**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
t=int(input())
while t:
    n=int(input())
    if n==1:
        print("Kosuke")
    elif ((n+1)/2)%2==0:
        print("Kosuke")
    else:
        print("Sakurako")
    t-=1

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

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

Think step by step.

Response:
```plaintext
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: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
import math
for _ in range(int(input())):
    n = int(input())
    if round(math.log(n, 2)) % 2 == 0:
        print("Kosuke")
    else:
        print("Sakurako")

**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 consider the following:

1. **Edge Cases**: Values of `n` that are close to the boundaries of the allowed range (1 to 100).
2. **Critical Points**: Values of `n` that are powers of 2, as the logarithm function in the buggy code might handle these values differently.
3. **Symmetry**: Values of `n` that are symmetric around the midpoint of the range to test the even and odd behavior of the logarithm.

Let's analyze the buggy code:
- The code uses `round(math.log(n, 2)) % 2` to determine the result. This means it is checking if the rounded logarithm base 2 of `n` is even or odd.
- This approach is incorrect because the game's outcome depends on the sequence of moves, not just the logarithm of `n`.

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

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

Explanation:
- **n = 1**: The game ends immediately after the first move by Sakurako. The correct result is "Kosuke".
- **n = 2**: The game will end after Sakurako's second move. The correct result is "Sakurako".
- **n = 3**: The game will end after Kosuke's second move. The correct result is "Kosuke".
- **n = 4**: The game will end after Sakurako's third move. The correct result is "Sakurako".

This test input includes values that are powers of 2 and values that are not, which should help in exposing the bug in the logarithm-based decision.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t=sc.nextInt();
        int n=t;
        for (int i=0;i<t;i++) {
            if (n%2==0)
                System.out.println("Kosuke");
            else
                System.out.println("Sakurako");
            n = sc.nextInt();
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
import java.math.*;
import java.text.DecimalFormat;
import java.time.*;
import java.util.*;
import java.util.Map.Entry;
import java.util.function.Function;
import java.util.stream.*;
import java.util.regex.*;

public class App {

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

        while (tc-- > 0) {
            int n = sc.nextInt();

            int sum = 0;
            for (int i = 1; i <= 100; i++) {
                sum += (i % 2 == 0) ? (2 * i - 1) : -(2 * i - 1);
                if (Math.abs(sum) > n) {
                    System.out.println((i % 2 == 0) ? "Kosuke" : "Sakurako");
                    break;
                }
            }
        }
    }
}

**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
50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define int long long
#define endl "\n"
#define Endl "\n"
const LL mod = 1e9 + 7;
const LL sz = 1e7 + 5;

vector<LL> prefix_function(string &s)
{
    LL n = (LL)s.length();
    vector<LL> pi(n);
    for (LL i = 1; i < n; i++)
    {
        LL j = pi[i - 1];
        while (j > 0 && s[i] != s[j])
            j = pi[j - 1];
        if (s[i] == s[j])
            j++;
        pi[i] = j;
    }
    return pi;
}

LL bin(LL a, LL p)
{
    if (p == 0)
        return 1;
    LL res = bin(a, p / 2ll);
    res = (res * res) % mod;
    if (p & 1ll)
        res = (res * a) % mod;
    return res;
}

LL mmi(LL a)
{
    return bin(a, mod - 2);
}

vector<LL> primes;
void sieve()
{
    vector<bool> isp(sz, true);
    isp[0] = false, isp[1] = false;
    for (LL i = 2; i * i <= sz; i++)
        if (isp[i])
            for (LL j = i * i; j < sz; j += i)
                isp[j] = false;
    for (int i = 2; i < sz; i++)
        if (isp[i])
            primes.push_back(i);
}

bool cmp(const pair<int, int> &p1, const pair<int, int> &p2)
{
    // ip / op
    return (p1.second * p2.first) >= (p1.first * p2.second);
}
void solve()
{
    int n, p, x, pt = 0;
    cin >> n >> p;
    vector<pair<int, int>> vp(n);
    multiset<int> ms;
    for (int i = 0; i < n; i++)
        cin >> vp[i].first;
    for (int i = 0; i < n; i++)
        cin >> vp[i].second;
    for (int i = 0; i < p; i++)
    {
        cin >> x;
        ms.insert(x);
    }
    sort(vp.begin(), vp.end(), cmp);
    for (int i = 0; i < n;)
    {
        auto val = vp[i];
        if (ms.size() == 0)
            break;
        int cost = val.first;
        int last = *prev(ms.end());
        if (cost > last)
        {
            i++;
            continue;
        }
        ms.erase(prev(ms.end()));
        int ways = last / cost;
        ms.insert((last % cost) + (ways * val.second));
        pt += (2ll * ways);
    }
    cout << pt << endl;
}
signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
2 2
3 5
2 4
10 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 2e6 + 1003;
#define ll long long int
ll n, m, c[N];
struct red {
	int a, b, c;
}node[N];
bool cmp(red x, red y) {
	return x.c < y.c;
}
int main() {
	std::ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n >> m;
	for (int i = 1; i <= n; i++) cin >> node[i].a;
	for (int i = 1; i <= n; i++) {
		cin >> node[i].b;
		node[i].c = node[i].a - node[i].b;
	}
	sort(node + 1, node + 1 + n, cmp);
	ll sum = 0;
	for (int i = 1; i <= m; i++) { cin >> c[i]; }
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			if (c[j] >= node[i].a) { sum += 2; c[j] -= (node[i].c); }
		}
	}
	cout << sum;
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
def solve():
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    c = list(map(int, input().split()))

    mi = min(a)
    d = sorted([(x, x - y) for x, y in zip(a, b)])
    dp = [0] * 1000005
    dp[mi] = 1
    for i in range(mi + 1, 1000005):
        l, r = 0, n - 1
        idx = n
        while l <= r:
            mid = (l + r) // 2
            if d[mid][0] > i:
                idx = mid
                r = mid - 1
            else:
                l = mid + 1
        ai, diff = d[idx-1]
        pro = ((i - ai) // diff) + 1
        res = i - (pro * diff)
        dp[i] = dp[res] + pro

    ans = 0
    for i in c:
        ans += dp[i]
    print(2 * ans)

solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
def finnMinsteDiff(mengde, ordbok, differanser, start, slutt):
    mid = int((start + slutt) / 2)
    if mengde >= ordbok[differanser[mid]]:
        if mid == start or mengde < ordbok[differanser[mid + 1]]:
            return differanser[mid], ordbok[differanser[mid]]
        else:
            return finnMinsteDiff(mengde, ordbok, differanser, start, mid)
    else:
        return finnMinsteDiff(mengde, ordbok, differanser, mid + 1, slutt)

info = input().split()
antallVaapen = int(info[0])
antallMetaller = int(info[1])
vaapenKrav = [int(tall) for tall in input().split()]
faarTilbake = [int(tall) for tall in input().split()]
inventoryMetaller = [int(tall) for tall in input().split()]
differanse = []
xp = 0
minsteKrav = min(vaapenKrav)

for i in range(antallVaapen):
    differanse.append(vaapenKrav[i] - faarTilbake[i])

sortedDifferanse = differanse.copy()
sortedDifferanse.sort()

ordbok = {}

for i in range(len(vaapenKrav)):
    if differanse[i] in ordbok:
        if vaapenKrav[i] < ordbok[differanse[i]]:
            ordbok[differanse[i]] = vaapenKrav[i]
    else:
        ordbok[differanse[i]] = vaapenKrav[i]

i = 0
while i < antallVaapen:
    fjern = []
    for k in range(i+1, len(sortedDifferanse)):
        if ordbok[sortedDifferanse[i]] < ordbok[sortedDifferanse[k]]:
            fjern.append(sortedDifferanse[k])
    fjern = list(set(fjern))

    for objekt in fjern:
        ordbok.pop(objekt)
        antallVaapen -= 1
    i += 1

differanser = list(ordbok.keys())
differanser.sort()

for mengde in inventoryMetaller:
    igjen = mengde
    while igjen >= minsteKrav:
        minstediff, kravet = finnMinsteDiff(igjen, ordbok, differanser, 0, len(differanser))
        rest = int(kravet / minstediff)
        if rest == 0:
            nyxp = int(igjen / minstediff)
            igjen = igjen - nyxp * minstediff
        else:
            nyxp = int(igjen / minstediff) - rest
            igjen = igjen - nyxp * minstediff
            if igjen >= kravet and igjen >= minstediff:
                igjen -= minstediff
                nyxp += 1

        xp += nyxp * 2

print(xp)

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

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

Think step by step.

Response:
```plaintext
3 3
10 20 30
9 19 29
10 20 30
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
def finnMinsteDiff(mengde, sortertDifferanse, ordbok):
    for diff in sortertDifferanse:
        if mengde >= ordbok[diff]:
            return diff, ordbok[diff]

info = input().split()
antallVaapen = int(info[0])
antallMetaller = int(info[1])
vaapenKrav = [int(tall) for tall in input().split()]
faarTilbake = [int(tall) for tall in input().split()]
inventoryMetaller = [int(tall) for tall in input().split()]
differanse = []
xp = 0
minsteKrav = min(vaapenKrav)

for i in range(antallVaapen):
    differanse.append(vaapenKrav[i] - faarTilbake[i])

sortedDifferanse = differanse.copy()
sortedDifferanse.sort()

ordbok = {}

for i in range(len(vaapenKrav)):
    if differanse[i] in ordbok:
        if vaapenKrav[i] < ordbok[differanse[i]]:
            ordbok[differanse[i]] = vaapenKrav[i]
    else:
        ordbok[differanse[i]] = vaapenKrav[i]

fjern = []
for i in range(len(sortedDifferanse)):
    for k in range(i+1, len(sortedDifferanse)):
        if ordbok[sortedDifferanse[i]] < ordbok[sortedDifferanse[k]]:
            fjern.append(sortedDifferanse[k])

for objekt in fjern:
    ordbok.pop(objekt)

for mengde in inventoryMetaller:
    igjen = mengde
    while igjen >= minsteKrav:
        minstediff, kravet = finnMinsteDiff(igjen, sortedDifferanse, ordbok)
        rest = int(kravet / minstediff)
        if rest == 0:
            nyxp = int(igjen / minstediff)
            igjen = igjen - nyxp * minstediff
        else:
            nyxp = int(igjen / minstediff) - rest
            igjen = igjen - nyxp * minstediff
            if igjen >= kravet and igjen >= minstediff:
                igjen -= minstediff
                nyxp += 1

        xp += nyxp * 2

print(xp)

**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 2
5 10 15
4 5 10
20 30
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
import java.text.DecimalFormat;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        int M = in.nextInt();
        int[] a = new int[N];
        int[] b = new int[N];
        int[] c = new int[M];
        for(int i = 0; i< N; i++) a[i] = in.nextInt();
        for (int i = 0; i < N; i++) b[i] = in.nextInt();
        for (int i = 0; i < M; i++) c[i] = in.nextInt();
        in.close();

        List<int[]> list = new ArrayList<>();// (实际消耗， 制造消耗)
        for(int i = 0; i < N; i++){
            list.add(new int[]{a[i] - b[i], a[i]});
        }
        list.sort((x, y) -> { //排序，实际消耗升序，相同则制造消耗升序
            if (x[0] == y[0]) return x[1] - y[1];
            return x[0] - y[0];
        });

        double exe = 0;
        for (int ignot: c) {
            for(int[] item: list){
                if (ignot > item[1]) {
                    int ans = (ignot - item[1]) / item[0];
                    ignot -= ans*item[0];
                    exe += ans;
                }
                if (ignot == item[1]) {
                    ignot -= item[0];
                    exe++;
                }
            }
        }
        System.out.println(new DecimalFormat("#").format(exe*2));
    }

}

**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 2
5 4
1 1
10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
#include <iostream>
#include <vector>
#include <string>
#include <string.h>
#include <queue>
#include <bitset>
#include<numeric>
#include <iomanip>
#include <map>
#include <stack>
#include <set>
#include <algorithm>
#include <math.h>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define MOD 998244353
#define N 2000007
int f[N], g[N];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    int a[n],b[n],c[m];
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    int p = INF, id = -1;
    int minv = INF;
    for(int i=0;i<N;i++)
    {
        f[i] = INF;
        g[i] = -1;
    }
    for(int i=0;i<n;i++)
    {
        scanf("%d",&b[i]);
        minv = min(minv, a[i]);
        if(a[i]-b[i]<p)
        {
            p=a[i]-b[i];
            id = i;
        }
        f[a[i]] = a[i]-b[i];
        g[a[i]] = i;
    }
    for(int i=1;i<N;i++)
        if(f[i]>f[i-1])
        {
            f[i]=f[i-1];
            g[i]=g[i-1];
        }
    ll ans = 0;
    priority_queue<int>que;
    for(int i=0;i<m;i++)
    {
        scanf("%d",&c[i]);
        if(c[i] >= minv)
        {
            if(c[i] >= a[id])
            {
                int y = c[i] - a[id];
                if(y % p == 0)
                {
                    int z = y / p + 1;
                    ans += z * 2;
                    que.push(c[i] - z * p);
                }
                else
                {
                    int z = y / p + 1;
                    ans += z * 2;
                    que.push(c[i] - z * p);
                }
            }
            else
                que.push(c[i]);
        }
    }
    while(!que.empty())
    {
        int x = que.top();que.pop();
        if(x<minv) break;
        int y = x - a[g[x]];
        if(y % f[x] == 0)
        {
            int z = y / f[x] +1;
            ans += z * 2;
            que.push(x - z * f[x]);
        }
        else
        {
            int z = y / f[x] + 1;
            ans += z * 2;
            que.push(x - z * f[x]);
        }
    }
    printf("%ld\n",ans);
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3 2
5 3 2
1 1 1
10 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
import java.util.*;
public class smithing_skill {
    // public static int sort rowWise(int mtx[][], int n){
    //     for(int i=0; i<n; i++){
    //         Arrays.sort(mtx[1][i]);
    //     }

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

            long n = sc.nextLong();
            long m = sc.nextLong();

            long mtx[][] = new long[(int)n][3];

            long min = Long.MAX_VALUE;
            for(long i=0; i<n; i++){
                mtx[(int)i][0] = sc.nextLong();
                min = (long)Math.min(min, mtx[(int)i][0]);
            }

            for(long i=0; i<n; i++){
                mtx[(int)i][2] = sc.nextLong();
                mtx[(int)i][1] = (mtx[(int)i][0] - mtx[(int)i][2]);

                // mtx[1][-1] = Integer.MIN_VALUE;
            }

            // for(int i=0; i<n; i++){
            //     System.out.print(mtx[1][i]+" ");
            // }
            // System.out.println();

            Arrays.sort(mtx, Comparator.comparingInt(a -> (int)(a[0] - a[2])));

            // for(int i=0; i<n; i++){
            //     System.out.print(mtx[i][0]+" ");
            // }
            // System.out.println();
            // for(int i=0; i<n; i++){
            //     System.out.print(mtx[i][2]+" ");
            // }
            // System.out.println();
            // for(int i=0; i<n; i++){
            //     System.out.print(mtx[i][1]+" ");
            // }

            // for(long i=1; i<n; i++){

            //     long temp[][] = new long[3][(int)n];

            //     if(mtx[1][(int)i] < mtx[1][(int)(i-1)]){
            //         for(int j=0; j<3; j++){
            //             temp[j][(int)i] = mtx[j][(int)i];
            //             mtx[j][(int)i] = mtx[j][(int)(i-1)];
            //             mtx[j][(int)(i-1)] = temp[j][(int)i];
            //         }

            //     }

            //     if(mtx[1][(int)i] == mtx[1][(int)(i-1)]){
            //         if(mtx[0][(int)i] > mtx[0][(int)(i-1)])
            //         for(int j=0; j<3; j++){
            //             temp[j][(int)i] = mtx[j][(int)i];
            //             mtx[j][(int)i] = mtx[j][(int)(i-1)];
            //             mtx[j][(int)(i-1)] = temp[j][(int)i];
            //         }

            //     }
            // }

            // for(int i=0; i<n; i++){
            //     System.out.print(mtx[0][i]+" ");
            // }
            // System.out.println();
            // for(int i=0; i<n; i++){
            //     System.out.print(mtx[2][i]+" ");
            // }
            // System.out.println();
            // for(int i=0; i<n; i++){
            //     System.out.print(mtx[1][i]+" ");
            // }

            Long types[] = new Long[(int)m];
            for(long i=0; i<m; i++){
                types[(int)i] = sc.nextLong();
            }
            Arrays.sort(types, Collections.reverseOrder());
            long max_type = types[0];

            long exper = 0;
                if(n==1){
                    if(types[0]>=mtx[0][0]){
                        exper = 1 + ((types[0]-mtx[0][0])/mtx[0][1]);
                    }
                }else{
                    for(long i=0; i<n; i++){
                        long excess = 0;
                        long diff = 0;
                        long use = 0;
                        long type_1 = types[1];

                        if(types[0]>=mtx[(int)i][0]){
                            excess = (types[0] - mtx[(int)i][0]);
                            diff = mtx[(int)i][1];
                            use = 1 + (excess/diff);
                            types[0] -= use*diff;
                            // System.out.println("use:"+use);
                            // System.out.println("types:"+types[0]);
                            exper += use;
                            // System.out.println("exper="+exper);
                        }else if(types[0]==mtx[(int)i][0]){
                            exper ++;
                            types[0] = mtx[(int)i][1];
                            // System.out.println("types:"+types[0]);
                            // System.out.println("exper="+exper);
                        }
                        Arrays.sort(types, Collections.reverseOrder());
                        // System.out.print("i:"+i+" ");
                        // for(int j=0; j<m; j++){
                        //     System.out.print(types[j]+" ");
                        // }
                        max_type = types[0];
                        // System.out.println("type_1:"+type_1);
                        // System.out.println("max_type:"+max_type);
                        if(type_1==max_type){
                            i = -1;
                        }

                        // System.out.println();
                        if(types[0] < min){
                            i = n;
                        }

                        // System.out.println("exp"+exper);
                    }
                }

            System.out.println(exper*2);

        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3 1
10 20 30
0 0 0
25
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
n,m=list(map(int,input().split(' ')))
a=list(map(int,input().split(' ')))
b=list(map(int,input().split(' ')))
c=list(map(int,input().split(' ')))
ans=0
Req=[]
for i in range(n):
    Req.append([a[i]-b[i],a[i]])
Req.sort()
req=[Req[0]]
for i in range(1,n):
    if(Req[i][0]!=Req[i-1][0]):
        req.append(Req[i])
bar=0
c.sort(reverse=True)
for i in range(m):
    pre=c[i]
    while(bar<len(req) and pre<a[bar]):
        bar+=1
    for j in range(bar,len(req)):
        ans+=2*(max(0,(pre-req[j][1])//req[j][0]))
        pre-=req[j][0]*max(0,((pre-req[j][1])//req[j][0]))
        if(pre>=req[j][1]):
            pre-=req[j][0]
            ans+=2
print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
# ------------------------------- ESSENTIALS -------------------------------
import sys
import platform

def is_online_judge():
    if platform.system()=='Windows' and platform.win32_ver()[0]=='10' and platform.win32_ver()[2]=='SP0' and platform.node()=='DESKTOP-07FV8E1':
        return False
    else:
        return True
is_judge : bool = is_online_judge()

if is_judge:
    input = sys.stdin.readline
else:
    inp = open('./InpOut/input.txt','r')
    out = open('./InpOut/output.txt','w')
    input = inp.readline
    if str(sys.stdout)[1:29] == 'ipykernel.iostream.OutStream':
        og_stdout = sys.stdout
        sys.stdout = out

def readList():
    return list(map(int, input().split()))
def readLists(n):
    return [readList() for _ in range(n)]
def readInt():
    return int(input())
def readInts():
    return map(int, input().split())
def readStr():
    return input().strip()
def print_array(arr):
    print(" ".join(map(str, arr)))
def printer(is_True):
    print('YES') if is_True else print('NO')

# ------------------------------- __MAIN__ -------------------------------

def solve(n, m, a, b, c):
    cnt = 0
    c.sort(reverse=True)
    duh = dict()
    zero = float('inf')

    for i in range(n):
        if b[i]==0:
            zero = min(zero, a[i])
        elif a[i] in duh:
            if duh[a[i]]>b[i]:
                duh[a[i]] = b[i]
        else:
            duh[a[i]] = b[i]

    cla = sorted([[x,duh[x]] for x in duh], key=lambda x:(x[0]/x[1], x[0]))
    f = len(cla)
    # print(cla)
    useless =(zero!=float('inf'))
    low_lim = min(a)

    if f==0 and useless:
        for met in c:
            cnt += met//zero
        print(2*cnt)
        return

    first = True
    noah = 0

    for i,met in enumerate(c):
        # print('for',met)
        k= noah

        while met>0 and k<f:
            # print('while', met, k, cla[k][0])
            if met<low_lim:
                break
            if met<cla[k][0]:
                k+=1
                continue

            if first:
                noah = k
                first = False
            di = cla[k][0]-cla[k][1]
            tim = (met-cla[k][0])//di + 1
            # print(tim)
            cnt += tim
            met -= tim*di
            k+=1
        first = True

        if useless: cnt += met//zero

    print(2*cnt)
    return

t = 1
# t = readInt()
for _ in range(t):
    n, m, = readInts()
    a = readList()
    b = readList()
    c = readList()
    solve(n, m, a, b, c)

if not is_judge:
    sys.stdout = og_stdout
    out.close()

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

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

Think step by step.

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

1. **Identify Potential Bugs**:
   - The code uses a dictionary `duh` to store the minimum `b[i]` for each `a[i]`.
   - It sorts the classes based on the ratio `a[i] / b[i]` and the value of `a[i]`.
   - It handles the case where `b[i] == 0` separately.
   - It iterates over the metal ingots and tries to maximize the experience points.

2. **Edge Cases to Consider**:
   - **Case with `b[i] == 0`**: If there are multiple classes with `b[i] == 0`, the code should handle them correctly.
   - **Case with `a[i] > c[j]`**: If the number of ingots required to forge a weapon is greater than the available ingots, the code should not count this as a valid option.
   - **Case with `a[i] == b[i] + 1`**: This is a boundary case where the number of ingots returned by melting is just one less than the number required to forge.
   - **Case with very large numbers**: The code should handle large numbers without integer overflow or incorrect calculations.

3. **Constructing the Test Input**:
   - We will create a test case with a small number of weapon classes and metal types to keep the input small.
   - We will include a class with `b[i] == 0` to test the handling of zero returns.
   - We will include a class with `a[i] == b[i] + 1` to test the boundary condition.
   - We will include a class with `a[i] > c[j]` to test the case where forging is not possible.

Here is the test input:

```plaintext
3 2
5 4 6
0 3 5
10 15
```

**Explanation**:
- **n = 3**: There are 3 weapon classes.
- **m = 2**: There are 2 types of metal ingots.
- **a = [5, 4, 6]**: The number of ingots required to forge weapons of each class.
- **b = [0, 3, 5]**: The number of ingots returned by melting weapons of each class.
- **c = [10, 15]**: The number of ingots available of each metal type.

This test case includes:
- A class with `b[i] == 0` (class 1).
- A class with `a[i] == b[i] + 1` (class 2).
- A class with `a[i] > c[j]` (class 3, for metal type 2).

This input should expose potential issues in the code's handling of these edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
import java.util.*;

public class D {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt();
        int[][] a = new int[n][2];
        for (int i = 0; i < n; i++) {
            a[i][0] = sc.nextInt();
        }
        for (int i = 0; i < n; i++) {
            a[i][1] = sc.nextInt();
        }
        Arrays.sort(a, (o1, o2) -> {
            if ((o1[0] - o1[1]) != o2[0] - o2[1]) return (o1[0] - o1[1]) - (o2[0] - o2[1]);
            return o1[0] - o2[0];
        });
        int N = (int) 1e6 + 1;
        int[] cnt = new int[N];
        long ans = 0;
        for (int i = 0; i < m; i++) {
            int d = sc.nextInt();
            if (d >= a[0][0]) {
                int c = ((d - a[0][0]) / (a[0][0] - a[0][1])) + 1;
                ans += c * 2;
                cnt[d - c * (a[0][0] - a[0][1])]++;
            } else {
                cnt[d]++;
            }

        }
        int pt = 0;
        for (int i = N - 1; i > 0; i--) {
            if (cnt[i] > 0) {
                while (pt < n && a[pt][0] > i) pt++;
                if (pt == n) break;
                int c = ((i - a[pt][0]) / (a[pt][0] - a[pt][1])) + 1;
                ans += (long) c * 2 * cnt[i];
                cnt[i - c * (a[pt][0] - a[pt][1])] += cnt[i];
            }
        }

        System.out.println(ans);
    }

}

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

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

Think step by step.

Response:
```plaintext
2 2
5 3
4 1
10 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * @Author congge
 * @Date 2024/6/28 0:09
 * @description https://codeforces.com/contest/1989/problem/D
 */
public class Main {
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static int N = 1000010;
    static int[][] a = new int[N][2];
    static int[] c = new int[N];
    static int[] stack = new int[N], r = new int[N];
    static int top = 0;
    static int n, m;

    public static void main(String[] args) {
        n = sc.nextInt();
        m = sc.nextInt();
        int minV = N;
        for (int i = 1; i <= n; i++) {
            a[i][0] = sc.nextInt();
            minV = Math.min(minV, a[i][0]);
        }

        for (int i = 1; i <= n; i++) a[i][1] = sc.nextInt();
        for (int i = 1; i <= m; i++) c[i] = sc.nextInt();
        Arrays.sort(a, 1, n + 1, ((o1, o2) -> {
            if (o1[0] - o1[1] == o2[0] - o2[1]) {
                return o1[0] - o2[0];
            } else {
                return o1[0] - o1[1] - (o2[0] - o2[1]);
            }
        }));
        long res = 0;
        a[n + 1][0] = 0;
        // 单调栈
        stack[++top] = n + 1;
        for (int i = n; i >= 1; i--) {
            while (top >= 1 && a[stack[top]][0] >= a[i][0]) top--;
            r[i] = stack[top];
            stack[++top] = i;
        }

        for (int i = 1; i <= m; i++) {
            int j = 1;
            while (j <= n) {
                if (c[i] >= a[j][0]) {
                    long x = (c[i] - a[j][0]) / (a[j][0] - a[j][1]) + 1;
                    res += x * 2;
                    c[i] -= x * (a[j][0] - a[j][1]);
                }
                if (c[i] < minV) break;
                j = r[j];
            }
        }
        out.println(res);
        out.flush();
    }
}

class FastReader {
    StringTokenizer st;
    BufferedReader br;

    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());
    }

    String nextLine() {
        String s = "";
        try {
            s = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return s;
    }

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

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

    // 是否由下一个
    boolean hasNext() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = br.readLine();
                if (line == null)
                    return false;
                st = new StringTokenizer(line);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return true;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or scenarios that can maximize the failure probability. The code sorts the weapon classes based on the net cost of forging and melting, and then uses a monotonic stack to find the optimal sequence of forging and melting. However, there might be issues with how it handles the remaining ingots and the calculation of the number of weapons that can be forged and melted.

### Step-by-Step Reasoning:

1. **Edge Case: Minimal Return from Melting**:
   - If the return from melting is very low (e.g., `b_i` is very close to 0), the number of times you can forge and melt a weapon will be limited. This can expose issues in the code's handling of the remaining ingots.

2. **Edge Case: Large Initial Ingots**:
   - If you have a large number of initial ingots, the code should be able to handle the repeated forging and melting without running into integer overflow or incorrect calculations.

3. **Edge Case: Mixed Ingots and Weapon Classes**:
   - If there are multiple types of ingots and weapon classes with varying costs, the code should correctly handle the optimal sequence of forging and melting.

### Test Input:
Let's create a small-scale test input that combines these edge cases:

- **Number of weapon classes (n)**: 3
- **Number of metal types (m)**: 2
- **Forge costs (a_i)**: 10, 20, 30
- **Melt returns (b_i)**: 1, 2, 3
- **Initial ingots (c_j)**: 100, 200

This input will test the code's ability to handle minimal returns from melting and large initial ingots.

### Test Input:
```plaintext
3 2
10 20 30
1 2 3
100 200
```

### Explanation:
- **Weapon Class 1**: Costs 10 ingots to forge, returns 1 ingot when melted.
- **Weapon Class 2**: Costs 20 ingots to forge, returns 2 ingots when melted.
- **Weapon Class 3**: Costs 30 ingots to forge, returns 3 ingots when melted.
- **Initial Ingots**:
  - 100 ingots of type 1
  - 200 ingots of type 2

This input will test the code's ability to correctly calculate the maximum experience points by forging and melting weapons multiple times, especially when the return from melting is minimal.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

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

#define debug(n1) cout << n1 << endl
#define SORT(array, initial, final) sort(&array[initial], &array[final]);
#define MOD 1000000007

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

#define F first
#define S second
#define PB push_back
#define MP make_pair

/*
HELP:
Mod of Division:

Method 1 -

    IF a Overflows OR/AND b overflows use ALWAYS (EVEN if 'a')

    For (a/b)%c = ((a%c)*((1/b)%c))%c = ((a%c)*(modInverse(b, MOD)%c))%c ;
    For b=x*y which overflows - take mod where required -
         WRONG : (a/(x*y))%c != ((a%c)*(modInverse(x, MOD)%c)*(modInverse(y, MOD)%c))%c   - (NOT EQUAL)
         CORRECT: (a/(x*y))%c = ((a%c)*(modInverse(modMul(x,y,MOD), MOD)%c))%c

Method 2 -

    For (a/b)%c = ((a%c)*((1/b)%c))%c = ((a%c)*(modExpo(b, MOD-2, MOD)%c))%c ;

Here modExpo with exponent = MOD-2 is same modInverse.

HELP END;
*/

// To compute (a * b) % mod when a*b overflows (i.e when mod > INT_MAX so mod*mod > ll_MAX)
ll modMul(ll a, ll b, ll mod){
    ll res = 0; // Initialize result
    a = a % mod;
    b = b % mod;
    while (b > 0){
        // If b is odd, add 'a' to result
        if(b % 2 == 1){
            res = (res + a) % mod;
        }
        // Multiply 'a' with 2
        a = (a * 2) % mod;

        // Divide b by 2
        b /= 2;
    }

    // Return result
    return res % mod;
}

ll modExpo(ll x, ll n, ll modulo){ 		//pass MOD for modulo usually
//	int modulo = 1000000007;
    ll product = 1;
    ll p = x%modulo;
    while (n != 0) {
        if ((n & 1) == 1) {
            product = product * p % modulo;
        }
        p = (p * p % modulo);
        n >>= 1;
    }
    return product;
}

// Function for extended Euclidean Algorithm
int gcdExtended(ll a, ll b, int* x, int* y){
    // Base Case
    if (a == 0) {
        *x = 0, *y = 1;
        return b;
    }
    // To store results of recursive call
    int x1, y1;
    int gcd = gcdExtended(b % a, a, &x1, &y1);

    // Update x and y using results of recursive
    // call
    *x = y1 - (b / a) * x1;
    *y = x1;
    return gcd;
}

ll modInverse(ll A, int M){ 		//pass MOD for M usually
    int x, y;
    ll g = gcdExtended(A, M, &x, &y);
    if (g != 1){
//		cout << "Inverse doesn't exist";
        return -1;
    }else {
//		m is added to handle negative x
        ll res = (x % M + M) % M;
//		cout << "Modular multiplicative inverse is " << res;
        return res;
    }
}

ll gcd(ll a, ll b){
    if(b%a==0)return a;
    else return gcd(b%a,a);
}

ll fac(ll n, ll p){
    if(n==0 || n==1)return 1;
    return (fac(n-1,p)*n)%p;
}

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

void readArr(vector<ll> &array, ll n){
    for(ll i=0; i<n; i++){
        cin>>array[i];
    }
}

void printArr(vector<ll> &array, ll n){
    cout<<"Array : ";
    for(ll i=0; i<n-1; i++){
        cout<<array[i]<<", ";
    }
    cout << array[n-1] << endl;
}

void solve(){
    ll ans = 0;
    ll sum = 0;

    ll n, m;
    cin >> n >> m;
    vector<ll> a(n, 0);

    ll mn = 1000000000;

    for(int i=0; i<n; i++){
        cin>>a[i];
        mn = min(a[i], mn);
    }
    vector<ll> b(n, 0);

    for(int i=0; i<n; i++){
        cin>>b[i];
    }

    vector<ll> c(m, 0);

    priority_queue<ll> pq;

    map <ll, ll> cnt;
    for(int i=0; i<m; i++){
        cin>>c[i];
        cnt[c[i]]++;
    }

    for(auto x: cnt){
        pq.push(x.first);
    }
    vector<tuple<ll, ll, ll>> d(n);

    for(int i=0; i<n; i++){
        d[i] = {a[i]-b[i], a[i], b[i]};
    }
    sort(d.begin(), d.end());

    ll curr = 0;

    while(!pq.empty() && curr < n){

        ll tempAns = 0;
        ll mx;
        while(!pq.empty() && pq.top() >= get<1>(d[curr])){
            mx = pq.top();
            ll diff = get<0>(d[curr]);
            ll steps = (((mx - get<1>(d[curr]))/diff)+1);
            tempAns += 2*steps;
            pq.pop();
            ll left = mx - diff*steps;
            if(left >= mn){
                cnt[left] = cnt[mx];
                pq.push(left);
            }
        }
        tempAns *= cnt[mx];
        ans += tempAns;
        curr++;
    }
    cout << ans << "\n";
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t;
    t=1;
    // cin >> t;

    while(t-->0){
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. Here are some key considerations:

1. **Small Values for `n` and `m`**: Since the problem constraints allow for very large values of `n` and `m` (up to \(10^6\)), a small-scale test input can still be effective in exposing bugs.

2. **Special Cases for `a_i` and `b_i`**: Consider cases where `a_i` and `b_i` are such that the algorithm might miscalculate the number of steps or the amount of experience points.

3. **Large Values for `c_j`**: Large values for `c_j` can help expose issues in handling large numbers and ensuring the algorithm does not overflow or underflow.

4. **Edge Cases for `a_i - b_i`**: Values where `a_i - b_i` is very small or very large can cause issues in the algorithm's logic.

5. **Multiple Metal Types and Weapon Classes**: Ensure the test case includes multiple metal types and weapon classes to test the algorithm's ability to handle different combinations.

Given these considerations, here is a small-scale test input that should be challenging:

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

### Explanation:
- **`n = 5` and `m = 3`**: Small values to ensure the test case is manageable but still challenging.
- **`a_i = [9, 6, 7, 5, 5]`**: Different values to test various crafting costs.
- **`b_i = [8, 4, 5, 1, 2]`**: Values that are less than `a_i` but close to them to test the algorithm's ability to handle small differences.
- **`c_j = [10, 4, 7]`**: Different amounts of metal ingots to test the algorithm's ability to handle different starting materials.

This test input should help expose any issues in the algorithm's logic for handling small differences in `a_i` and `b_i`, as well as different combinations of metal types and weapon classes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Smithing Skill
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get $1$ experience point, and every time you melt an item, you also get $1$ experience point.

There are $n$ classes of weapons you can forge and $m$ types of metal ingots.

You can create one weapon of the $i$-th class, spending $a_i$ ingots of metal of the same type. Melting a weapon of the $i$-th class (which you crafted earlier) returns you $b_i$ ingots of the type of metal it was made of.

You have $c_j$ metal ingots of the $j$-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.

What is the maximum total amount of experience you can earn by crafting and melting weapons?

Input: The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^6$) — the number of weapon classes and metal types.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the number of ingots you need to forge one weapon of the $i$-th class.

The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ($0 \le b_i < a_i$), where $b_i$ is the number of ingots you return by melting one weapon of the $i$-th class you forged earlier.

The fourth line contains $m$ integers $c_1, c_2, \dots, c_m$ ($1 \le c_j \le 10^9$) — the number of ingots you have of the corresponding metal type.

Output: Print one integer — the maximum total experience points you can gain by repeatedly forging and melting weapons.

Examples:
input:
5 3
9 6 7 5 5
8 4 5 1 2
10 4 7
output:
12

input:
3 4
10 20 20
0 0 0
9 10 19 20
output:
8

input:
1 5
3
1
1000000000 1000000000 1000000000 1000000000 1000000000
output:
4999999990

Note: In the first example, you can do the following:

  1. craft one weapon of the $1$-st class from the $1$-st type of metal, spending $9$ ingots;    2. melt that weapon, returning $8$ ingots of the $1$-st metal type;    3. again, craft and melt one weapon of the $1$-st class from the $1$-st metal type;    4. craft and melt one weapon of the $3$-rd class from the $1$-st metal type;    5. craft and melt one weapon of the $3$-rd class from the $3$-rd metal type;    6. craft and melt one weapon of the $4$-th class from the $1$-st metal type;    7. craft and melt one weapon of the $5$-th class from the $3$-rd metal type;

In the end you'll have $c = [2, 4, 2]$ ingots left. In total, you've crafted $6$ weapons and melted $6$ weapons, gaining $12$ experience points in total.

Buggy Code:
#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;
#define The_Flash_mode ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define ll long long
#define ull unsigned long long
#define dd double
#define all(v) ((v).begin()),((v).end())
#define allr(v) ((v).rbegin()),((v).rend())
#define pll pair<ll,ll>
#define pii pair<int,int>
#define dpp(arr,val) memset(arr,val,sizeof(arr))
#define F first
#define S second
void File() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}
ll gcd(ll x, ll y) { return ((!y) ? x : gcd(y, x % y)); }
ll lcm(ll x, ll y) { return ((x / gcd(x, y)) * y); }
const int dx[8] = {0,-1,0,1,1,-1,1,-1 };
const int dy[8] = {-1,0,1,0,1,1,-1,-1 };
const int N = 2e5 + 5, mod = 1e9 + 7, md = 998244353;
ll fpow (ll x, ll y) {
    ll res = 1;
    while (y) {
        if (y & 1)res *= x;
        y >>= 1, x *= x;
    }
    return res;
}
bool cmp(pll p, pll p2) {
    return (p.F - p.S) > (p2.F - p2.S);
}
void Solve() {
    int n, m;
    cin >> n >> m;
    vector<pll > v(n);
    for (auto &it: v)cin >> it.F;
    for (auto &it: v)cin >> it.S;
    set<ll> s;
    unordered_map<ll, ll> mp;
    while (m--) {
        int x;
        cin >> x;
        mp[x]++;
        s.insert(x);
    }
    ll res = 0;
    for (int i = 0; i < n; i++) {
        if (s.empty())break;
        auto it = s.end();
        it--;
        if (v[i].F > *it)continue;
        ll diff = v[i].F - v[i].S, num = *it;
        res += ((((num - v[i].F) / diff) + 1) * 2) * mp[num];
        s.erase(it);
        if (num - (((num - v[i].F) / diff) + 1)) {
            s.insert(num - (((num - v[i].F) / diff) + 1) * diff);
            mp[num - (((num - v[i].F) / diff) + 1) * diff] = mp[num];
        }
        i--;
    }
    cout << res << '\n';
}
signed main() {
    The_Flash_mode;
    File();
    int t = 1; //cin >> t;
    while (t--) {
        Solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
2 2
3 5
2 4
5 7
```