**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
"""
Capitalization

word=input()
if word[0].isupper()==True:
    print(word)
elif word[0].islower()==True:
    print(word[0].upper()+word[1:])
    """
"""
Petia And The String
def String_Compare(string1,string2,comparesion):
            if string1<string2:
                comparesion = -1
            elif string2<string1:
                comparesion = 1
            else:
                comparesion= 0
            return comparesion

print(String_Compare(input().lower(),input().lower(),0))
"""
"""
dominoes
import math
m,n=(int(i)for i in input().split())
total_space_of_board=m*n
no_Dominos_can_be_placed=total_space_of_board/2
math.ceil(no_Dominos_can_be_placed)
print(int(no_Dominos_can_be_placed))
"""
"""
boy or girl
name=input()
c=''
for character in name :
    if character not in c:
        c=c+character
if len(c)%2==0:
    print("CHAT WITH HER!")
else:
    print("IGNORE HIM!")
"""
"""
helpful maths
xenia_input=input()
lista1=[]
lista2=[]
lista3=[]
nums=xenia_input.split("+")
for i in nums:
    if i=='1':
        lista1.append(i)
    elif i=='2':
        lista2.append(i)
    elif i== '3':
        lista3.append(i)
print("+".join(lista1+lista2+lista3))
"""
"""
string task
input_string=list(input())
vouls=["a","A","e","E","i","I","o","O","u","U","y","Y","e"]
for j in input_string:
    for i in vouls*5:
        if i in input_string:
            input_string.remove(i)
input_string=".".join(input_string)
print(str("."+input_string.lower()))
"""
'''
Elephant
import math
position=int(input())
required_steps=0
if position in range(1,6):
    required_steps=1
elif position>5:
    x=position/5
    required_steps=math.ceil(x)

print(required_steps)
'''
'''
banana
price_of_banana,dollars_i_have,no_bananas=(int(i)for i in input().split())
price=price_of_banana
num_of_ban=1
x=0
for i in range(no_bananas):
    price_of_banana=price*num_of_ban
    num_of_ban +=1
    x+=price_of_banana
if x>dollars_i_have:
    print(x-dollars_i_have)
elif x <dollars_i_have:
    print(0)
else:
    print(0)
    '''

def yougert(test_cases):
    for i in range(test_cases):
        n,a,b=(int(i) for i in input().split())
        if n%2==0:
            if n*(b/2)<n*a:
                return b
            else:
                return n*a
        if n%2!=0:
            if b+a<n*a:
                return b+a
            else:
                return n*a
test_cases=int(input())
for i in range(test_cases)   :
    print(yougert(test_cases))
'''
def yougert(test_cases):
    results = []  # Initialize an empty list to store results for each test case
    for _ in range(test_cases):
        n, a, b = map(int, input().split())
        if n % 2 == 0:
            results.append(min(n * (b / 2), n * a))
        else:
            results.append(min(b + a, n * a))
    return results

test_cases = int(input())
for result in yougert(test_cases):
    print(result)

'''

'''
no_test_cases=int(input())
for case in range(no_test_cases):
    n,a,b=(int(i)for i in input().split())
    if n%2!=0 and :
        print(b)
    else:
        print(n*a)
    if n%2==0 and

    else:
        print()
'''

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

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

Think step by step.

Response:
```plaintext
5
1 10 20
2 10 15
3 10 25
4 10 30
5 10 40
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
tt = int(input())
for t in range(tt):
    n,a,b = map(int, input().split())
    print(min(n*a, (n%2)*a+(n//2)*b, (n//2+1)*b))

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

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

Think step by step.

Response:
```plaintext
1
5 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

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();
		for (int i=0;i<t;i++) {
			int a=sc.nextInt();
			int b=sc.nextInt();
			int c=sc.nextInt();
			if (b*2<=c) {
				System.out.println(a*b);;
			}
			else if(a>2) {
				int n=a/2;
				System.out.println(n*c+b);
			}
			else if(a==2) {
				System.out.println(c);
			}
			else System.out.println(a*b);
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
1
5 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
import java.util.*;
public class Main{
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner reader=new Scanner(System.in);
		int n=reader.nextInt();
		int x,y,z;
		for(int i=0;i<n;i++) {
			x=reader.nextInt();
			y=reader.nextInt();
			z=reader.nextInt();
			int num1=x*y;
			int num2;
			int t=x%2;
			if(t==0) {
				num2=(x/2)*z;
			}else {
				num2=(x%2)*z+(x%2)*y;
			}

			if(num1>num2) {
				System.out.println(num2);
			}else {
				System.out.println(num1);
			}
		}
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the logic in the code might fail. Specifically, the code calculates the cost of buying yogurts in two ways: buying all yogurts at the regular price and buying as many pairs as possible at the promotion price with the remainder at the regular price. The bug seems to be in the calculation of the promotion price when there is an odd number of yogurts.

Let's break down the potential issues:
1. **Regular Price Calculation**: `num1 = x * y` (This is correct).
2. **Promotion Price Calculation**:
   - If the number of yogurts is even: `num2 = (x / 2) * z` (This is correct).
   - If the number of yogurts is odd: `num2 = (x % 2) * z + (x % 2) * y` (This is incorrect. It should be `num2 = (x / 2) * z + y`).

Given this, let's create a test case where the number of yogurts is odd and the promotion price is significantly cheaper than the regular price.

### Test Input
```plaintext
1
3 5 10
```

### Explanation
- **Number of test cases**: 1
- **Number of yogurts (n)**: 3
- **Price of one yogurt (a)**: 5
- **Price of two yogurts on promotion (b)**: 10

### Expected Output
- **Minimum cost**: 15

### Reasoning
- **Regular Price**: `3 * 5 = 15`
- **Promotion Price**: `1 * 10 + 5 = 15` (1 pair of yogurts at 10 burles and 1 yogurt at 5 burles)

In this case, the correct minimum cost is 15 burles. The buggy code incorrectly calculates the promotion price for an odd number of yogurts, which should be `15` but might be calculated incorrectly as `10` due to the bug.

This test case should expose the bug in the promotion price calculation for an odd number of yogurts.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

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

#define pb push_back
int const p = 1e9+7;
using namespace std;

int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    int t;
    cin >> t;
    while (t--){
     int n;
       double b, c; cin >> n >> b >> c;
       double d = b, e = c/2;
       if(n%2==0){

            cout << min((n/2)*c , n*d) << '\n';
       }
       else{
            cout  <<  min(((n/2)*c + min(b,c)), n*d) << '\n';
       }

    }

}

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

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

Think step by step.

Response:
```plaintext
3
1 5 9
2 5 9
3 5 11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
void solve()
{
  int n, a, b;
  cin >> n >> a >> b;
  int sum1 = (n * a);
  double dam = (double)b / 2;
  int sum2 = ceil(dam * n);
  if (sum1 > sum2)
  {
    cout << sum2 << '\n';
    return;
  }
  else
  {
    cout << sum1 << '\n';
    return;
  }
}

int32_t main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--)
  {
    solve();
  }

  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can highlight the issues in the logic. The primary issue in the buggy code is the incorrect calculation of the promotion price. The code incorrectly calculates the promotion price using a double and then applies a ceiling function, which can lead to incorrect results.

### Key Considerations:
1. **Edge Cases for `n`**:
   - When `n` is odd, the promotion price calculation should consider the extra yogurt.
   - When `n` is even, the promotion price should be straightforward.

2. **Edge Cases for `a` and `b`**:
   - When `b` is exactly twice `a`, the promotion price should be the same as buying individual yogurts.
   - When `b` is slightly more than twice `a`, the promotion should not be used.
   - When `b` is much less than twice `a`, the promotion should be used as much as possible.

### Challenging Test Input:
```plaintext
5
1 5 9
2 5 9
3 5 9
4 5 9
5 5 9
```

### Explanation:
1. **Test Case 1**: `n = 1, a = 5, b = 9`
   - Maxim should buy 1 yogurt for 5 burles.
   - Expected output: `5`

2. **Test Case 2**: `n = 2, a = 5, b = 9`
   - Maxim should buy 2 yogurts for 9 burles using the promotion.
   - Expected output: `9`

3. **Test Case 3**: `n = 3, a = 5, b = 9`
   - Maxim should buy 2 yogurts for 9 burles and 1 yogurt for 5 burles.
   - Expected output: `14`

4. **Test Case 4**: `n = 4, a = 5, b = 9`
   - Maxim should buy 4 yogurts, 2 at 9 burles each.
   - Expected output: `18`

5. **Test Case 5**: `n = 5, a = 5, b = 9`
   - Maxim should buy 4 yogurts, 2 at 9 burles each, and 1 yogurt for 5 burles.
   - Expected output: `23`

### Why This Test Input is Challenging:
- **Test Case 1** and **Test Case 2** are straightforward and should work correctly.
- **Test Case 3** and **Test Case 4** involve both individual and promotion prices, which can expose the issue in the promotion price calculation.
- **Test Case 5** involves an odd number of yogurts, which can further test the logic for handling the extra yogurt.

This test input should help identify the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
t = int(input())
for i in range(t):
    n, a, b = map(int, input().split())
    if a * 2 > b and n % 2 == 0 and n != 2:
        print(b * n)
    elif n == 2 and a * 2 > b and n % 2 == 0:
        print(b)
    elif a * 2 > b and n % 2 != 0 and n != 3:
        print((a + b)*n)
    elif n == 3 and a * 2 > b and n % 2 != 0:
        print(a+b)
    else:
        print(a * n)

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

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

Think step by step.

Response:
```plaintext
5
1 5 9
2 5 9
3 5 9
4 5 9
5 5 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
def solve(n, a, b):

    res = 0
    while n:

        if n==1:
            res += min(a,b)
            break
        res+=min(2*a, b)
        n-=2

    return res

Nt = int(input())

for _ in range(Nt):
    n, a, b = [int(num) for num in input().split(' ')]
    print(solve(n,a,b))

# 4
# 2 5 9
# 3 5 9
# 3 5 11
# 4 5 11

# 9
# 14
# 15
# 20

**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 where the code's logic might fail. The main issue in the provided code is how it handles the last yogurt when `n` is odd. The code always adds `min(a, b)` when `n` is 1, which is not always the optimal solution.

Let's consider the following scenarios:

1. **Odd `n` where the promotion price is significantly cheaper than the regular price:**
   - If `n` is odd and the promotion price `b` is significantly cheaper than `2a`, the optimal solution would be to buy as many pairs of yogurts as possible and then buy the last yogurt at the regular price. The code might fail if it incorrectly handles the last yogurt.

2. **Even `n` where the promotion price is slightly more expensive than the regular price:**
   - If `n` is even and the promotion price `b` is slightly more expensive than `2a`, the optimal solution would be to buy all yogurts at the regular price. The code might fail if it incorrectly uses the promotion price.

3. **Edge cases with `n = 1` or `n = 2`:**
   - These are simple cases but important to verify the correctness of the code.

Let's create a test input that covers these scenarios:

```plaintext
5
1 10 5
3 10 15
4 5 11
5 5 9
2 5 10
```

Explanation:
1. **Test Case 1:**
   - `n = 1`, `a = 10`, `b = 5`
   - The optimal solution is to buy one yogurt at the regular price, which is `10` burles.
   - The code should return `10`.

2. **Test Case 2:**
   - `n = 3`, `a = 10`, `b = 15`
   - The optimal solution is to buy two yogurts at the promotion price and one at the regular price, which is `15 + 10 = 25` burles.
   - The code should return `25`.

3. **Test Case 3:**
   - `n = 4`, `a = 5`, `b = 11`
   - The optimal solution is to buy four yogurts at the regular price, which is `4 * 5 = 20` burles.
   - The code should return `20`.

4. **Test Case 4:**
   - `n = 5`, `a = 5`, `b = 9`
   - The optimal solution is to buy two pairs of yogurts at the promotion price and one at the regular price, which is `2 * 9 + 5 = 23` burles.
   - The code should return `23`.

5. **Test Case 5:**
   - `n = 2`, `a = 5`, `b = 10`
   - The optimal solution is to buy two yogurts at the regular price, which is `2 * 5 = 10` burles.
   - The code should return `10`.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        int arr[] = new int[4];
        for(int i = 0 ; i<tc ; i++){
            int n = sc.nextInt();
            int a = sc.nextInt();
            int b = sc.nextInt();
            if(n==2){
                if(a*2<b)
                arr[i]=a*2;
            }
                else
                arr[i] = b;
            if(n%2==0){
                if(a*2<b){
                    arr[i]=a*n;
                }
                else{
                    arr[i]=b*(n/2);
                }
            }
            if(n%2!=0){
                if(a*2<b){
                    arr[i]=a*n;
                }
                else{
                    arr[i]=(b*(n/2))+a;
                }
            }
        }
        for(int j = 0; j<tc; j++){
            System.out.println(arr[j]);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
5
1 5 9
2 5 9
3 5 9
4 5 9
5 5 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
import java.util.Scanner;

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

        while (t-- > 0) {
            int n = in.nextInt();
            int a = in.nextInt();
            int b = in.nextInt();

            long costOption1 = n * a;
            long costOption2 = (long)Math.ceil((double) n / 2 * b);

            System.out.println(Math.min(costOption1, costOption2));
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
5
1 10 20
2 10 20
3 10 20
4 10 20
5 10 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
/*###############################################################################################################
##                                    Author : Kim Tae Yoon (Serendipity__)                                    ##
###############################################################################################################*/

#include <bits/stdc++.h>
#define fastio std::ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define prntyes cout<<"Yes\n";
#define prntno cout<<"No\n";
using namespace std;
#ifndef ONLINE_JUDGE
#include "___debug.h"
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
// mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count()); // random int64 generator
typedef long long ll;
typedef unsigned long ul;
typedef unsigned long long ull;
typedef __int128 ll128;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pii;
typedef complex<double> inum;
// Macros from KACTL pdf
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef vector<int> vi;
typedef vector<double> vd;
const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
const ll LLINF = 1000000000000000000LL;
const ll MAX = 200005; // depending on the problem
const ll MOD = 998244353; // depending on the problem

void solve() {
    int n,a,b; cin>>n>>a>>b;

    int ans = n*a;
    ans = min(ans, (n+1)/2*b);

    if (n%2==0){
        ans = min(ans, n/2*b);
    }else{
        ans = min(ans, n/2*b+a);
    }
    cout<<ans<<"\n";

}

int main() {
    fastio;
    int tc = 1;
    cin >> tc;
    while (tc--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 6 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

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

int main()
{
  int tc;
  cin >> tc;

  while (tc--)
  {

    int n, a, b;
    cin >> n >> a >> b;

    int dp[n + 1];
    memset(dp, -1, sizeof(dp));

    function<int(int)> fn = [&](int n)
    {
      if (n <= 0)
        return 0;
      if (dp[n] != -1)
        return dp[n];
      return dp[n] = min(fn(n - 2) + b, fn(n - 1) + a);
    };

    cout << fn(n) << endl;
  }

  return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where `n` is at the boundary of the allowed range (1 and 100).
2. **Promotion Price vs. Regular Price**: Test cases where the promotion price `b` is significantly higher or lower than the regular price `a`.
3. **Odd and Even Values of `n`**: Test cases where `n` is odd or even, as the algorithm should handle both scenarios correctly.
4. **Corner Cases**: Test cases where `b` is exactly twice `a` or where `b` is very close to `2a`.

Let's construct a test input that includes these scenarios:

```plaintext
6
1 5 9
2 5 9
3 5 11
4 5 11
100 5 9
100 5 10
```

Explanation:
1. **`1 5 9`**: This tests the case where `n` is 1, and the promotion price is higher than the regular price.
2. **`2 5 9`**: This tests the case where `n` is 2, and the promotion price is higher than the regular price.
3. **`3 5 11`**: This tests the case where `n` is 3, and the promotion price is higher than the regular price.
4. **`4 5 11`**: This tests the case where `n` is 4, and the promotion price is higher than the regular price.
5. **`100 5 9`**: This tests the case where `n` is 100, and the promotion price is lower than the regular price.
6. **`100 5 10`**: This tests the case where `n` is 100, and the promotion price is exactly twice the regular price.

This test input should help expose any issues in the buggy code, especially in handling different values of `n` and the relationship between `a` and `b`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
t = int(input())
data = []
for i in range(t):
    s = input()

    s = s.split()
    if len(s) == 3 and 1<= t <=10**3:
        n = int(s[0])
        a = int(s[1])
        b = int(s[2])
        normal_price = n *a
        if n % 2 != 0:
            even_num = int(n/2)
            remain_num = 1
        else:
            even_num = int(n/2)
            remain_num = 0

        promotion = (even_num * b) + remain_num * a

        data.append(min(normal_price,promotion))

for s in data:
    print(s)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The main issues in the buggy code are:

1. The condition `if len(s) == 3 and 1<= t <=10**3:` is incorrect because it checks `t` instead of `n`.
2. The calculation of `even_num` and `remain_num` is not correctly handling the case where `n` is odd.

Let's consider the following scenarios:

1. **When `n` is odd and `b` is very high**: The code might incorrectly choose the promotion price even though it is more expensive.
2. **When `n` is even and `b` is very low**: The code should always choose the promotion price.
3. **When `n` is 1**: The code should only consider the single yogurt price.
4. **When `n` is 2**: The code should compare the single yogurt price and the promotion price.

Based on these scenarios, here is a small-scale test input that can expose the bug:

```plaintext
4
1 5 10
2 5 9
3 5 11
4 5 11
```

Explanation:
- **Test Case 1**: `n = 1, a = 5, b = 10`
  - Since `n` is 1, only the single yogurt price should be considered.
  - Expected output: 5
- **Test Case 2**: `n = 2, a = 5, b = 9`
  - Since `n` is 2, the promotion price should be considered.
  - Expected output: 9
- **Test Case 3**: `n = 3, a = 5, b = 11`
  - Since `n` is 3, the code should correctly handle the odd case.
  - Expected output: 15 (2 * 5 + 1 * 5)
- **Test Case 4**: `n = 4, a = 5, b = 11`
  - Since `n` is 4, the code should correctly handle the even case.
  - Expected output: 20 (2 * 11 + 0 * 5)

This test input should help in identifying the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
import sys

def intinput():
    return int(sys.stdin.readline().strip())

def solution():
    n, a, b = map(int, input().split())

    if b >= 2 * a:
        print(n * a)
    elif b >= a:
        print(b * (n // 2) + a * (n % 2))
    else:
        print(b * (n // 2) + b)

t = intinput()
for _ in range(t):
    solution()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

Buggy Code:
import java.util.*;

import java.io.*;

public class Sol{

    static PrintWriter pw = new PrintWriter(System.out);
    static FastReader fr = new FastReader(System.in);
    private static long MOD = 1000000007;
    static int maxGlobalReach = Integer.MAX_VALUE;

    public static boolean res = false;
    public static void main(String[] args) throws Exception {
        int t = fr.nextInt();
        while (t-->0) {
            int n = fr.nextInt();
            int a = fr.nextInt();
            int b = fr.nextInt();

            if(b<a){
                int pairNeeded = (n+1)/2;
                pw.println(pairNeeded*b);
            }else if((a*2)>b){
                int maxDouble = n/2;
                int remain = n%2;
                pw.println((maxDouble*b)+(remain*a));
            }else{
                pw.println(a*n);
            }

        }
        pw.close();
    }

    public static long dpKnapSack(int index,int containerSize,long[][] dp,int n,Pair[] vals){
        if(containerSize<0 || index<0){
            return 0;
        }
        if(dp[index][containerSize]!=-1){
            return dp[index][containerSize];
        }

        long possibility = 0;
        if(containerSize>=vals[index].opNeeded){
            possibility = dpKnapSack(index-1, containerSize-vals[index].opNeeded, dp, n, vals)+vals[index].profit;
        }
        possibility = Math.max(possibility,dpKnapSack(index-1, containerSize, dp, n, vals));
        dp[index][containerSize] = possibility;
        return dp[index][containerSize];
    }

    public static int opNeeded(long val){
        int start = 1;
        int opNeeded = 0;

        while (start+start<=val) {
            start = start+start;
            opNeeded++;
        }

        if(start!=val){
            long neededVal = val-start;
            long x = start/neededVal;
            long sum = start/x;

            if(sum+start==val){
                opNeeded++;
            }else{
                opNeeded+=2;
            }
        }

        return opNeeded;
    }

    static class Pair {
        long val;
        int opNeeded;
        long profit;

        public Pair(long val,int opNeeded){
            this.val = val;
            this.opNeeded = opNeeded;
        }
    }

    static long[] factorial = new long[200005];
    public  static void instantiateFactorial(int max){
        factorial = new long[200005];
        factorial[0] = 1;

        for (int i = 1; i < factorial.length; i++) {
            factorial[i] = (factorial[i-1] * i) % MOD;
        }
    }

    static long nPr(int n,int r){
        long num = factorial[n];
        long denom = factorial[n-r];

        long inv = modPow(denom,MOD-2);
        long res = (num * inv) % MOD;

        return res;

    }

    static long modPow(long a,long b){
        long result = 1;
        while (b>0) {

            //if odd
            if((b&1)>0){
                result =(long) ((result* 1l * a)%MOD);
            }
            a = (long)(((a%MOD)*(a%MOD))%MOD);
            //right shift
            b>>=1;
        }
        return result;
    }

    /////////////////////
    //DSU
    static int[] parent;
    static int[] power;
    public static int findParent(int x){
        if(parent[x]==x){
            return x;
        }
        parent[x] = findParent(parent[x]);
        return parent[x];
    }

    public static void union(int a,int b){
        int parA = findParent(a);
        int parB = findParent(b);

        if(power[parA]<power[parB]){
            parent[parA] = parB;
        }else if(power[parB]<power[parA]){
            parent[parB] = parA;
        }else{
            parent[parB] = parA;
            power[parA]++;
        }
    }

    /////////////////////////////

    static int isPerfectSquare(int vals){
        int lastPow=1;
        while(lastPow*lastPow<vals){
            lastPow++;
        }
        if(lastPow*lastPow==vals){
            return lastPow*lastPow;
        }else{
            return -1;
        }
    }

    static class FastReader{

        byte[] buf = new byte[2048];
        int index, total;
        InputStream in;

        FastReader(InputStream is) {
            in = is;
        }

        int scan() throws IOException {
            if (index >= total) {
                index = 0;
                total = in.read(buf);
                if (total <= 0) {
                    return -1;
                }
            }
            return buf[index++];
        }

        String next() throws IOException {
            int c;
            for (c = scan(); c <= 32; c = scan());
            StringBuilder sb = new StringBuilder();
            for (; c > 32; c = scan()) {
                sb.append((char) c);
            }
            return sb.toString();
        }

        int nextInt() throws IOException {
            int c, val = 0;
            for (c = scan(); c <= 32; c = scan());
            boolean neg = c == '-';
            if (c == '-' || c == '+') {
                c = scan();
            }
            for (; c >= '0' && c <= '9'; c = scan()) {
                val = (val << 3) + (val << 1) + (c & 15);
            }
            return neg ? -val : val;
        }

        long nextLong() throws IOException {
            int c;
            long val = 0;
            for (c = scan(); c <= 32; c = scan());
            boolean neg = c == '-';
            if (c == '-' || c == '+') {
                c = scan();
            }
            for (; c >= '0' && c <= '9'; c = scan()) {
                val = (val << 3) + (val << 1) + (c & 15);
            }
            return neg ? -val : val;
        }
    }

    public static long GCD(long numA, long numB){
        if(numA==0){
            return numB;
        }else if(numB==0){
            return numA;
        }else{
            if(numA>numB){
                return GCD(numA%numB,numB);
            }else{
                return GCD(numA,numB%numA);
            }
        }
    }
}

//Things to remember
//Important XOR property
//:- X^X = 0
//:- X^X^X = X
//: A^B = C same as C^B = A

//https://www.geeksforgeeks.org/explore

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

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

Think step by step.

Response:
```plaintext
1
5 5 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yogurt Sale
time_limit_ms: 1000
memory_limit_mb: 256
Description: The price of one yogurt at the "Vosmiorochka" store is $$$a$$$ burles, but there is a promotion where you can buy two yogurts for $$$b$$$ burles.

Maxim needs to buy exactly $$$n$$$ yogurts. When buying two yogurts, he can choose to buy them at the regular price or at the promotion price.

What is the minimum amount of burles Maxim should spend to buy $$$n$$$ yogurts?

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

The first and only line of each test case contains three integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \le n \le 100$$$, $$$1 \le a, b \le 30$$$) — the number of yogurts Maxim wants to buy, the price for one yogurt, and the price for two yogurts on promotion.

Output: For each test case, print in a separate line the minimum cost of buying $$$n$$$ yogurts at "Vosmiorochka".

Examples:
input:
4
2 5 9
3 5 9
3 5 11
4 5 11
output:
9
14
15
20

Note: In the third test case of the example, it is more advantageous to buy three yogurts for $$$15$$$ burles than two for $$$11$$$ and one for $$$5$$$.

In the fourth test case of the example, you need to buy four yogurts, each for $$$5$$$ burles.

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

public class Main {
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = sc.nextInt();
        while(t-->0) {
            int n = sc.nextInt(); // number of yogurts maxim wants to buy
            int a = sc.nextInt(); // price of one yogurt
            int b  = sc.nextInt(); // price of two yogurts on promotion
            int minPrice = Math.min((b*n/2) + (a*n%2), a*n);
            out.println(minPrice);
        }
        out.close();

    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

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

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

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

        return gcd(b , a % b);
    }
}

**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 edge cases where the logic in the code might fail. The main issue in the buggy code is the way it calculates the minimum price. Specifically, it uses a formula that might not always give the correct result, especially when the promotion price is more advantageous than buying individual yogurts.

### Step-by-Step Analysis:

1. **Identify the Bug:**
   - The code calculates the minimum price using the formula: `Math.min((b * n / 2) + (a * n % 2), a * n)`.
   - This formula assumes that the best strategy is either to buy all yogurts at the promotion price or to buy all yogurts individually, which is not always true. The correct approach should consider a combination of both strategies.

2. **Edge Cases to Consider:**
   - **Case 1: Promotion price is much lower than the individual price.**
     - Example: `n = 5`, `a = 10`, `b = 15`. Here, the promotion is a better deal, and the correct answer should be `3 * 15 + 1 * 10 = 55`, but the buggy code might not handle this correctly.
   - **Case 2: Promotion price is higher than the individual price.**
     - Example: `n = 6`, `a = 5`, `b = 11`. Here, the promotion is not a good deal, and the correct answer should be `6 * 5 = 30`, but the buggy code might not handle this correctly.
   - **Case 3: Mixed scenarios where both promotion and individual prices are used.**
     - Example: `n = 7`, `a = 5`, `b = 9`. Here, the correct answer should be `3 * 9 + 1 * 5 = 32`, but the buggy code might not handle this correctly.

3. **Generate the Test Input:**
   - We will create a test input with a few test cases that cover the above scenarios.

### Challenging Test Input:
```plaintext
3
5 10 15
6 5 11
7 5 9
```

### Explanation:
- **Test Case 1:**
  - `n = 5`, `a = 10`, `b = 15`
  - Correct answer: `3 * 15 + 1 * 10 = 55`
- **Test Case 2:**
  - `n = 6`, `a = 5`, `b = 11`
  - Correct answer: `6 * 5 = 30`
- **Test Case 3:**
  - `n = 7`, `a = 5`, `b = 9`
  - Correct answer: `3 * 9 + 1 * 5 = 32`

This test input should expose the bug in the given buggy code by challenging the logic used to calculate the minimum price.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

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

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#include <iostream>
#include <cstring>
#include <vector>
#include <map>
#include <cmath>
#include <unordered_map>
using namespace std;
#define int long long
int n;
const int N=2e5+10;
int a[N];
int st[N];
int huan[N];
int cc[N];
int dist[N];
int du[N];
void hh(int u){
    int j=a[u];
    huan[u]=1;
    if(!huan[j]) hh(j);
}
void dfs(int u){
    st[u]=1;
    cc[u]=1;
    int j=a[u];
    if(!st[j]) dfs(j);
    else if(cc[j]) hh(j);
    if(huan[u]){
        dist[u]=2;
        cc[u]=0;
        return;
    }
    if(huan[j]){
        dist[u]=3;
    }
    else{
        dist[u]=dist[j]+du[j];
    }
    cc[u]=0;
}
void solve(){
    cin>>n;
    for(int i=1;i<=n;i++) {
        st[i]=0;
        dist[i]=1;
        huan[i]=0;
        cc[i]=0;
        du[i]=0;
    }
    for(int i=1;i<=n;i++){
        cin>>a[i];
        du[a[i]]++;
    }
    for(int i=1;i<=n;i++){
        if(!st[i]){
            dfs(i);
        }
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        ans=max(ans,dist[i]);
    }
    cout<<ans<<'\n';
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases or scenarios that can cause the algorithm to fail. Here are the steps to think through the problem:

1. **Understanding the Bug**: The provided code seems to be trying to identify cycles and calculate the distance of each spider from the cycle. However, it might not correctly handle cases where there are multiple cycles or where the cycles are not immediately apparent.

2. **Edge Cases**:
   - **Single Cycle**: A test case where all spiders form a single cycle.
   - **Multiple Cycles**: A test case with multiple disjoint cycles.
   - **Self-loops**: A test case where some spiders give plushies to themselves (though this is not allowed, it's good to check for robustness).
   - **Long Chains**: A test case with a long chain of spiders, where the last spider in the chain gives to the first spider, forming a cycle.
   - **Complex Graphs**: A test case with a more complex graph structure, including multiple cycles and chains.

3. **Challenging Test Case**:
   - Let's create a test case with a small number of spiders, but with a structure that can challenge the algorithm. For example, a test case with a small cycle and a chain leading into the cycle.

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

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

This test case has 5 spiders, and the plushies are exchanged in a cycle: spider 1 gives to spider 2, spider 2 gives to spider 3, spider 3 gives to spider 4, spider 4 gives to spider 5, and spider 5 gives to spider 1. This forms a single cycle, and the process should become stable after 2 years.

This input is designed to test the algorithm's ability to correctly identify and handle a simple cycle, which should help in exposing any bugs related to cycle detection and distance calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

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

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#include <iostream>
#include <set>
#include <vector>
#include <algorithm>
int r[200005];
int n, numTest;
std::set <int> in[200005];
std::set <int> listNotIn;
int d[200005];
bool visited[200005];

int main () {
  std::ios_base::sync_with_stdio(false);
  std::cin >> numTest;
  for (int testCase = 1; testCase <= numTest; testCase++) {
    std::cin >> n;
    listNotIn.clear();
    for (int i = 1; i <= n; i++) {
      in[i].clear();
      d[i] = 1;
      visited[i] = false;
    }
    for (int i = 1; i <= n; i++) {
      std::cin >> r[i];
      in[r[i]].insert(i);
    }
    for (int i = 1; i <= n; i++) {
      if (in[i].empty() == true) {
        listNotIn.insert(i);
      }
    }
    int ans = 1;
    while (listNotIn.empty() == false) {
      std::vector <int> newNotIn;
      for (std::set <int>::iterator it = listNotIn.begin(); it != listNotIn.end(); it++) {
        int u = *it;
        int next = r[u];
        d[next] += d[u];
        in[next].erase(in[next].find(u));
        if (in[next].empty() == true) {
          newNotIn.push_back(next);
        }
      }
      listNotIn.clear();
      while (newNotIn.empty() == false) {
        listNotIn.insert(newNotIn.back());
        newNotIn.pop_back();
      }
    }
    for (int i = 1; i <= n; i++) {
      if (visited[i] == false && in[i].empty() == false) {
        int p = i;
        int lenCycle = 0;
        bool isEqual = true;
        while (visited[p] == false) {
          ans = std::max(ans, d[p]);
          visited[p] = true;
          p = r[p];
        }
      }
    }
    std::cout << ans + 1 << std::endl;
  }
  return 0;
}

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

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

Think step by step.

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

1. **Cycles and Loops**: Ensure that the input creates complex cycles and loops that the algorithm might not handle correctly.
2. **Multiple Test Cases**: Include multiple test cases to ensure the algorithm handles different scenarios.
3. **Special Edge Cases**: Consider scenarios where the number of spiders is minimal or maximal within the given constraints.

Let's construct a test input that includes these aspects:

1. A simple cycle of 2 spiders.
2. A more complex cycle of 5 spiders.
3. A scenario with a large number of spiders to test the limits.
4. A scenario where the process becomes stable after a few years.

Here is the test input:

```plaintext
4
2
2 1
5
2 3 4 5 1
10
2 3 4 5 1 2 3 4 5 1
20
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1
```

Explanation:
1. **Test Case 1**: A simple cycle of 2 spiders. This should stabilize in year 2.
2. **Test Case 2**: A cycle of 5 spiders. This should stabilize in year 2.
3. **Test Case 3**: A cycle of 10 spiders. This should stabilize in year 2.
4. **Test Case 4**: A large cycle of 20 spiders. This should stabilize in year 2.

This test input is designed to ensure that the algorithm correctly handles different cycle lengths and large numbers of spiders. The output for this test input should be:

```plaintext
2
2
2
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

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

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#include <cstdio>
#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <cassert>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
using namespace std;

#define ll long long
#define ld long double
#define ull unsigned long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pdd pair<double, double>
#define vi vector<int>
#define vll vector<ll>
#define vd vector<double>
#define vpii vector<pii>
#define vpll vector<pll>
#define vpd vector<pdd>
#define st set<int>
#define mset(a, b) memset(a, b, sizeof(a))
#define sz(a) int((a).size())
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) a.begin(), a.end()
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep1(i, n) for (int i = 1; i <= (n); i++)
#define repr(i, n) for (int i = (n) - 1; i >= 0; i--)
#define repi(i, l, r) for (int i = (l); i <= (r); i++)
#define repir(i, l, r, c) for (int i = (l); i <= (r); i += (c))
#define per(i, l, r) for (int i = (l); i >= (r); i--)
#define perir(i, l, r, c) for (int i = (l); i >= (r); i -= (c))
#define debug(x) cerr << #x << '=' << (x) << endl
#define repdebug(i, n) cerr << #i << '=' << (i) << endl
#define debugarr(a, n)              \
    cerr << #a << "=[";             \
    rep1(i, n) cerr << a[i] << ' '; \
    cerr << ']' << endl
#define debugall(a)            \
    cerr << #a << '=' << endl; \
    for (auto x : a)           \
        cerr << x << ' ';      \
    cerr << endl
#define IOS                      \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0)
#define endl '\n'
#define endll endl << '\n'
#define endi endl << ' '
#define endd endl << ' '
#define endall(a)         \
    for (auto x : a)      \
        cerr << x << ' '; \
    cerr << endl
#define INF 0x3f3f3f3f3f3f3f3f

const int maxn = 2e5 + 5, maxm = 1e3 + 5, maxq = 1e7 + 5;
const int mod = 1e9 + 7;

ll n, m, k, t, u, v, i, j, ans, res, mid, cnt, tmp, sum;
ll arr[maxn], c[maxn], s[maxn];

ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll power(ll a, ll b)
{
    ll res = 1;
    while (b > 0)
    {
        if (b & 1)
            res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}
ll pre_fnd(ll *a, ll x, ll len)
{
    ll l = 1, r = len, rs = -1;
    while (l <= r)
    {
        mid = (l + r) >> 1;
        if (a[mid] <= x)
        {
            rs = mid;
            l = mid + 1;
        }
        else
            r = mid - 1;
    }
    return rs;
}
ll suf_fnd(ll *a, ll x, ll len)
{
    ll l = 1, r = len, rs = -1;
    while (l <= r)
    {
        mid = (l + r) >> 1;
        if (a[mid] >= x)
        {
            rs = mid;
            r = mid - 1;
        }
        else
            l = mid + 1;
    }
    return rs;
}
inline void read(int &a)
{
    int s = 0, w = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            w = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        s = (s << 3) + (s << 1) + ch - '0';
        ch = getchar();
    }
    a = s * w;
}

struct treeNode
{
    ll lef, rig, val, stag, mtag;
} tr[maxn << 2];

void build(ll p, ll l, ll r)
{
    tr[p] = {l, r, 0, 0};
    if (l == r)
    {
        tr[p].val = s[l];
        tr[p].stag = 0;
        tr[p].mtag = 1;
        return;
    }
    ll mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    tr[p].val = tr[p << 1].val + tr[p << 1 | 1].val;
    tr[p].stag = 0;
    tr[p].mtag = 1;
    return;
}

void push_down(ll p)
{
    tr[p << 1].val = tr[p << 1].val * tr[p].mtag + tr[p].stag * (tr[p << 1].rig - tr[p << 1].lef + 1);
    tr[p << 1 | 1].val = tr[p << 1 | 1].val * tr[p].mtag + tr[p].stag * (tr[p << 1 | 1].rig - tr[p << 1 | 1].lef + 1);
    tr[p << 1].stag = tr[p << 1].stag * tr[p].mtag + tr[p].stag;
    tr[p << 1 | 1].stag = tr[p << 1 | 1].stag * tr[p].mtag + tr[p].stag;
    tr[p << 1].mtag = tr[p << 1].mtag * tr[p].mtag;
    tr[p << 1 | 1].mtag = tr[p << 1 | 1].mtag * tr[p].mtag;
    tr[p].stag = 0;
    tr[p].mtag = 1;
}

void add(ll p, ll l, ll r, ll k)
{
    if (tr[p].lef >= l && tr[p].rig <= r)
    {
        tr[p].val += k * (tr[p].rig - tr[p].lef + 1);
        tr[p].stag += k;
        return;
    }
    push_down(p);
    if (l <= tr[p << 1].rig)
        add(p << 1, l, r, k);
    if (r >= tr[p << 1 | 1].lef)
        add(p << 1 | 1, l, r, k);
    tr[p].val = tr[p << 1].val + tr[p << 1 | 1].val;
    return;
}

void mul(ll p, ll l, ll r, ll k)
{
    if (tr[p].lef >= l && tr[p].rig <= r)
    {
        tr[p].val *= k;
        tr[p].stag *= k;
        tr[p].mtag *= k;
        return;
    }
    push_down(p);
    if (l <= tr[p << 1].rig)
        mul(p << 1, l, r, k);
    if (r >= tr[p << 1 | 1].lef)
        mul(p << 1 | 1, l, r, k);
    tr[p].val = tr[p << 1].val + tr[p << 1 | 1].val;
    return;
}

ll search(ll p, ll l, ll r)
{
    if (tr[p].lef >= l && tr[p].rig <= r)
        return tr[p].val;
    if (tr[p].lef > r || tr[p].rig < l)
        return 0;
    push_down(p);
    ll sr = 0;
    if (tr[p << 1].rig >= l)
        sr += search(p << 1, l, r);
    if (tr[p << 1 | 1].lef <= r)
        sr += search(p << 1 | 1, l, r);
    return sr;
}

struct node
{
    ll val, ste;
} nodes[maxn];

vector<ll> G[maxn];
ll in[maxn];
queue<pair<ll, ll>> q;

void build_Edge(ll u, ll v)
{
    G[u].push_back(v);
    in[v]++;
}

void bfs()
{
    rep1(i, n) if (in[i] == 0) q.push({i, 2});
    while (!q.empty())
    {
        auto [tu, step] = q.front();
        ans = max(ans, step);
        q.pop();
        for (auto tv : G[tu])
        {
            if (step == nodes[tv].ste)
                nodes[tv].val++;
            else
                nodes[tv].ste = step, nodes[tv].val = 1;
            in[tv]--;
            if (in[tv] == 0)
                q.push({tv, step + 1});
        }
    }
}

int main()
{
    IOS;
    cin >> t;
    while (t--)
    {
        cin >> n;
        rep1(i, n) G[i].clear();
        while (!q.empty())
            q.pop();
        ans = 1;
        rep1(i, n) in[i] = 0, nodes[i].val = 1, nodes[i].ste = 0;
        rep1(i, n) cin >> tmp, build_Edge(i, tmp);
        bfs();
        k = 0;
        rep1(i, n) k = max(k, nodes[i].val);
        cout << ans + k << endl;
    }
    return 0;
}

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code seems to be using a BFS approach to determine the first stable year. However, it does not correctly handle the simultaneous nature of the plushie exchanges. The BFS approach might not accurately simulate the simultaneous exchanges, leading to incorrect results.

2. **Identify Edge Cases**:
   - **Single Cycle**: A single cycle where each spider gives a plushie to another in a loop.
   - **Multiple Cycles**: Multiple independent cycles of spiders giving plushies.
   - **Complex Graph**: A more complex graph with multiple cycles and paths, including dead ends.

3. **Create a Test Case**:
   - A test case with a single cycle will help ensure the algorithm correctly handles the simultaneous nature of the exchanges.
   - A test case with multiple cycles will test the algorithm's ability to handle multiple stable cycles.
   - A test case with a complex graph will test the algorithm's ability to handle more intricate scenarios.

4. **Generate the Test Input**:
   - For a single cycle, we can use a small number of spiders, say 5, where each spider gives a plushie to the next spider in a loop.
   - For multiple cycles, we can use a small number of spiders, say 10, with two separate cycles of 5 spiders each.
   - For a complex graph, we can use a small number of spiders, say 7, with a mix of cycles and dead ends.

Given the constraints and the need for a small-scale test input, let's create a test case with a single cycle of 5 spiders:

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

This test case will help expose any issues with the algorithm's handling of simultaneous exchanges in a single cycle. If the algorithm is incorrect, it will not correctly determine that the first stable year is 2.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

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

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import sys
from heapq import heappop, heappush
input = sys.stdin.readline

T = int(input().strip())
for _ in range(T):
    N = int(input().strip())
    graph = [-1] + list(map(int, input().split()))
    deg = [0]*(N+1)
    visited = [False]*(N+1)
    qu = []
    for i in graph[1:]:
        deg[i] += 1
    for i in range(1, N+1):
        if deg[i] == 0:
            heappush(qu, (2, i))
    plush = deg[:]
    res = 0
    flag = 1
    mid = []
    if not qu:
        flag = 0
        res = 2
    while qu:
        time, t = heappop(qu)
        if time > res:
            res = time
        visited[t] = True
        if deg[t] == 0:
            deg[graph[t]] -= 1
            if deg[graph[t]] == 0 and not visited[graph[t]]:
                if time+plush[graph[t]] > res:
                    res = time+plush[graph[t]]
                qu.append((time+plush[graph[t]], graph[t]))
    if flag:
        print(res+1)
    else:
        print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

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

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
from itertools import permutations, combinations, accumulate
from collections import defaultdict, deque, Counter
from heapq import heapify, heappush, heappop, merge
from bisect import bisect_right, bisect_left
from io import StringIO,BytesIO
from math import gcd,inf,isqrt
from random import randint
_MOD = 998244353
MOD = 10**9 + 7
import sys,os

# sys.stdin=BytesIO(os.read(0,os.fstat(0).st_size))

odd=lambda N:N&1
input=sys.stdin.readline
even=lambda N:not(odd(N))
getone=lambda:int(input())
lcm=lambda x,y:x*y//gcd(x,y)
getonestr=lambda:input().strip()
get=lambda:map(int,input().split())
getstr=lambda:list(input().split())
CEIL_DIV=lambda a,b:(a//b)+bool(a%b)
SINGLE_TESTCASE=CALL=lambda func:func()
write=lambda obj:sys.stdout.write(f"{obj}")
getarr=lambda:list(map(int,input().split()))
writeln=lambda obj='':sys.stdout.write(f"{obj}\n")
MULTIPLE_TESTCASE=lambda func:exec('for _ in range(int(input())):func()')
_2D_ARRAY_SPAWN=lambda i,j,T=None:[[T for _j in range(j)] for _i in range(i)]
_3D_ARRAY_SPAWN=lambda i,j,k,T=None:[[[T for _k in range(k)] for _j in range(j)] for _i in range(i)]
dp5state = [[[[[0 for i in range(1)] for j in range(1)] for k in range(1)] for l in range(1)] for z in range(1)]

def solve() :
    ...
    n = getone()
    arr = getarr()

    pred = defaultdict(int)
    for x in arr :
        pred[x] += 1

    bfs = []
    for u in range(1, n + 1) :
        if pred[u] == 0 :
            bfs += [(u, 0)]

    val = defaultdict(lambda: 1)
    for u, d in bfs :
        pred[arr[u - 1]] -= 1
        if pred[arr[u - 1]] == 0 :
            bfs += (arr[u - 1], d + 1),

        val[arr[u - 1]] += val[u]

    if not val :
        return 2

    return 1 + max(val.values())

@MULTIPLE_TESTCASE
def main() :

    # stress test
    # while 1  :
    #     l = randint(1, 100)
    #     r = randint(l + 1, 500)
    #     G = randint(1, 20)
    #     solve(l, r, G)
    #     print(f"checked {l},{r},{G}")

    ans=solve()
    if ans is not None: writeln(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

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

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
def solve(vs):

	ins = [[] for _ in range(len(vs)+1)]
	for v,u in enumerate(vs):
		ins[u].append(v+1)

	seeng = set([])
	def find_cycle(v):
		seen = set([])
		while True:
			if v in seeng:
				seeng.update(seen)
				return None
			if v in seen:
				seeng.update(seen)
				return v
			seen.add(v)
			v = vs[v-1]

	m = 0
	for v in range(1,len(vs)+1):
		v = find_cycle(v)
		if not v: continue

		def vlen(v):
			vl = sum(map(vlen,ins[v])) + 1
			return vl

		u = vs[v-1]
		p = v
		while True:
			ws = [w for w in ins[u] if w != p]
			m = max(m, sum(map(vlen,ws)))
			p = u
			u = vs[u-1]
			if p == v: break

	return m + 2

for _ in range(int(input())):
	input()
	print(solve([*map(int,input().split())]))

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

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

Think step by step.

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

**Explanation:**
This test case is designed to challenge the buggy code by creating a simple cycle among the spiders. Here's how the process would unfold:

- **Initial State (Year 1):** Each spider has 1 plushie.
  - Spider 1: 1 plushie
  - Spider 2: 1 plushie
  - Spider 3: 1 plushie
  - Spider 4: 1 plushie

- **Year 2:**
  - Spider 1 gives 1 plushie to Spider 2.
  - Spider 2 gives 1 plushie to Spider 3.
  - Spider 3 gives 1 plushie to Spider 4.
  - Spider 4 gives 1 plushie to Spider 1.
  - State after Year 2:
    - Spider 1: 1 plushie
    - Spider 2: 1 plushie
    - Spider 3: 1 plushie
    - Spider 4: 1 plushie

Since the state after Year 2 is the same as the initial state, the process becomes stable in Year 2. This test case should expose any issues in the buggy code that might incorrectly handle cycles or the stability condition.