**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can't choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$ and $i$ wasn't chosen before, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
18
1281
4
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 1$ and set $x$ to $\text{mex}(x, a_{1, 1}, a_{1, 2}, a_{1, 3}, a_{1, 4}, a_{1, 5}) = \text{mex}(1, 0, 2, 0, 4, 11) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(1) = 3$.

It can be seen that $f(0) = 4$, $f(1) = 3$, $f(2) = 4$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 3 + 4 + 3 + 4 = 18$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 1$. So $f(0) + f(1) = 3 + 1 = 4$.

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

typedef pair<int,int> PII;

const int N = 200010;
const int inf=1e18;

set<int> s[N];
int n,m;

void solve()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)s[i].clear();
    map<int,vector<int>> g;
    map<int,int> mp;

    for(int i=1;i<=n;i++)
    {
        int cnt;cin>>cnt;
        while(cnt--)
        {
            int x;cin>>x;
            s[i].insert(x);
        }
    }

    int mx=0,mxu=0,mxv=0;
    for(int i=1;i<=n;i++)
    {
        int u,v;
        int mex=0;
        while(s[i].count(mex))mex++;
        u=mex;
        mxu=max(mxu,mex);
        s[i].insert(mex);
        while(s[i].count(mex))mex++;
        v=mex;
        mxv=max(mxv,mex);
        g[u].push_back(v);
    }
    mx=mxu;

    for(int u=mxv;u>=0;u--)
    {
        mp[u]=u;
        if(!g.count(u))continue;
        for(auto t:g[u])mp[u]=max(mp[u],mp[t]);
        if(g[u].size()>1)mx=max(mx,mp[u]);
    }

    int ans=0;

    for(int i=0;i<=m;i++)
    {
        if(!mp[i])ans+=max(i,mx);
        else ans+=max({i,mx,mp[i]});
    }
    cout<<ans<<'\n';
    return;
}

signed main()
{
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;

    while(T --)
    {
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that can maximize the failure probability. The key aspects to focus on are:

1. **Handling of MEX calculations**: Ensure that the MEX calculation is correct, especially when sequences have repeated elements or are very large.
2. **Edge cases for the initial value of \( x \)**: Test with different initial values of \( x \) to see if the code correctly handles the MEX operations.
3. **Large values for \( m \)**: Test with large values of \( m \) to ensure the code handles large sums efficiently.
4. **Sequences with very large elements**: Test with sequences containing very large elements to ensure the code can handle such inputs.

Given these considerations, let's construct a small-scale test input that can expose potential bugs:

```plaintext
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
```

This input covers a variety of scenarios, including:
- Sequences with repeated elements.
- Sequences with large elements.
- Different values of \( m \) to test the sum calculation.
- Different lengths of sequences to test the MEX calculation.

By running this input through the provided code, we can observe if the code handles these edge cases correctly and identify any potential bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can't choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$ and $i$ wasn't chosen before, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
18
1281
4
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 1$ and set $x$ to $\text{mex}(x, a_{1, 1}, a_{1, 2}, a_{1, 3}, a_{1, 4}, a_{1, 5}) = \text{mex}(1, 0, 2, 0, 4, 11) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(1) = 3$.

It can be seen that $f(0) = 4$, $f(1) = 3$, $f(2) = 4$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 3 + 4 + 3 + 4 = 18$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 1$. So $f(0) + f(1) = 3 + 1 = 4$.

Buggy Code:
import java.util.*;

public class Main {
	static int max(int a, int b) { return a > b ? a : b; }
	static int min(int a, int b) { return a < b ? a : b; }

	static class Pair {
		public int first = 0, second = 0;
		Pair(int f, int s) {
			this.first = f;
			this.second = s;
		}
	}

	static int[] mx;
	static boolean[] used;
	static ArrayList<Integer>[] g;

	static int dfs(int v) {
		if (used[v]) return mx[v];
		used[v] = true;
		mx[v] = v;
		for (int to : g[v]) mx[v] = max(mx[v], dfs(to));
		return mx[v];
	}

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

		int t = sc.nextInt();
		for (int test = 0; test < t; test++) {
			int n = sc.nextInt(), m = sc.nextInt();

			Pair[] mex = new Pair[n];
			int[] cnt = new int[n + 5];
			int _mexmx = n;

			for (int i = 0; i < n; i++) {
				mex[i] = new Pair(-1, -1);
				int l = sc.nextInt();
				int[] arr = new int[l];
				for (int j = 0; j < l; j++) arr[j] = sc.nextInt();
				Arrays.sort(arr);

				int _mex = 0;
				for (int j = 0; j < l;) {
					if (arr[j] == _mex) _mex++;
					else if (arr[j] == _mex - 1) {}
					else if (mex[i].first == -1) {
						mex[i].first = _mex++;
						continue;
					}
					else {
						mex[i].second = _mex;
						break;
					}
					j++;
				}
				if (mex[i].first == -1) mex[i].first = _mex++;
				if (mex[i].second == -1) mex[i].second = _mex;
				cnt[mex[i].first]++;
				_mexmx = max(_mexmx, mex[i].second);
			}
			int _mx = 0;

			g = new ArrayList[_mexmx + 5];
			for (int i = 0; i < _mexmx + 5; i++) g[i] = new ArrayList<>();
			mx = new int[_mexmx + 5];
			used = new boolean[_mexmx + 5];

			for (Pair p : mex) g[p.first].add(p.second);
			for (Pair p : mex) if (cnt[p.first] > 1) _mx = max(_mx, dfs(p.first)); else _mx = max(_mx, p.first);

			long ans = (m * 1L * (m + 1L)) / 2;
			for (int i = 0; i <= min(m, _mexmx + 1); i++) ans += max(max(i, _mx), dfs(i)) - i;
			System.out.println(ans);
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
1
3 4
2 0 1
2 1 2
2 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and a MEX Problem (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. In this version of the problem, you can't choose the same integer twice or more. You can make hacks only if both versions are solved.

One day, Turtle was playing with $n$ sequences. Let the length of the $i$-th sequence be $l_i$. Then the $i$-th sequence was $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$.

Piggy gave Turtle a problem to solve when Turtle was playing. The statement of the problem was:

  * There was a non-negative integer $x$ at first. Turtle would perform an arbitrary number (possibly zero) of operations on the integer.    * In each operation, Turtle could choose an integer $i$ such that $1 \le i \le n$ and $i$ wasn't chosen before, and set $x$ to $\text{mex}^{\dagger}(x, a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i})$.    * Turtle was asked to find the answer, which was the maximum value of $x$ after performing an arbitrary number of operations.

Turtle solved the above problem without difficulty. He defined $f(k)$ as the answer to the above problem when the initial value of $x$ was $k$.

Then Piggy gave Turtle a non-negative integer $m$ and asked Turtle to find the value of $\sum\limits_{i = 0}^m f(i)$ (i.e., the value of $f(0) + f(1) + \ldots + f(m)$). Unfortunately, he couldn't solve this problem. Please help him!

$^{\dagger}\text{mex}(c_1, c_2, \ldots, c_k)$ is defined as the smallest non-negative integer $x$ which does not occur in the sequence $c$. For example, $\text{mex}(2, 2, 0, 3)$ is $1$, $\text{mex}(1, 2)$ is $0$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n, m$ ($1 \le n \le 2 \cdot 10^5, 0 \le m \le 10^9$).

Each of the following $n$ lines contains several integers. The first integer $l_i$ ($1 \le l_i \le 2 \cdot 10^5$) represents the length of the $i$-th sequence, and the following $l_i$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, l_i}$ ($0 \le a_{i, j} \le 10^9$) represent the elements of the $i$-th sequence.

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

Output: For each test case, output a single integer — the value of $\sum\limits_{i = 0}^m f(i)$.

Examples:
input:
6
3 4
2 0 2
3 2 3 3
4 7 0 1 5
3 4
5 0 2 0 4 11
1 1
5 1 3 0 3 3
2 50
2 1 2
2 1 2
1 1
7 1 2 4 1 4 9 5
4 114514
2 2 2
5 7 3 6 0 3
3 0 1 1
5 0 9 2 1 5
5 1919810
1 2
2 324003 0
3 1416324 2 1460728
4 1312631 2 0 1415195
5 1223554 192248 2 1492515 725556
output:
16
18
1281
4
6556785365
1842836177961

Note: In the first test case, when $x$ is initially $2$, Turtle can choose $i = 3$ and set $x$ to $\text{mex}(x, a_{3, 1}, a_{3, 2}, a_{3, 3}, a_{3, 4}) = \text{mex}(2, 7, 0, 1, 5) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(2) = 3$.

It can be seen that $f(0) = 3$, $f(1) = 3$, $f(2) = 3$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 3 + 3 + 3 + 3 + 4 = 16$.

In the second test case, when $x$ is initially $1$, Turtle can choose $i = 1$ and set $x$ to $\text{mex}(x, a_{1, 1}, a_{1, 2}, a_{1, 3}, a_{1, 4}, a_{1, 5}) = \text{mex}(1, 0, 2, 0, 4, 11) = 3$. It can be proved that Turtle can't make the value of $x$ greater than $3$, so $f(1) = 3$.

It can be seen that $f(0) = 4$, $f(1) = 3$, $f(2) = 4$, $f(3) = 3$, and $f(4) = 4$. So $f(0) + f(1) + f(2) + f(3) + f(4) = 4 + 3 + 4 + 3 + 4 = 18$.

In the fourth test case, it can be seen that $f(0) = 3$ and $f(1) = 1$. So $f(0) + f(1) = 3 + 1 = 4$.

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.TreeMap;

public class Main {

	private static long cem(long x) {
		return x * (x + 1) / 2;
	}

	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in), 200000 * 20);
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out), 50000000);

		int T = Integer.parseInt(in.readLine());
		List<Integer>[] graph = new ArrayList[2000000];
		boolean[] flag = new boolean[2000000];
		int[] dp = new int[2000000];

		for (int ts=1; ts<=T; ts++) {
			String[] line = in.readLine().split(" ");
			int N = Integer.parseInt(line[0]);
			int M = Integer.parseInt(line[1]);
			List<Integer> firsts = new ArrayList<>();

			//out.write(M + "\n");

			int maxSecond = -1;
			int maxFirst = -1;
			for (int i=1; i<=N; i++) {
				line = in.readLine().split(" ");
				int L = Integer.parseInt(line[0]);

				//out.write("******\n");
				for (int j=1; j<=L; j++) {
					//out.write(line[j] + "\n");
					//out.flush();
					int num = Integer.parseInt(line[j]);
					flag[num] = true;
				}
				//out.write("******\n");

				int first = -1;
				int second = -1;
				for (int idx=0; second < 0; idx++) {
					if (flag[idx]) {
						continue;
					}

					if (first == -1) {
						first = idx;
					} else {
						second = idx;
					}
				}
				//out.write(first + " " + second + "\n");
				maxSecond = Math.max(maxSecond, second);
				maxFirst = Math.max(maxFirst, first);

				if (graph[first] == null) {
					graph[first] = new ArrayList<>();
				}
				graph[first].add(second);
				firsts.add(first);

				for (int j=1; j<=L; j++) {
					//out.write(line[j] + "\n");
					//out.flush();
					int num = Integer.parseInt(line[j]);
					flag[num] = false;
				}
			}

			dp[maxSecond] = maxSecond;
			for (int i=maxSecond - 1; i>=0; i--) {
				dp[i] = i;
				if (graph[i] == null) {
					continue;
				}

				for (final int w : graph[i]) {
					dp[i] = Math.max(dp[i], dp[w]);
				}
				//out.write(i + " " + dp[i] + "\n");
			}

			long resetMax = -1;
			for (int i=maxSecond - 1; i>=0; i--) {
				if (graph[i] != null && graph[i].size() > 1) {
					resetMax = Math.max(resetMax, dp[i]);
				}
			}

			long ans = 0;
			for (int i=0; i<=Math.min(maxSecond, M); i++) {
				long maxValue = i;
				if (graph[i] != null) {
					maxValue = Math.max(maxValue, dp[i]);
				}
				maxValue = Math.max(maxValue, resetMax);
				maxValue = Math.max(maxValue, maxFirst);
				ans += maxValue;
				//out.write(i + " " + maxValue + "\n");
			}

			//out.write(maxSecond + " " + M + "\n");
			if (M > maxSecond) {
				ans += cem(M) - cem(maxSecond);
			}

			for (final int first : firsts) {
				graph[first] = null;
			}

			out.write(ans + "\n");

		}

		in.close();
		out.close();
	}

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can highlight potential issues in the algorithm. Here are the steps to create such a test input:

1. **Small Number of Test Cases**: Start with a small number of test cases to ensure the input is manageable.
2. **Special Cases for `n` and `m`**: Use values of `n` and `m` that can push the algorithm to its limits.
3. **Sequences with Specific Patterns**: Create sequences that can lead to incorrect MEX calculations or edge cases in the dynamic programming (DP) logic.

### Test Case 1
- **Description**: A single sequence with a large `m` and a sequence that contains all integers up to a certain point, but skips one specific integer.
- **Purpose**: To test the MEX calculation and ensure the algorithm correctly identifies the missing integer.

### Test Case 2
- **Description**: Multiple sequences with overlapping and non-overlapping ranges.
- **Purpose**: To test the interaction between different sequences and ensure the algorithm correctly handles multiple operations.

### Test Case 3
- **Description**: A sequence with a large number of unique integers and a small `m`.
- **Purpose**: To test the performance and correctness of the algorithm with a large input size.

### Test Case 4
- **Description**: A sequence with repeated integers and a small `m`.
- **Purpose**: To test the algorithm's ability to handle repeated integers and ensure it correctly identifies the MEX.

### Test Case 5
- **Description**: A sequence with a single integer and a large `m`.
- **Purpose**: To test the algorithm's ability to handle a single integer and ensure it correctly handles the sum calculation.

### Test Case 6
- **Description**: A sequence with a large `m` and a sequence that contains all integers up to `m`.
- **Purpose**: To test the algorithm's ability to handle a large `m` and ensure it correctly identifies the maximum value.

### Chosen Test Input
```plaintext
6
1 5
5 0 1 2 3 4
3 2
2 0 1
2 1 2
2 0 2
3 10
3 0 1 2
3 1 2 3
3 2 3 4
1 0
5 0 0 0 0 0
1 10
10 0 1 2 3 4 5 6 7 8 9
1 100
100 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
```

This test input includes a variety of scenarios that can help identify potential bugs in the algorithm, such as incorrect MEX calculations, handling of repeated integers, and performance with large input sizes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
using ll = int64_t;

struct Event {
    ll cord, type;

    bool operator<(const Event& a) const {
        if (cord != a.cord) return cord < a.cord;
        return cord < a.type;
    }
};

void solve() {
    ll n, k;
    cin >> n >> k;
    vector<ll> a(n), b(n);
    for (auto& i : a) cin >> i;
    for (auto& i : b) cin >> i;
    vector<Event> ev(2 * n);
    for (ll i = 0; i < n; i++) {
        ev[2 * i] = {a[i] + 1, 1};
        ev[2 * i + 1] = {b[i] + 1, 2};
    }
    sort(ev.begin(), ev.end());
    ll ans = n;
    ll cnt = n, bad = 0;
    for (auto [c, t] : ev) {
        ans = max(ans, cnt * (c - 1));
        if (t == 1) {
            bad++;
        } else {
            cnt--;
        }
        if (bad > k) break;
    }
    cout << ans << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 2
1 2 3 4 5
6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>

using namespace std;

void solve() {
    long long n, k;

    vector<pair<long long, long long>> things;

    cin >> n >> k;

    for (int i = 0; i < n; i++) {
        long long a;
        cin >> a;
        things.push_back({a, 0});
    }

    for (int i = 0; i < n; i++) {
        long long b;
        cin >> b;
        things.push_back({b, 1});
    }

    sort(things.begin(), things.end());

    long long maxS = 0, badReviews = 0, buyingCustomers = n;

    for (int i = 0; i < 2 * n; i++) {
        if (maxS < things[i].first * buyingCustomers && badReviews <= k) {
            maxS = things[i].first * buyingCustomers;
        }

        if (things[i].second == 0) {
            badReviews++;
        }
        else {
            buyingCustomers--;
            badReviews--;
        }
    }

    cout << maxS << endl;
}

int main() {

    int t;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 0
1
2
2 1
1 2
3 4
3 1
2 3 4
5 6 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
#include <iostream>
#include <algorithm>
#include <bitset>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <sstream>
#include <iomanip>
#include <stdexcept>
#include <utility>
#include <ranges>
#include <deque>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef set<long long> sll;
typedef long double ld;
typedef vector<long long> vll;
ll MOD = 1e9 + 7;

void meow() {
    ll n,k;cin>>n>>k;
    vll a(n), b(n); for(auto &k: a) cin>>k; for(auto &k:b) cin>>k;
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());

    ll bc = n;
    ll ic = 0;
    ll maxs = 0;
    ll ai = 0, bi = 0;

    while(ai < n) {
        if(a[ai] <= b[bi]) {
            if(ic<=k && ((ai+1) == n || (a[ai] != a[ai+1]))) maxs = max(maxs, bc*a[ai]);
            ic++;
            ai++;
        } else {
            if(ic<=k && (bi+1 == n || (b[bi] != b[bi+1]))) maxs = max(maxs, bc*b[bi]);
            ic--;
            bc--;
            bi++;
        }
    }

    while(bi < n) {
        if(ic<=k) maxs = max(maxs, bc*b[bi]);
        ic--;
        bc--;
        bi++;
    }

    cout<<maxs<<"\n";

}

int main() {
    //these first two lines speed up input / output significantly
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    ll t; cin>>t;

    while(t--) {
        meow();
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

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

while(t):
    n,k = map(int,input().split(' '))
    a = list(map(int,input().split(' ')))
    b = list(map(int,input().split(' ')))
    d = []
    for i in range(n):
        d.append((b[i],a[i]))
    d.sort(reverse=True)
    profit = 0
    for i in a:
        nr = 0
        tp = 0
        for j in d:
            if(i<=j[0]):
                if(i>j[1]):
                    nr+=1
                tp+=1
            else:
                new_profit = i*tp
                if(profit<new_profit):
                    profit = new_profit
                break
            if(nr>k):
                break
    else:
        new_profit = i*tp
        if(profit<new_profit):
            profit = new_profit

    for i in b:
        nr = 0
        tp = 0
        for j in d:
            if(i<=j[0]):
                if(i>j[1]):
                    nr+=1
                tp+=1
            else:
                new_profit = i*tp
                if(profit<new_profit):
                    profit = new_profit
                break
            if(nr>k):
                break
        else:
            new_profit = i*tp
            if(profit<new_profit):
                profit = new_profit
    ans.append(profit)
    t-=1
for i in ans:
    print(i)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
# t=int(input())
# for i in range(t):
#   n,k=map(int,input().split())
#   a=list(map(int,input().split()))
#   b=list(map(int,input().split()))

#   pos_array=[0]*(max(a)+1)
#   neg_array=[0]*(max(b)+1)
#   pos_len=(max(a)+1)
#   neg_len=max(b)+1
#   for i in range(n):
#     pos_array[1]+=1
#     if(a[i]+1<pos_len):
#       pos_array[a[i]+1]-=1
#     neg_array[a[i]+1]+=1
#     if(b[i]+1<neg_len):
#       neg_array[b[i]+1]-=1
#   curr_ans=0
#   for i in range(1,pos_len):
#     curr_pos=pos_array[i-1]+pos_array[i]
#     curr_neg=neg_array[i-1]+neg_array[i]
#     if(curr_neg<=k):
#       curr_ans=max(curr_ans,(curr_pos+curr_neg)*i)

#     pos_array[i]=curr_pos
#     neg_array[i]=curr_neg
#   for i in range(pos_len,neg_len):
#     curr_neg=neg_array[i-1]+neg_array[i]
#     if(curr_neg<=k):
#         curr_ans=max(curr_ans,curr_neg*i)

#     neg_array[i]=curr_neg
#   #print(pos_array,neg_array)
#   print(curr_ans)

from collections import defaultdict
t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    # Dictionaries to store increments and decrements
    pos_updates = defaultdict(int)
    neg_updates = defaultdict(int)

    for i in range(n):
        # Update for positive array
        pos_updates[1] = pos_updates.get(1, 0) + 1
        if(a[i]+1<max(a)+1):
            pos_updates[a[i]+1]-=1

        # Update for negative array
        neg_updates[a[i] + 1] = neg_updates.get(a[i] + 1, 0) + 1
        if(b[i]+1<max(b)+1):
            neg_updates[b[i]+1]-=1

    curr_pos = 0
    curr_neg = 0
    curr_ans = 0

    max_value = max(max(a), max(b)) + 1

    for i in range(1, max(a) + 1):
        # Update prefix sums for current value
        curr_pos =curr_pos+ pos_updates.get(i, 0)
        curr_neg += neg_updates.get(i, 0)

        # Check if the current value can lead to a better result
        if curr_neg <= k:
            curr_ans = max(curr_ans, (curr_pos + curr_neg) * i)
    for i in range(max(a)+1,max(b)+1):
      curr_neg += neg_updates.get(i - 1, 0)

        # Check if the current value can lead to a better result
      if curr_neg <= k:
            curr_ans = max(curr_ans, curr_neg * i)

    print(curr_ans)

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

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

Think step by step.

Response:
```plaintext
1
5 2
1 2 3 4 5
6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
for _ in range(int(input())):
    n,k = map(int,input().split())
    a = list(map(int,input().split()))
    b = list(map(int,input().split()))
    if (k==n):
        ans = 0
        b.sort()
        for i in range(n):
            ans = max(ans,b[i]*(n-i))
        print(ans)
        continue

    lst = [ ]
    for i in range(n):
        lst.append([a[i],b[i]])

    lst.sort()
    # print(lst)

    lst1 = [ ]
    for i in range(k):
        lst1.append(lst[i][1])

    lst1.sort()
    ans = 0
    # print(lst1)

    for i in range(k):
        ans = max(ans,min(lst1[i],lst[k][0])*(n-i))

    ans1 = lst[0][0]*n
    for i in range(1,n):
        if (lst[i][0]>lst[i-1][1]):
            ans1 = max(ans1,lst[i][0]*(n-i))

    print(max(ans1,ans))

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

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

Think step by step.

Response:
```plaintext
5
2 1
1 2
3 4
1 1
2
5
3 3
1 2 3
4 5 6
4 3
1 2 3 4
5 6 7 8
3 1
1 2 3
4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc= new Scanner(System.in);
        int tes=sc.nextInt();
        label:while(tes-->0){
           int n=sc.nextInt(),k=sc.nextInt(),i,max=0;
           int a[]=new int[n];
           int b[]=new int[n];
           for(i=0;i<n;i++) { a[i]=sc.nextInt(); max=Math.max(max,a[i]); }
           for(i=0;i<n;i++) { b[i]=sc.nextInt(); max=Math.max(max,b[i]); }
           long low=1,high=(long)max,ans=0;
           while(high>=low){
                long mid=(low+high)>>1;
                int cnt=0,c=0;
                for(i=0;i<n;i++){
                    if(b[i]<mid) continue;
                    if(mid>a[i] && mid<=b[i]) cnt++;
                    c++;
                }
                if(cnt<=k){
                    //System.out.println(c+" "+mid);
                    ans=Math.max(ans,c*mid);
                    low=mid+1;
                }
                else high=mid-1;
           }
           System.out.println(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 expose the bug in the provided code, we need to focus on scenarios where the code's logic for counting positive and negative reviews might fail. Specifically, the code seems to have an issue with correctly counting the number of customers who will buy a tree and the number of negative reviews.

Here are the steps to create a challenging test input:

1. **Edge Case with All Customers Leaving Negative Reviews:**
   - Set all `a_i` values to a very low number, and all `b_i` values to a higher number.
   - Ensure that the maximum price `mid` can be set such that all customers leave negative reviews, but the code incorrectly counts the number of customers who will buy the tree.

2. **Edge Case with No Customers Buying the Tree:**
   - Set all `a_i` and `b_i` values such that no customer will buy the tree at any price.
   - This should result in zero earnings, but the code might incorrectly count some customers as buying the tree.

3. **Edge Case with Exactly k Negative Reviews:**
   - Set the prices such that exactly `k` customers will leave negative reviews.
   - Ensure that the code correctly counts the number of negative reviews and the maximum earnings.

4. **Mixed Case with Various `a_i` and `b_i` Values:**
   - Set a mix of `a_i` and `b_i` values to test the code's ability to handle a variety of customer behaviors.
   - Ensure that the code correctly identifies the optimal price to maximize earnings while respecting the limit on negative reviews.

Based on these considerations, here is a small-scale test input:

```plaintext
3
5 2
1 2 3 4 5
6 7 8 9 10
5 3
1 2 3 4 5
6 7 8 9 10
5 0
1 2 3 4 5
6 7 8 9 10
```

This input includes:
- A case where exactly 2 negative reviews are allowed.
- A case where up to 3 negative reviews are allowed.
- A case where no negative reviews are allowed.

These scenarios will help expose any issues in the code's logic for counting positive and negative reviews and determining the maximum earnings.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
// for string input, always use: String s=sc.readLine().trim();
// otherwise it will read '\n' at end of line.
// ArrayIndexOutOfBound Error Correction - readLine() method of reader class has byte[] buf array -> increase its size to upper bound of string length.
import java.util.*;
import java.io.*;

public class pr1 {
  public static void main(String[] args) throws IOException {
    Reader sc = new Reader();
    PrintWriter out = new PrintWriter(System.out);
    int t = sc.nextInt();
    while (t-- > 0) {
      int n = sc.nextInt(), k = sc.nextInt();
      long a[][] = new long[n][3];
      for (int i = 0; i < n; i++) {
        a[i][0] = sc.nextLong();
      }
      for (int i = 0; i < n; i++) {
        a[i][1] = sc.nextLong();
        a[i][2] = a[i][1] - a[i][0];
      }
      Arrays.sort(a, (aa, bb) -> Long.compare(bb[2], aa[2]));
      long low = 1, high = (long) 1e10, ans = 0;
      while (low <= high) {
        long mid = low + (high - low) / 3;
        long tans = 0;
        int cc = 0, chk = 0;
        for (int i = 0; i < n; i++) {
          if (mid <= a[i][0]) {
            tans += mid;
          } else if (mid <= a[i][1]) {
            if (cc < k) {
              tans += mid;
              cc++;
            } else {
              chk = 1;
            }
          }
        }
        //
        long mid2 = high - (high - low) / 3;
        long tans2 = 0;
        int cc2 = 0, chk2 = 0;
        for (int i = 0; i < n; i++) {
          if (mid2 <= a[i][0]) {
            tans2 += mid2;
          } else if (mid2 <= a[i][1]) {
            if (cc2 < k) {
              tans2 += mid2;
              cc2++;
            } else {
              chk2 = 1;
            }
          }
        }
        if (chk == 1)
          tans = 0;
        if (chk2 == 1)
          tans2 = 0;
        if (tans == 0) {
          high = mid - 1;
        } else if (tans > tans2) {
          high = mid2 - 1;
        } else if (tans < tans2) {
          low = mid + 1;
        } else {
          high = mid2 - 1;
          low = mid + 1;
        }
        ans = Math.max(ans, Math.max(tans, tans2));
        // System.out.println(ans + " " + tans + " " + tans2 + " " + mid + " " + mid2);
      }

      out.println(ans);
    }
    out.close();
  }

  // ------------------------------------reader class-----------------------------
  static class Reader {
    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader() {
      din = new DataInputStream(System.in);
      buffer = new byte[BUFFER_SIZE];
      bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException {
      din = new DataInputStream(
          new FileInputStream(file_name));
      buffer = new byte[BUFFER_SIZE];
      bufferPointer = bytesRead = 0;
    }

    public String readLine() throws IOException {
      byte[] buf = new byte[64]; // line length
      int cnt = 0, c;
      while ((c = read()) != -1) {
        if (c == '\n') {
          if (cnt != 0) {
            break;
          } else {
            continue;
          }
        }
        buf[cnt++] = (byte) c;
      }
      return new String(buf, 0, cnt);
    }

    public int nextInt() throws IOException {
      int ret = 0;
      byte c = read();
      while (c <= ' ') {
        c = read();
      }
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');
      if (neg)
        return -ret;
      return ret;
    }

    public long nextLong() throws IOException {
      long ret = 0;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');
      if (neg)
        return -ret;
      return ret;
    }

    public double nextDouble() throws IOException {
      double ret = 0, div = 1;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');
      if (c == '.') {
        while ((c = read()) >= '0' && c <= '9') {
          ret += (c - '0') / (div *= 10);
        }
      }
      if (neg)
        return -ret;
      return ret;
    }

    private void fillBuffer() throws IOException {
      bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
      if (bytesRead == -1)
        buffer[0] = -1;
    }

    private byte read() throws IOException {
      if (bufferPointer == bytesRead)
        fillBuffer();
      return buffer[bufferPointer++];
    }

    public void close() throws IOException {
      if (din == null)
        return;
      din.close();
    }
  }
  // -----------------------------------------------
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - **No Negative Reviews Allowed (k = 0)**: This can test if the algorithm correctly maximizes earnings without any negative reviews.
   - **All Customers Have the Same a and b Values**: This can test if the algorithm handles uniform input correctly.
   - **High Number of Customers (n close to 200,000)**: This can test the algorithm's performance and correctness with large inputs.

2. **Specific Scenarios**:
   - **a_i and b_i Values are Very Close**: This can test the precision of the algorithm in handling small differences.
   - **a_i and b_i Values are Very Different**: This can test the algorithm's ability to handle a wide range of values.

3. **Combination of the Above**:
   - **Mixed Values with k = 1**: This can test the algorithm's ability to balance between maximizing earnings and minimizing negative reviews.

Let's create a test input that includes a mix of the above scenarios:

```plaintext
5
3 0
1 2 3
2 3 4
1 1
2
5
5 3
1 2 3 4 5
2 3 4 5 6
10 5
1 2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8 9 10 11
20 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
```

This test input includes:
- A case with 3 customers and 0 negative reviews allowed.
- A case with 1 customer and 1 negative review allowed.
- A case with 5 customers and 3 negative reviews allowed.
- A case with 10 customers and 5 negative reviews allowed.
- A case with 20 customers and 1 negative review allowed.

These scenarios should help expose any issues 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: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

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

#define fast_io cin.tie(0)->sync_with_stdio(0);
#define dbg(x) cout << #x << " = " << x << endl
#define all(x) x.begin(), x.end()
#define sz(a) ((int)((a).size()))
#define int long long
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define lb(vect, x) (lower_bound(all(vect), x) - vect.begin())
#define ub(vect, x) (upper_bound(all(vect), x) - vect.begin())

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fll;

#define ADD 0
#define RMV 1

int n, k;
set<pii> ev;
vi a;
vi b;

int qtNeg(int valor)
{
    return lb(a, valor);
}

int qtComprada(int valor)
{
    return n - lb(b, valor);
}

void solve()
{
    cin >> n >> k;

    a.resize(n);
    b.resize(n);

    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        ev.insert({a[i], ADD});
    }

    for (int i = 0; i < n; i++)
    {
        cin >> b[i];
        ev.insert({b[i], RMV});
    }

    sort(all(a));
    sort(all(b));

    int maior = 0;
    for (auto &p : ev)
    {
        if (qtNeg(p.f) > k)
        {
            break;
        }

        maior = max(maior, p.f * qtComprada(p.f));
    }

    cout << maior << endl;
}

int32_t main()
{
    // casas decimais
    // cout << fixed << setprecision(1);

    // horario
    // cout << setfill('0') << setw(2);

    fast_io;
    int t = 1;
    cin >> t;

    for (int i = 1; i <= t; i++)
    {
        solve();

        // #ifdef ONPC
        //         cout << "__________________________" << endl;
        // #endif
    }

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
// #define int long long
#define all(x) x.begin(), x.end()

#ifndef ONLINE_JUDGE
#include "dbg.cpp"
#else
#define debug(...)
#endif

const int INF = INT_MAX;

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

// find_by_order, order_of_key -> gives number of elements just greater/lesser in pbds
template<typename T, class C = std::less<T>>
using ordered_set = tree<T, null_type, C,
      rb_tree_tag, tree_order_statistics_node_update>;

void solve(){
    int n, k; cin >> n >> k;
    vector<int>a(n), b(n);
    for(int &i : a) cin >> i;
    for(int &i : b) cin >> i;
    // take all people, take starting a as ur main person
    int l = 0, h = INF, ans = 0;
    while(l <= h){
        int mid = (l + h) >> 1;
        int cnt = 0;
        for(int i = 0; i < n; i++){
            if(mid > b[i]){
                cnt = (int)1e9;
            }
            if(mid > a[i]) cnt++;
        }
        if(cnt > k){
            h = mid - 1;
        }
        else{
            l = mid + 1, ans = mid;
        }
    }
    ans *= n;
    vector<array<int, 2>>vec;
    for(int i = 0; i < n; i++){
        vec.push_back({b[i], a[i]});
    }
    sort(all(vec));
    for(auto &[x, y] : vec) swap(x, y);

    ordered_set<int, less_equal<int>>os;
    for(int &i : a) os.insert(i);

    int i = 0;
    while(i < n){
        int num = vec[i][1];
        while(i < n and vec[i][1] == num){
            os.erase(os.find_by_order(os.order_of_key(vec[i][0])));
            i++;
        }
        num++;
        if(i < n and k){
            int next = -1;
            next = (*os.find_by_order(min(k, (int)os.size()) - 1)) -  1;
            if(num <= next){
                ans = max(ans, (n - i) * (max(next, num)));
            }
        }
    }
    cout << ans << '\n';
}

signed main(){
    #ifndef ONLINE_JUDGE
    freopen("error.txt", "w", stderr);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    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
5
2 0
1 1
2 2
1 1
1
3 3
1 1 1
2 2 2
4 3
1 1 1 1
2 2 2 2
3 1
1 1 1
2 2 2
```

**Explanation:**

1. **Test Case 1:**
   - \( n = 2 \), \( k = 0 \)
   - \( a = [1, 1] \), \( b = [2, 2] \)
   - Both customers will buy the tree if the price is 1, and no one will leave a negative review.
   - Expected output: \( 2 \)

2. **Test Case 2:**
   - \( n = 1 \), \( k = 1 \)
   - \( a = [1] \), \( b = [1] \)
   - The customer will buy the tree at price 1, and it will count as a negative review.
   - Expected output: \( 1 \)

3. **Test Case 3:**
   - \( n = 3 \), \( k = 3 \)
   - \( a = [1, 1, 1] \), \( b = [2, 2, 2] \)
   - All customers will buy the tree at price 1, and all will leave negative reviews.
   - Expected output: \( 3 \)

4. **Test Case 4:**
   - \( n = 4 \), \( k = 3 \)
   - \( a = [1, 1, 1, 1] \), \( b = [2, 2, 2, 2] \)
   - All customers will buy the tree at price 1, and all will leave negative reviews.
   - Expected output: \( 4 \)

5. **Test Case 5:**
   - \( n = 3 \), \( k = 1 \)
   - \( a = [1, 1, 1] \), \( b = [2, 2, 2] \)
   - All customers will buy the tree at price 1, and only one can leave a negative review.
   - Expected output: \( 3 \)

These test cases are designed to cover edge cases and scenarios where the number of negative reviews is a critical factor. The input is small-scale and complete, ensuring that it can be used to validate the correctness of the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
import bisect
import sys

def main():
    input = sys.stdin.read
    data = input().split()
    idx = 0
    t = int(data[idx])
    idx += 1
    for _ in range(t):
        n = int(data[idx])
        k = int(data[idx + 1])
        idx += 2
        a = list(map(int, data[idx:idx + n]))
        idx += n
        b = list(map(int, data[idx:idx + n]))
        idx += n
        # Collect all possible price points
        P = sorted(set(a + b))
        # Sort a in descending order
        A = sorted(a, reverse=True)
        # Sort b in ascending order
        B = sorted(b)
        max_earnings = 0
        for p in P:
            # Number of positive buyers: a_i >= p
            pos = bisect.bisect_left(A, p)
            positive_buyers = n - pos
            # Number of buyers with b_i >= p
            neg = bisect.bisect_right(B, p - 1)
            total_buyers = n - neg
            # Negative buyers
            negative_buyers = total_buyers - positive_buyers
            if negative_buyers <= k:
                earnings = p * (positive_buyers + negative_buyers)
                if earnings > max_earnings:
                    max_earnings = earnings
        print(max_earnings)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the logic of the algorithm might fail. Here are some key points to consider:

1. **Edge Case with No Negative Reviews Allowed (k = 0):**
   - Ensure that the price is set such that no negative reviews are generated.
   - This can be a scenario where all `a_i` values are very low, and all `b_i` values are much higher.

2. **Edge Case with All Negative Reviews Allowed (k = n):**
   - Ensure that the price is set such that all customers buy the tree, even if it means receiving all negative reviews.
   - This can be a scenario where all `a_i` values are very high, and all `b_i` values are just slightly higher.

3. **Edge Case with Mixed Reviews:**
   - Ensure that the price is set such that the number of negative reviews is exactly equal to `k`.
   - This can be a scenario where `a_i` and `b_i` values are spread out to create a mix of positive and negative reviews.

4. **Edge Case with Duplicates:**
   - Ensure that the code handles duplicate values correctly.
   - This can be a scenario where some `a_i` and `b_i` values are the same.

5. **Edge Case with Large Values:**
   - Ensure that the code handles large values correctly.
   - This can be a scenario where the values of `a_i` and `b_i` are close to the upper limit of the range.

Based on these considerations, let's create a small-scale test input that can expose potential issues in the code:

```plaintext
4
5 0
1 1 1 1 1
2 2 2 2 2
5 5
1000000000 1000000000 1000000000 1000000000 1000000000
1000000001 1000000001 1000000001 1000000001 1000000001
5 2
1 2 3 4 5
6 7 8 9 10
5 3
1 2 3 4 5
6 7 8 9 10
```

Explanation:
- **First Test Case (k = 0):** All `a_i` values are 1, and all `b_i` values are 2. The price should be set to 1 to ensure no negative reviews.
- **Second Test Case (k = 5):** All `a_i` values are 1000000000, and all `b_i` values are 1000000001. The price should be set to 1000000001 to ensure all customers buy the tree with negative reviews.
- **Third Test Case (k = 2):** `a_i` values are 1, 2, 3, 4, 5, and `b_i` values are 6, 7, 8, 9, 10. The price should be set to 3 to ensure exactly 2 negative reviews.
- **Fourth Test Case (k = 3):** `a_i` values are 1, 2, 3, 4, 5, and `b_i` values are 6, 7, 8, 9, 10. The price should be set to 4 to ensure exactly 3 negative reviews.

This test input should help expose any issues in the logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
for _ in range(int(input())):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    if k == 0:
        best = min(a)
        print(best * n)
    elif k == n:
        best = min(b)
        print(best * n)
    else:
        res = []
        for i in range(max(b)):
            sk = 0
            ss = 0
            for j in range(n):
                if i <= a[j]:
                    ss += i
                elif i <= b[j] and sk < k:
                    ss += i
                    sk += 1
                elif sk == k:
                    ss = 0
                    break
            res.append(ss)
        print(max(res))

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

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

Think step by step.

Response:
```plaintext
1
5 2
1 2 3 4 5
6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Queue;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;

@SuppressWarnings("DuplicatedCode")
public class SolutionB {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
        List<Integer> nextIntList(int n) {
            List<Integer> list = new ArrayList<>();
            for (int i=0; i<n; i++) {
                list.add(nextInt());
            }
            return list;
        }
        List<Long> nextLongList(int n) {
            List<Long> list = new ArrayList<>();
            for (int i=0; i<n; i++) {
                list.add(nextLong());
            }
            return list;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

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

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

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

        public void printf(double num, int precision) throws IOException {
            print(String.format("%." + precision + "f", num));
        }

        public void printfln(double num, int precision) throws IOException {
            print(String.format("%." + precision + "f", num));
            bw.append("\n");
        }

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

    static final long MOD = (long) 1e9 + 7;

    private static class lPair {
        long first;
        long second;

        public lPair(long a, long b) {
            first = a;
            second = b;
        }
    }

    private static lPair isPossible(long price, int k, List<lPair> list) {
        long sum = 0;
        int count = 0;
        int n = list.size();

        for (int i=0; i<n; i++) {
            if (list.get(i).first >= price) {
                sum += price;
            } else if (list.get(i).second >= price) {
                sum += price;
                count++;
            }
        }
        return new lPair(sum, count);
    }
    public static void main(String[] args) {
        try {
            FastReader cin = new FastReader();
            FastWriter cout = new FastWriter();

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

                List<lPair> list = new ArrayList<>();
                long maxPrice = 0;
                for (int i=0; i<n; i++) {
                    list.add(new lPair(cin.nextLong(), -1));
                    maxPrice = Math.max(maxPrice, Math.max(list.get(i).first, list.get(i).second));
                }
                for (int i=0; i<n; i++) {
                    list.get(i).second = cin.nextLong();
                    maxPrice = Math.max(maxPrice, Math.max(list.get(i).first, list.get(i).second));
                }

                long low = 0;
                long high = maxPrice;
                long ans = 0;

                while (low <= high) {
                    long mid = low + ((high - low) >> 1);
                    lPair temp = isPossible(mid, k, list);
                    if (temp.second <= k) {
                        ans = Math.max(ans, temp.first);
                        low = mid+1;
                    } else {
                        high = mid-1;
                    }
                }
                cout.println(ans);
            }

            cout.close();
        } catch (Exception e) {
            e.printStackTrace();
            return;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Best Price
time_limit_ms: 2000
memory_limit_mb: 256
Description: A batch of Christmas trees has arrived at the largest store in Berland. $$$n$$$ customers have already come to the store, wanting to buy them.

Before the sales begin, the store needs to determine the price for one tree (the price is the same for all customers). To do this, the store has some information about each customer.

For the $$$i$$$-th customer, two integers $$$a_i$$$ and $$$b_i$$$ are known, which define their behavior:

- if the price of the product is at most $$$a_i$$$, the customer will buy a tree and leave a positive review;
- otherwise, if the price of the product is at most $$$b_i$$$, the customer will buy a tree but leave a negative review;
- otherwise, the customer will not buy a tree at all.

Your task is to calculate the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 2 \cdot 10^9$$$; $$$a_i < b_i$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — the maximum possible earnings for the store, given that it can receive no more than $$$k$$$ negative reviews.

Examples:
input:
5
2 0
2 1
3 4
1 1
2
5
3 3
1 5 2
3 6 4
4 3
2 3 2 8
3 7 3 9
3 1
2 9 5
12 14 9
output:
2
5
9
14
15

Note: Consider the example from the statement:

- In the first test case, the price should be set to $$$1$$$. Then both customers will buy one tree each and leave no negative reviews;
- In the second test case, the price should be set to $$$5$$$. Then the only customer will buy a tree and leave a negative review;
- In the third test case, the price should be set to $$$3$$$. Then all customers will buy one tree each, and the store will receive two negative reviews.
- In the fourth test case, the price should be set to $$$7$$$. Then two customers will buy one tree each, and the store will receive one negative review.

Buggy Code:
import java.util.*;

public class MaxEarnings {

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

        int t = sc.nextInt(); // Number of test cases

        while (t-- > 0) {
            int n = sc.nextInt(); // Number of elements in the arrays
            long k = sc.nextLong(); // The threshold value

            long[] arrA = new long[n];
            long[] arrB = new long[n];

            // Read elements into arrays arrA and arrB
            for (int i = 0; i < n; i++) arrA[i] = sc.nextLong();
            for (int i = 0; i < n; i++) arrB[i] = sc.nextLong();

            // Sort both arrays
            Arrays.sort(arrA);
            Arrays.sort(arrB);

            // Create a combined list and remove duplicates
            TreeSet<Long> combinedSet = new TreeSet<>();
            for (long val : arrA) combinedSet.add(val);
            for (long val : arrB) combinedSet.add(val);

            long maxEarnings = 0;

            // Iterate over all unique values in the combined set
            for (long value : combinedSet) {
                // Find the position of 'value' in arrA and arrB using binary search
                int lbA = Arrays.binarySearch(arrA, value);
                if (lbA < 0) lbA = -(lbA + 1);

                int lbB = Arrays.binarySearch(arrB, value);
                if (lbB < 0) lbB = -(lbB + 1);

                // Calculate D (difference between the number of elements before value in arrA and arrB)
                long D = lbA - lbB;

                if (D <= k) {
                    long buyers = n - lbB;
                    long earnings = value * buyers;
                    maxEarnings = Math.max(maxEarnings, earnings);
                }
            }

            System.out.println(maxEarnings); // Output the maximum earnings
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 2
1 2 3 4 5
6 7 8 9 10
```