)1e9;

    public static void main(String[] args) {

        int T = in.nextInt();

        while (T-- > 0) {

            a = in.next().toCharArray();
            b = in.next().toCharArray();
            c = in.next().toCharArray();

            n = a.length;
            m = b.length;

            int[][] dp = new int[n + m + 1][n + 1];

            for(int i = 0; i <= n + m; ++i) {
                Arrays.fill(dp[i], INF);
            }

            dp[0][0] = 0;

            for(int t = 1; t <= n; ++t) {
                dp[t][0] = dp[t - 1][0] + (b[t - 1] != c[t - 1] ? 1 : 0);
            }

            for(int t = 1; t <= n + m; ++t) {
                for(int j = 1; j <= n; ++j) {
                    dp[t][j] = min(dp[t][j], dp[t - 1][j - 1] + (a[j - 1] != c[t - 1] ? 1 : 0));
                    if(t - j - 1 >= 0 && t - j - 1 < m) {
                        dp[t][j] = min(dp[t][j], dp[t - 1][j] + (b[t - j - 1] != c[t - 1] ? 1 : 0));
                    }
                }
            }

            out.println(dp[n + m][n]);

        }

        out.close();

    }

    // Remember to initialize the whole adj list
    // with empty arraylist.
    // static void dfs(int node) {
    // visited[node] = true;
    // Just after entering the node.

    // for (int child : g[node]) {
    // Take action on each child
    // if (!visited[child])
    // dfs(child);
    // }

    // Take action after exiting all the
    // childs

    // }

    static long binPow(long a, long n, long mod) {
        long res = 1L;

        while (n > 0L) {
            if ((n & 1L) == 1L)
                res = ((a % mod) * (res % mod)) % mod;

            a = ((a % mod) * (a % mod)) % mod;
            n >>= 1;
        }

        return res;
    }

    // Strictly lesser than target if less than equal to target < := <=
    static int lowerBound(long[] a, int l, int r, long target) {
        while (l <= r) {
            int mid = (r - l) / 2 + l;

            if (a[mid] < target) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return r;
    }

    // Strictly greater than target if greater than equal to target <= := <
    static int upperBound(long[] a, int l, int r, long target) {
        while (l <= r) {
            int mid = (r - l) / 2 + l;

            if (a[mid] <= target) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    // Strictly greater than target if greater than equal to target <= := <
    static int upperBound(int[] a, int l, int r, int target) {
        while (l <= r) {
            int mid = (r - l) / 2 + l;

            if (a[mid] <= target) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    // Strictly lesser than target if less than equal to target < := <=
    static int lowerBound(int[] a, int l, int r, int target) {
        while (l <= r) {
            int mid = (r - l) / 2 + l;

            if (a[mid] < target) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return r;
    }

    static void push(Map<Long, Long> mp, long k, long v) {
        if (mp.containsKey(k)) {
            mp.put(k, mp.get(k) + v);
        } else {
            mp.put(k, v);
        }
    }

    static void pull(Map<Long, Long> mp, long k, long v) {
        long x = mp.get(k);

        if (v >= x)
            mp.remove(k);
        else
            mp.put(k, x - v);
    }

    static boolean isPrime(long n) {
        if (n < 2)
            return false;
        if (n == 2 || n == 3)
            return true;
        if (n % 2 == 0 || n % 3 == 0)
            return false;

        for (long i = 6; i * i <= n; i += 6) {
            if ((n % (i - 1) == 0) || (n % (i + 1) == 0)) {
                return false;
            }
        }

        return true;
    }

    static long gcd(long a, long b) {
        return b == 0L ? a : gcd(b, a % b);
    }

    static void reverse(int[] a, int l, int r) {
        while (l < r)
            swap(a, l++, r--);
    }

    static void reverse(long[] a, int l, int r) {
        while (l < r)
            swap(a, l++, r--);
    }

    static void swap(long[] a, int l, int r) {
        long temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }

    static void swap(int[] a, int l, int r) {
        int temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }

    static void ruffleSort(long[] a, int l, int r) {
        shuffleArray(a, l, r);
        Arrays.sort(a, l, r + 1);
    }

    static void ruffleSort(int[] a, int l, int r) {
        shuffleArray(a, l, r);
        Arrays.sort(a, l, r + 1);
    }

    static void shuffleArray(int[] arr, int l, int r) {
        Random rnd = new Random();
        for (int i = l; i <= r; ++i) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(r - i + 1);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
    }

    static void shuffleArray(long[] arr, int l, int r) {
        Random rnd = new Random();
        for (int i = l; i <= r; ++i) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(r - i + 1);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
    }

    static long[] readArrayL(int n) {
        long[] a = new long[n + 1];

        for (int i = 1; i <= n; ++i) {
            a[i] = in.nextLong();
        }

        return a;
    }

    static ArrayList<Integer> readList(int n) {
        ArrayList<Integer> list = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            list.add(in.nextInt());
        }

        return list;
    }

    static int[] readArray(int n) {
        int[] arr = new int[n + 1];

        for (int i = 1; i <= n; i++) {
            arr[i] = in.nextInt();
        }

        return arr;
    }

    static class FastScanner {
        // I don't understand how this works lmao
        private int BS = 1 << 16;
        private char NC = (char) 0;
        private byte[] buf = new byte[BS];
        private int bId = 0, size = 0;
        private char c = NC;
        private double cnt = 1;
        private InputStream in;

        public FastScanner() {
            in = new BufferedInputStream(System.in, BS);
        }

        public FastScanner(String s) {
            try {
                in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
            } catch (Exception e) {
                in = new BufferedInputStream(System.in, BS);
            }
        }

        private char nextChar() {
            while (bId == size) {
                try {
                    size = in.read(buf);
                } catch (Exception e) {
                    return NC;
                }
                if (size == -1)
                    return NC;
                bId = 0;
            }
            return (char) buf[bId++];
        }

        public int nextInt() {
            return (int) nextLong();
        }

        public long nextLong() {
            cnt = 1;
            boolean neg = false;
            if (c == NC)
                c = nextChar();
            for (; (c < '0' || c > '9'); c = nextChar()) {
                if (c == '-')
                    neg = true;
            }
            long res = 0;
            for (; c >= '0' && c <= '9'; c = nextChar()) {
                res = (res << 3) + (res << 1) + c - '0';
                cnt *= 10;
            }
            return neg ? -res : res;
        }

        public double nextDouble() {
            double cur = nextLong();
            return c != '.' ? cur : cur + nextLong() / cnt;
        }

        public String next() {
            StringBuilder res = new StringBuilder();
            while (c <= 32)
                c = nextChar();
            while (c > 32) {
                res.append(c);
                c = nextChar();
            }
            return res.toString();
        }

        public String nextLine() {
            StringBuilder res = new StringBuilder();
            while (c <= 32)
                c = nextChar();
            while (c != '\n') {
                res.append(c);
                c = nextChar();
            }
            return res.toString();
        }

        public boolean hasNext() {
            if (c > 32)
                return true;
            while (true) {
                c = nextChar();
                if (c == NC)
                    return false;
                else if (c > 32)
                    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 generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases that might cause the algorithm to fail. The code seems to be using dynamic programming to find the minimum number of character changes required to transform the string `c` into a valid sequence formed by strings `a` and `b`. Here are some potential edge cases to consider:

1. **Empty Strings**: Although the problem constraints guarantee that the strings are non-empty, it's good to test with minimal valid inputs.
2. **Equal Length Strings**: Strings `a` and `b` of equal length.
3. **Different Length Strings**: Strings `a` and `b` of different lengths.
4. **Random Character Changes**: Ensure that some characters in `c` are different from what would be expected from `a` and `b`.
5. **Maximum Length Strings**: Test with the maximum possible lengths for `a` and `b` within the constraints.

Given the constraints and the need for a small-scale test input, let's consider a case where `a` and `b` are of different lengths and some characters in `c` are changed.

### Test Input
```plaintext
1
abc
def
adbecf
```

### Explanation
- **a = "abc"**: Length 3
- **b = "def"**: Length 3
- **c = "adbecf"**: Length 6, which is the sum of the lengths of `a` and `b`.

In this case, `c` is a valid sequence that can be formed by interleaving `a` and `b`, but with some characters changed. The expected output should be 0 because `c` can be formed without any changes.

This test input is small and valid, and it should help in exposing any issues in the dynamic programming approach used in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

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

using namespace std;
#define ull unsigned long long
#define cy cout << "YES\n"
#define cn cout << "NO\n"
#define new cout << endl
#define ll  long long
#define pb push_back
#define vll vector<long long>
#define INF (long long)1e9
#define MOD 1000000007
#define mp make_pair
#define F first
#define S second
#define f(i, a, b) for(long long i = a; i < b; i++) // Loop from a to b
long long gcd(long long a, long long b) {
    while (b != 0) {
        long long temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}
long long lcm(long long a, long long b) {
    return (a * b) /gcd(a, b);
}
long long factorial(int n) {
    long long result = 1;
    for (int i = 1; i <= n; ++i) {
        result *= i;
    }
    return result;
}

long long combination(int n, int r) {
    if (r > n) return 0;
    if (r == 0 || r == n) return 1;

    if (r > n - r) r = n - r;

    long long result = 1;

    for (int i = 0; i < r; ++i) {
        result *= (n - i);
        result /= (i + 1);
    }

    return result;
}

bool damn(const pair<int, int>& a, const pair<int, int>& b) {
    return a.first<b.first;

};

vector<long long> sieve_of_eratosthenes(long long limit) {
    vector<bool> prime(limit + 1, true);
    prime[0] = prime[1] = false;

    for (long long num = 2; num * num <= limit; ++num) {
        if (prime[num]) {
            for (long long multiple = num * num; multiple <= limit; multiple += num) {
                prime[multiple] = false;
            }
        }
    }

    vector<long long> primes;
    for (long long i = 2; i <= limit; ++i) {
        if (prime[i]) {
            primes.push_back(i);
        }
    }

    return primes;
};
void merge(int arr[], int left, int right) {
    if (left >= right) return;
    int mid = left + (right - left) / 2;
    merge(arr, left, mid);
    merge(arr, mid + 1, right);

    int n1 = mid - left + 1;
    int n2 = right - mid;

    int L[n1], R[n2];
    for (int i = 0; i < n1; i++) L[i] = arr[left + i];
    for (int i = 0; i < n2; i++) R[i] = arr[mid + 1 + i];

    int i = 0, j = 0, k = left;
    while (i < n1 && j < n2) {
        if (L[i] <= R[j]) arr[k++] = L[i++];
        else arr[k++] = R[j++];
    }

    while (i < n1) arr[k++] = L[i++];
    while (j < n2) arr[k++] = R[j++];
};

void mergeSort(int arr[], int left, int right) {
    if (left < right) {
        int mid = left + (right - left) / 2;
        mergeSort(arr, left, mid);
        mergeSort(arr, mid + 1, right);
        merge(arr, left, right);
    }
};
struct item{
    ll m,c;
};
int count_trailing_zeros(ll n) {
    int count = 0;

    while ((n & 1) == 0 && n != 0) {
        count++;
        n >>= 1;
    }

    return count;
}

int popcount(ll n) {
    int count = 0;

    while (n) {

        n = n & (n - 1);
        count++;
    }

    return count;
}

struct segTree {
    ll size;
    vll sums;

    void init(int n) {
        size = 1;
        while (size < n) {
            size *= 2;
        }
        sums.assign(2 * size, 0LL);
    }

    void set(ll i, ll v, ll x, ll lx, ll rx) {
        if (rx - lx == 1) {
            sums[x] = v;
            return;
        }
        ll m = (lx + rx) / 2;
        if (i < m) {
            set(i, v, 2 * x + 1, lx, m);
        }
        else {
            set(i, v, 2 * x + 2, m, rx);
        }
        sums[x] = gcd(sums[2 * x + 1] , sums[2 * x + 2]);
    }

    void set(ll i, ll v) {
        set(i, v, 0, 0, size);
    }

    ll sum(ll l, ll r, ll x, ll lx, ll rx) {
        if (lx >= r || l >= rx) return 1e9;
        if (lx >= l && rx <= r) return sums[x];
        ll m = (lx + rx) / 2;
        ll s1 = sum(l, r, 2 * x + 1, lx, m);
        ll s2 = sum(l, r, 2 * x + 2, m, rx);
        return gcd(s1 , s2);
    }

    ll sum(ll l, ll r) {
        return sum(l, r, 0, 0, size);
    }
};
bool isPerfectPowerOf3(int n) {
    if (n <= 0) {
        return false;
    }
    while (n % 3 == 0) {
        n /= 3;
    }

    return n == 1;
}

int modInverse(int a, int m) {
    a = a % m;
    for (int x = 1; x < m; x++) {
        if ((a * x) % m == 1) {
            return x;
        }
    }
    return -1;
}

int chineseRemainderTheorem(const vector<int>& a, const vector<int>& n) {
    int k = a.size();
    int N = 1;
    for (int i = 0; i < k; i++) {
        N *= n[i];
    }

    int result = 0;
    for (int i = 0; i < k; i++) {
        int Ni = N / n[i];
        int yi = modInverse(Ni, n[i]);
        if (yi == -1) {
            return -1;
        }
        result += a[i] * Ni * yi;
    }

    return result % N;
}
ll dp[1005][1005];
ll sumi(string str1,string str2,string str3,ll ptr1,ll ptr2,ll count){
    if(ptr1==str1.length() && ptr2==str2.length()){
        return dp[ptr1][ptr2]=count;
    }
    if(count>=dp[ptr1][ptr2]) return dp[ptr1][ptr2];
    if(ptr1==str1.length()){
        if(str2[ptr2]!=str3[ptr2+ptr1]) return dp[ptr1][ptr2]=sumi(str1,str2,str3,ptr1,ptr2+1,count+1);
        else return dp[ptr1][ptr2]=sumi(str1,str2,str3,ptr1,ptr2+1,count);
    }
    if(ptr2==str2.length()){
        if(str1[ptr1]!=str3[ptr2+ptr1]) return dp[ptr1][ptr2]=sumi(str1,str2,str3,ptr1+1,ptr2,count+1);
        else return dp[ptr1][ptr2]=sumi(str1,str2,str3,ptr1+1,ptr2,count);
    }
    //if(dp[ptr1][ptr2]!=-1) return dp[ptr1][ptr2];
    if(str1[ptr1]==str3[ptr1+ptr2] && str2[ptr2]!=str3[ptr1+ptr2] ) return dp[ptr1][ptr2]=min(sumi(str1,str2,str3,ptr1+1,ptr2,count),sumi(str1,str2,str3,ptr1,ptr2+1,count+1));
    else if(str1[ptr1]!=str3[ptr1+ptr2] && str2[ptr2]==str3[ptr1+ptr2]) return dp[ptr1][ptr2]=min(sumi(str1,str2,str3,ptr1,ptr2+1,count),sumi(str1,str2,str3,ptr1+1,ptr2,count+1));
    else if(str1[ptr1]==str3[ptr1+ptr2] && str2[ptr2]==str3[ptr1+ptr2]) return dp[ptr1][ptr2]=min(sumi(str1,str2,str3,ptr1,ptr2+1,count),sumi(str1,str2,str3,ptr1+1,ptr2,count));
    else return dp[ptr1][ptr2]=min(sumi(str1,str2,str3,ptr1,ptr2+1,count+1),sumi(str1,str2,str3,ptr1+1,ptr2,count+1));
}

int main() {

    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ll test_cases;
    cin>>test_cases;
    auto start = chrono::high_resolution_clock::now();
    //vll vec=sieve_of_eratosthenes(10000);
    f(j,0,test_cases){
        memset(dp,10000,sizeof(dp));
        ll n,m,l1,l2,r1,r2,k,x=0,maxi2=0,maxi=0;ll maxi_pos=0;
        bool change=false, sorted=true;
        ll sum=0;ll sum1=0;string str;
        //cout<<dp[0][0]<<endl;
        string str1, str2,str3;
        cin>>str1;
        cin>>str2;
        cin>>str3;

        ll pointer_1=0;
        ll pointer_2=0;
        ll count=0;

        cout<<sumi(str1,str2,str3,0,0,0)<<endl;

    }

    auto end = chrono::high_resolution_clock::now();
    chrono::duration<double> duration = end - start;
    //cout << "Time taken: " << duration.count() << " seconds" << endl;
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
abc
def
abcdef
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

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

ll f(ll ind1, ll ind2, ll &n, ll &m, string &s1, string &s2, string &s, vector<vector<ll>> &dp)
{
    if (ind1 + ind2 == n + m)
        return 0;
    ll mn = 1e9;
    if (ind1 < n)
        mn = min(mn, (s1[ind1] != s[ind1 + ind2]) + f(ind1 + 1, ind2, n, m, s1, s2, s, dp));
    if (ind2 < m)
        mn = min(mn, (s2[ind2] != s[ind1 + ind2]) + f(ind1, ind2 + 1, n, m, s1, s2, s, dp));
    return dp[ind1][ind2] = mn;
}

void Ankur()
{
    string s1, s2, s;
    cin >> s1 >> s2 >> s;
    ll n = s1.size(), m = s2.size();
    vector<vector<ll>> dp(n + m, vector<ll>(n + m, -1));
    cout << f(0, 0, n, m, s1, s2, s, dp) << "\n";
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t = 1;
    cin >> t;
    while (t--)
    {
        Ankur();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
abc
def
abcde
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

Input: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^3$$$) — the number of test cases.

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
import sys
import os
import time
import math
import random

def min_changes_required(a, b, c):
    n = len(a)
    m = len(b)
    total = n + m

    def get_possible_strings(a, b):
        if not a:
            return {b}
        if not b:
            return {a}

        # Try taking from a first
        result1 = set()
        for s in get_possible_strings(a[1:], b):
            result1.add(a[0] + s)

        # Try taking from b first
        result2 = set()
        for s in get_possible_strings(a, b[1:]):
            result2.add(b[0] + s)

        return result1.union(result2)

    # Get all possible strings
    possible = get_possible_strings(a, b)

    # Find minimum differences
    min_diff = total
    for p in possible:
        diff = sum(1 for x, y in zip(p, c) if x != y)
        min_diff = min(min_diff, diff)

    return min_diff

def solve():
    t = int(input())
    for _ in range(t):
        a = input()
        b = input()
        c = input()
        print(min_changes_required(a, b, c))

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
3
a
a
aa
ab
ba
abab
abc
cba
abcabc
```