**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.time.chrono.MinguoEra;
import java.util.*;
import java.io.PrintWriter;

public class A2027 {
    final static FastReader reader = new FastReader();
    final static PrintWriter pw = new PrintWriter(System.out);
    final static int MOD = (int) 1e9 + 7;

    public static void main(String[] args) {
        int t = reader.nextInt();
        while (t-- > 0) {
            int n = reader.nextInt();
            Map<Long, List<Long>> surplus = new HashMap<>();
            for (int i = 0; i < n; i++) {
                long a = reader.nextLong();
                long diff = a - (n - i);
                if (diff >= 0 && diff <= 600000) {
                    List<Long> integers = surplus.get(diff);
                    if (integers == null) {
                        List<Long> l = new ArrayList<>();
                        l.add((long) (i + 1));
                        surplus.put(diff, l);
                    } else {
                        integers.add((long) (i + 1));
                    }
                }
            }

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

            long[] dp = new long[350000];
            Arrays.fill(dp, 0);
            solve(dp, 0, surplus);

            pw.println(n + dp[0]);
            pw.flush();
        }
    }

    private static long solve(long[] dp, long extra, Map<Long, List<Long>> surplus) {
        if (surplus.get(extra) == null) {
            return 0;
        }

        if (dp[(int) extra] != 0) {
            return dp[(int) extra];
        }

        long max = 0;
        List<Long> integers = surplus.get(extra);

        for (Long integer : integers) {
            if (integer != 1) {
                max = Math.max(solve(dp, extra + integer - 1, surplus) + integer - 1, max);
            }
        }
        dp[(int) extra] = max;
        return max;

    }

    static long fastpow(long n, long ti) {
        if (ti == 0) return 1l;
        if (ti % 2 == 0) {
            long y = fastpow(n, ti / 2);
            long k = y * y;
            k %= MOD;
            return k;
        } else {
            long y = fastpow(n, ti / 2);
            long k = ((n * y) % MOD) * y;
            k %= MOD;
            return k;
        }
    }

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

    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

        String next() {
            while (st == null || !st.hasMoreElements()) {
                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 {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the code seems to be related to the condition `if (diff >= 0 && diff <= 600000)` and the way it handles the `surplus` map. This condition might be too restrictive or incorrectly handling some edge cases.

2. **Edge Cases**:
   - **Single Element Array**: The code already handles this case correctly, so it's not a good candidate.
   - **Array with Elements Close to the Condition**: Elements close to the boundary of `diff` can expose issues.
   - **Array with Elements That Satisfy the Condition Multiple Times**: This can help in testing the recursive logic.
   - **Array with Elements That Do Not Satisfy the Condition**: This can help in testing the fallback logic.

3. **Small-Scale Test Input**:
   - We need a small input that can be easily verified and is likely to expose the bug.

Let's create a test case with an array that has elements close to the boundary and elements that satisfy the condition multiple times.

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

### Explanation:
- **Test Case**: `1`
  - **Array Length**: `5`
  - **Array**: `[6, 5, 4, 3, 2]`

### Reasoning:
- For `i = 1`, `a_1 = 6` and `diff = 6 - (5 - 1) = 2` (satisfies the condition).
- For `i = 2`, `a_2 = 5` and `diff = 5 - (5 - 2) = 2` (satisfies the condition).
- For `i = 3`, `a_3 = 4` and `diff = 4 - (5 - 3) = 2` (satisfies the condition).
- For `i = 4`, `a_4 = 3` and `diff = 3 - (5 - 4) = 2` (satisfies the condition).
- For `i = 5`, `a_5 = 2` and `diff = 2 - (5 - 5) = 2` (satisfies the condition).

This input should expose any issues with the condition `if (diff >= 0 && diff <= 600000)` and the way the `surplus` map is handled. If the code is not correctly handling these repeated conditions, it might fail to compute the correct maximum possible length of the array.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Add Zeros
time_limit_ms: 3000
memory_limit_mb: 256
Description: You're given an array $a$ initially containing $n$ integers. In one operation, you must do the following:

  * Choose a position $i$ such that $1 < i \le |a|$ and $a_i = |a| + 1 - i$, where $|a|$ is the current size of the array.    * Append $i - 1$ zeros onto the end of $a$.

After performing this operation as many times as you want, what is the maximum possible length of the array $a$?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). The description of the test cases follows.

The first line of each test case contains $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^{12}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible length of $a$ after performing some sequence of operations.

Examples:
input:
4
5
2 4 6 2 5
5
5 4 4 5 1
4
6 8 2 3
1
1
output:
10
11
10
1

Note: In the first test case, we can first choose $i = 4$, since $a_4 = 5 + 1 - 4 = 2$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0]$. We can then choose $i = 3$ since $a_3 = 8 + 1 - 3 = 6$. After this, the array becomes $[2, 4, 6, 2, 5, 0, 0, 0, 0, 0]$, which has a length of $10$. It can be shown that no sequence of operations will make the final array longer.

In the second test case, we can choose $i=2$, then $i=3$, then $i=4$. The final array will be $[5, 4, 4, 5, 1, 0, 0, 0, 0, 0, 0]$, with a length of $11$.

Buggy Code:
import javax.print.DocFlavor;
import java.util.*;

public class sep27 {
    public static long dprec(long[] a, long l, Map<Long, Long> m){
        if(m.containsKey(l)) return m.get(l);

        long maxL = l;
        for(int i = 1; i < a.length; i++){
            if(a[i] == l - i){
                long nl = l + i;
                maxL = Math.max(maxL, dprec(a, nl, m));
            }
        }
        m.put(l, maxL);
        return maxL;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0) {
            long n = sc.nextLong();
            long[] a = new long[(int)n];
            for(int i = 0; i < n; i++) a[i] = sc.nextLong();
            Map<Long, Long> m = new HashMap<>();
            System.out.println(dprec(a, n, m));
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
// HỒ LY IS DA BEST
// YOASOBI - 「Halzion」 2:08
//
// ..............................................................-...................=......
// ..#######*=.*#*:::#*=-....:################..###############=............##=..=##*:-###..
// ..###.....-.--*#*:==..::............-...-=#+.:-.........-....:.......###-:-+*+:--:.=###..
// ..###.....*::.--=*=-...........#.##+=:......*#-.........:+*#--:*****#=:*-+----.**#..###..
// ..###......-.=.--=:-+--.-...*#*--......=####:..=:..-..=.....-##.-----***-:..=..-....###..
// ..###......#:....:-..:=---=#-:...::.:##......:##.:###...::..:..+#...**=**-=.=**.....###..
// ..###.......#**.-*+*+===+**....:-..#...:-.=:..::....####..+...:.-#-*+==**.-----.....###..
// ..###........##...:::.-:..:.......#..:-..........-++:-###..::...-#+===-+*.+*-.......###..
// ..###...........:...-...#.:..#--.=..........:..--#=*##.--##.:=..-#==##=+*.--........###..
// .-###.........#**#+-...#....#+=#-=+#*=*.+---..---##.......##.:..-##=+*++*...........###..
// ..###............-...-*#:..#+###=--:::=*+##-..-#.#..+..--..-..=..-##%.**...**.......###..
// ..###.............-.+#+:..#*---:+*#########-#....:.:#-....:.#....--#%-**.***........###..
// ..###.............-#..=::#--.................-..:-.-..:--+:....:..-*--.*****........###..
// ..###...........--....##--...:..+.....-=-....-.:-..#..-..--.#.=....--#*.%%*.........###..
// ..###..........-*....*-....-..:.......:........-...=#-.....+#.#...#-.**.#%..........###..
// ..###.........-.....#-...:..+-....:....-....-...-.:-........:#+.:###****............###..
// ..###.......-......-...:.:#=..+-#:#:###....-...--.-..........-:..##+-#**##..........###..
// ..###......-.........:=*=:....###.:.......-....#:..........:..:..--##=**:#-.........###..
// ..###.....-..-=######+....:-:.......=....+...#-......-##=:...-....-##=*#:.=.........###.=
// ..###.............*.....:=.........=...+-...-.....-#@@%@@@####=.....##*..+..........###..
// ..###.....##-++=--::-+##...:-.....:..++..........+..*%%%%%...:...:-.#:...+.:........###..
// ..###............+#............+#+##=:.#.+...........-.:*##.::.=.=:....=..*.-.:.....###..
// ..###.=..........#############@#####-##..............---:...-.:.:+..#..=:..:.+#--...#*#..
// ..###...........###*##-#-####..:--:.........-:................::-:.#.:.#...=....##-:###..
// ..###.........*####----#.###*#.....:.......................:=:.....#.:.-#.:::.......:.:..
// ..###........--:-......###+=-##..........................-#.....##.#....*.::::.-....###..
// ..###.......:.....=:...##=++=###.......................#.+...####.##..:.*-.-::..*...###..
// ..###......#.::#.:-:...##+++=#*##.........#.....##.....=.#.:###.-###..:..*..:-...#..###..
// ..###.....-.::=-.+:..#-.#++++#*###*..........-.-......-+..#**#.#####..-..+....-:.:#=..#..
// ..###.......:#..::..##=.#+==#*#*+####...................###:.###=#:#......#.-...:.:###=..
// .......-:--#.=:.-..####.##=####-##+####%..............#-##-#=-+#++##.:....-*.....-.::.#..
// ..###-.#=...##..-.=#*:##.####:----*=#.-###*........##+-.###.#:--#+##.-:....-=.......:###.
// ..#**.....:##...-:####*#####:====#-##-=:#-----#+#----+*%+.-#=#---##..:.#....-=:...-.:.#..
// ..###....+#.-...:#+--*+=##-*==+==*=#+-=-%%%%%%%%%%%%%%%%*:--=##-**#....#..:..:#......::=.
// ..###-.-###.#...##+=+=##+-*===+=*:##---#%%%%%%%%##%##%#.#-..--=##....#.#...-..-....:..-:.
// ..+...##....#.-.:######+#+==----.=+..-#+.:---------:-...###..+*.....##.-#.:.::.-...:.=.:.
// .-..#-.:..###.-.-....:+#.#**+##=########*.............*##*###..-..-#+=#.#....:.-....:.::.
// ...##.-..##.#.:..-.....-#...*.**#*#***#..............*##**..##...*=-#-#.#-.::.-.....:.#..
// .-###:..+##.-.::..:#.....-###*#**####**##.=........##=#**###..##########.#..:.#.....:.#..
// ..##...:#*#...::#...:=..:*=..#=#*###########:-:..####***.:.###*=#-#...:##-#...:...:.:.#..
// ..#:...#+#..=:..#=--...=-#.-..####:.............##=#*:.##*******###..........#....-.:..#.
// ..-....#=#..-:=....#-:##=...-.........#####.:.####*******+###--=#:+.............:.-..#.*.
// .#....#*-#.==-.:-+...--............-########.##+***###----=**##--.-...........:-.-..#.#-.
// .+....#-##..-++:..#.########+.-............#####---:=****#--=#..+#-........-.--.:..:.#--.
// ..::...###...-+##-.........--#................:##**#-----#......-#:.....=.:-+:...:.###-..
// ..:..#..#.:...............---+....................---=.........*--..:#-+###...:=##..#.-..
// ..::.##-..--=............###=..........:=####-::...............-###-----+...:##---..##-..
// .......##..............:=-..........:............####..........-:-.................**+-..
#include <iostream>
#include <stack>
#include <algorithm>
#include <map>
#include <numeric>
#include <vector>
#include <string>
#include <deque>
#include <climits>
#include <functional>
#include <cassert>
#include <math.h>
#include <set>
#include <queue>
#define ll long long
#define ull unsigned ll
using namespace std;
typedef vector<vector<int>> vvi;
typedef vector<int> vi;

int r = 0;
#define fori(st,end,step) for(int i = st ;i < end;i+= step)
#define fort(tenbien,st,end,step) for(int tenbien = st ;tenbien < end;tenbien+= step)
template <typename T>
void PrintVector(vector<T> &a)
{
    for (auto &x : a)
    {
        cout << x << ' ';
    }

}
template <typename T>
void ReadVector(vector<T> &a)
{
    for (auto &x : a)
    {
        cin >> x;
    }
}
#include <cstring>
#define tc
const int mod = 1e9  + 7;
#define int long long
struct ev
{
    int start;
    int objend = -1;
    vi val;
};

auto TestCase(int test)
{
    int n,m,k;
    cin >> n >> m >> k;
    int h[n];
    int x[n];
    for (int i = 0;i < n;i++)
        cin >> h[i];
    for (int i =0 ;i < n;i++)
        cin >> x[i];
    int lp = 0,rp = 0;
    int sum  = 0;
    int minhp = INFINITY;
    int lpm,rpm;
    auto checkmin = [&](int newmin)
    {
        if (x[rp] - x[lp] <=  2 * (m - 1) && rp - lp + 1 >= k)
            {
                if (minhp > newmin)
                {

                    minhp = newmin;
                    lpm = lp;
                    rpm = rp;
                }
            }
    };
    for (rp = 0;rp < n;rp++)
    {

        if (rp - lp + 1 <= k)
            sum += h[rp],
            checkmin(sum);
        else
        {
            sum -= h[lp++];
            sum += h[rp];
            checkmin(sum);

        }
    }
    if (minhp == LLONG_MAX)
    {
        cout << -1 << '\n';
        return;
    }
    int p = (x[rpm] + x[lpm]) / 2;
    int res = 0;
    for (int i = lpm;i <= rpm;i++)
    {
        int damage  = m - abs(p - x[i]);
        res = max(res ,h[i]  / damage + (h[i] % damage != 0) );
    }

    cout << res << endl;
}
main()
{
    int t;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    #ifdef local
    freopen("input","r",stdin);
    #endif
    int i;
    #ifdef tc
    cin >> t;
    for (i = 0;i < t;i++)
    #endif
        TestCase(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:
To expose the bug in the given code, we need to focus on edge cases where the algorithm might fail. The code is supposed to find the minimum number of attacks required to defeat at least `k` enemies by positioning Xilonen at an optimal point. The key points to consider are:

1. **Positioning Xilonen at the correct point**: The code might not correctly handle the optimal positioning of Xilonen.
2. **Damage calculation**: The code might not correctly calculate the damage dealt to enemies, especially when the positions of enemies are far apart.
3. **Edge cases**: The code might fail when `k` is equal to `n` or when the enemies are positioned in a way that makes it impossible to defeat `k` enemies with the given attack damage.

Let's generate a small-scale test input that can expose these issues:

### Test Input
```plaintext
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
```

### Explanation
1. **First Test Case**:
   - `n = 5`, `m = 5`, `k = 3`
   - Enemy health: `7 7 7 7 7`
   - Enemy positions: `1 2 3 4 5`
   - Optimal position: `p = 3`
   - Damage calculation: Each attack, the first enemy takes `4` damage, the second enemy takes `5` damage, the third enemy takes `4` damage, the fourth enemy takes `3` damage, and the fifth enemy takes `2` damage.
   - Expected output: `2` (since it takes 2 attacks to defeat the first three enemies).

2. **Second Test Case**:
   - `n = 9`, `m = 5`, `k = 9`
   - Enemy health: `2 4 6 8 10 8 6 4 2`
   - Enemy positions: `1 2 3 4 5 6 7 8 9`
   - Optimal position: `p = 5`
   - Damage calculation: All nine enemies will be defeated in 2 attacks.
   - Expected output: `2`.

3. **Third Test Case**:
   - `n = 2`, `m = 10`, `k = 2`
   - Enemy health: `1 1`
   - Enemy positions: `1 20`
   - No position can damage both enemies at the same time.
   - Expected output: `-1`.

4. **Fourth Test Case**:
   - `n = 2`, `m = 10`, `k = 1`
   - Enemy health: `69696969 420420420`
   - Enemy positions: `1 20`
   - Optimal position: `p = 1`
   - Damage calculation: The first enemy will be defeated in 6969697 attacks.
   - Expected output: `6969697`.

5. **Fifth Test Case**:
   - `n = 2`, `m = 10`, `k = 2`
   - Enemy health: `10 15`
   - Enemy positions: `1 19`
   - Optimal position: `p = 10`
   - Damage calculation: Both enemies will be defeated in 15 attacks.
   - Expected output: `15`.

6. **Sixth Test Case**:
   - `n = 2`, `m = 2`, `k = 2`
   - Enemy health: `1000000000 1`
   - Enemy positions: `1 3`
   - Optimal position: `p = 2`
   - Damage calculation: The first enemy will take 1 damage per attack, and the second enemy will take 0 damage per attack.
   - Expected output: `1000000000`.

This test input covers a variety of scenarios, including edge cases where the enemies are positioned far apart, where `k` is equal to `n`, and where it is impossible to defeat `k` enemies. These cases should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

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

using namespace std;

typedef long long ll;
typedef pair<ll, ll> ii;

constexpr ll INF = 1e9 + 5;

struct Enemy{
    ll h;
    ll x;
};

ll my_abs(ll n){
    if(n > 0) return n;
    return -n;
}

ll n_strikes(ll health, ll power){
    if(power == 0) return INF;
    return (health / power) + (health % power != 0);
}

ll attack_f(ll m, ll p, ll x){
    // cout << "m: " << m << " p: " << p << " x: " << x << " res: " << m - my_abs(p - x) << '\n';
    return m - my_abs(p - x);
}

ll calculate_minimum(vector<Enemy> enemies, int m, int p, int k){
    priority_queue<ll> queue;
    for(int i = 0 ; i < enemies.size(); i++){
        ll damage = max(0LL, attack_f(m, p, enemies[i].x));
        // cout << "the attack is " << attack(he[i], damage) << " damage " << damage  << '\n';
        ll ns = n_strikes(enemies[i].h, damage);
        // cout << "the n strikes is " << ns << " with damage " << damage << '\n';
        queue.push(-ns);
    }
    int counter = 0;
    ll ans = 0;
    while(!queue.empty()){
        if(counter == k) break;

        ll value = -queue.top(); queue.pop();
        if (value == INF){
            return INF;
        }

        counter++;
        ans = max(ans, value);
    }
    return ans;
}

ll search_p(vector<Enemy> &enemies, int minxx, int maxx, int &m, int &k){
    ll low = minxx;
    ll hi = maxx;
    ll mid;
    ll ans = INF;
    while (low <= hi){
        mid = (low + hi) >> 1;
        // cout << "Low: " << low << " Hi: " << hi << " Mid: " << mid << '\n';
        ll tmp = calculate_minimum(enemies, m, mid, k);
        // cout << "Ans: " << tmp << '\n';
        if(tmp <= ans && tmp != -1){
            hi = mid - 1;
            ans = tmp;
        } else {
            low = mid + 1;
        }
    }
    // cout << "Ans for " << startp << " and " << endp << " is " << ans << '\n';
    return ans;
}

ll search2_p(vector<Enemy> &enemies, int minx, int maxx, int &m, int &k){
    ll low = minx;
    ll hi = maxx;
    ll mid;
    ll ans = INF;
    while (low <= hi){
        mid = (low + hi) >> 1;
        // cout << "Low: " << low << " Hi: " << hi << " Mid: " << mid << '\n';
        ll tmp = calculate_minimum(enemies, m, mid, k);
        // cout << "Ans: " << tmp << '\n';
        if(tmp <= ans && tmp != -1){
            low = mid + 1;
            ans = tmp;
        } else {
            hi = mid - 1;
        }
    }
    // cout << "Ans for " << startp << " and " << endp << " is " << ans << '\n';
    return ans;
}

bool cmp(Enemy o, Enemy s) {
    if(o.h < s.h) return true;
    else if(o.h > s.h) return false;
    return o.x < s.x;
}

ii get_range(Enemy e, int m){
    ll first_side = max(0LL, (-1 + m + e.x));
    ll second_side = max(0LL, (1 - m + e.x));
    return {min(first_side, second_side), max(first_side, second_side)};
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(nullptr); cout.tie(nullptr);
    int t, n, m, k;
    cin >> t;
    while(t--){
        cin >> n >> m >> k;
        vector<Enemy> enemies(n);
        ll minimum_x = INF, maximum_x = 0;

        for(int i = 0 ; i < n ; i++) cin >> enemies[i].h;
        for(int i = 0 ; i < n ; i++){
            cin >> enemies[i].x;
            minimum_x = min(minimum_x, enemies[i].x);
            maximum_x = max(maximum_x, enemies[i].x);
        }

        // for(int i = 0 ; i < n ; i++){
        //     ii rng = get_range(enemies[i], m);
        //     cout << rng.first << ' ' << rng.second << '\n';
        // }

        ll ans1 = search_p(enemies, minimum_x, maximum_x, m, k);
        ll ans2 = search2_p(enemies, minimum_x, maximum_x, m, k);

        ll ans = min(ans1, ans2);

        if(ans == INF) cout << "-1";
        else cout << ans;
        cout << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
#include <iostream>
#include <iomanip>
#include <vector>
#include <array>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <numeric>
#include <string>
#include <cstring>
#include <bitset>
#include <random>
#include <tuple>
using namespace std;

//_-----------------------------
/*
#define mp make_pair
#define  isz(x) int((x).size())
#define ins insert
#define pb push_back
#define ft front()
#define bk back()
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define sortlargetosmall(x) sort(all(x), greater<>())
#define sortsmalltolarge(x) sort(all(x), less<>())
#define fin(x) for (auto &it: x) cin >> it
#define fout(x) for (auto &it: x) cout << it << ' '; cout << en
#define finpairs(x) for (auto &it: x) cin >> it.first >> it.second
#define foutpairs(x) for (auto &it: x) cout << it.first << ' '<<it.second << ' '; cout << en
#define cno cout << "NO" << en
#define cyes cout << "YES" << en
#define PI acos(-1.0L)
#define fac(x, col) for(auto x: col)
#define facl(x, col) for(auto &x: col)
#define forn(start, end) for(int i = start; i < end; ++i)
#define cmo cout << -1 << en
#define ifyn(x) x ? cyes : cno
#define xout(x) cout << x << en
#define xin(x) cin >> x
#define nline cout << en
#define cen cerr <<"\n"
*/
//_---------_-
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void print(const pair<T, V> &x) {cerr << '{'; print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}

#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

#define all(v) ((v).begin()), ((v).end())
#define rall(v) ((v).rbegin()), ((v).rend())
#define sz(v) ((int)((v).size()))
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define eb emplace_back
#define ef emplace_front

#define _sum(a) accumulate(all(a), 0ll)
#define _min(a) *min_element(all(a))
#define _max(a) *max_element(all(a))
#define _cnt(a, b) count(all(a), b)
#define _iota(a, b) iota(all(a), b)
#define _fill(a, b) fill(all(a), b)

#define rep(n) for(int i=0;i<(n);i++)
#define repr(n) for(int i=(n)-1;i>=0;i--)

#define reps(n,s) for(int i=0;i<(n);i+=s)
#define reprs(n,s) for(int i=(n)-1;i>=0;i-=s)

#define repi(s,n) for(int i=s;i<(n);i++)
#define repri(s,n) for(int i=(n)-1;i>=s;i--)

#define rrep(n) for(int j=0;j<(n);j++)
#define rrepr(n) for(int j=(n)-1;j>=0;j--)
#define rrepi(s,n) for(int j=s;j<(n);j++)
#define rrepri(s,n) for(int j=(n)-1;j>=s;j--)

#define rrrep(n) for(int k=0;k<(n);k++)
#define rrrepr(n) for(int k=(n)-1;k>=0;k--)
#define rrrepi(s,n) for(int k=s;k<(n);k++)
#define rrrepri(s,n) for(int k=(n)-1;k>=s;k--)
#define _even(n) for(int i=0;i<(n);i+=2)
#define _odd(n) for(int i=1;i<(n);i+=2)

#define _evenr(n) for(int i=n-1;i>=0;i-=2)
#define _oddr(n) for(int i=n-1;i>=1;i-=2)

#define fir first
#define sec second
#define pb push_back

#define ll long long
//-----
void removeIdx(vector<int>& v,int idx){
	 v.erase(v.begin()+idx);
}
void removeAll(vector<int>& v,int num){
	v.erase(remove(all(v), num),v.end());
}
bool removeFirst(vector<int> &v,int num){

	auto it=find(all(v),num);

    if (it!=v.end()){
        v.erase(it);
        return 1;
    }
    return 0;
}
bool removeLast(vector<int> &v,int num){

	auto it=find(rall(v),num);

    if (it!=v.rend()){
        v.erase((it+1).base());
        return 1;
    }
    return 0;
}
//------

//-----------------------------

template<typename T>
void printV(vector<T> &v){
    int n=v.size();
    for(int i=0;i<n;i++)
        cout<<v[i]<<" ";
    cout<<endl;
}

template<typename T,typename F>
void printP(vector<pair<T,F>> &v){
    int n = v.size();
    for(int i=0;i<n;i++)
        cout<<"{ "<<v[i].fir<<" , "<<v[i].sec<< " } ";
    cout<<endl;
}
template<typename T,typename F,typename H>
void printT(vector<tuple<T,F,H>> &v){
    int n = v.size();
    for(int i=0;i<n;i++)
        cout<<"{ "<<get<0>(v[i])<<" , "<<get<1>(v[i])<<" , "<<get<2>(v[i])<< " } ";
    cout<<endl;
}
template<typename T,typename F>
void printM(map<T,F> &mp){
    for(auto p:mp)
        cout<<"{ "<<p.fir<<" , "<<p.sec<< " } ";
    cout<<endl;
}
template<typename T>
void printSt(stack<T> stk){
     while(!stk.empty()){
     	cout<<stk.top()<<" ";
         stk.pop();
     }
     cout<<endl;
}
void makeSuff(vector<vector<int>> &a, string s){
    int n=s.size();
    for(int i=0;i<n;i++){
        a[s[i]-'a'][i]++;
        for(int j=0;j<26;j++){
            if(i>0){
                a[j][i]+=a[j][i-1];
            }
        }
    }
}
int numberOfChar(int l,int r,vector<vector<int>> &v){
        l--; r--;
        int diff=0;
        for(int i=0;i<26;i++){
            int cnt=v[i][r]-(l>0?v[i][l-1]:0);
            diff+=cnt;
        }
        return diff;
}
bool isPrime(int x) {
	if (x <= 4 || x % 2 == 0 || x % 3 == 0) return x == 2 || x == 3;
	for (int i = 5; i * i <= x; i += 6) if (x % i == 0 || x % (i + 2) == 0) return 0; return 1;
}
ll fact(ll num){
	if(num<=1) return num;
	return num*fact(num-1);
}
ll C(ll a, ll b) {
    if (b > a) return 0;
    if (b == 0 || b == a) return 1;
    ll res = 1;
    for (ll i = 1; i <= b; ++i) {
        res = res * (a - i + 1) / i;
    }
    return res;
}

int sums(ll s, ll e){
    return (e-s+1)*(s+e)/2;
}

void format(int num,int size){
	cout<<fixed<<setprecision(size)<<num<<endl;

}

//const ll MN=INT_MAX,MX=INT_MIN,mod = 1e9 + 7;

int bs(vector<int> &v,int l,int r,int k){
	if(l>r) return -(r+1);

	int mid=l+(r-l)/2;
	if(v[mid]>k)
	  return bs(v,l,mid-1,k);
	else if(v[mid]<k)
	  return bs(v,mid+1,r,k);
	else return mid;
}

// أول عنصر أكبر او يساوي
int lbs(vector<int> &v,int l,int r,int k){
	if(l>r) return l;

    int mid=l+(r-l)/2;
	if(v[mid]>=k)
	  return lbs(v,l,mid-1,k);
	else
	  return lbs(v,mid+1,r,k);
}
// اول عنصر أصغر أو يساوي
int lbsr(vector<int> &v,int l,int r,int k,int result=-1){
    if (l>r) return result;

    int mid=l+(r-l)/2;

    if(v[mid]<=k){
        result=mid;
        return lbsr(v,l,mid-1,k,result);
    } else {
        return lbsr(v,mid+1,r,k,result);
    }
}

// أول عنصر أكبر فقط
int rbs(vector<int> &v,int l,int r,int k){
	if(l>r) return l;

	int mid=l+(r-l)/2;
	if(v[mid]>k)
	  return rbs(v,l,mid-1,k);
	else
	  return rbs(v,mid+1,r,k);
}
bool isSorted(deque<int> d){
	for(int i=0;i<d.size()-1;i++){
		if(d[i]>d[i+1])
		   return 0;
	}
	return 1;
}

ll can(vector<ll> &v,vector<ll> &p,ll mid,ll m,ll k){
    vector<ll> attack;

    rep(sz(v)){
        ll damage=m-abs(mid-p[i]);
        if(damage>0){
            attack.eb((v[i]+damage-1)/damage);
        }
    }

    if(k>sz(attack)) return -1;
    sort(all(attack));
    return attack[k-1];
}

void solve(){
    ll n,m,k;
    cin>>n>>m>>k;

    vector<ll> v(n),p(n);
    rep(n) cin>>v[i];
    rep(n) cin>>p[i];

    ll l=p[0],r=p[n-1];
    ll mn=1e18;

    while(l<=r){
        ll mid=l+(r-l)/2;
        ll r1=can(v,p,mid,m,k);

        if(r1!=-1){
            mn=min(mn,r1);
            r=mid-1;
        } else l=mid+1;

    }

    cout<<(mn==1e18?-1:mn)<<endl;
}
int main(){
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
   int t=1;
   cin>>t;
   while(t--){
        solve();
    }
}
/*

6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3

*/

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

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

Think step by step.

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

1. **Edge Case with High Health Values and Low Damage:**
   - This scenario tests the algorithm's ability to handle large health values and low damage per attack, which can lead to a high number of attacks required.
   - Example: \( n = 2 \), \( m = 2 \), \( k = 2 \), \( h = [1000000000, 1000000000] \), \( x = [1, 2] \).

2. **Edge Case with Distant Enemies:**
   - This scenario tests the algorithm's ability to handle enemies that are far apart, where the damage distribution is minimized.
   - Example: \( n = 2 \), \( m = 1 \), \( k = 2 \), \( h = [1, 1] \), \( x = [1, 1000000000] \).

3. **Edge Case with Minimum Health and Maximum Damage:**
   - This scenario tests the algorithm's ability to handle the simplest case where the health of each enemy is 1 and the damage is high.
   - Example: \( n = 2 \), \( m = 10 \), \( k = 2 \), \( h = [1, 1] \), \( x = [1, 2] \).

4. **Edge Case with Single Enemy:**
   - This scenario tests the algorithm's ability to handle a single enemy, which should be straightforward but can still expose issues.
   - Example: \( n = 1 \), \( m = 5 \), \( k = 1 \), \( h = [10] \), \( x = [1] \).

5. **Edge Case with Impossible Scenario:**
   - This scenario tests the algorithm's ability to handle a case where it is impossible to defeat the required number of enemies.
   - Example: \( n = 2 \), \( m = 1 \), \( k = 2 \), \( h = [10, 10] \), \( x = [1, 2] \).

Given the constraints and the need for a small-scale test input, let's combine some of these scenarios into a single test case:

```plaintext
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
1 1 1
10
1
2 2 2
1 1
1 1000000000
1 2
```

This test input includes a mix of scenarios to ensure the algorithm is thoroughly tested for various edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
def minimum_attacks(t, test_cases):
    results = []

    for n, m, k, health, positions in test_cases:
        min_attacks = float('inf')
        possible = False

        # Iterate over a range of possible positions to place Xilonen
        for p in range(positions[0], positions[-1] + 1):
            damages = []

            # Calculate the damage each enemy takes for position p
            for i in range(n):
                dist = abs(positions[i] - p)
                damage = max(0, m - dist)
                if damage > 0:
                    damages.append((damage, health[i]))

            # Sort by damage in descending order to prioritize higher damage to enemies
            damages.sort(reverse=True, key=lambda x: x[0])

            # Calculate the number of attacks required to defeat at least k enemies
            total_attacks = 0
            enemies_defeated = 0
            for damage, hp in damages:
                attacks = -(-hp // damage)  # Equivalent to math.ceil(hp / damage)
                total_attacks = max(total_attacks, attacks)
                enemies_defeated += 1
                if enemies_defeated >= k:
                    min_attacks = min(min_attacks, total_attacks)
                    possible = True
                    break

        if possible:
            results.append(min_attacks)
        else:
            results.append(-1)

    return results

def main():
    t = int(input())
    test_cases = []
    for _ in range(t):
        n, m, k = map(int, input().split())
        health = list(map(int, input().split()))
        positions = list(map(int, input().split()))
        test_cases.append((n, m, k, health, positions))

    results = minimum_attacks(t, test_cases)
    for result in results:
        print(result)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
import math

def can_defeat(n, m, k, h, x, attacks):
    for p in range(x[0] - m + 1, x[-1] + 1):
        enemies = []
        for i in range(n):
            damage = max(0, m - abs(p - x[i]))
            if damage > 0:
                enemies.append(math.ceil(h[i] / damage))
        enemies.sort()
        count = 0
        for i in range(min(k, len(enemies))):
            if enemies[i] > attacks:
                break
            count += 1
        if count >= k:
            return True
    return False

def minimum_attacks(t, test_cases):
    results = []
    for case in test_cases:
        n, m, k, h, x = case
        left, right, result = 1, 10**9, -1
        while left <= right:
            mid = (left + right) // 2
            if can_defeat(n, m, k, h, x, mid):
                result = mid
                right = mid - 1
            else:
                left = mid + 1
        results.append(result)
    return results

# Input Reading
t = int(input())
test_cases = []
for _ in range(t):
    n, m, k = map(int, input().split())
    h = list(map(int, input().split()))
    x = list(map(int, input().split()))
    test_cases.append((n, m, k, h, x))

# Processing
results = minimum_attacks(t, test_cases)

# Output Results
for res in results:
    print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
import sys

read_all = sys.stdin.read

class Point:
    def __init__(self, position, type_):
        self.position = position
        self.type = type_

    def __lt__(self, other):
        if self.position != other.position:
            return self.position < other.position
        return self.type > other.type

def split_data(data):
    return data.split()

def get_integer(tokens, pointer):
    value = int(tokens[pointer])
    return value, pointer + 1

def get_list(tokens, pointer, count):
    lst = list(map(int, tokens[pointer:pointer + count]))
    return lst, pointer + count

def parse_test_case(tokens, pointer):
    n, pointer = get_integer(tokens, pointer)
    m, pointer = get_integer(tokens, pointer)
    L, pointer = get_integer(tokens, pointer)
    health, pointer = get_list(tokens, pointer, n)
    positions, pointer = get_list(tokens, pointer, n)
    return n, m, L, health, positions, pointer

def calculate_required_hits(health, mid):
    return [(h + mid - 1) // mid for h in health]

def calculate_flexibility(required_hits, m, n, health, positions, mid):
    flexibility = []
    for i in range(n):
        if required_hits[i] > m:
            flexibility.append(None)
            continue
        delta = m - required_hits[i]
        start = positions[i] - delta
        end = positions[i] + delta
        flexibility.append((start, end))
    return flexibility

def generate_points(flexibility, n):
    points = []
    for i in range(n):
        if flexibility[i] is None:
            continue
        start, end = flexibility[i]
        points.append(Point(start, 1))
        points.append(Point(end + 1, -1))
    return points

def sort_points(points):
    points.sort()

def sweep_points(points, L):
    count = 0
    for i in range(len(points)):
        count += points[i].type
        if i + 1 < len(points) and points[i].position == points[i + 1].position:
            continue
        if count >= L:
            return True
    return False

def is_valid_mid(points, L):
    if not points:
        return False
    sort_points(points)
    return sweep_points(points, L)

def binary_search(n, m, L, health, positions):
    low, high = 1, int(1e9)
    answer = -1
    while low <= high:
        mid = (low + high) // 2
        required_hits = calculate_required_hits(health, mid)
        flexibility = calculate_flexibility(required_hits, m, n, health, positions, mid)
        points = generate_points(flexibility, n)
        valid = is_valid_mid(points, L)
        if valid:
            answer = mid
            high = mid - 1
        else:
            low = mid + 1
    return answer

def process_cases(tokens, pointer, total_cases):
    results = []
    for _ in range(total_cases):
        n, m, L, health, positions, pointer = parse_test_case(tokens, pointer)
        min_mid = binary_search(n, m, L, health, positions)
        results.append(str(min_mid))
    return results

def write_output(results):
    sys.stdout.write("\n".join(results) + "\n")

def run():
    data = read_all()
    tokens = split_data(data)
    pointer = 0
    total_cases, pointer = get_integer(tokens, pointer)
    results = process_cases(tokens, pointer, total_cases)
    write_output(results)

run()

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that could cause the algorithm to fail. Here are some potential edge cases to consider:

1. **Single Enemy Test Case**: This can help verify the basic functionality.
2. **Multiple Enemies with High Health**: This can test the algorithm's ability to handle high health values.
3. **Multiple Enemies with High Damage**: This can test the algorithm's ability to handle high damage values.
4. **Enemies with Equal Health and Positions**: This can test the algorithm's ability to handle uniform conditions.
5. **Enemies with Varying Health and Positions**: This can test the algorithm's ability to handle diverse conditions.
6. **Impossible to Defeat k Enemies**: This can test the algorithm's ability to handle cases where it is impossible to defeat the required number of enemies.

Let's create a small-scale test input that includes these edge cases:

```plaintext
6
1 10 1
100
10
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
```

This test input includes:
1. A single enemy with high health.
2. Multiple enemies with equal health and positions.
3. Multiple enemies with varying health and positions.
4. A case where it is impossible to defeat the required number of enemies.
5. High damage and health values to test the algorithm's limits.
6. A case with very high health values to test the algorithm's ability to handle large numbers.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
//package com.aakash.Contest.C988;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class F{
    static BufferedReader br = new BufferedReader(
            new InputStreamReader(System.in));

    public static void main(String[] args) throws IOException {

        StringTokenizer s1
                = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(s1.nextToken());
        while (t-- > 0) {
            StringTokenizer s2
                    = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(s2.nextToken());
            int m = Integer.parseInt(s2.nextToken());
            int k = Integer.parseInt(s2.nextToken());

            int[] pow = new int[n];
            int[] ind=new int[n];
            StringTokenizer s3
                    = new StringTokenizer(br.readLine());
            StringTokenizer s4
                    = new StringTokenizer(br.readLine());
            for (int i = 0; i < n; i++) {
                pow[i] = Integer.parseInt(s3.nextToken());
                ind[i] = Integer.parseInt(s4.nextToken());
            }
            solve(pow,ind,n,m,k);
        }

    }

    static void solve(int[] pow,int[] ind,int n,int m,int k) {
        long s=1,e=(long)1e18;
        while(s<=e) {
            long mid=s+(e-s)/2;
            if(is(pow,ind,n,m,k,mid)) e=mid-1;
            else s=mid+1;
        }
        if(s>(long)1e18) s=-1;
        System.out.println(s);
    }
    static boolean is(int[] pow,int[] ind,int n,long m,long k,long a) {
        PriorityQueue<Long> pq=new PriorityQueue<>(Collections.reverseOrder());
        ArrayList<long[]> list=new ArrayList<>();
        for(int i=n-1;i>=0;i--) {
            while(!pq.isEmpty() &&pq.peek()>ind[i]) {
                list.add(new long[]{pq.peek(),pq.size()});
                pq.remove();
            }
            long curr=ind[i]+(pow[i]/a)-m;
            if(pow[i]%a!=0) curr++;
            if(curr<=ind[i]) pq.add(curr);
            list.add(new long[]{ind[i],pq.size()});
        }
        pq=new PriorityQueue<>();
        int p=0;
//        for(int i=list.size()-1;i>=0;i--) {
//            System.out.println(Arrays.toString(list.get(i)));
//        }
        for(int i=list.size()-1;i>=0;i--) {
            long[] temp=list.get(i);
            long index=temp[0],cnt=temp[1];
            while(!pq.isEmpty() &&pq.peek()<index) pq.remove();
            if(pq.size()+cnt>=k) return true;
            if(p>=n ||ind[p]!=index) continue;
            long curr=m+ind[p]-(pow[p]/a);
            if(pow[p]%a!=0) curr--;

            if(curr>index) pq.add(curr);
            p++;
        }
        return false;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 3 3
1 2 3 4 5
1 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: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

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

public class Main {
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);

    static class Pair{
        Integer x ;
        Integer y ;

        public 	Pair(Integer i, Integer j) {
            this.x = i;
            this.y = j;
        }
    }

    public static void solve() throws IOException{
        int n = sc.nextInt(), m = sc.nextInt(), k = sc.nextInt();
        int a [] = sc.nextIntArray(n);
        int b [] = sc.nextIntArray(n);
        Queue<Pair> qu = new LinkedList<>();
        qu.add(new Pair(1, b[n - 1]));
        int ans = Integer.MAX_VALUE;
        while(!qu.isEmpty()) {
            int low = qu.peek().x;
            int high = qu.peek().y;
            qu.poll();
            if (low <= high) {
                int mid1 = low + (high - low) / 3;
                int mid2 = high - (high - low) / 3;
                int cnt1 = Integer.MAX_VALUE, cnt2 = Integer.MAX_VALUE;
                ArrayList<Integer> list1 = new ArrayList<>();
                ArrayList<Integer> list2 = new ArrayList<>();
                for (int i = 0; i < n; i++) {
                    if (Math.abs(b[i] - mid1) < m) {
                        int power = m - Math.abs(b[i] - mid1);
                        list1.add((a[i] + power - 1) / power);
                    }
                }
                Collections.sort(list1);
                for (int i = 0; i < n; i++) {
                    if (Math.abs(b[i] - mid2) < m) {
                        int power = m - Math.abs(b[i] - mid2);
                        list2.add((a[i] + power - 1) / power);
                    }
                }
                Collections.sort(list2);
                if (list1.size() >= k) {
                    cnt1 = list1.get(k - 1);
                }
                if (list2.size() >= k) {
                    cnt2 = list2.get(k - 1);
                }
                ans = Math.min(ans, Math.min(cnt1, cnt2));
                if (cnt1 > cnt2) {
                    qu.add(new Pair(mid2 + 1, high));
                } else if (cnt1 < cnt2) {
                    qu.add(new Pair(low, mid1 - 1));
                } else {
                    qu.add(new Pair(mid2 + 1, high));
                    qu.add(new Pair(low, mid1 - 1));
                    qu.add(new Pair(mid1 + 1, mid2 - 1));
                }
            }
        }
        pw.println(ans == Integer.MAX_VALUE ? -1 : ans);
    }
    public static void main(String[] args) throws IOException {
        int t = sc.nextInt();
        while(t-- > 0)
            solve();
        pw.flush();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongArray(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
/*
  Author: Pranava
  scholar id -> 2312177
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <class T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define pranav                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);
#define endl '\n'
#define mp make_pair
#define pb push_back
#define ll long long
#define ld long double
#define be begin()
#define en end()
#define rbe rbegin()
#define ren rend()
#define asort(vec) sort(vec.begin(), vec.end())
#define dsort(vec) sort(vec.begin(), vec.end(), greater<long long>())
#define vll vector<long long>
#define vpll vector<pair<long long, long long>>
#define forn(i, n) for (long long i = 0; i < n; i++)
#define rep(var, start, end) for (long long var = start; var < end; var++)
#define ff first
#define ss second
#define vi vector<int>
#define apriority priority_queue<long long, vector<long long>, greater<long long>>
#define dpriority priority_queue<long long>
#define input(vec, n)                 \
    for (long long i = 0; i < n; i++) \
    {                                 \
        cin >> vec[i];                \
    }
#define lcm(a, b) ((a * b) / (__gcd(a, b)))
#define lpow(a, b) (ll) pow(a, b)
#define minel *min_element
#define maxel *max_element

// Debugging helpers
#define debug(x) cerr << #x << " = " << (x) << endl;

// Template for modular arithmetic (modular addition, subtraction, multiplication)
const ll MOD = 1e9 + 7;
ll mod_add(ll a, ll b, ll m = MOD) { return ((a % m) + (b % m)) % m; }
ll mod_sub(ll a, ll b, ll m = MOD) { return (((a % m) - (b % m)) + m) % m; }
ll mod_mul(ll a, ll b, ll m = MOD) { return ((a % m) * (b % m)) % m; }

// Function to perform modular exponentiation
ll mod_exp(ll base, ll exp, ll mod)
{
    ll result = 1;
    while (exp > 0)
    {
        if (exp % 2 == 1)
        {
            result = mod_mul(result, base, mod);
        }
        base = mod_mul(base, base, mod);
        exp /= 2;
    }
    return result;
}

// Function to compute modular inverse
ll mod_inv(ll a, ll mod)
{
    return mod_exp(a, mod - 2, mod);
}

bool isPrime(long long n)
{
    return n > 1 && (n == 2 || (n % 2 && none_of(begin(array<int, 5>{3, 5, 7, 11, 13}), end(array<int, 5>{3, 5, 7, 11, 13}), [n](int i)
                                                 { return i * i <= n && n % i == 0; })));
}

vector<bool> primesieve(ll n)
{
    vector<bool> ans(n + 1, true);
    ans[0] = false;
    ans[1] = false;
    for (ll i = 2; i * i <= n; i++)
    {
        if (ans[i])
        {
            for (ll j = i * i; j <= n; j += i)
            {
                ans[j] = false;
            }
        }
    }
    return ans;
}

vector<ll> spf(ll n)
{
    vector<ll> ret(n + 1, 0);
    for (ll i = 1; i <= n; i++)
    {
        ret[i] = i;
    }

    for (ll i = 2; i * i <= n; i++)
    {
        if (ret[i] == i)
        {
            for (ll j = i * i; j <= n; j += i)
            {
                if (ret[j] == j)
                {
                    ret[j] = i;
                }
            }
        }
    }
    return ret;
}

void factorisation(ll n)
{
    vll factors = spf(n);
    while (n != 1)
    {
        cout << factors[n] << ' ';
        n /= factors[n];
    }
    cout << endl;
}

void print(vll a)
{
    for (auto pos : a)
    {
        cout << pos << ' ';
    }
}

pair<ll, ll> prod(ll val)
{
    if ((val & 1) == 0)
    {
        return mp((val >> 1), (val >> 1));
    }
    return mp((val >> 1), ((val >> 1) + 1));
}

class DisjointSets {
  private:
	vector<int> parents;
	vector<int> sizes;

  public:
	DisjointSets(int size) : parents(size), sizes(size, 1) {
		for (int i = 0; i < size; i++) { parents[i] = i; }
	}

	/** @return the "representative" node in x's component */
	int find(int x) { return parents[x] == x ? x : (parents[x] = find(parents[x])); }

	/** @return whether the merge changed connectivity */
	bool unite(int x, int y) {
		int x_root = find(x);
		int y_root = find(y);
		if (x_root == y_root) { return false; }

		if (sizes[x_root] < sizes[y_root]) { swap(x_root, y_root); }
		sizes[x_root] += sizes[y_root];
		parents[y_root] = x_root;
		return true;
	}

	/** @return whether x and y are in the same connected component */
	bool connected(int x, int y) { return find(x) == find(y); }
};

/*************TEMPLATE ENDS******************/

void solve(){
ll n,m,k;cin>>n>>m>>k;
vll a(n);input(a,n);
vll b(n);input(b,n);

ll low=1;
ll high=1e9;
ll ans=-1;
while(low<=high){
	ll mid=(low+high)>>1;
	bool check=true;
	vpll temp;
	forn(i,n){
		ll req=(a[i]/mid)+(a[i]%mid!=0);
		if(req<=m){
			ll diff=m-req;
			ll mini=1;
			mini=max(mini,(b[i]-diff));
			ll maxi=b[i]+diff;
			temp.pb({mini,maxi});
		}
	}
	asort(temp);
	ll coun=0;
	map<ll,ll>co;
	ordered_multiset<ll>jus;
	for(auto pos:temp){
		if(pos.ff==pos.ss){
			co[pos.ff]++;
		}
		else{
			co[pos.ff]++;
			co[pos.ss]++;
		}
	}

	for(auto pos:temp){
		ll cur=co[pos.ff];
		cur+=((ll)jus.size()-jus.order_of_key(pos.ff));
		coun=max(coun,cur);
		cur=co[pos.ss];
		cur+=((ll)jus.size()-jus.order_of_key(pos.ss));
		coun=max(coun,cur);
		jus.insert(pos.ss);
	}
	if(coun>=k){
		ans=mid;
		high=mid-1;
	}
	else{
		low=mid+1;
	}
}
cout<<ans<<endl;

}

int main()
{
	pranav
    ll tt=1; cin>>tt;
    while(tt--){
    solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

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

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
/// idk
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while(t--){
        int n, m, k;
        cin >> n >> m >> k;
        vector<long long>h(n), x(n);
        for(int i = 0;i < n;++i){
            cin >> h[i];
        }
        for(int i = 0;i < n;++i){
            cin >> x[i];
        }
        long long l = 1, r = 1e9, ans = -1;
        while(l <= r){
            long long mid = (l + r) >> 1;
            vector<pair<long long, long long>>v(n);
            for(int i = 0;i < n;++i){
                v[i] = {((h[i] + mid - 1) / mid + x[i] - m), m + x[i] - (h[i] + mid - 1) / mid};
            }
            multiset<long long>L, R;
            bool found = false;
            for(int i = 0;i < k;++i){
                L.insert(v[i].first);
                R.insert(v[i].second);
            }
            if(*(--L.end()) <= *R.begin())found = true;
            for(int i = k;i < n;++i){
                L.insert(v[i].first);
                R.insert(v[i].second);
                L.erase(L.find(v[i - k].first));
                R.erase(R.find(v[i - k].second));
                if(*(--L.end()) <= *R.begin())found = true;
            }
            if(found){
                r = mid - 1;
                ans = mid;
            }
            else{
                l = mid + 1;
            }
        }
        cout << ans << '\n';
    }
}

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

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

Think step by step.

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