**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
import java.util.*;

public class PreparingForTheExam {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for(int i=0;i<t;i++){
            int l1[]=new int[3];
            for(int j=0;j<3;j++){
                l1[j]=sc.nextInt();
            }
            int m[]=new int[l1[1]];
            for(int j=0;j<l1[1];j++){
                m[j]=sc.nextInt();
            }
            boolean know[]=new boolean[l1[1]+1];
            for(int j=0;j<l1[2];j++){
                int a=sc.nextInt();
                know[a]=true;
            }
            List<Character> obj=new ArrayList<>();
            String w="";
            if(l1[2]==l1[0]){
                for(int j=0;j<l1[1];j++)
                    w+='1';
            }
            else if(l1[2]==l1[0]-1){
                recur(obj,m,know,l1[1],l1[0]-1,0,0,0);
                for(char j:obj){
                    w=j+w;
                }
                for(int j=0;j<l1[1]-obj.size();j++)
                    w+='0';
            }
            else{
                for(int j=0;j<l1[1];j++)
                    w+='0';
            }
            System.out.println(w);
        }
    }
    public static void recur(List<Character> obj,int m[],boolean k[],
                             int ml,int l,int count,int index,int taken){
        if(taken==l){
            if(count==l)
                obj.add('1');
            else
                obj.add('0');
            return;
        }
        if(index==ml) return;
        int c=0;
        if(k[m[index]]){
            c=count+1;
        }
        recur(obj,m,k,ml,l,c,index+1,taken+1);
        recur(obj,m,k,ml,l,count,index+1,taken);
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

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

int main(){
	int tc;
	cin >> tc;
	while(tc--){
		int n,m,k;
		cin >> n >> m >> k;
		vector<int> v1;
		vector<int> v2;
		int sum = 0;
		for(int i = 0 ; i < m ; i++){
			int el;
			cin >> el;
			v1.push_back(el);
		}
		for(int i = 0 ; i < k ; i++){
			int el;
			cin >> el;
			v2.push_back(el);
			sum+=el;
		}
		if(n == k){
			string ans = "";
			for(int i = 0 ; i < m ; i++){
				ans = ans + '1';
			}
			cout << ans << endl;
			continue;
		}
		if(n - k > 1){
			string ans = "";
			for(int i = 0 ; i < m ; i++){
				ans = ans + '0';
			}
			cout << ans << endl;
			continue;
		}
		else{
			int missing = ((n*(n+1))/2) - sum;
			string ans = "";
			for(auto num : v1){
				if(num==missing){
					ans = ans + '1';
				}
				else{
					ans = ans + '0';
				}
			}
			cout << ans << 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
5 5 4
1 2 3 4 5
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

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

int main() {
    int t;cin >> t;
    while (t--) {
        int n,m,k;cin >> n>>m>>k;
        vector<int>a(m),q(k);
        for (int i = 0; i < m; i++) {
            cin >> a[i];
        }
        for (int i = 0; i < k; i++) {
            cin >> q[i];
        }

        if (n-k>1) {
            for (int i = 0; i < m; i++) {
                cout<<0;
            }
            cout<<endl;
        }

        else if (n-k==0) {
            for (int i = 0; i < m; i++) {
                cout<<1;
            }
            cout<<endl;
        }

        else {
            int num=0;
            for (int i = 0; i < m; i++) {
                bool ac=false;
                for (int j = 0; j < k; j++) {
                    if (a[i]<q[j]) {break;}
                    if (a[i]==q[j]) {
                        ac=true;
                    }
                }
                if (ac) {cout<<0;}
                else {cout<<1;}
                // if (a[i]==q[i]) {
                //     cout<<0;
                //     num++;
                // }
                // else {
                //     cout<<1;
                //     num++;
                //     break;
                // }
            }
            // for (int i = num; i < m; i++) {
            //     cout<<0;
            // }
            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:
```plaintext
2
4 4 2
1 2 3 4
1 2
5 4 2
1 2 3 4
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
#!/usr/bin/python
# _*_ coding:utf-8 _*_

t = int(input())

for _ in range(t):
    n, m, k = map(int, input().split(" "))
    al = list(map(int, input().split(" ")))
    ql = list(map(int, input().split(" ")))
    problem_list = list(range(1, n + 1))
    for i in range(m):
        p = [] + problem_list
        # print(p, al[i])
        p.remove(al[i])
        # print(ql, p)
        if set(p) <= set(ql):
            print(1, end="")
        else:
            print(0, end="")
    print()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
x = int(input())
ans = []

for i in range(x) :
    n , m , k = map(int, input().split())
    lm = list(map(int, input().split()))
    lk = list(map(int, input().split()))
    if n-k > 1 :
        a1 = "0" * m
        ans.append(a1)
    elif n-k == 0 :
        a3 = "1" * m
        ans.append(a3)
    else :
        a2 = ""
        for s in range(m) :
            if lm[s] in lk :
               a2 += "0"
            else :
               a2 += "1"
        ans.append(a2)

for num in ans :
    print(num)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();

            int[] a = new int[m];
            for (int i = 0; i < m; i++) {
                a[i] = sc.nextInt();
            }

            int[] b = new int[k];
            for (int i = 0; i < k; i++) {
                b[i] = sc.nextInt();
            }

            HashSet<Integer> st1 = new HashSet<>();
            for (int i = 1; i <= n; i++) {
                st1.add(i);
            }

            HashSet<Integer> st2 = new HashSet<>();
            for (int value : b) {
                st2.add(value);
            }

            ArrayList<Integer> list = new ArrayList<>();

            for (int i = 0; i < m; i++) {
                st1.remove(a[i]);
                if (st2.containsAll(st1)) {
                    list.add(1);
                } else {
                    list.add(0);
                }
                st1.add(a[i]);
            }

            // Print results
            for (int result : list) {
                System.out.print(result);
            }
            System.out.println();
        }

        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
5 3 2
1 3 5
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
import java.util.*;
import java.io.*;
public class codeforces995d3p3 {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st;

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

        for (int i = 0; i < t; i++) {
            int n = readInt();
            int m = readInt();
            int k = readInt();
            ArrayList<Integer> mValues = new ArrayList<Integer>();
            ArrayList<Integer> kValues = new ArrayList<Integer>();

            for (int j = 0; j < m; j++) {
                mValues.add(readInt());
            }
            for (int l = 0; l < k; l++) {
                kValues.add(readInt());
            }

            if (n > k + 1) {
                for (int b = 0; b < m; b++) {
                    System.out.print("0");
                }
            }
            else if (n == k) {
                for (int c = 0; c < m; c++) {
                    System.out.print("1");
                }
            }
            else {
                for (int a = 0; a < m; a++) {
                    if (mValues.get(a) == 1) {
                        if (mValues.get(a) == kValues.get(mValues.get(a) - 1)) {
                            System.out.print("0");
                        }
                        else {
                            System.out.print("1");
                        }
                    }
                    else if (mValues.get(a) > k) {
                        if (mValues.get(a) == kValues.get(mValues.get(a) - 2)) {
                            System.out.print("0");
                        }
                        else {
                            System.out.print("1");
                        }
                    }
                    else {
                        if (mValues.get(a) == kValues.get(mValues.get(a) - 1) || mValues.get(a) == kValues.get(mValues.get(a) - 2)) {
                            System.out.print("0");
                        }
                        else {
                            System.out.print("1");
                        }
                    }

                }
            }

            System.out.println();
        }
    }

    static String next () throws IOException {
        while (st == null || ! st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    static long readLong () throws IOException {
        return Long.parseLong(next());
    }
    static int readInt () throws IOException {
        return Integer.parseInt(next());
    }
    static double readDouble () throws IOException {
        return Double.parseDouble(next());
    }
    static char readCharacter () throws IOException {
        return next().charAt(0);
    }
    static String readLine () throws IOException {
        return br.readLine().trim();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
t = int(input())
for i in range(t):
    n , m , k = map(int,input().split(" "))
    que = list(map(int,input().split(" ")))
    know = list(map(int,input().split()))
    if k < n-1:
        print('0'*m)
    elif k == n:
        print('1'*m)
    else:
        know.sort()
        index = k+1
        for j in range(k):
            if know[j] != j+1:
                index = j+1
                break

        ans = ''
        for ele in que:
            if ele == index:
                ans += '1'
            else:
                ans += '0'

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
n = int(input())

for _ in range(n):
    n, m, k = map(int, input().split())
    l = list(map(int, input().split()))
    c = set(map(int, input().split()))

    result = []
    for missing in l:
        questions = set(range(1, n + 1)) - {missing}
        if questions.issubset(c):
            result.append('1')
        else:
            result.append('0')

    print(''.join(result))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

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

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

    while (t-- > 0) {
      int n = sc.nextInt();
      int m = sc.nextInt();
      int k = sc.nextInt();

      int[] a = new int[m];
      for (int i = 0; i < m; i++) {
        a[i] = sc.nextInt();
      }

      HashSet<Integer> q = new HashSet<>();
      for (int i = 0; i < k; i++) {
        q.add(sc.nextInt());
      }

      if (k == n) {
        for (int i = 0; i < m; i++) {
          System.out.print('1');
        }
        System.out.println();
      }

      else if (k == n - 1) {
        int mis = 1;
        for (int x : q) {
          if (x == mis) {
            mis++;
          } else {
            break;
          }
        }
        if (mis > n) {
          mis = n;
        }

        StringBuilder res = new StringBuilder();
        for (int x : a) {
          if (x == mis) {
            res.append('1');
          } else {
            res.append('0');
          }
        }
        System.out.println(res);
      }

      else {
        for (int i = 0; i < m; i++) {
          System.out.print('0');
        }
        System.out.println();
      }
    }

    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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the code's logic might fail. The buggy code has issues when handling the case where \( k = n - 1 \). Specifically, it incorrectly determines the missing question and checks if the missing question is in the list of questions that Monocarp knows.

Let's consider a test case where \( k = n - 1 \) and the missing question is not the last one in the sequence. This will expose the bug in the code's logic.

Here's a small-scale test input:

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

### Explanation:
- \( t = 1 \): One test case.
- \( n = 5 \): There are 5 questions.
- \( m = 3 \): There are 3 lists of questions.
- \( k = 4 \): Monocarp knows 4 questions.
- \( a = [1, 3, 5] \): The lists of questions are:
  - List 1: [2, 3, 4, 5] (missing 1)
  - List 2: [1, 2, 4, 5] (missing 3)
  - List 3: [1, 2, 3, 4] (missing 5)
- \( q = [1, 2, 4, 5] \): Monocarp knows questions 1, 2, 4, and 5.

### Expected Output:
- For List 1: [2, 3, 4, 5] (Monocarp knows all questions except 3, so he won't pass).
- For List 2: [1, 2, 4, 5] (Monocarp knows all questions, so he will pass).
- For List 3: [1, 2, 3, 4] (Monocarp doesn't know 3, so he won't pass).

The expected output is:
```plaintext
010
```

This test case should expose the bug in the code's logic for handling the case where \( k = n - 1 \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
import java.util.*;

public class MonocarpExam {

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

        int t = scanner.nextInt();

        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            int k = scanner.nextInt();

            int[] a = new int[m];
            for (int j = 0; j < m; j++) {
                a[j] = scanner.nextInt();
            }

            int[] b = new int[k];
            for (int j = 0; j < k; j++) {
                b[j] = scanner.nextInt();
            }

            String result = solve(n, m, a, b, k);
            System.out.println(result);
        }

        scanner.close();
    }

    public static String solve(int n, int m, int[] a, int[] b, int k) {
        StringBuilder result = new StringBuilder();

        for (int i = 0; i < m; i++) {
            HashSet<Integer> knownQuestions = new HashSet<>();
            for (int j = 0; j < k; j++) {
                knownQuestions.add(b[j]);
            }

            boolean canPass = true;
            for (int j = 1; j <= n; j++) {
                if (j != a[i] && !knownQuestions.contains(j)) {
                    canPass = false;
                    break;
                }
            }

            if (canPass) {
                result.append("1");
            } else {
                result.append("0");
            }
        }

        return result.toString();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that can cause the algorithm to fail. The primary issue in the code is the logic used to determine if Monocarp can pass the exam. Specifically, the code checks if each question in the list is known by Monocarp, but it does not correctly handle the case where Monocarp knows all the questions in the list.

Here are the key points to consider:
1. **Monocarp knows all questions except the one missing in the list**: If Monocarp knows all questions except the one that is missing in the list, he should pass.
2. **Monocarp knows a subset of the questions**: If Monocarp knows a subset of the questions, he should fail if the missing question is one he doesn't know.

Let's construct a test case that maximizes the failure probability:

### Test Case:
- **n = 4**: There are 4 questions.
- **m = 4**: There are 4 lists, each missing one question.
- **k = 3**: Monocarp knows 3 questions.

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

### Explanation:
- **List 1**: Missing question 1, contains questions [2, 3, 4]. Monocarp knows [1, 3, 4], so he fails (0).
- **List 2**: Missing question 2, contains questions [1, 3, 4]. Monocarp knows [1, 3, 4], so he passes (1).
- **List 3**: Missing question 3, contains questions [1, 2, 4]. Monocarp knows [1, 3, 4], so he fails (0).
- **List 4**: Missing question 4, contains questions [1, 2, 3]. Monocarp knows [1, 3, 4], so he fails (0).

### Expected Output:
```plaintext
0100
```

### Test Case:
- **n = 5**: There are 5 questions.
- **m = 5**: There are 5 lists, each missing one question.
- **k = 4**: Monocarp knows 4 questions.

### Input:
```plaintext
1
5 5 4
1 2 3 4 5
1 2 4 5
```

### Explanation:
- **List 1**: Missing question 1, contains questions [2, 3, 4, 5]. Monocarp knows [1, 2, 4, 5], so he fails (0).
- **List 2**: Missing question 2, contains questions [1, 3, 4, 5]. Monocarp knows [1, 2, 4, 5], so he fails (0).
- **List 3**: Missing question 3, contains questions [1, 2, 4, 5]. Monocarp knows [1, 2, 4, 5], so he passes (1).
- **List 4**: Missing question 4, contains questions [1, 2, 3, 5]. Monocarp knows [1, 2, 4, 5], so he fails (0).
- **List 5**: Missing question 5, contains questions [1, 2, 3, 4]. Monocarp knows [1, 2, 4, 5], so he fails (0).

### Expected Output:
```plaintext
00100
```

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;cin >>t;
    while(t--){
        int n,m,k;
        cin >>n>>m>>k;
        int a[m];
        int s[n+1]={0};
        int ans[m];
        for(int i=0;i<m;i++){
            cin >>a[i];
        }
        for(int i=0;i<k;i++){
            int l;
            cin >>l;
            s[l]=1;
        }

        for(int i=0;i<m;i++){
            ans[i]=1;
        }
        cout<<"\n";
        for(int i=1;i<n+1;i++){
            if(s[i]==0){
                for(int o=0;o<m;o++){
                    if(a[o]!=i){
                        ans[o]=0;
                    }
                }
            }
        }
        for(int i=0;i<m;i++){
            cout<<ans[i];
        }
        cout<<"\n";
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Preparing for the Exam
time_limit_ms: 1500
memory_limit_mb: 256
Description: Monocarp is preparing for his first exam at the university. There are $$$n$$$ different questions which can be asked during the exam, numbered from $$$1$$$ to $$$n$$$. There are $$$m$$$ different lists of questions; each list consists of exactly $$$n-1$$$ different questions. Each list $$$i$$$ is characterized by one integer $$$a_i$$$, which is the index of the only question which is not present in the $$$i$$$-th list. For example, if $$$n = 4$$$ and $$$a_i = 3$$$, the $$$i$$$-th list contains questions $$$[1, 2, 4]$$$.

During the exam, Monocarp will receive one of these $$$m$$$ lists of questions. Then, the professor will make Monocarp answer all questions from the list. So, Monocarp will pass only if he knows all questions from the list.

Monocarp knows the answers for $$$k$$$ questions $$$q_1, q_2, \dots, q_k$$$. For each list, determine if Monocarp will pass the exam if he receives that list.

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

Each test case consists of three lines:

- the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$2 \le n \le 3 \cdot 10^5$$$; $$$1 \le m, k \le n$$$);
- the second line contains $$$m$$$ distinct integers $$$a_1, a_2, \dots, a_m$$$ ($$$1 \le a_i \le n$$$; $$$a_i < a_{i+1}$$$);
- the third line contains $$$k$$$ distinct integers $$$q_1, q_2, \dots, q_k$$$ ($$$1 \le q_i \le n$$$; $$$q_i < q_{i+1}$$$).

Additional constraints on the input:

- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a string of $$$m$$$ characters. The $$$i$$$-th character should be 1 if Monocarp passes the exam if he receives the $$$i$$$-th question list, 0 if Monocarp won't pass.

Examples:
input:
4
4 4 3
1 2 3 4
1 3 4
5 4 3
1 2 3 4
1 3 4
4 4 4
1 2 3 4
1 2 3 4
2 2 1
1 2
2
output:
0100
0000
1111
10

Note: In the first test case, Monocarp knows the questions $$$[1, 3, 4]$$$. Let's consider all the question lists:

- the first list consists of questions $$$[2, 3, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the second list consists of questions $$$[1, 3, 4]$$$. Monocarp knows all these questions, so he will pass;
- the third list consists of questions $$$[1, 2, 4]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass;
- the fourth list consists of questions $$$[1, 2, 3]$$$. Monocarp doesn't know the $$$2$$$-nd question, so he won't pass.

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

using namespace std ;

//#define x first
//#define y second

typedef long long ll ;

typedef pair<int , int>PII ;

const int N = 3e5 + 10 ;
int a[N] , q[N] ;

map<int , int>mp ;

void solve()
{
	int n , m , k ;
	cin >> n >> m >> k ;
	for(int i = 1 ; i <= m ; i++){
		cin >> a[i] ;
	}
	for(int i = 1 ; i <= k ; i++){
		cin >> q[i] ;
		mp[q[i]]++ ;
	}
	string s ;
	if(n - k > 1){
		for(int i = 0 ; i < m ; i++){
			s += '0' ;
		}
		cout << s << '\n' ;
		mp.clear() ;
		return ;
	}
	if(n == k){
		for(int i = 0 ; i < m ; i++){
			s += '1' ;
		}
		cout << s << '\n' ;
		mp.clear() ;
		return ;
	}
	ll tt = 0 ;
	for(int i = 1 ; i <= n ; i++){
		if(mp[i] == 0){
			tt = i ;  //记录所缺的那个数是什么
		}
	}
	for(int i = 0 ; i < m ; i++){
		s += '0' ;
	}
	s[tt - 1] = '1' ;
	cout << s << '\n' ;
	mp.clear() ;
}

int main()
{
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;
	int t ;
	cin >> t ;
	while(t--) solve() ;
	return 0 ;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Message Spread
time_limit_ms: 12000
memory_limit_mb: 1024
Description: Given is an undirected graph with $n$ vertices and $m$ edges. Each edge connects two vertices $(u, v)$ and has a probability of $\frac{p}{q}$ of appearing each day.

Initially, vertex $1$ has a message. At the end of the day, a vertex has a message if and only if itself or at least one of the vertices adjacent to it had the message the day before. Note that each day, each edge chooses its appearance independently.

Calculate the expected number of days before all the vertices have the message, modulo $998\,244\,353$.

Input: The first line contains two integers $n$ and $m$ ($1\leq n\leq 21$, $n-1\leq m\leq\frac{n(n-1)}{2}$).

Then $m$ lines follow, each containing four integers $u$, $v$, $p$, and $q$ ($1\leq u\neq v\leq n$, $1\leq p<q<998\,244\,353$, $\gcd(p,q)=1$) — there is an undirected edge between $u$ and $v$, and it has a probability of appearance of $\frac{p}{q}$ each day.

It is guaranteed that there are no self-loops or multiple-edges in the graph and that the graph is connected if all of the edges appear.

Additional constraint in the input: Let $g_{i,j}$ be the probability of appearance of the edge between $i$ and $j$ ($g_{i,j}=0$ if there is no edge between $i$ and $j$). It is guaranteed that for any $S\subseteq\\{1,2,\ldots,n\\}$ ($|S|\ge 1$), $$ \prod_{i\in S}\left(\prod_{j\in\\{1,2,\ldots,n\\}\setminus S}(1-g_{i,j})\right)\not\equiv1\pmod{998\,244\,353}. $$

Output: Print a single integer in the only line of the output — the expected number of days, modulo $998\,244\,353$.

Formally, let $M = 998\,244\,353$. It can be shown that the exact answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Examples:
input:
2 1
1 2 1 10
output:
10

input:
3 3
1 2 1 2
1 3 1 2
2 3 1 2
output:
887328316

input:
1 0
output:
0

input:
5 8
1 2 1 11
1 3 2 11
1 4 3 11
1 5 4 11
2 4 5 11
2 5 6 11
3 4 7 11
4 5 8 11
output:
469993557

input:
21 22
1 2 3 4
2 3 4 5
3 4 5 6
5 6 7 8
6 7 8 9
7 8 9 10
8 9 2 3
9 10 3 4
10 11 4 5
11 12 5 6
12 13 6 7
13 14 7 8
14 15 8 9
15 16 9 10
16 17 2 3
17 18 3 4
18 19 4 5
19 20 5 6
20 21 6 7
1 10 100 1001
15 4 147 220
4 11 1 998244352
output:
299529765

Note: In the first test, the answer is equal to the expected number of days before the only edge in the graph first appears, and that is $\frac{1}{0.1}=10$.

In the second test, the answer is equal to $\frac{20}{9}$ before it is taken modulo $998\,244\,353$.

In the third test, the only vertex already has the message, so the answer is $0$.

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

typedef long long ll;
const int MOD = 998244353;

ll modpow(ll base, ll exp) {
    ll res = 1;
    while (exp > 0) {
        if (exp % 2 == 1) res = res * base % MOD;
        base = base * base % MOD;
        exp /= 2;
    }
    return res;
}

ll modinv(ll x) {
    return modpow(x, MOD - 2);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m;
    cin >> n >> m;
    vector<vector<pair<int, pair<int, int>>>> adj(n);
    for(int i=0; i<m; i++){
        int u, v, p, q;
        cin >> u >> v >> p >> q;
        u--; v--;
        adj[u].push_back({v, {p, q}});
        adj[v].push_back({u, {p, q}});
    }

    vector<double> dp((1<<n), 1e18);
    dp[1] = 0;

    for(int mask=1; mask<(1<<n); mask++){
        for(int u=0; u<n; u++){
            if(mask & (1<<u)){
                for(auto [v, pq]: adj[u]){
                    int p = pq.first, q = pq.second;
                    int next_mask = mask | (1<<v);
                    double prob = 1.0 * p / q;
                    dp[next_mask] = min(dp[next_mask], dp[mask] + 1.0 / prob);
                }
            }
        }
    }

    // Convert the expected number of days to a fraction
    double expected_days = dp[(1<<n)-1];
    ll numerator = round(expected_days * MOD);
    ll denominator = MOD;

    // Simplify the fraction and find the modular inverse
    ll gcd = __gcd(numerator, denominator);
    numerator /= gcd;
    denominator /= gcd;
    ll result = numerator * modinv(denominator) % MOD;

    cout << result;
}

**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 scenarios that can maximize the failure probability. The buggy code uses a double-precision floating-point number to store the expected number of days, which can lead to precision issues. Additionally, the code assumes that the expected number of days can be directly converted to a fraction, which might not be accurate due to the nature of floating-point arithmetic.

Here are the steps to generate a challenging test input:

1. **Small-Scale Input**: Since the input needs to be small-scale, we can use a graph with a small number of vertices and edges.
2. **Edge Case**: Use a graph where the probabilities of edge appearances are very small, which can lead to a large expected number of days. This can cause precision issues when using floating-point arithmetic.
3. **Connected Graph**: Ensure the graph is connected to satisfy the problem constraints.

Let's create a test input with 3 vertices and 2 edges, where the probabilities of edge appearances are very small:

```plaintext
3 2
1 2 1 1000000
2 3 1 1000000
```

This input represents a graph with 3 vertices and 2 edges. Each edge has a very small probability of appearing (1/1000000), which should cause the expected number of days to be very large. This can expose the precision issues in the floating-point arithmetic used in the buggy code.

Here is the test input:

```plaintext
3 2
1 2 1 1000000
2 3 1 1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Message Spread
time_limit_ms: 12000
memory_limit_mb: 1024
Description: Given is an undirected graph with $n$ vertices and $m$ edges. Each edge connects two vertices $(u, v)$ and has a probability of $\frac{p}{q}$ of appearing each day.

Initially, vertex $1$ has a message. At the end of the day, a vertex has a message if and only if itself or at least one of the vertices adjacent to it had the message the day before. Note that each day, each edge chooses its appearance independently.

Calculate the expected number of days before all the vertices have the message, modulo $998\,244\,353$.

Input: The first line contains two integers $n$ and $m$ ($1\leq n\leq 21$, $n-1\leq m\leq\frac{n(n-1)}{2}$).

Then $m$ lines follow, each containing four integers $u$, $v$, $p$, and $q$ ($1\leq u\neq v\leq n$, $1\leq p<q<998\,244\,353$, $\gcd(p,q)=1$) — there is an undirected edge between $u$ and $v$, and it has a probability of appearance of $\frac{p}{q}$ each day.

It is guaranteed that there are no self-loops or multiple-edges in the graph and that the graph is connected if all of the edges appear.

Additional constraint in the input: Let $g_{i,j}$ be the probability of appearance of the edge between $i$ and $j$ ($g_{i,j}=0$ if there is no edge between $i$ and $j$). It is guaranteed that for any $S\subseteq\\{1,2,\ldots,n\\}$ ($|S|\ge 1$), $$ \prod_{i\in S}\left(\prod_{j\in\\{1,2,\ldots,n\\}\setminus S}(1-g_{i,j})\right)\not\equiv1\pmod{998\,244\,353}. $$

Output: Print a single integer in the only line of the output — the expected number of days, modulo $998\,244\,353$.

Formally, let $M = 998\,244\,353$. It can be shown that the exact answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Examples:
input:
2 1
1 2 1 10
output:
10

input:
3 3
1 2 1 2
1 3 1 2
2 3 1 2
output:
887328316

input:
1 0
output:
0

input:
5 8
1 2 1 11
1 3 2 11
1 4 3 11
1 5 4 11
2 4 5 11
2 5 6 11
3 4 7 11
4 5 8 11
output:
469993557

input:
21 22
1 2 3 4
2 3 4 5
3 4 5 6
5 6 7 8
6 7 8 9
7 8 9 10
8 9 2 3
9 10 3 4
10 11 4 5
11 12 5 6
12 13 6 7
13 14 7 8
14 15 8 9
15 16 9 10
16 17 2 3
17 18 3 4
18 19 4 5
19 20 5 6
20 21 6 7
1 10 100 1001
15 4 147 220
4 11 1 998244352
output:
299529765

Note: In the first test, the answer is equal to the expected number of days before the only edge in the graph first appears, and that is $\frac{1}{0.1}=10$.

In the second test, the answer is equal to $\frac{20}{9}$ before it is taken modulo $998\,244\,353$.

In the third test, the only vertex already has the message, so the answer is $0$.

Buggy Code:
#include<cstdio>
#define TY ll
#define MAXN 2097152
#define FOR(i,a,b) for(TY i=(a);i<=(b);i=-~i)
#define fOR(i,a,b) for(TY i=(a);i<(b);i=-~i)
#define ROF(i,a,b) for(TY i=(a);i>=(b);i=~-i)
#define rOF(i,a,b) for(TY i=(a);i>(b);i=~-i)
#define EDG(i,u) for(TY i=hed[u];i;i=nxt[i])
using namespace std;
typedef long long ll;
const TY M=998244353;
typedef unsigned long long ull;
TY _abs(TY a){return a<0?-a:a;}
TY maxn(TY a,TY b){return a>b?a:b;}
TY minn(TY a,TY b){return a<b?a:b;}
inline void updmx(TY &x,TY y){if(x<y)x=y;}
inline void updmn(TY &x,TY y){if(x>y)x=y;}
inline void add(TY &x,TY y){if((x+=y)>=M)x-=M;}
TY gcd(TY a,TY b){return b?gcd(b,a%b):a;}
TY qp(TY a,TY b){TY ans=1;do{if(1&b)ans=ans*a%M;a=a*a%M;}while(b>>=1);return ans;}
char getc(){char ch=getchar();while(ch==' '||ch=='\n'||ch=='\r')ch=getchar();return ch;}
TY qr(){
	char ch=getchar();TY s=0,x=1;
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')x=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())s=s*10+ch-'0';return x*s;
}void qw(TY a){if(a>9)qw(a/10);putchar(a%10+'0');}
void qw(TY a,char ch){
	if(a<0){a=-a;putchar('-');}
	if(a>9)qw(a/10);putchar(a%10+'0');if(ch)putchar(ch);
}TY n,m,S,u,vl,ml[MAXN],ct[MAXN],dp[MAXN],per[MAXN],ar[MAXN],A[MAXN],B[20][MAXN],C[22][MAXN],ans;
int main(){
	n=qr();m=qr();S=(1<<n)-1;
	FOR(i,0,S){ml[i]=1;ct[i]=ct[i>>1]+(1&i);}
	while(m--){
		u=1<<qr()-1;u|=1<<qr()-1;vl=qr();
		ml[u]=M+1-vl*qp(qr(),M-2)%M;
	}FOR(i,7,S)if(ct[i]>2){
		fOR(j,0,n)if(1&(i>>j)){u=j;break;}ml[i]=ml[i^(1<<u)];
		fOR(j,u+1,n)if(1&(i>>j))ml[i]=ml[i]*ml[(1<<u)|(1<<j)]%M;
	}FOR(i,1,S)per[i]=ml[S]*qp(ml[i]*ml[S^i]%M,M-2)%M;
	dp[1]=1;vl=ml[S]*qp((M+1-per[1])*ml[S^1]%M,M-2)%M;
	for(TY i=0;i<S;i+=2)ar[i|1]=ml[i]%M*vl%M;
	fOR(x,1,n-1){
		FOR(j,0,S)A[j]=0;
		for(TY i=2;i<S;i+=2)if(ct[i]==x)A[i]=ml[i];
		fOR(k,0,n)FOR(j,1<<k,S)if(1&(j>>k))
			add(A[j],A[j^(1<<k)]);
		FOR(i,0,S)B[x][i]=A[i];
	}fOR(x,2,n){
		qw(x,'\n');
		FOR(i,0,S)A[i]=C[x][i];
		fOR(k,0,n)FOR(i,1<<k,S)if(1&(i>>k))
			add(A[i],M-A[i^(1<<k)]);
		FOR(i,0,S)if(ct[i]==x)add(ar[i],A[i]);
		for(TY i=1;i<=S;i+=2)A[i]=ar[i]*qp(ml[i],M-2)%M;
		fOR(k,1,n)for(TY i=S;i>=(1<<k);i-=2)
			if(1&(i>>k))add(A[i],M-A[i^(1<<k)]);
		for(TY i=1;i<=S;i+=2)if(ct[i]==x){
			dp[i]=A[i];A[i]=A[i]*qp(ml[S^i]*(M+1-per[i])%M,M-2)%M*ml[S]%M;
			add(ar[i],A[i]);
		}else A[i]=0;
		fOR(k,0,n)FOR(j,1<<k,S)if(1&(j>>k))
			add(A[j],A[j^(1<<k)]);
		ROF(i,n-x,1)FOR(j,0,S)add(C[i+x][j],A[j]*B[i][j]%M);
	}fOR(i,1,S)add(ans,dp[i]*qp(M+1-per[i],M-2)%M);
	qw(ans);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, we need to consider edge cases and scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Graph Size and Connectivity**:
   - Use a small number of vertices and edges to keep the input manageable.
   - Ensure the graph is connected, but not fully connected, to test the edge appearance probabilities.

2. **Edge Probabilities**:
   - Use probabilities that are not trivial (e.g., not 0 or 1).
   - Ensure the probabilities are not too high or too low to avoid easy or hard edge cases.

3. **Special Cases**:
   - Consider a graph where the message spread is not straightforward, such as a graph with a bottleneck or a long path.

Given these considerations, let's create a test input with 5 vertices and 6 edges. We will use probabilities that are not trivial and ensure the graph is connected but not fully connected.

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

This input describes a graph with 5 vertices and 6 edges. The edges have a probability of 1/3 of appearing each day. The graph is connected but not fully connected, and the probabilities are chosen to be non-trivial. This should expose any issues in the buggy code, particularly in how it handles the message spread and the expected number of days calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Message Spread
time_limit_ms: 12000
memory_limit_mb: 1024
Description: Given is an undirected graph with $n$ vertices and $m$ edges. Each edge connects two vertices $(u, v)$ and has a probability of $\frac{p}{q}$ of appearing each day.

Initially, vertex $1$ has a message. At the end of the day, a vertex has a message if and only if itself or at least one of the vertices adjacent to it had the message the day before. Note that each day, each edge chooses its appearance independently.

Calculate the expected number of days before all the vertices have the message, modulo $998\,244\,353$.

Input: The first line contains two integers $n$ and $m$ ($1\leq n\leq 21$, $n-1\leq m\leq\frac{n(n-1)}{2}$).

Then $m$ lines follow, each containing four integers $u$, $v$, $p$, and $q$ ($1\leq u\neq v\leq n$, $1\leq p<q<998\,244\,353$, $\gcd(p,q)=1$) — there is an undirected edge between $u$ and $v$, and it has a probability of appearance of $\frac{p}{q}$ each day.

It is guaranteed that there are no self-loops or multiple-edges in the graph and that the graph is connected if all of the edges appear.

Additional constraint in the input: Let $g_{i,j}$ be the probability of appearance of the edge between $i$ and $j$ ($g_{i,j}=0$ if there is no edge between $i$ and $j$). It is guaranteed that for any $S\subseteq\\{1,2,\ldots,n\\}$ ($|S|\ge 1$), $$ \prod_{i\in S}\left(\prod_{j\in\\{1,2,\ldots,n\\}\setminus S}(1-g_{i,j})\right)\not\equiv1\pmod{998\,244\,353}. $$

Output: Print a single integer in the only line of the output — the expected number of days, modulo $998\,244\,353$.

Formally, let $M = 998\,244\,353$. It can be shown that the exact answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Examples:
input:
2 1
1 2 1 10
output:
10

input:
3 3
1 2 1 2
1 3 1 2
2 3 1 2
output:
887328316

input:
1 0
output:
0

input:
5 8
1 2 1 11
1 3 2 11
1 4 3 11
1 5 4 11
2 4 5 11
2 5 6 11
3 4 7 11
4 5 8 11
output:
469993557

input:
21 22
1 2 3 4
2 3 4 5
3 4 5 6
5 6 7 8
6 7 8 9
7 8 9 10
8 9 2 3
9 10 3 4
10 11 4 5
11 12 5 6
12 13 6 7
13 14 7 8
14 15 8 9
15 16 9 10
16 17 2 3
17 18 3 4
18 19 4 5
19 20 5 6
20 21 6 7
1 10 100 1001
15 4 147 220
4 11 1 998244352
output:
299529765

Note: In the first test, the answer is equal to the expected number of days before the only edge in the graph first appears, and that is $\frac{1}{0.1}=10$.

In the second test, the answer is equal to $\frac{20}{9}$ before it is taken modulo $998\,244\,353$.

In the third test, the only vertex already has the message, so the answer is $0$.

Buggy Code:
#include <iostream>
#define ewrewwrewa int

using namespace std;

int main()
{
    cout << 10;

    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, we need to consider the following points:

1. **Input Validation**: The code simply outputs `10` without processing any input. This suggests that the code is not even reading the input, which is a major bug.
2. **Edge Cases**: We should focus on creating an input that is valid but not trivial, to ensure that the code's behavior is clearly incorrect.
3. **Small Scale**: The input should be small enough to be manageable and to fit within the constraints provided.

Given these points, let's create a test input that includes a small, valid graph with a non-trivial number of vertices and edges. We will use a simple graph with 3 vertices and 3 edges, which is a connected graph and satisfies the problem constraints.

Here is the test input:

```plaintext
3 3
1 2 1 2
1 3 1 2
2 3 1 2
```

This input describes a complete graph with 3 vertices and 3 edges, where each edge has a probability of 1/2 of appearing each day. This is a non-trivial case that should clearly expose the bug in the code, as the code will output `10` regardless of the input, which is incorrect for this graph.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Message Spread
time_limit_ms: 12000
memory_limit_mb: 1024
Description: Given is an undirected graph with $n$ vertices and $m$ edges. Each edge connects two vertices $(u, v)$ and has a probability of $\frac{p}{q}$ of appearing each day.

Initially, vertex $1$ has a message. At the end of the day, a vertex has a message if and only if itself or at least one of the vertices adjacent to it had the message the day before. Note that each day, each edge chooses its appearance independently.

Calculate the expected number of days before all the vertices have the message, modulo $998\,244\,353$.

Input: The first line contains two integers $n$ and $m$ ($1\leq n\leq 21$, $n-1\leq m\leq\frac{n(n-1)}{2}$).

Then $m$ lines follow, each containing four integers $u$, $v$, $p$, and $q$ ($1\leq u\neq v\leq n$, $1\leq p<q<998\,244\,353$, $\gcd(p,q)=1$) — there is an undirected edge between $u$ and $v$, and it has a probability of appearance of $\frac{p}{q}$ each day.

It is guaranteed that there are no self-loops or multiple-edges in the graph and that the graph is connected if all of the edges appear.

Additional constraint in the input: Let $g_{i,j}$ be the probability of appearance of the edge between $i$ and $j$ ($g_{i,j}=0$ if there is no edge between $i$ and $j$). It is guaranteed that for any $S\subseteq\\{1,2,\ldots,n\\}$ ($|S|\ge 1$), $$ \prod_{i\in S}\left(\prod_{j\in\\{1,2,\ldots,n\\}\setminus S}(1-g_{i,j})\right)\not\equiv1\pmod{998\,244\,353}. $$

Output: Print a single integer in the only line of the output — the expected number of days, modulo $998\,244\,353$.

Formally, let $M = 998\,244\,353$. It can be shown that the exact answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Examples:
input:
2 1
1 2 1 10
output:
10

input:
3 3
1 2 1 2
1 3 1 2
2 3 1 2
output:
887328316

input:
1 0
output:
0

input:
5 8
1 2 1 11
1 3 2 11
1 4 3 11
1 5 4 11
2 4 5 11
2 5 6 11
3 4 7 11
4 5 8 11
output:
469993557

input:
21 22
1 2 3 4
2 3 4 5
3 4 5 6
5 6 7 8
6 7 8 9
7 8 9 10
8 9 2 3
9 10 3 4
10 11 4 5
11 12 5 6
12 13 6 7
13 14 7 8
14 15 8 9
15 16 9 10
16 17 2 3
17 18 3 4
18 19 4 5
19 20 5 6
20 21 6 7
1 10 100 1001
15 4 147 220
4 11 1 998244352
output:
299529765

Note: In the first test, the answer is equal to the expected number of days before the only edge in the graph first appears, and that is $\frac{1}{0.1}=10$.

In the second test, the answer is equal to $\frac{20}{9}$ before it is taken modulo $998\,244\,353$.

In the third test, the only vertex already has the message, so the answer is $0$.

Buggy Code:
MOD = 998244353

def mod_inv(a, m=MOD):
    return pow(a, m-2, m)

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

    n = int(data[0])
    m = int(data[1])
    edges = []
    idx = 2
    for _ in range(m):
        u = int(data[idx]) - 1  # 0-indexed
        v = int(data[idx+1]) - 1  # 0-indexed
        p = int(data[idx+2])
        q = int(data[idx+3])
        edges.append((u, v, p, q))
        idx += 4

    adj = [{} for _ in range(n)]
    for u, v, p, q in edges:
        adj[u][v] = p * mod_inv(q)
        adj[v][u] = p * mod_inv(q)

    all_mask = (1 << n) - 1
    dp = [0] * (1 << n)

    for mask in range(all_mask, -1, -1):
        if mask == all_mask:
            continue  # already initialized to 0
        informed = []
        uninformed = []
        for i in range(n):
            if mask & (1 << i):
                informed.append(i)
            else:
                uninformed.append(i)

        prob_stay = 1
        for u in uninformed:
            prob_u_informs = 0
            for v in informed:
                if v in adj[u]:
                    prob_u_informs += adj[u][v]
            prob_u_informs %= MOD
            prob_stay *= (1 - prob_u_informs)
            prob_stay %= MOD

        dp[mask] = (1 + dp[mask]) % MOD
        if prob_stay != 1:
            dp[mask] = (dp[mask] * mod_inv(1 - prob_stay)) % MOD

    print(dp[1])  # only vertex 1 is informed initially

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Edge Cases**: Test the boundary conditions of the input constraints, such as the minimum and maximum values for `n` and `m`.
2. **Graph Structure**: Ensure the graph is connected and test different types of graphs (e.g., complete graph, linear chain, star graph).
3. **Probabilities**: Use probabilities that are not trivial (e.g., 0 or 1) to ensure the code handles floating-point arithmetic correctly.
4. **Modulo Operations**: Ensure the probabilities and their inverses are well-defined modulo 998244353.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
3 3
1 2 1 3
1 3 1 3
2 3 1 3
```

### Explanation:
- **Graph Structure**: This input represents a complete graph with 3 vertices (a triangle). Each vertex is connected to every other vertex.
- **Probabilities**: Each edge has a probability of 1/3 of appearing each day.
- **Expected Output**: The expected number of days for all vertices to have the message should be calculated.

This test case is designed to:
- Ensure the graph is connected.
- Test the handling of non-trivial probabilities.
- Verify the correctness of the modulo operations and probability calculations.

By using this input, we can check if the code correctly handles the propagation of the message through a complete graph with non-trivial edge probabilities.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Message Spread
time_limit_ms: 12000
memory_limit_mb: 1024
Description: Given is an undirected graph with $n$ vertices and $m$ edges. Each edge connects two vertices $(u, v)$ and has a probability of $\frac{p}{q}$ of appearing each day.

Initially, vertex $1$ has a message. At the end of the day, a vertex has a message if and only if itself or at least one of the vertices adjacent to it had the message the day before. Note that each day, each edge chooses its appearance independently.

Calculate the expected number of days before all the vertices have the message, modulo $998\,244\,353$.

Input: The first line contains two integers $n$ and $m$ ($1\leq n\leq 21$, $n-1\leq m\leq\frac{n(n-1)}{2}$).

Then $m$ lines follow, each containing four integers $u$, $v$, $p$, and $q$ ($1\leq u\neq v\leq n$, $1\leq p<q<998\,244\,353$, $\gcd(p,q)=1$) — there is an undirected edge between $u$ and $v$, and it has a probability of appearance of $\frac{p}{q}$ each day.

It is guaranteed that there are no self-loops or multiple-edges in the graph and that the graph is connected if all of the edges appear.

Additional constraint in the input: Let $g_{i,j}$ be the probability of appearance of the edge between $i$ and $j$ ($g_{i,j}=0$ if there is no edge between $i$ and $j$). It is guaranteed that for any $S\subseteq\\{1,2,\ldots,n\\}$ ($|S|\ge 1$), $$ \prod_{i\in S}\left(\prod_{j\in\\{1,2,\ldots,n\\}\setminus S}(1-g_{i,j})\right)\not\equiv1\pmod{998\,244\,353}. $$

Output: Print a single integer in the only line of the output — the expected number of days, modulo $998\,244\,353$.

Formally, let $M = 998\,244\,353$. It can be shown that the exact answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Examples:
input:
2 1
1 2 1 10
output:
10

input:
3 3
1 2 1 2
1 3 1 2
2 3 1 2
output:
887328316

input:
1 0
output:
0

input:
5 8
1 2 1 11
1 3 2 11
1 4 3 11
1 5 4 11
2 4 5 11
2 5 6 11
3 4 7 11
4 5 8 11
output:
469993557

input:
21 22
1 2 3 4
2 3 4 5
3 4 5 6
5 6 7 8
6 7 8 9
7 8 9 10
8 9 2 3
9 10 3 4
10 11 4 5
11 12 5 6
12 13 6 7
13 14 7 8
14 15 8 9
15 16 9 10
16 17 2 3
17 18 3 4
18 19 4 5
19 20 5 6
20 21 6 7
1 10 100 1001
15 4 147 220
4 11 1 998244352
output:
299529765

Note: In the first test, the answer is equal to the expected number of days before the only edge in the graph first appears, and that is $\frac{1}{0.1}=10$.

In the second test, the answer is equal to $\frac{20}{9}$ before it is taken modulo $998\,244\,353$.

In the third test, the only vertex already has the message, so the answer is $0$.

Buggy Code:
MOD = 998244353

def modinv(a): # Function to compute modular inverse using Fermat's little theorem
    return pow(a, MOD - 2, MOD)

def solve():
    n, m = map(int, input().split())
    edges = [[] for _ in range(n + 1)]
    probs = [[0] * (n + 1) for _ in range(n + 1)]

    for _ in range(m):
        u, v, p, q = map(int, input().split())
        prob = p * modinv(q) % MOD
        edges[u].append((v, prob))
        edges[v].append((u, prob))
        probs[u][v] = probs[v][u] = prob

    dp = [[0] * (1 << n) for _ in range(n + 1)]
    for i in range(1, n + 1):
        dp[i][1 << (i - 1)] = 1

    for mask in range(1, 1 << n):
        for node in range(1, n + 1):
            if not ((mask >> (node - 1)) & 1): continue

            for next_node, prob in edges[node]:
                if mask & (1 << (next_node - 1)): continue

                new_mask = mask | (1 << (next_node - 1))
                wait_time = dp[node][mask] * modinv(probs[node][next_node]) % MOD
                dp[next_node][new_mask] += wait_time
                dp[next_node][new_mask] %= MOD

    print(dp[1][(1 << n) - 1])

solve()

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

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

Think step by step.

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

1. **Small-scale Test Input**: Since the problem constraints allow for up to 21 vertices, we can start with a smaller graph to ensure the test input is manageable and complete.

2. **Edge Cases**:
   - **Single Vertex**: The simplest case where there is only one vertex.
   - **Fully Connected Graph**: A graph where every vertex is connected to every other vertex.
   - **Linear Graph**: A graph where vertices are connected in a linear fashion.
   - **Graph with High Edge Probabilities**: A graph where the edges have very high probabilities of appearing, which can cause the modular inverse to be computed incorrectly.
   - **Graph with Low Edge Probabilities**: A graph where the edges have very low probabilities of appearing, which can also cause issues in the computation.

3. **Challenging Scenario**:
   - **Graph with Specific Edge Probabilities**: A graph where the edge probabilities are chosen to create a complex dependency structure, making the dynamic programming (DP) approach more likely to fail.

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

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

This test input represents a small graph with 4 vertices and 5 edges, where each edge has a probability of 1/3 of appearing each day. This setup is designed to be non-trivial and to test the correctness of the modular inverse and DP calculations.

### Explanation:
- **Vertices and Edges**: The graph is small enough to be manageable but complex enough to test the algorithm.
- **Edge Probabilities**: All edges have the same probability (1/3), which can create a scenario where the modular inverse and DP calculations need to be accurate.
- **Graph Structure**: The graph is connected, and each vertex is connected to at least two other vertices, creating a non-linear structure.

This test input should help expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Message Spread
time_limit_ms: 12000
memory_limit_mb: 1024
Description: Given is an undirected graph with $n$ vertices and $m$ edges. Each edge connects two vertices $(u, v)$ and has a probability of $\frac{p}{q}$ of appearing each day.

Initially, vertex $1$ has a message. At the end of the day, a vertex has a message if and only if itself or at least one of the vertices adjacent to it had the message the day before. Note that each day, each edge chooses its appearance independently.

Calculate the expected number of days before all the vertices have the message, modulo $998\,244\,353$.

Input: The first line contains two integers $n$ and $m$ ($1\leq n\leq 21$, $n-1\leq m\leq\frac{n(n-1)}{2}$).

Then $m$ lines follow, each containing four integers $u$, $v$, $p$, and $q$ ($1\leq u\neq v\leq n$, $1\leq p<q<998\,244\,353$, $\gcd(p,q)=1$) — there is an undirected edge between $u$ and $v$, and it has a probability of appearance of $\frac{p}{q}$ each day.

It is guaranteed that there are no self-loops or multiple-edges in the graph and that the graph is connected if all of the edges appear.

Additional constraint in the input: Let $g_{i,j}$ be the probability of appearance of the edge between $i$ and $j$ ($g_{i,j}=0$ if there is no edge between $i$ and $j$). It is guaranteed that for any $S\subseteq\\{1,2,\ldots,n\\}$ ($|S|\ge 1$), $$ \prod_{i\in S}\left(\prod_{j\in\\{1,2,\ldots,n\\}\setminus S}(1-g_{i,j})\right)\not\equiv1\pmod{998\,244\,353}. $$

Output: Print a single integer in the only line of the output — the expected number of days, modulo $998\,244\,353$.

Formally, let $M = 998\,244\,353$. It can be shown that the exact answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Examples:
input:
2 1
1 2 1 10
output:
10

input:
3 3
1 2 1 2
1 3 1 2
2 3 1 2
output:
887328316

input:
1 0
output:
0

input:
5 8
1 2 1 11
1 3 2 11
1 4 3 11
1 5 4 11
2 4 5 11
2 5 6 11
3 4 7 11
4 5 8 11
output:
469993557

input:
21 22
1 2 3 4
2 3 4 5
3 4 5 6
5 6 7 8
6 7 8 9
7 8 9 10
8 9 2 3
9 10 3 4
10 11 4 5
11 12 5 6
12 13 6 7
13 14 7 8
14 15 8 9
15 16 9 10
16 17 2 3
17 18 3 4
18 19 4 5
19 20 5 6
20 21 6 7
1 10 100 1001
15 4 147 220
4 11 1 998244352
output:
299529765

Note: In the first test, the answer is equal to the expected number of days before the only edge in the graph first appears, and that is $\frac{1}{0.1}=10$.

In the second test, the answer is equal to $\frac{20}{9}$ before it is taken modulo $998\,244\,353$.

In the third test, the only vertex already has the message, so the answer is $0$.

Buggy Code:
MOD = 998244353

def modinv(a, p):
    return pow(a, p - 2, p)

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

    idx = 0
    n = int(data[idx])
    idx += 1
    m = int(data[idx])
    idx += 1

    edges = []
    prob = [[0] * n for _ in range(n)]

    for _ in range(m):
        u = int(data[idx]) - 1
        idx += 1
        v = int(data[idx]) - 1
        idx += 1
        p = int(data[idx])
        idx += 1
        q = int(data[idx])
        idx += 1

        inv_q = modinv(q, MOD)
        prob[u][v] = prob[v][u] = p * inv_q % MOD
        edges.append((u, v, prob[u][v]))

    dp = [float('inf')] * (1 << n)
    dp[1] = 0  # Starting with vertex 1 having the message

    for mask in range(1 << n):
        if dp[mask] == float('inf'):
            continue

        for i in range(n):
            if not (mask & (1 << i)):
                continue

            for j in range(n):
                if mask & (1 << j):
                    continue

                if prob[i][j]:
                    new_mask = mask | (1 << j)
                    transition_prob = prob[i][j]

                    current_expected_days = dp[mask]
                    additional_days = 1

                    # Calculate the expected additional days needed
                    dp[new_mask] = min(dp[new_mask], current_expected_days + additional_days * modinv(transition_prob, MOD))

    full_mask = (1 << n) - 1
    print(dp[full_mask] % MOD)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Message Spread
time_limit_ms: 12000
memory_limit_mb: 1024
Description: Given is an undirected graph with $n$ vertices and $m$ edges. Each edge connects two vertices $(u, v)$ and has a probability of $\frac{p}{q}$ of appearing each day.

Initially, vertex $1$ has a message. At the end of the day, a vertex has a message if and only if itself or at least one of the vertices adjacent to it had the message the day before. Note that each day, each edge chooses its appearance independently.

Calculate the expected number of days before all the vertices have the message, modulo $998\,244\,353$.

Input: The first line contains two integers $n$ and $m$ ($1\leq n\leq 21$, $n-1\leq m\leq\frac{n(n-1)}{2}$).

Then $m$ lines follow, each containing four integers $u$, $v$, $p$, and $q$ ($1\leq u\neq v\leq n$, $1\leq p<q<998\,244\,353$, $\gcd(p,q)=1$) — there is an undirected edge between $u$ and $v$, and it has a probability of appearance of $\frac{p}{q}$ each day.

It is guaranteed that there are no self-loops or multiple-edges in the graph and that the graph is connected if all of the edges appear.

Additional constraint in the input: Let $g_{i,j}$ be the probability of appearance of the edge between $i$ and $j$ ($g_{i,j}=0$ if there is no edge between $i$ and $j$). It is guaranteed that for any $S\subseteq\\{1,2,\ldots,n\\}$ ($|S|\ge 1$), $$ \prod_{i\in S}\left(\prod_{j\in\\{1,2,\ldots,n\\}\setminus S}(1-g_{i,j})\right)\not\equiv1\pmod{998\,244\,353}. $$

Output: Print a single integer in the only line of the output — the expected number of days, modulo $998\,244\,353$.

Formally, let $M = 998\,244\,353$. It can be shown that the exact answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Examples:
input:
2 1
1 2 1 10
output:
10

input:
3 3
1 2 1 2
1 3 1 2
2 3 1 2
output:
887328316

input:
1 0
output:
0

input:
5 8
1 2 1 11
1 3 2 11
1 4 3 11
1 5 4 11
2 4 5 11
2 5 6 11
3 4 7 11
4 5 8 11
output:
469993557

input:
21 22
1 2 3 4
2 3 4 5
3 4 5 6
5 6 7 8
6 7 8 9
7 8 9 10
8 9 2 3
9 10 3 4
10 11 4 5
11 12 5 6
12 13 6 7
13 14 7 8
14 15 8 9
15 16 9 10
16 17 2 3
17 18 3 4
18 19 4 5
19 20 5 6
20 21 6 7
1 10 100 1001
15 4 147 220
4 11 1 998244352
output:
299529765

Note: In the first test, the answer is equal to the expected number of days before the only edge in the graph first appears, and that is $\frac{1}{0.1}=10$.

In the second test, the answer is equal to $\frac{20}{9}$ before it is taken modulo $998\,244\,353$.

In the third test, the only vertex already has the message, so the answer is $0$.

Buggy Code:
MOD = 998244353

def modinv(a, m):
    return pow(a, m - 2, m)

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

    index = 0
    def next_int():
        nonlocal index
        r = int(data[index])
        index += 1
        return r

    n = next_int()
    m = next_int()

    edges = []
    for _ in range(m):
        u = next_int() - 1
        v = next_int() - 1
        p = next_int()
        q = next_int()
        prob = p * modinv(q, MOD) % MOD
        edges.append((u, v, prob))

    dp = [float('inf')] * (1 << n)
    dp[1] = 0

    for mask in range(1, 1 << n):
        for u, v, prob in edges:
            if (mask >> u) & 1 == 1 and (mask >> v) & 1 == 0:
                new_mask = mask | (1 << v)
                not_appear_prob = (1 - prob + MOD) % MOD
                dp[new_mask] = min(dp[new_mask], dp[mask] + (dp[mask] * not_appear_prob + 1) * modinv(prob, MOD) % MOD)

    print(dp[(1 << n) - 1] % MOD)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

This test input is designed to expose issues in the given buggy code. Here's why:

1. **Fully Connected Graph**: The graph is fully connected with 3 vertices, which means every pair of vertices is connected by an edge. This can help in identifying issues with the way the code handles fully connected graphs.

2. **Equal Probabilities**: Each edge has a probability of $\frac{1}{2}$. This is a common and simple probability that can help in verifying the correctness of the code's probability calculations.

3. **Small Scale**: The graph is small (3 vertices and 3 edges), making it easy to manually verify the expected output and the correctness of the code's logic.

4. **Expected Output**: The expected number of days for all vertices to have the message in this case should be $\frac{20}{9}$, which, when taken modulo $998\,244\,353$, is $887\,328\,316$. This can help in verifying the correctness of the code's final output.

By using this input, you can test whether the code correctly handles fully connected graphs and probability calculations, and whether it correctly computes the expected number of days.