**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 areeaerae 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 a scenario where the code's fixed output of `10` is clearly incorrect. The buggy code simply prints `10` without considering the input, which is a clear indication of a flawed solution.

### Step-by-Step Reasoning:
1. **Identify the Bug**: The buggy code always prints `10`, regardless of the input. This means it fails for any input that does not result in an expected output of `10`.

2. **Choose a Simple but Effective Input**:
   - **Vertices and Edges**: Choose a small number of vertices and edges to keep the input manageable.
   - **Edge Probabilities**: Ensure the edge probabilities are such that the expected number of days is not `10`.

3. **Expected Output Calculation**:
   - For a simple graph with a small number of vertices and edges, calculate the expected number of days using the given formula or a known result.
   - Ensure the expected number of days is not `10`.

4. **Example Input**:
   - Consider a graph with 3 vertices and 2 edges.
   - Set the probabilities such that the expected number of days is different from `10`.

### Chosen Input:
- **Vertices**: 3
- **Edges**: 2
- **Edge 1**: Between vertices 1 and 2 with probability 1/2
- **Edge 2**: Between vertices 2 and 3 with probability 1/2

### Calculation:
- The expected number of days for a graph with 3 vertices and 2 edges, each with a probability of 1/2, can be calculated using the formula for the expected number of days for a connected graph with edge probabilities.
- The expected number of days for this configuration is not `10`.

### Expected Output:
- The expected number of days for this configuration is 4.

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

This input will expose the bug in the buggy code because the expected output is `4`, but the code will output `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:
#include<iostream>
#include<cstdio>
#define mod 998244353
using namespace std;
int n,m,p,q,ans;
int e[25];
int f[1<<21];
int pc[1<<21];
int P[25][25];
inline int add(int x,int y){
	if((x+=y)>=mod) x-=mod;
	return x;
}
inline int sub(int x,int y){
	if((x-=y)<0) x+=mod;
	return x;
}
inline int qp(int x,int y=mod-2){
	int res=1;
	while(y){
		if(y&1) res=1ll*res*x%mod;
		x=1ll*x*x%mod; y>>=1;
	}
	return res;
}
int solve(int s){
	if(s&1) return f[s]=0;
	if(~f[s]) return f[s];
	int t=0;
	for(int i=0;i<n;i+=1){
		if(s>>i&1){
			t|=e[i];
		}
	}
	t&=((1<<n)-1-s);
	int p=1;
	for(int i=0;i<n;i+=1){
		if(s>>i&1){
			for(int j=0;j<n;j+=1){
				if(t>>j&1){
					if(e[i]>>j&1){
						p=1ll*p*P[i][j]%mod;
					}
				}
			}
		}
	}
	f[s]=add(solve(t),qp(sub(1,p)));
	return f[s];
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i+=1){
		int x,y; scanf("%d%d",&x,&y);
		scanf("%d%d",&p,&q);
		p=sub(1,1ll*p*qp(q)%mod);
		x-=1; y-=1;
		e[x]|=(1<<y);
		e[y]|=(1<<x);
		P[x][y]=P[y][x]=p;
	}
	for(int i=1;i<(1<<n);i+=1) f[i]=-1;
	for(int i=1;i<(1<<n);i+=1){
		pc[i]=pc[i>>1]+(i&1);
		solve(i);
		if(pc[i]&1) ans=add(ans,f[i]);
		else ans=sub(ans,f[i]);
	}
	printf("%d\n",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:
```plaintext
3 3
1 2 1 10
1 3 1 10
2 3 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:
#include <iostream>
#define eafnfajnkankjanknnknknknk 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:
```plaintext
21 210
1 2 1 2
1 3 1 2
1 4 1 2
1 5 1 2
1 6 1 2
1 7 1 2
1 8 1 2
1 9 1 2
1 10 1 2
1 11 1 2
1 12 1 2
1 13 1 2
1 14 1 2
1 15 1 2
1 16 1 2
1 17 1 2
1 18 1 2
1 19 1 2
1 20 1 2
1 21 1 2
2 3 1 2
2 4 1 2
2 5 1 2
2 6 1 2
2 7 1 2
2 8 1 2
2 9 1 2
2 10 1 2
2 11 1 2
2 12 1 2
2 13 1 2
2 14 1 2
2 15 1 2
2 16 1 2
2 17 1 2
2 18 1 2
2 19 1 2
2 20 1 2
2 21 1 2
3 4 1 2
3 5 1 2
3 6 1 2
3 7 1 2
3 8 1 2
3 9 1 2
3 10 1 2
3 11 1 2
3 12 1 2
3 13 1 2
3 14 1 2
3 15 1 2
3 16 1 2
3 17 1 2
3 18 1 2
3 19 1 2
3 20 1 2
3 21 1 2
4 5 1 2
4 6 1 2
4 7 1 2
4 8 1 2
4 9 1 2
4 10 1 2
4 11 1 2
4 12 1 2
4 13 1 2
4 14 1 2
4 15 1 2
4 16 1 2
4 17 1 2
4 18 1 2
4 19 1 2
4 20 1 2
4 21 1 2
5 6 1 2
5 7 1 2
5 8 1 2
5 9 1 2
5 10 1 2
5 11 1 2
5 12 1 2
5 13 1 2
5 14 1 2
5 15 1 2
5 16 1 2
5 17 1 2
5 18 1 2
5 19 1 2
5 20 1 2
5 21 1 2
6 7 1 2
6 8 1 2
6 9 1 2
6 10 1 2
6 11 1 2
6 12 1 2
6 13 1 2
6 14 1 2
6 15 1 2
6 16 1 2
6 17 1 2
6 18 1 2
6 19 1 2
6 20 1 2
6 21 1 2
7 8 1 2
7 9 1 2
7 10 1 2
7 11 1 2
7 12 1 2
7 13 1 2
7 14 1 2
7 15 1 2
7 16 1 2
7 17 1 2
7 18 1 2
7 19 1 2
7 20 1 2
7 21 1 2
8 9 1 2
8 10 1 2
8 11 1 2
8 12 1 2
8 13 1 2
8 14 1 2
8 15 1 2
8 16 1 2
8 17 1 2
8 18 1 2
8 19 1 2
8 20 1 2
8 21 1 2
9 10 1 2
9 11 1 2
9 12 1 2
9 13 1 2
9 14 1 2
9 15 1 2
9 16 1 2
9 17 1 2
9 18 1 2
9 19 1 2
9 20 1 2
9 21 1 2
10 11 1 2
10 12 1 2
10 13 1 2
10 14 1 2
10 15 1 2
10 16 1 2
10 17 1 2
10 18 1 2
10 19 1 2
10 20 1 2
10 21 1 2
11 12 1 2
11 13 1 2
11 14 1 2
11 15 1 2
11 16 1 2
11 17 1 2
11 18 1 2
11 19 1 2
11 20 1 2
11 21 1 2
12 13 1 2
12 14 1 2
12 15 1 2
12 16 1 2
12 17 1 2
12 18 1 2
12 19 1 2
12 20 1 2
12 21 1 2
13 14 1 2
13 15 1 2
13 16 1 2
13 17 1 2
13 18 1 2
13 19 1 2
13 20 1 2
13 21 1 2
14 15 1 2
14 16 1 2
14 17 1 2
14 18 1 2
14 19 1 2
14 20 1 2
14 21 1 2
15 16 1 2
15 17 1 2
15 18 1 2
15 19 1 2
15 20 1 2
15 21 1 2
16 17 1 2
16 18 1 2
16 19 1 2
16 20 1 2
16 21 1 2
17 18 1 2
17 19 1 2
17 20 1 2
17 21 1 2
18 19 1 2
18 20 1 2
18 21 1 2
19 20 1 2
19 21 1 2
20 21 1 2
```


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

public class Solution {
    static final int MOD = 998244353;

    static class Edge {
        int u, v;
        long p, q;
        Edge(int u, int v, long p, long q) {
            this.u = u;
            this.v = v;
            this.p = p;
            this.q = q;
        }
    }

    static long modPow(long base, long exp) {
        if (exp == 0) return 1;
        long half = modPow(base, exp / 2);
        long result = (half * half) % MOD;
        if (exp % 2 == 1) {
            result = (result * base) % MOD;
        }
        return result;
    }

    static long modInverse(long x) {
        return modPow(x, MOD - 2);
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());

        if (n == 1) {
            System.out.println(0);
            return;
        }

        List<Edge> edges = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            st = new StringTokenizer(br.readLine());
            int u = Integer.parseInt(st.nextToken()) - 1;
            int v = Integer.parseInt(st.nextToken()) - 1;
            long p = Long.parseLong(st.nextToken());
            long q = Long.parseLong(st.nextToken());
            edges.add(new Edge(u, v, p, q));
        }

        // dp[mask] = expected number of days to reach all vertices from state mask
        long[] dp = new long[1 << n];
        Arrays.fill(dp, -1);
        dp[(1 << n) - 1] = 0;  // all vertices have message

        // Process states from all vertices to just vertex 1
        for (int mask = (1 << n) - 2; mask >= 1; mask--) {
            if ((mask & 1) == 0) continue;  // must include vertex 1

            // Calculate probability of transitioning to each next state
            long num = 0;
            long den = 1;

            // For each edge
            for (Edge e : edges) {
                if ((mask & (1 << e.u)) != 0 && (mask & (1 << e.v)) == 0) {
                    // Edge can spread message
                    long prob = (e.p * modInverse(e.q)) % MOD;
                    long nextMask = mask | (1 << e.v);

                    // Add contribution of this transition
                    long contribution = (prob * dp[(int)nextMask]) % MOD;
                    num = (num + contribution) % MOD;
                    den = (den * (1 - prob + MOD)) % MOD;
                }
            }

            if (den == 1) continue;  // No possible transitions

            // Calculate expected value
            dp[mask] = (num * modInverse(1 - den + MOD)) % MOD;
            dp[mask] = (dp[mask] + modInverse(1 - den + MOD)) % MOD;
        }

        System.out.println(dp[1]);  // State with just vertex 1
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios where the code might fail. Here are the steps to create such a test input:

1. **Edge Cases:**
   - A graph with a single vertex.
   - A graph with a single edge.
   - A graph where the message can spread quickly.
   - A graph where the message spreads very slowly.

2. **Scenarios:**
   - A graph where the edges have very low probabilities of appearing.
   - A graph where the edges have very high probabilities of appearing.
   - A graph with a specific structure that can lead to complex transitions.

Given the constraints and the need for a small-scale test input, let's focus on a graph with a specific structure that can lead to complex transitions and low probabilities of edges appearing.

### Test Input
```plaintext
3 3
1 2 1 1000
1 3 1 1000
2 3 1 1000
```

### Explanation
- **Graph Structure:**
  - Vertices: 1, 2, 3
  - Edges: (1, 2), (1, 3), (2, 3)
- **Edge Probabilities:**
  - Each edge has a very low probability of appearing (1/1000).

### Why This Test Input is Challenging
- **Low Edge Probabilities:** The low probabilities of edges appearing (1/1000) will make it very unlikely for the message to spread quickly. This can lead to a large number of days required for the message to reach all vertices, which can expose issues in the calculation of expected values.
- **Complete and Valid:** The graph is connected, and the input adheres to the constraints provided.
- **Complex Transitions:** The structure of the graph with multiple edges can lead to complex state transitions, which can help in identifying bugs in the state transition logic.

This test input should help in exposing any issues in the code's handling of low probability edges and state transitions.


**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 afafdaaffaafsdafeeeeeeeee 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:
```plaintext
3 3
1 2 1 3
1 3 1 3
2 3 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 <iostream>
#define uuauysyysya 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 create a valid and small-scale input that forces the code to produce an incorrect result. The buggy code simply outputs `10` without considering any input, which is clearly incorrect for most valid inputs.

Let's consider the following steps:

1. **Choose a small graph**: Since the input constraints allow up to 21 vertices and the graph must be connected, we can choose a small number of vertices to ensure the input is manageable and easy to validate.

2. **Ensure the graph is connected**: We need to ensure that the graph is connected if all edges appear, as per the problem constraints.

3. **Set edge probabilities**: We need to set edge probabilities such that the expected number of days to spread the message is not trivially 10.

4. **Check the additional constraint**: The additional constraint must be satisfied, ensuring that the product of probabilities does not equal 1 modulo \(998\,244\,353\).

Let's construct a test input with 3 vertices and 3 edges, which is a simple connected graph:

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

### Explanation:
- **Vertices**: 3
- **Edges**: 3
- **Edge Probabilities**:
  - Edge between 1 and 2 has a probability of \(\frac{1}{2}\)
  - Edge between 1 and 3 has a probability of \(\frac{1}{2}\)
  - Edge between 2 and 3 has a probability of \(\frac{1}{2}\)

This input ensures that the graph is connected and the probabilities are non-trivial. The expected number of days to spread the message to all vertices is not 10, which will expose the bug in the buggy code.

### Output:
```plaintext
3 3
1 2 1 2
1 3 1 2
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: 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;

// Fast exponentiation
inline int power_pow(int a, int b, int MOD) {
    ll res = 1;
    ll base = a;
    while(b > 0){
        if(b & 1){
            res = res * base % MOD;
        }
        base = base * base % MOD;
        b >>=1;
    }
    return res;
}

// Structure to hold edges
struct Edge {
    int u;
    ll one_minus_g; // (1 - g_e) mod MOD
};

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    // Nodes are 0-based
    vector<vector<Edge>> node_edges(n, vector<Edge>());
    for(int i=0;i<m;i++){
        int u, v, p, q;
        cin >> u >> v >> p >> q;
        u -=1; v -=1;
        ll inv_q = power_pow(q, MOD-2, MOD);
        ll g_e = (1LL * p * inv_q) % MOD;
        ll one_minus_g = (1 - g_e + MOD) % MOD;
        // Since the graph is undirected, add to both nodes
        node_edges[u].push_back(Edge{v, one_minus_g});
        node_edges[v].push_back(Edge{u, one_minus_g});
    }
    int size = 1 << n;
    vector<int> E(size, 0);
    // Iterate from full set to empty set
    for(int mask = (1<<n)-1; mask >=0; mask--){
        if(mask == (1<<n)-1){
            E[mask] = 0;
            continue;
        }
        // List of nodes not in mask
        vector<int> nodes_not_S;
        vector<ll> p_v_list;
        for(int v=0; v<n; v++){
            if( (mask & (1<<v)) ==0 ){
                nodes_not_S.push_back(v);
                // Compute p_v
                ll prod =1;
                for(auto &e: node_edges[v]){
                    if( (mask & (1<<e.u)) ){
                        prod = (prod * e.one_minus_g) % MOD;
                    }
                }
                ll p_v = (1 - prod + MOD) % MOD;
                p_v_list.push_back(p_v);
            }
        }
        // Compute Q = 1 - product(1 - p_v)
        ll Q_product =1;
        for(auto &p_v: p_v_list){
            ll term = (1 - p_v + MOD) % MOD;
            Q_product = (Q_product * term) % MOD;
        }
        ll Q = (1 - Q_product + MOD) % MOD;
        // Compute inverse of Q
        ll inv_Q = power_pow(Q, MOD-2, MOD);
        // Compute sum_p_E = sum(p_v * E[S union {v}])
        ll sum_p_E =0;
        for(int i=0; i< nodes_not_S.size(); i++){
            int v = nodes_not_S[i];
            ll p_v = p_v_list[i];
            int new_mask = mask | (1<<v);
            sum_p_E = (sum_p_E + (p_v * 1LL * E[new_mask]) % MOD ) % MOD;
        }
        // E[mask] = (1 + sum_p_E) * inv_Q mod MOD
        ll numerator = (1 + sum_p_E) % MOD;
        ll E_val = (numerator * inv_Q) % MOD;
        E[mask] = E_val;
    }
    // Initial mask has only vertex 0
    int initial_mask = 1<<0;
    cout << E[initial_mask] << "\n";
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Small Scale**: Since the problem constraints allow up to 21 vertices, we should use a small number of vertices to ensure the input is manageable and can be verified easily.
2. **Connected Graph**: The graph must be connected if all edges appear, and it should have no self-loops or multiple edges.
3. **Probability Values**: Use probabilities that are non-trivial and can cause issues in the computation, especially when dealing with modular arithmetic.
4. **Edge Cases**: Consider scenarios where the probabilities are very low or very high to test the limits of the algorithm.

Given these considerations, let's create a test input with 4 vertices and 4 edges. This small-scale input will allow us to verify the correctness of the algorithm more easily.

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

### Explanation:
- **Vertices and Edges**: We have 4 vertices (1, 2, 3, 4) and 4 edges.
- **Probabilities**: Each edge has a probability of 1/3 of appearing each day.
- **Graph Connectivity**: The graph is a complete graph (K4), which is connected if all edges appear.
- **Challenges**:
  - The probabilities are not trivial (1/3), which can cause issues in the modular arithmetic.
  - The complete graph structure ensures that the message can spread quickly, but the probabilities need to be handled correctly to compute the expected number of days.

This test input should be challenging enough to expose any issues in the algorithm, especially in the handling of modular arithmetic and the computation of 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:
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int MOD = 998244353;

// Fast exponentiation
ll power_pow(ll a, ll b) {
    ll res = 1;
    a %= MOD;
    while(b > 0){
        if(b & 1) res = res * a % MOD;
        a = a * a % MOD;
        b >>=1;
    }
    return res;
}

// Modular inverse
ll inv(ll a){
    return power_pow(a, MOD-2);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    // Adjacency list with probabilities
    vector<vector<pair<int, pair<ll, ll>>>> adj(n+1, vector<pair<int, pair<ll, ll>>>());
    for(int i=0;i<m;i++){
        int u, v;
        ll p, q;
        cin >> u >> v >> p >> q;
        adj[u].emplace_back(v, make_pair(p, q));
        adj[v].emplace_back(u, make_pair(p, q));
    }
    int size = 1<<n;
    // Initialize E
    vector<ll> E(size, 0); // E[S}
    // E[full] = 0 already
    // Precompute for each vertex, the list of neighbors and their (q - p)/q
    // To compute p_v = 1 - product_{u in S connected to v} (1 - g_{u,v})
    // where g_{u,v} = p/q
    // 1 - g_{u,v} = (q - p)/q
    // Precompute for each vertex, the list of connected edges with (q - p)/q
    vector<vector<pair<int, ll>>> vertex_probs(n+1, vector<pair<int, ll>>());
    for(int v=1; v<=n; v++){
        for(auto &[u, pq]: adj[v]){
            ll p = pq.first;
            ll q = pq.second;
            ll term = (q - p) * inv(q) % MOD;
            vertex_probs[v].emplace_back(u, term);
        }
    }
    // Precompute for each vertex v, all possible products for subsets of neighbors
    // To speed up p_v computation
    // Since n <=21, and degree <=20, it's manageable
    // For each v, map from mask to product
    // The order of neighbors matters, assign an index to each neighbor
    vector<vector<ll>> pre_product(n+1, vector<ll>());
    vector<vector<ll>> pre_inv_term(n+1, vector<ll>());
    vector<vector<ll>> pre_pv(n+1, vector<ll>());
    // To handle up to degree 20, use dynamic programming
    // But n=21, degrees up to 20 are manageable
    // Instead, we compute on the fly
    // Iterate S from high to low
    for(int S = (1<<n)-1; S >=0; S--){
        // Compute E[S]
        // If S is full, E[S] = 0
        if(S == (1<<n)-1){
            E[S] = 0;
            continue;
        }
        // For each v not in S, compute p_v
        // p_v = 1 - product_{u in S connected to v} (1 - g_{u,v})
        // which is 1 - product of terms from vertex_probs[v]
        vector<ll> p_v(n+1, 0);
        for(int v=1; v<=n; v++){
            if( (S & (1<<(v-1))) ) continue;
            ll prod = 1;
            for(auto &[u, term]: vertex_probs[v]){
                if(S & (1<<(u-1))){
                    prod = prod * term % MOD;
                }
            }
            p_v[v] = (1 - prod + MOD) % MOD;
        }
        // Compute P(S stays) = product_{v not in S} (1 - p_v)
        ll P_stay = 1;
        for(int v=1; v<=n; v++){
            if( (S & (1<<(v-1))) ) continue;
            P_stay = P_stay * ((1 - p_v[v] + MOD) % MOD) % MOD;
        }
        // Compute sum = sum_{U superset S} [ product_{v in U \ S} p_v * product_{v not in U} (1 - p_v) ] * E[U}
        // This is equivalent to sum_{T subset not S} [ product_{v in T} p_v * product_{v not in T} (1 - p_v) ] * E[S | T}
        // Which is sum_{U superset S} [ product_{v in U \ S} p_v * product_{v not in U} (1 - p_v) ] * E[U}
        // To compute this efficiently, iterate over all subsets T of not S
        // and accumulate E[S | T} ] * product terms
        // But n=21 is too slow, need optimizations
        // Instead, iterate over U superset S
        ll sum = 0;
        // Get the list of vertices not in S
        int notS = (~S) & ((1<<n)-1);
        // Iterate over all supersets U of S
        // U ranges from S to (1<<n)-1, where U has all bits of S
        // To iterate efficiently, iterate T subset notS and set U = S | T
        // Enumerate T from 0 to notS
        // To optimize, use fast subset enumeration
        for(int T = notS; T >=1; T = (T-1) & notS){
            // U = S | T
            int U = S | T;
            // Compute product_{v in T} p_v * product_{v not in U} (1 - p_v)
            ll coef = 1;
            // product_{v in T} p_v
            // and product_{v not in U} (1 - p_v)
            // Note that U = S | T, so not U = notS & ~T
            for(int v=1; v<=n; v++){
                if( (T & (1<<(v-1))) ){
                    coef = coef * p_v[v] % MOD;
                }
            }
            ll notU = (~U) & ((1<<n)-1);
            for(int v=1; v<=n; v++){
                if( (notU & (1<<(v-1))) ){
                    coef = coef * ((1 - p_v[v] + MOD) % MOD) % MOD;
                }
            }
            sum = (sum + coef * E[U] ) % MOD;
        }
        // Now, E[S} } = (1 + sum) / (1 - P_stay)
        // Compute denominator inverse
        ll denom = (1 - P_stay + MOD) % MOD;
        ll denom_inv = inv(denom);
        E[S] = (1 + sum) * denom_inv % MOD;
    }
    // The initial state is S = 1 (only vertex 1 has the message)
    int initial = 1;
    cout << E[initial] << "\n";
}

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

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

Think step by step.

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

1. **Small Scale**: The input should be small enough to fit within the constraints but complex enough to challenge the algorithm.
2. **Edge Cases**: Consider cases where the graph is connected but has specific edge probabilities that might cause issues.
3. **Critical Path**: Focus on graphs where the message spread is not straightforward, such as graphs with low probability edges or specific connectivity patterns.

Given these considerations, let's create a small-scale test input:

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

### Explanation
- **Graph Structure**: The graph is a simple path from vertex 1 to vertex 4, with an additional edge between vertex 1 and vertex 4.
- **Edge Probabilities**: Each edge has a probability of 1/3 of appearing each day. This low probability increases the complexity of the message spread calculation.
- **Initial State**: Only vertex 1 has the message initially.

### Why This Input is Challenging
- **Low Probability Edges**: The low probability (1/3) of each edge appearing each day makes the expected number of days to spread the message to all vertices more complex to calculate. This can expose issues in the calculation of the expected number of days.
- **Multiple Paths**: The additional edge between vertex 1 and vertex 4 creates multiple paths for the message to spread, which can complicate the computation of the probabilities and the expected number of days.

This input should help in exposing any bugs in the algorithm, especially in the computation of the expected number of days and the handling of low probability edges.


**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>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll p = 998244353;
ll fp(ll a, ll b) {
    ll s = 1;
    for(; b; b >>=1, a = a * a % p) {
        if (b&1) {
            s = s * a % p;
        }
    }
    return s;
}
int n, m;
ll a[30][30];
ll pr[1 << 21];
ll ex[1 << 21];
int occur[1 << 21][22];
ll any[1 << 21];
ll any_inv[1 << 21];
ll g[1 << 21];
int main() {
    scanf("%d%d", &n, &m);
    int x, y;
    for (int i = 1; i <= m; i++) {
        int u, v, x, y;
        scanf("%d%d%d%d", &u, &v, &x, &y);
        ll prob = x * fp(y, p-2) % p;
        a[u][v] = a[v][u] = prob;
    }
    for (int i = 1; i < 1 << n; i++) {
        any[i] = 1;
        for (int j = 1; j <= n; j++) {
            if (!((i >> (j-1))&1)) {
                occur[i][j] = 1;
                for (int k = 1; k <= n; k++) {
                    if ((i >> (k-1))&1) {
                        occur[i][j] = occur[i][j] * (1 - a[j][k]) % p;
                    }
                }
                occur[i][j] = 1 - occur[i][j];
                any[i] = any[i] * (1 - occur[i][j]) % p;
            }
        }
        any[i] = 1 - any[i];
        any_inv[i] = fp(any[i], p-2);
        g[i] = fp(1 - any[i], p-2);
    }

    pr[1] = 1;
    ex[1] = 0;
    for (int i = 3; i < 1 << n; i += 2) {
        for (int j = (i - 2) & i;; j = (j - 2) & i) {
            ll temp_in = 1;
            ll temp_out = 1;
            for (int k = 2; k <= n; k++) {
                if (!((j >> (k-1))&1)) {
                    if ((i >> (k-1))&1) {
                        temp_in = temp_in * occur[j][k] % p;
                    } else {
                        temp_out = temp_out * (1 - occur[j][k]) % p;
                    }
                }
            }
            ll prob = temp_in * temp_out % p * any_inv[j] % p;
            ll prob2 = prob * pr[j] % p;
            pr[i] = (pr[i] + prob2);
            ex[i] = (ex[i] + ex[j] * prob + prob2 * any_inv[j]) % p;
            if (j <= 2) {
                break;
            }
        }
        pr[i] %= p;
    }
    ll ans = 0;
    for (int i = 1; i < (1 << n) - 1; i += 2) {
        ans = (ans + pr[i] * any_inv[i]) % p;
    }
    ans = (ans % p + p) % p;
    printf("%lld\n", 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:
```plaintext
3 3
1 2 1 3
1 3 1 3
2 3 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 <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
// typedef tree<int, null_type, std::less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

#include <bits/stdc++.h>

#define f first
#define s second
#define pb push_back

typedef long long int ll;
// typedef __int128 lll;
typedef unsigned long long int ull;
using namespace std;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

template<typename T> int die(T x) { cout << x << endl; return 0; }

#define mod_fft 998244353
#define mod_nfft 1000000007
#define INF 100000000000000
#define LNF 1e15
#define LOL 12345678912345719ll

struct LL {

    static const ll m = mod_fft;
    long long int val;

    LL(ll v) {val=reduce(v);};
    LL() {val=0;};
    ~LL(){};
    LL(const LL& l) {val=l.val;};
    LL& operator=(int l) {val=l; return *this;}
    LL& operator=(ll l) {val=l; return *this;}
    LL& operator=(LL l) {val=l.val; return *this;}

    static long long int reduce(ll x, ll md = m) {
        x %= md;
        while (x >= md) x-=md;
        while (x < 0) x+=md;
        return x;
    }

    bool operator<(const LL& b) { return val<b.val; }
    bool operator<=(const LL& b) { return val<=b.val; }
    bool operator==(const LL& b) { return val==b.val; }
    bool operator>=(const LL& b) { return val>=b.val; }
    bool operator>(const LL& b) { return val>b.val; }

    LL operator+(const LL& b) { return LL(val+b.val); }
    LL operator+(const ll& b) { return (*this+LL(b)); }
    LL& operator+=(const LL& b) { return (*this=*this+b); }
    LL& operator+=(const ll& b) { return (*this=*this+b); }

    LL operator-(const LL& b) { return LL(val-b.val); }
    LL operator-(const ll& b) { return (*this-LL(b)); }
    LL& operator-=(const LL& b) { return (*this=*this-b); }
    LL& operator-=(const ll& b) { return (*this=*this-b); }

    LL operator*(const LL& b) { return LL(val*b.val); }
    LL operator*(const ll& b) { return (*this*LL(b)); }
    LL& operator*=(const LL& b) { return (*this=*this*b); }
    LL& operator*=(const ll& b) { return (*this=*this*b); }

    static LL exp(const LL& x, const ll& y){
        ll z = y;
        z = reduce(z,m-1);
        LL ret = 1;
        LL w = x;
        while (z) {
            if (z&1) ret *= w;
            z >>= 1; w *= w;
        }
        return ret;
    }
    LL& operator^=(ll y) { return (*this=LL(val^y)); }

    LL operator/(const LL& b) { return ((*this)*exp(b,-1)); }
    LL operator/(const ll& b) { return (*this/LL(b)); }
    LL operator/=(const LL& b) { return ((*this)*=exp(b,-1)); }
    LL& operator/=(const ll& b) { return (*this=*this/LL(b)); }

}; ostream& operator<<(ostream& os, const LL& obj) { return os << obj.val; }

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int T = 1; // cin >> T;
    while (T--) {
        int N, M; cin >> N >> M;
        vector<vector<LL>> W(N, vector<LL>(N, 1)), w(N, vector<LL>(N, 1));
        for (int m = 0; m < M; m++) {
            int u, v, p, q; cin >> u >> v >> p >> q; --u; --v;
            W[u][v] = W[v][u] = LL(1) - LL(p)/q;
            w[u][v] = w[v][u] = LL(1) / W[u][v];
        }
        vector<vector<LL>> row(N, vector<LL>(1<<N, 1));
        for (int a = 0; a < N; a++) {
            for (int z = 0; z < (1<<N); z++) {
                for (int b = 0; b < N; b++) if ((1 << b) & z) {
                    row[a][z] *= W[a][b];
                }
            }
        }
        vector<vector<LL>> irow(N, vector<LL>(1<<N, 1));
        for (int a = 0; a < N; a++) {
            for (int z = 0; z < (1<<N); z++) {
                irow[a][z] = (LL(1) - row[a][z]) / row[a][z];
            }
        }

        vector<LL> P(1<<N, 0), E(1<<N, 0), Z(1<<N);
        P[1] = 1;
        E[1] = 0;
        for (int y = 1; y < (1<<N) - 1; y += 2) {
            E[y] /= P[y];
            if (y == (1<<N) - 1) break;
            Z[y] = 1;
            for (int a = 1; a < N; a++) if ((1 << a) & y) {} else {
                Z[y] *= row[a][y];
            }
            LL IZ = LL(1) / (LL(1) - Z[y]);
            Z[y] *= IZ;
            for (int a = 1; a < N; a++) if ((1 << a) & y) {} else {
                for (int x = ((1<<a) | y); x < y+(1<<a<<1); x = (x + 2) | y) {
                    Z[x] = Z[x^(1<<a)] * irow[a][y];
                }
            }
            for (int x = (y + 2) | y; x < (1<<N); x = (x + 2) | y) {
                P[x] += P[y] * Z[x];
            }
            for (int x = (y + 2) | y; x < (1<<N); x = (x + 2) | y) {
                E[x] += (E[y] + IZ) * P[y] * Z[x];
            }
        }

        cout << E[(1<<N) - 1] << endl;
    }
}

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

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

Think step by step.

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

1. **Small-Scale Test Input**: Since the input needs to be small, we should focus on a graph with a small number of vertices and edges.
2. **Edge Cases**: Consider graphs that have specific properties such as:
   - A single vertex with no edges.
   - A graph with a single edge.
   - A complete graph (all vertices are connected to each other).
   - A graph with a specific structure that might cause the algorithm to fail, such as a star graph or a linear graph.
3. **Probability Values**: Use probabilities that can cause the algorithm to fail, such as very low or very high probabilities.

Given these considerations, let's create a test input for a small graph that can expose potential bugs in the code.

### Test Input: Linear Graph with Low Probability Edges

A linear graph with low probability edges can be a good candidate because it will test the algorithm's ability to handle low probabilities and the propagation of the message through a chain of vertices.

```plaintext
5 4
1 2 1 1000
2 3 1 1000
3 4 1 1000
4 5 1 1000
```

### Explanation:
- **Vertices and Edges**: The graph has 5 vertices and 4 edges, forming a linear structure.
- **Edge Probabilities**: Each edge has a very low probability of appearance (1/1000). This will test the algorithm's ability to handle very low probabilities and the expected number of days for the message to spread to all vertices.

This test input is small and valid, and it should help expose any issues in the buggy code related to handling low probabilities and the propagation of the message through a linear structure.


**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:
//A tree without skin will surely die.
//A man without face will be alive.
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,j,k) for(int i=j;i<=k;++i)
#define per(i,j,k) for(int i=j;i>=k;--i)
#define add(x,y) (x=((x+y>=mod)?(x+y-mod):(x+y)))
int const mod=998244353,iv2=(mod+1)/2;
int n,m,S,w[21][21],sm[1<<21],B[1<<21],P[1<<21],Q[1<<21],R[1<<21],f[1<<21],g[1<<21],h[1<<21];
inline int qpow(int a,int b=mod-2){int res=1;while (b){if (b&1) res*=a,res%=mod;a*=a,a%=mod,b>>=1;}return res;}
int a[21][1<<21],b[21][1<<21];
inline void FWT(int *f,int S){
	rep(i,0,20) rep(j,0,S) if (j>>i&1) add(f[j],f[j^(1<<i)]);
}
inline void IFWT(int *f,int n){
	rep(i,0,20) rep(j,0,S) if (j>>i&1) add(f[j],mod-f[j^(1<<i)]);
}
inline void zjzj(int p,int *B){
	rep(i,0,p) rep(j,0,S) b[i][j]=0;
	rep(i,0,S) b[__builtin_popcount(i)][i]=B[i];
	rep(i,0,p) FWT(b[i],S);
	rep(j,0,p) rep(k,0,S) add(h[k],a[j][k]*b[p-j][k]%mod);
	IFWT(h,S);
}
inline void solve(){
	cin>>n>>m,S=(1<<n)-1;
	rep(i,0,S) P[i]=R[i]=1;
	while (m--){
		int u,v,p,q;cin>>u>>v>>p>>q,--u,--v,p*=qpow(q),p%=mod;
		w[u][v]=w[v][u]=p;rep(i,0,S)
			if ((i>>u&1) && (i>>v&1)) P[i]*=(1+mod-p),P[i]%=mod;
			else if ((i>>u&1) || (i>>v&1)) R[i]*=(1+mod-p),R[i]%=mod;
	}
	rep(i,0,S) Q[i]=qpow(P[i],mod-2),g[i]=qpow((1+mod-R[i])%mod,mod-2);
	rep(i,0,S) a[__builtin_popcount(i)][i]=P[i];
	rep(i,0,20) FWT(a[i],S);
	rep(p,1,n){
		for (int i=1;i<S;i+=2) sm[i]=(f[i]+R[i]*g[i]%mod*f[i]%mod)%mod;
		rep(j,0,n-1) rep(i,0,S)
			if (i>>j&1) add(sm[i],sm[i^(1<<j)]);
		rep(i,0,S) B[i]=g[i]*f[i]%mod*Q[S-i]%mod,h[i]=0;
		zjzj(p,B);
		for (int i=1;i<S;i+=2){
			if (__builtin_popcount(i)!=p) continue;
			h[i]*=P[S],h[i]%=mod,h[i]*=Q[i],h[i]%=mod;
			++h[i],h[i]%=mod;
			f[i]=(h[i]+mod-sm[i])%mod;
		}
	}
	int ans=0;
	for (int i=1;i<S;i+=2) ans+=f[i]*g[i]%mod,ans%=mod;
	cout<<ans<<'\n';
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int t=1;
	// 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
3 3
1 2 1 3
1 3 1 3
2 3 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:
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <cstdio>
#include <map>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <queue>
#include <stack>
#include <vector>
#include <random>
#include <cstring>
#include <ctime>
#include <cmath>
#include <assert.h>
#include <unordered_map>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define LL long long
#define pp pair<LL,int>
#define mp make_pair
#define ull unsigned long long
namespace IO{
	const int sz=1<<22;
	char a[sz+5],b[sz+5],*p1=a,*p2=a,*t=b,p[105];
	inline char gc(){
	//	return p1==p2?(p2=(p1=a)+fread(a,1,sz,stdin),p1==p2?EOF:*p1++):*p1++;
		return getchar();
	}
	template<class T> void gi(T& x){
		x=0; int f=1;char c=gc();
		if(c=='-')f=-1;
		for(;c<'0'||c>'9';c=gc())if(c=='-')f=-1;
		for(;c>='0'&&c<='9';c=gc())
			x=x*10+(c-'0');
		x=x*f;
	}
	inline void flush(){fwrite(b,1,t-b,stdout),t=b; }
	inline void pc(char x){*t++=x; if(t-b==sz) flush(); }
	template<class T> void pi(T x,char c='\n'){
		if(x<0)pc('-'),x=-x;
		if(x==0) pc('0'); int t=0;
		for(;x;x/=10) p[++t]=x%10+'0';
		for(;t;--t) pc(p[t]); pc(c);
	}
	struct F{~F(){flush();}}f;
}
using IO::gi;
using IO::pi;
using IO::pc;
const int mod=998244353;
inline int add(int x,int y){
	return x+y>=mod?x+y-mod:x+y;
}
inline int dec(int x,int y){
	return x-y<0?x-y+mod:x-y;
}
inline int mul(int x,int y){
	return 1ll*x*y%mod;
}
inline int qkpow(int a,int b){
	if(b<0)return 0;
	int ans=1,base=a%mod;
	while(b){
		if(b&1)ans=1ll*ans*base%mod;
		base=1ll*base*base%mod;
		b>>=1;
	}
	return ans;
}
int fac[3000005],inv[3000005],Invn[3000005];
inline int binom(int n,int m){
	if(n<m||m<0)return 0;
	return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
void init_C(int n){
	fac[0]=1;
	for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
	inv[0]=1;
	inv[n]=qkpow(fac[n],mod-2);
	for(int i=n-1;i>=1;i--)inv[i]=1ll*inv[i+1]*(i+1)%mod;
	Invn[0]=1;
	for(int i=1;i<=n;i++)Invn[i]=1ll*inv[i]*fac[i-1]%mod;
}
int t,n,m,wp[23][23],w[23],f[23],g[(1<<21)+5],len;
int dp[(1<<21)+5],tmp[(1<<21)+5],f2[(1<<21)+5];
inline void solve(){
	gi(n),gi(m);
	for(int i=1;i<=m;i++){
		int u,v,p,q;
		gi(u),gi(v),gi(p),gi(q);
		u--,v--;
		wp[u][v]=wp[v][u]=1ll*p*qkpow(q,mod-2)%mod;
	}
	dp[(1<<n)-1]=0;
	__int128 Sum=0;
	for(int i=(1<<n)-2;i>=1;i--){
		if(!(i&1))continue;
		int val=1;
		for(int j=0;j<n;j++){
			if(!((1<<j)&i)){
				f[j]=1;
				for(int k=0;k<n;k++){
					if((1<<k)&i){
						if(wp[j][k])f[j]=1ll*f[j]*dec(1,wp[j][k])%mod;
					}
				}
				val=1ll*val*f[j]%mod;
				g[(1<<j)]=1ll*dec(1,f[j])*qkpow(f[j],mod-2)%mod;
			}
		}
		int xs=qkpow(dec(1,val),mod-2);
		int S=((1<<n)-1)^i;
		len=0;
		for(int j=S;j;j=(j-1)&S)tmp[++len]=j;
		f2[0]=val;
		Sum=0;
		for(int j=len;j;j--){
			int x=tmp[j];
			f2[x]=1ll*f2[x^(x&-x)]*g[x&-x]%mod;
			Sum+=(LL)f2[x]*dp[i|x];
		}
		dp[i]=add(dp[i],Sum%mod);
		dp[i]=add(dp[i],1);
		dp[i]=1ll*xs*dp[i]%mod;

	}
	pi(dp[1]);
}
signed main(){
	solve();
	return 0;
}
/*
*/

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

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

Think step by step.

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

1. **Understand the Algorithm**: The algorithm calculates the expected number of days for all vertices to have a message in an undirected graph with probabilistic edges.

2. **Identify Potential Bugs**: Common issues in such problems include:
   - Incorrect handling of edge probabilities.
   - Incorrect state transitions in the dynamic programming (DP) array.
   - Modulo arithmetic errors.
   - Edge cases with very high or very low probabilities.

3. **Create a Test Case**:
   - **Small Scale**: Use a small number of vertices and edges to ensure the test case is manageable.
   - **Probabilities**: Use probabilities that are neither too high nor too low to ensure the algorithm must handle intermediate values correctly.
   - **Graph Structure**: Create a graph that is connected but has a structure that could expose the bug, such as a linear chain or a star graph.

4. **Example Test Case**:
   - **Vertices and Edges**: Use 4 vertices and 3 edges to form a linear chain.
   - **Probabilities**: Use probabilities that are not trivial (e.g., 1/2, 1/3).

Here is a small-scale test input that fits these criteria:

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

Explanation:
- **Vertices**: 4
- **Edges**: 3
- **Probabilities**:
  - Edge (1, 2) has a probability of 1/3.
  - Edge (2, 3) has a probability of 1/4.
  - Edge (3, 4) has a probability of 1/5.

This test case ensures that the algorithm must correctly handle intermediate probabilities and the transition of the message through the vertices in a linear chain. This can help expose any issues in the handling of edge probabilities and state transitions in the DP array.


**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>
#define pi pair<int,int>
#define pb push_back
#define vec vector<int>
#define int long long
#define M 998244353
using namespace std;
void in(int &x){
    int f=1;x=0;
    char c=getchar();
    while(!isdigit(c)&&c!='-')c=getchar();
    if(c=='-')f=-1,c=getchar();
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x*=f;
}
int ksm(int n,int m){
    int ans=1;
    while(m){
        if(m&1)ans=ans*n%M;
        n=n*n%M;m/=2;
    }return ans;
}
int t,n,m,k,x,y,z,r,U,ans;
int u[1005],v[1005],w[1005];
int p[2097152],q[2097152],h[2097152];
int f[2097152],F[2097152],g[2097152];
int pm[2097152],pf[2097152],pcnt[2097152];
int A[22][2097152],B[22][2097152],C[2097152];
int lowbit(int x){return x&(-x);}
void upd(int &x,int y){x=(x+y)%M;}
void FWT(int n,int *f,int opt){
    for(int k=1;k*2<=n;k*=2){
        for(int i=0;i<n;i+=2*k){
            for(int j=0;j<k;j++){
                f[i+j+k]=(f[i+j+k]+f[i+j]*opt+M)%M;
            }
        }
    }
}
signed main(){
    in(n);in(m);
    for(int i=1;i<=m;i++){
        in(x);in(y);in(z);in(r);
        x--;y--;
        z=z*ksm(r,M-2)%M;
        u[i]=x;v[i]=y;w[i]=z;
    }
    for(int i=0;i<(1<<n);i++){
        pcnt[i]=__builtin_popcount(i);
        p[i]=h[i]=1;
        for(int j=1;j<=m;j++){
            if((i>>u[j])%2&&(i>>v[j])%2)p[i]=p[i]*(1-w[j]+M)%M;
            if((i>>u[j])%2+(i>>v[j])%2==1)h[i]=h[i]*(1-w[j]+M)%M;
        }
        q[i]=ksm(p[i],M-2);
        g[i]=ksm(1-h[i]+M,M-2);
        A[pcnt[i]][i]=p[i];
    }
    for(int i=0;i<=n;i++)FWT((1<<n),A[i],1);
    f[1]=F[1]=1;U=(1<<n)-1;
    for(int c=2;c<=n;c++){
        for(int i=1;i<=U;i++){
            if(i>1)pf[i]=f[i];
            pm[i]=f[i]*g[i]%M*h[i]%M;
        }
        for(int i=0;i<n;i++)for(int j=0;j<=U;j++)if((j>>i)&1)upd(pm[j],pm[j^(1<<i)]),upd(pf[j],pf[j^(1<<i)]);
        for(int i=0;i<(1<<n);i++)if(pcnt[i]==c-1)B[c-1][i]=q[U-i]*g[i]%M*f[i]%M;
        FWT((1<<n),B[c-1],1);
        memset(C,0,sizeof C);
        for(int i=1;i<c;i++)for(int j=0;j<(1<<n);j++)C[j]=(C[j]+A[i][j]*B[c-i][j])%M;
        FWT((1<<n),C,-1);
        for(int s=1;s<=U;s+=2){
            if(pcnt[s]!=c)continue;
            F[s]=C[s]*p[U]%M*q[s]%M;
            F[s]=(F[s]-pm[s]+M)%M;
            f[s]=(F[s]-pf[s]+M)%M;
        }
    }
    for(int i=0;i<U;i++)ans=(ans+g[i]*f[i])%M;
    printf("%lld\n",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:
To create 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 think through this:

1. **Understand the Bug**: The buggy code might have issues with:
   - Handling edge cases where the graph is very simple.
   - Incorrectly calculating probabilities.
   - Issues with the Fast Walsh-Hadamard Transform (FWT) or modular arithmetic.

2. **Simple Graphs**: Start with a small, simple graph to ensure the input is manageable and easy to verify.

3. **Edge Cases**:
   - A graph with a single vertex.
   - A graph with two vertices and one edge.
   - A graph with a small number of vertices and edges, ensuring the graph is connected and the probabilities are non-trivial.

4. **Non-Trivial Probabilities**: Use probabilities that are not too simple (e.g., 1/2) to ensure the code handles more complex calculations.

5. **Modulo Considerations**: Ensure the probabilities are chosen such that the modulo operations do not trivialize the problem.

Given these considerations, let's create a test input for a graph with 4 vertices and 5 edges, with non-trivial probabilities:

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

This input has:
- 4 vertices.
- 5 edges with probabilities that are not too simple.
- The graph is connected and the probabilities are chosen to ensure the problem is non-trivial.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Bento Box Adventure
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Image generated by ChatGPT 4o.

Boxed meals in Taiwan are very common, offering convenient and affordable nutrition-balanced choices for students and office workers. These meals typically include a variety of vegetables, protein, and rice, providing a well-rounded diet. With numerous options available at local self-service restaurants, they are a popular choice for those looking for a quick, healthy lunch.

There are five Taiwanese self-service restaurants numbered from 1 to 5. You plan to visit a different restaurant each day from Monday to Friday to pack a boxed meal for lunch. You've already visited one restaurant from Monday to Thursday, each time visiting a different one. Now, it's Friday, and you want to visit the last remaining restaurant that you haven't been to yet this week.

Write a program that takes as input the four restaurants you've already visited (one for each day from Monday to Thursday) and outputs the restaurant you should visit on Friday.

Input: A single line of input containing four integers $$$a, b, c, d$$$, each between 1 and 5 (inclusive), representing the restaurant numbers you visited from Monday to Thursday, in order.

- $$$1\le a, b, c, d\le 5$$$
- All four numbers will be different.

Output: Output the restaurant number you should visit on Friday.

Examples:
input:
1 3 2 5
output:
4

input:
2 5 4 3
output:
1

Note:

Buggy Code:
#include <iostream>
using namespace std;
int main()
{
    int arr[4];
    for(int i=0; i<4; i++)
    {
        cin >> arr[i];
    }
    int arr_2[5]={1,2,3,4,5};
    for(int i=0; i<5; i++)
    {
        bool isfound=false;
        for(int j=0; j<4; j++)
        {
            if (arr[i]==arr_2[j])
            {
                isfound=true;
                break;
            }
        }
        if (isfound==false)
        {
            cout << arr_2[i] << endl;
            break;
        }
    }
    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 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Bento Box Adventure
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Image generated by ChatGPT 4o.

Boxed meals in Taiwan are very common, offering convenient and affordable nutrition-balanced choices for students and office workers. These meals typically include a variety of vegetables, protein, and rice, providing a well-rounded diet. With numerous options available at local self-service restaurants, they are a popular choice for those looking for a quick, healthy lunch.

There are five Taiwanese self-service restaurants numbered from 1 to 5. You plan to visit a different restaurant each day from Monday to Friday to pack a boxed meal for lunch. You've already visited one restaurant from Monday to Thursday, each time visiting a different one. Now, it's Friday, and you want to visit the last remaining restaurant that you haven't been to yet this week.

Write a program that takes as input the four restaurants you've already visited (one for each day from Monday to Thursday) and outputs the restaurant you should visit on Friday.

Input: A single line of input containing four integers $$$a, b, c, d$$$, each between 1 and 5 (inclusive), representing the restaurant numbers you visited from Monday to Thursday, in order.

- $$$1\le a, b, c, d\le 5$$$
- All four numbers will be different.

Output: Output the restaurant number you should visit on Friday.

Examples:
input:
1 3 2 5
output:
4

input:
2 5 4 3
output:
1

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int restaurant[5] = {0, 0, 0, 0, 0};
    for (int i = 0; i < 3; i++)
    {
        int visited;
        cin >> visited;
        restaurant[visited - 1] = 1;
    }
    for (int i = 0; i < 4; i++)
    {
        if (restaurant[i] == 0)
        {
            cout << i + 1;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
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: The Bento Box Adventure
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Image generated by ChatGPT 4o.

Boxed meals in Taiwan are very common, offering convenient and affordable nutrition-balanced choices for students and office workers. These meals typically include a variety of vegetables, protein, and rice, providing a well-rounded diet. With numerous options available at local self-service restaurants, they are a popular choice for those looking for a quick, healthy lunch.

There are five Taiwanese self-service restaurants numbered from 1 to 5. You plan to visit a different restaurant each day from Monday to Friday to pack a boxed meal for lunch. You've already visited one restaurant from Monday to Thursday, each time visiting a different one. Now, it's Friday, and you want to visit the last remaining restaurant that you haven't been to yet this week.

Write a program that takes as input the four restaurants you've already visited (one for each day from Monday to Thursday) and outputs the restaurant you should visit on Friday.

Input: A single line of input containing four integers $$$a, b, c, d$$$, each between 1 and 5 (inclusive), representing the restaurant numbers you visited from Monday to Thursday, in order.

- $$$1\le a, b, c, d\le 5$$$
- All four numbers will be different.

Output: Output the restaurant number you should visit on Friday.

Examples:
input:
1 3 2 5
output:
4

input:
2 5 4 3
output:
1

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main(){
    vector <int> vec(4);
    for(auto &i :vec){
        cin>>i;
    }
    sort(vec.begin(),vec.end());
    int ans=0;
    for(int i=1;i<4;++i){
        if(vec[i]-vec[i-1]>1){
            ans=vec[i]-1;
        }
    }
    cout<<ans<<"\n";

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Bento Box Adventure
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Image generated by ChatGPT 4o.

Boxed meals in Taiwan are very common, offering convenient and affordable nutrition-balanced choices for students and office workers. These meals typically include a variety of vegetables, protein, and rice, providing a well-rounded diet. With numerous options available at local self-service restaurants, they are a popular choice for those looking for a quick, healthy lunch.

There are five Taiwanese self-service restaurants numbered from 1 to 5. You plan to visit a different restaurant each day from Monday to Friday to pack a boxed meal for lunch. You've already visited one restaurant from Monday to Thursday, each time visiting a different one. Now, it's Friday, and you want to visit the last remaining restaurant that you haven't been to yet this week.

Write a program that takes as input the four restaurants you've already visited (one for each day from Monday to Thursday) and outputs the restaurant you should visit on Friday.

Input: A single line of input containing four integers $$$a, b, c, d$$$, each between 1 and 5 (inclusive), representing the restaurant numbers you visited from Monday to Thursday, in order.

- $$$1\le a, b, c, d\le 5$$$
- All four numbers will be different.

Output: Output the restaurant number you should visit on Friday.

Examples:
input:
1 3 2 5
output:
4

input:
2 5 4 3
output:
1

Note:

Buggy Code:
def find_missing_restaurant(visited):
    # Define the set of all possible restaurant numbers
    all_restaurants = {1, 2, 3, 4, 5}

    # Convert the input list to a set
    visited_set = set(visited)

    # Find the missing restaurant
    missing_restaurant = all_restaurants - visited_set

    # Return the missing restaurant number
    return missing_restaurant.pop()

# Example usage:
input_example = [1, 3, 2, 5]
print(find_missing_restaurant(input_example))  # Output: 4

# atSJQYUq0FXETKjgfB9

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Bento Box Adventure
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Image generated by ChatGPT 4o.

Boxed meals in Taiwan are very common, offering convenient and affordable nutrition-balanced choices for students and office workers. These meals typically include a variety of vegetables, protein, and rice, providing a well-rounded diet. With numerous options available at local self-service restaurants, they are a popular choice for those looking for a quick, healthy lunch.

There are five Taiwanese self-service restaurants numbered from 1 to 5. You plan to visit a different restaurant each day from Monday to Friday to pack a boxed meal for lunch. You've already visited one restaurant from Monday to Thursday, each time visiting a different one. Now, it's Friday, and you want to visit the last remaining restaurant that you haven't been to yet this week.

Write a program that takes as input the four restaurants you've already visited (one for each day from Monday to Thursday) and outputs the restaurant you should visit on Friday.

Input: A single line of input containing four integers $$$a, b, c, d$$$, each between 1 and 5 (inclusive), representing the restaurant numbers you visited from Monday to Thursday, in order.

- $$$1\le a, b, c, d\le 5$$$
- All four numbers will be different.

Output: Output the restaurant number you should visit on Friday.

Examples:
input:
1 3 2 5
output:
4

input:
2 5 4 3
output:
1

Note:

Buggy Code:
def find_remaining_restaurant(visited):
    # Define all possible restaurant numbers
    all_restaurants = set([1, 2, 3, 4, 5])

    # Convert the list of visited restaurants to a set
    visited_set = set(visited)

    # Find the difference between all restaurants and the visited ones
    remaining_restaurant = all_restaurants - visited_set

    # Return the remaining restaurant as an integer
    return remaining_restaurant.pop()

# Example usage:
visited_restaurants = [1, 3, 2, 5]
print(find_remaining_restaurant(visited_restaurants))  # Output: 4

# sSDhicZm0T8xWPUYwFv

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

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

Think step by step.

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