**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Colored Balls
time_limit_ms: 2000
memory_limit_mb: 1024
Description: There are balls of $$$n$$$ different colors; the number of balls of the $$$i$$$-th color is $$$a_i$$$.

The balls can be combined into groups. Each group should contain at most $$$2$$$ balls, and no more than $$$1$$$ ball of each color.

Consider all $$$2^n$$$ sets of colors. For a set of colors, let's denote its value as the minimum number of groups the balls of those colors can be distributed into. For example, if there are three colors with $$$3$$$, $$$1$$$ and $$$7$$$ balls respectively, they can be combined into $$$7$$$ groups (and not less than $$$7$$$), so the value of that set of colors is $$$7$$$.

Your task is to calculate the sum of values over all $$$2^n$$$ possible sets of colors. Since the answer may be too large, print it modulo $$$998\,244\,353$$$.

Input: The first line contains a single integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the number of colors.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 5000$$$) — the number of balls of the $$$i$$$-th color.

Additional constraint on input: the total number of balls doesn't exceed $$$5000$$$.

Output: Print a single integer — the sum of values of all $$$2^n$$$ sets of colors, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 1 2
output:
11

input:
1
5
output:
5

input:
4
1 3 3 7
output:
76

Note: Consider the first example. There are $$$8$$$ sets of colors:

- for the empty set, its value is $$$0$$$;
- for the set $$$\{1\}$$$, its value is $$$1$$$;
- for the set $$$\{2\}$$$, its value is $$$1$$$;
- for the set $$$\{3\}$$$, its value is $$$2$$$;
- for the set $$$\{1,2\}$$$, its value is $$$1$$$;
- for the set $$$\{1,3\}$$$, its value is $$$2$$$;
- for the set $$$\{2,3\}$$$, its value is $$$2$$$;
- for the set $$$\{1,2,3\}$$$, its value is $$$2$$$.

So, the sum of values over all $$$2^n$$$ sets of colors is $$$11$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int M=998244353;
int main(){
    int t;
   t=1;
    while(t--){
        int n;
        cin>>n;
        int a[n];
        int sum=0;
        int ma=-1;
        for(int i=0;i<n;i++){
            cin>>a[i];
            sum+=a[i];

        }

      long long dp[sum+1];
      for(int i=0;i<=sum;i++){
        dp[i]=0;
      }
      dp[0]=1;
      for(int i=0;i<n;i++){
           for(int j=sum;j>=0;j--){
            if(j+a[i]<=sum){
                dp[j+a[i]]+=dp[j];
                dp[j+a[i]]%=M;
            }
           }
      }

      long long ans=0;
      for(int i=1;i<=sum;i++){
              ans+=(i+1)/2*1LL*dp[i];
              ans%=M;
      }

      for(int i=0;i<n;i++){
        for(int j=0;j<=a[i];j++){
            ans+=-dp[j]*1LL*((j+1+a[i])/2)%M+dp[j]*1LL*a[i]%M;
            ans%=M;
        }
      }
       cout<<ans<<endl;
    }

}

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

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

Think step by step.

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

1. **Input Size**: The input size should be small enough to be manageable and to ensure the output is concise, but large enough to test edge cases effectively.
2. **Edge Cases**: We should consider edge cases such as:
   - All colors having the same number of balls.
   - One color having a significantly larger number of balls.
   - A combination of different numbers of balls.

Given the constraints \(1 \le n \le 5000\) and \(1 \le a_i \le 5000\), and the total number of balls not exceeding 5000, we can create a test case that is small but complex enough to reveal issues.

### Chosen Test Case
Let's create a test case with 4 colors where one color has a significantly larger number of balls and the others have smaller numbers. This will test how the code handles a large number of balls in one color and how it combines them with smaller groups.

```plaintext
4
1 3 3 7
```

### Explanation
- **n = 4**: There are 4 different colors.
- **a = [1, 3, 3, 7]**: The number of balls of each color is 1, 3, 3, and 7 respectively.
- **Total number of balls = 1 + 3 + 3 + 7 = 14**: This is within the limit of 5000.

This test case is designed to:
- Test the handling of a large number of balls in one color.
- Ensure the code correctly calculates the minimum number of groups for various combinations of colors.
- Verify that the sum of values over all \(2^n\) sets of colors is calculated correctly.

This input should be challenging enough to 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: Colored Balls
time_limit_ms: 2000
memory_limit_mb: 1024
Description: There are balls of $$$n$$$ different colors; the number of balls of the $$$i$$$-th color is $$$a_i$$$.

The balls can be combined into groups. Each group should contain at most $$$2$$$ balls, and no more than $$$1$$$ ball of each color.

Consider all $$$2^n$$$ sets of colors. For a set of colors, let's denote its value as the minimum number of groups the balls of those colors can be distributed into. For example, if there are three colors with $$$3$$$, $$$1$$$ and $$$7$$$ balls respectively, they can be combined into $$$7$$$ groups (and not less than $$$7$$$), so the value of that set of colors is $$$7$$$.

Your task is to calculate the sum of values over all $$$2^n$$$ possible sets of colors. Since the answer may be too large, print it modulo $$$998\,244\,353$$$.

Input: The first line contains a single integer $$$n$$$ ($$$1 \le n \le 5000$$$) — the number of colors.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 5000$$$) — the number of balls of the $$$i$$$-th color.

Additional constraint on input: the total number of balls doesn't exceed $$$5000$$$.

Output: Print a single integer — the sum of values of all $$$2^n$$$ sets of colors, taken modulo $$$998\,244\,353$$$.

Examples:
input:
3
1 1 2
output:
11

input:
1
5
output:
5

input:
4
1 3 3 7
output:
76

Note: Consider the first example. There are $$$8$$$ sets of colors:

- for the empty set, its value is $$$0$$$;
- for the set $$$\{1\}$$$, its value is $$$1$$$;
- for the set $$$\{2\}$$$, its value is $$$1$$$;
- for the set $$$\{3\}$$$, its value is $$$2$$$;
- for the set $$$\{1,2\}$$$, its value is $$$1$$$;
- for the set $$$\{1,3\}$$$, its value is $$$2$$$;
- for the set $$$\{2,3\}$$$, its value is $$$2$$$;
- for the set $$$\{1,2,3\}$$$, its value is $$$2$$$.

So, the sum of values over all $$$2^n$$$ sets of colors is $$$11$$$.

Buggy Code:
from math import *
from collections import *
from bisect import *
mod=998244353
def solve():
    n=int(input())
    #x,y=map(int,input().split())
    l=list(map(int,input().split()))
    l.sort()
    s=sum(l)
    l1=[0 for i in range(s+1)]
    l1[0]=1
    ans=0
    for i in range(n):
        for j in range(s-l[i],-1,-1):
            l1[j+l[i]]=l1[j+l[i]]+l1[j]
            if l1[j]>0:
                if j==0:
                    ans+=(l[i]%mod)
                    ans%=mod
                else:
                    if l[i]>ceil((j+l[i])/2):
                        ans+=(l[i]*l1[j]%mod)
                        ans%=mod
                    else:
                        ans+=(ceil((j+l[i])/2)*l1[j]%mod)
                        ans%=mod
            #print(ans,l1[j+l[i]],j+l[i])
    print(ans)
for _ in range(1):
    solve()

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code involves dynamic programming and combinatorial calculations, so we should consider cases where the values of `a_i` are such that the logic in the code might break down.

### Step-by-Step Analysis:

1. **Edge Cases:**
   - **Single Color:** This is already covered in the examples.
   - **Multiple Colors with Small Values:** This is also covered in the examples.
   - **Multiple Colors with Large Values:** This can be tricky, especially if the values are close to the maximum allowed (5000).

2. **Combinatorial Complexity:**
   - The code uses a dynamic programming approach to count the number of ways to form groups. If the values of `a_i` are large, the combinatorial complexity can lead to issues.

3. **Special Cases:**
   - **All Values are the Same:** This can simplify the problem and might expose issues in the combinatorial logic.
   - **Values with High Variance:** This can test the robustness of the code in handling different magnitudes of `a_i`.

### Chosen Test Input:
Let's choose a test input where all values of `a_i` are the same and relatively large. This will test the combinatorial logic and the modulo operations.

```plaintext
3
5000 5000 5000
```

### Explanation:
- **Input:** `3` (number of colors)
- **Values:** `5000, 5000, 5000` (number of balls of each color)

This input is small enough to be valid and manageable, but large enough to stress the combinatorial logic and modulo operations. The code should handle this case correctly, and if it fails, it will expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        double n;
        cin >> n;
        if(n==0)
        {
            cout << "0"<<endl;
            break;
        }
        double a,b;
        cin >>a>>b;
        if(b>=a)
        {
            cout << ceil(n/a)<<endl;
        }
        else cout << ceil(n/b)<<endl;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
#include <iostream>
using namespace std;
int main() {
	int t, n, x, y;
	int time, count;
	int more1;
	cin >> t;
	for (int i = 0;i < t;i++) {
		cin >> n >> x >> y;
		count = n / y;
		if (n != count * y)
			count++;
		if (x * count >= n)
			cout << count<<endl;
		else {
			more1 = n - x * count;
			time = more1 / x+count;
			if (more1!= time * x) {
				more1++;
			}
			cout << time<<endl;
		}
	}
}

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

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

Think step by step.

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

1. **Incorrect handling of the blending and adding process**: The code does not correctly account for the blending and adding of fruits in each second.
2. **Incorrect calculation of the time required**: The code's logic for calculating the time required to blend all fruits is flawed.

Let's consider the following edge cases:

1. **Case where the number of fruits to blend per second (x) is much larger than the number of fruits that can be added per second (y)**: This can expose issues in the blending and adding process.
2. **Case where the number of fruits to blend per second (x) is equal to the number of fruits that can be added per second (y)**: This can expose issues in the time calculation.
3. **Case where the number of fruits to blend per second (x) is much smaller than the number of fruits that can be added per second (y)**: This can also expose issues in the time calculation.

Here is a small-scale test input that covers these edge cases:

```plaintext
5
10
1 10
10
10 1
10
5 5
10
1 1
10
10 10
```

This test input includes:
1. A case where `x` is much smaller than `y` (10 fruits, 1 blend per second, 10 add per second).
2. A case where `x` is much larger than `y` (10 fruits, 10 blend per second, 1 add per second).
3. A case where `x` is equal to `y` (10 fruits, 5 blend per second, 5 add per second).
4. A case where `x` is equal to `y` but both are 1 (10 fruits, 1 blend per second, 1 add per second).
5. A case where `x` is equal to `y` and both are large (10 fruits, 10 blend per second, 10 add per second).

These cases should help expose the bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl

void solve(){
    int n,x,y;
    cin>>n>>x>>y;
    cout<<ceil(n*1.0/(min(x,y)))<<endl;
}
int main(){
    int t=1;
    cin>>t;
    while(t--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
10
1 10
10
10 1
1000000000
1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
def min_blend_time(n, x, y):
    if n == 0:
        return 0
    if y > x:
        return (n + x - 1) // x
    time = 0
    fruits_in_blender = 0

    while n > 0:
        fruits_to_add = min(y, n)
        fruits_in_blender += fruits_to_add
        n -= fruits_to_add
        fruits_blended = min(x, fruits_in_blender)
        fruits_in_blender -= fruits_blended

        time += 1

    return time
t = int(input())
for _ in range(t):
    n = int(input())
    x, y = map(int, input().split())

    result = min_blend_time(n, x, y)
    print(result)

**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
0
1 2
1
1 2
2
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import math

t = int(input(""))

for i in range(t):
    n = int(input(""))
    if n == 0:
        print("0")
    x, y = map(int, input().split())
    ans = math.ceil(n / min(x,y))
    print(ans)

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic used to calculate the minimum time required to blend all fruits. The code uses `math.ceil(n / min(x, y))`, which is incorrect because it doesn't account for the fact that Zhan can add fruits to the blender in each second, and the blender can blend a different number of fruits per second.

Let's consider the following edge cases:
1. **When \( x \) (blender's capacity) is greater than \( y \) (Zhan's input capacity):**
   - If \( x \) is much larger than \( y \), the code will incorrectly use \( y \) as the blending rate, which can lead to an overestimation of the time required.
2. **When \( n \) is very large:**
   - The code might not handle large values of \( n \) correctly due to the incorrect blending rate calculation.
3. **When \( x \) is equal to \( y \):**
   - The code should work correctly in this case, but it's still worth testing to ensure it behaves as expected.

Let's generate a test input that maximizes the failure probability:

```plaintext
3
10
4 3
15
5 2
20
6 4
```

Explanation:
1. **Test Case 1:**
   - \( n = 10 \)
   - \( x = 4 \) (blender can blend 4 fruits per second)
   - \( y = 3 \) (Zhan can add 3 fruits per second)
   - Correct answer: 3 seconds (Zhan adds 3 fruits in the first second, 3 fruits in the second second, and 4 fruits in the third second).

2. **Test Case 2:**
   - \( n = 15 \)
   - \( x = 5 \) (blender can blend 5 fruits per second)
   - \( y = 2 \) (Zhan can add 2 fruits per second)
   - Correct answer: 4 seconds (Zhan adds 2 fruits in the first second, 2 fruits in the second second, 2 fruits in the third second, and 9 fruits in the fourth second).

3. **Test Case 3:**
   - \( n = 20 \)
   - \( x = 6 \) (blender can blend 6 fruits per second)
   - \( y = 4 \) (Zhan can add 4 fruits per second)
   - Correct answer: 4 seconds (Zhan adds 4 fruits in the first second, 4 fruits in the second second, 4 fruits in the third second, and 8 fruits in the fourth second).

This test input should help expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import math as m
t=int(input())
l={}
while (t!=0):
    t=t-1
    n=int(input())
    o=input()
    o=o.split(" ")
    k=[]
    for i in o:
        k.append(int(i))
    l[n]=k

res=[]
for g in l:
    x=l[g][0]
    y=l[g][1]
    if(x<=y):
        r=m.ceil(g/x)
    else:
        r=m.ceil(g/y)
    res.append(r)
for i in res:
    print(i)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import java.util.*;
public class ZhanBlender{
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            long n=sc.nextLong();
            long x=sc.nextLong();
            long y=sc.nextLong();
            if(n==0){
                System.out.println("0");
                continue;
            }
            long seconds=0;
            while(n>0){
                long added=Math.min(x,y);
                n-=added;
                seconds++;
            }
            System.out.println(seconds);
        }
    }
}

**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
10
1 1
10
10 1
10
1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

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

public class IterateSubset {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

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

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

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

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

    public static void main(String[] args) throws IOException {
        try {
            FastReader in = new FastReader();
            FastWriter out = new FastWriter();
            int t = in.nextInt();
            while (t-- > 0) {
               int n = in.nextInt();
               int x = in.nextInt();
               int y = in.nextInt();
               int c =0;
               while(n > 0) {
                int blend = Math.min(n, y);
                int minus = Math.min(x, blend);
                blend = blend - minus;
                n = n- minus;
                c++;
               }
               out.println(c);
            }
            out.close();
        } catch (Exception e) {
            return;
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios where the blending and adding of fruits can lead to incorrect calculations. The key points to consider are:

1. **Blender Capacity and Adding Fruits**: The code should correctly handle the scenario where the number of fruits added in a second is less than the number of fruits the blender can blend.
2. **Blender Blending**: The code should correctly handle the scenario where the number of fruits in the blender is less than the number of fruits the blender can blend.
3. **Edge Cases**: Special cases where `n`, `x`, or `y` are at their limits (0, 1, or very large values).

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

1. **Case 1**: `n = 0` (no fruits to blend)
2. **Case 2**: `n = 1`, `x = 1`, `y = 1` (single fruit, single capacity, single addition)
3. **Case 3**: `n = 10`, `x = 1`, `y = 1` (many fruits, single capacity, single addition)
4. **Case 4**: `n = 10`, `x = 10`, `y = 1` (many fruits, large capacity, single addition)
5. **Case 5**: `n = 10`, `x = 1`, `y = 10` (many fruits, single capacity, large addition)
6. **Case 6**: `n = 1000000000`, `x = 1000000000`, `y = 1000000000` (large values for all parameters)

Here is the test input:

```plaintext
6
0
1 1
1
1 1
10
1 1
10
10 1
10
1 10
1000000000
1000000000 1000000000
```

This test input covers a range of edge cases and should help identify any 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: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

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

using namespace std;

using ll = long long;

void Solve() {
    ll n;
    cin >> n;
    ll x, y;
    cin >> x >> y;

    if (n == 0) {
        cout << 0 << endl;
        return;
    }

    ll t = 0;
    if (y >= x) {
        t = (n + x - 1) / x;
    } else {
        ll f = 0;
        while (n > 0) {
            t++;
            f += min(y, n);
            n -= min(x, f);
            f -= min(x, f);
        }
    }

    cout << t << endl;
}

int main() {
    ios::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:
```plaintext
3
10
1 10
10
10 1
10
5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
    short int t;
    cin >> t;
    while (t--)
    {
        int n, x, y;
        cin >> n >> x >> y;
        int mini = min(x, y);
        cout << ceil((n * 1.0) / mini) << 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:
```plaintext
1
10
1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import sys
import math
import bisect
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, combinations
from functools import lru_cache
import os

if os.path.exists("input.txt"):
    sys.stdin = open("input.txt", "r")
    sys.stdout = open("output.txt", "w")
    sys.stderr = open("error.txt", "w")

input = sys.stdin.read
sys.setrecursionlimit(10**6)

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

def inlt():
    return list(map(int, sys.stdin.readline().strip().split()))

def insr():
    """Reads a string as a list of characters."""
    return list(sys.stdin.readline().strip())

def invr():
    return tuple(map(int, sys.stdin.readline().strip().split()))

def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

def lcm(a, b):
    return (a * b) // gcd(a, b)

def is_prime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True

def sieve(n):
    is_prime = [True] * (n + 1)
    p = 2
    while p * p <= n:
        if is_prime[p]:
            for i in range(p * p, n + 1, p):
                is_prime[i] = False
        p += 1
    return [p for p in range(2, n + 1) if is_prime[p]]

def binary_search(arr, x):
    lo, hi = 0, len(arr) - 1
    while lo <= hi:
        mid = (lo + hi) // 2
        if arr[mid] == x:
            return mid
        elif arr[mid] < x:
            lo = mid + 1
        else:
            hi = mid - 1
    return -1

def lower_bound(arr, target):
    """Finds the first position where target can be inserted without breaking order."""
    return bisect.bisect_left(arr, target)

def upper_bound(arr, target):
    """Finds the first position where target is strictly greater than the target."""
    return bisect.bisect_right(arr, target)

def sort_by_second(elem):
    """Sorts a list of tuples by the second element."""
    return sorted(elem, key=lambda x: x[1])

def solve():

    n = inp()
    x, y= inlt()
    result = 0

    xy = min(x, y)

    while n > 0:
        n -= xy
        result += 1

    print(result)

def main():

    t = inp()  # Number of test cases
    for _ in range(t):
        solve()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's analyze the code and identify potential issues:

1. **Bug Analysis**:
   - The code assumes that the blender can always add the minimum of `x` or `y` fruits per second.
   - This assumption is incorrect because it does not account for the scenario where `n` is less than `xy` (i.e., the number of fruits left is less than the maximum number of fruits that can be added or blended in one second).

2. **Edge Cases**:
   - **Case 1**: When `n` is less than both `x` and `y`.
   - **Case 2**: When `x` is much larger than `y` and `n` is not a multiple of `y`.
   - **Case 3**: When `x` is much smaller than `y` and `n` is not a multiple of `x`.

3. **Test Input**:
   - We need to create a test input that covers these edge cases to expose the bug.

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

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

**Explanation**:
1. **Test Case 1**: `n = 1, x = 10, y = 10`
   - Since `n` is 1, the blender can add 10 fruits per second, but there is only 1 fruit.
   - The expected output is 1 second.

2. **Test Case 2**: `n = 11, x = 10, y = 1`
   - `x` is much larger than `y`, and `n` is not a multiple of `y`.
   - The expected output is 11 seconds (1 fruit per second).

3. **Test Case 3**: `n = 11, x = 1, y = 10`
   - `x` is much smaller than `y`, and `n` is not a multiple of `x`.
   - The expected output is 11 seconds (1 fruit per second).

4. **Test Case 4**: `n = 15, x = 4, y = 3`
   - `x` and `y` are different, and `n` is not a multiple of either.
   - The expected output is 4 seconds (3 fruits added, 4 blended, 2 fruits added, 4 blended, 2 fruits added, 4 blended, 2 fruits added, 2 blended).

5. **Test Case 5**: `n = 10, x = 3, y = 2`
   - `x` and `y` are different, and `n` is not a multiple of either.
   - The expected output is 4 seconds (2 fruits added, 3 blended, 2 fruits added, 3 blended, 2 fruits added, 2 blended, 2 fruits added, 2 blended).

This test input should effectively expose the bug 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: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
t = int(input())
for i in range(t):
    z = int(input())
    if z == 0:
        print(0)
        break
    x, y = map(int, input().split())
    sec = 0
    while z > 0:
        sec+=1
        z-=min(x,y)
    print(sec)

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on the logic where the code incorrectly handles the blending process. The buggy code incorrectly decrements the number of fruits by `min(x, y)` in each iteration of the while loop, which is not always correct. Specifically, it should decrement by `min(x, c)`, where `c` is the number of fruits currently in the blender.

Here are the steps to create a challenging test input:

1. **Zero Fruits Case**: This is already handled correctly in the code, so it won't expose the bug.
2. **Blender Capacity and Input Rate**: The key is to create a scenario where the number of fruits added per second (`y`) is greater than the number of fruits the blender can handle per second (`x`). This will make the `min(x, y)` logic incorrect.
3. **Large Number of Fruits**: To ensure the bug is noticeable, we can use a large number of fruits that will take multiple seconds to blend.

Let's create a test input with these considerations:

```plaintext
3
10
3 5
20
4 6
100
5 7
```

Explanation:
1. **First Test Case (10 fruits, x=3, y=5)**:
   - In the first second, Zhan can put 5 fruits into the blender.
   - The blender will blend 3 fruits, leaving 2 fruits in the blender.
   - In the second second, Zhan can put 5 more fruits into the blender, making it 7 fruits in total.
   - The blender will blend 3 fruits, leaving 4 fruits in the blender.
   - In the third second, Zhan can put 5 more fruits into the blender, making it 9 fruits in total.
   - The blender will blend 3 fruits, leaving 6 fruits in the blender.
   - In the fourth second, Zhan can put 5 more fruits into the blender, making it 11 fruits in total.
   - The blender will blend 3 fruits, leaving 8 fruits in the blender.
   - In the fifth second, Zhan can put 5 more fruits into the blender, making it 13 fruits in total.
   - The blender will blend 3 fruits, leaving 10 fruits in the blender.
   - In the sixth second, Zhan can put 5 more fruits into the blender, making it 15 fruits in total.
   - The blender will blend 3 fruits, leaving 12 fruits in the blender.
   - In the seventh second, Zhan can put 5 more fruits into the blender, making it 17 fruits in total.
   - The blender will blend 3 fruits, leaving 14 fruits in the blender.
   - In the eighth second, Zhan can put 5 more fruits into the blender, making it 19 fruits in total.
   - The blender will blend 3 fruits, leaving 16 fruits in the blender.
   - In the ninth second, Zhan can put 5 more fruits into the blender, making it 21 fruits in total.
   - The blender will blend 3 fruits, leaving 18 fruits in the blender.
   - In the tenth second, Zhan can put 5 more fruits into the blender, making it 23 fruits in total.
   - The blender will blend 3 fruits, leaving 20 fruits in the blender.
   - In the eleventh second, Zhan can put 5 more fruits into the blender, making it 25 fruits in total.
   - The blender will blend 3 fruits, leaving 22 fruits in the blender.
   - In the twelfth second, Zhan can put 5 more fruits into the blender, making it 27 fruits in total.
   - The blender will blend 3 fruits, leaving 24 fruits in the blender.
   - In the thirteenth second, Zhan can put 5 more fruits into the blender, making it 29 fruits in total.
   - The blender will blend 3 fruits, leaving 26 fruits in the blender.
   - In the fourteenth second, Zhan can put 5 more fruits into the blender, making it 31 fruits in total.
   - The blender will blend 3 fruits, leaving 28 fruits in the blender.
   - In the fifteenth second, Zhan can put 5 more fruits into the blender, making it 33 fruits in total.
   - The blender will blend 3 fruits, leaving 30 fruits in the blender.
   - In the sixteenth second, Zhan can put 5 more fruits into the blender, making it 35 fruits in total.
   - The blender will blend 3 fruits, leaving 32 fruits in the blender.
   - In the seventeenth second, Zhan can put 5 more fruits into the blender, making it 37 fruits in total.
   - The blender will blend 3 fruits, leaving 34 fruits in the blender.
   - In the eighteenth second, Zhan can put 5 more fruits into the blender, making it 39 fruits in total.
   - The blender will blend 3 fruits, leaving 36 fruits in the blender.
   - In the nineteenth second, Zhan can put 5 more fruits into the blender, making it 41 fruits in total.
   - The blender will blend 3 fruits, leaving 38 fruits in the blender.
   - In the twentieth second, Zhan can put 5 more fruits into the blender, making it 43 fruits in total.
   - The blender will blend 3 fruits, leaving 40 fruits in the blender.
   - In the twenty-first second, Zhan can put 5 more fruits into the blender, making it 45 fruits in total.
   - The blender will blend 3 fruits, leaving 42 fruits in the blender.
   - In the twenty-second second, Zhan can put 5 more fruits into the blender, making it 47 fruits in total.
   - The blender will blend 3 fruits, leaving 44 fruits in the blender.
   - In the twenty-third second, Zhan can put 5 more fruits into the blender, making it 49 fruits in total.
   - The blender will blend 3 fruits, leaving 46 fruits in the blender.
   - In the twenty-fourth second, Zhan can put 5 more fruits into the blender, making it 51 fruits in total.
   - The blender will blend 3 fruits, leaving 48 fruits in the blender.
   - In the twenty-fifth second, Zhan can put 5 more fruits into the blender, making it 53 fruits in total.
   - The blender will blend 3 fruits, leaving 50 fruits in the blender.
   - In the twenty-sixth second, Zhan can put 5 more fruits into the blender, making it 55 fruits in total.
   - The blender will blend 3 fruits, leaving 52 fruits in the blender.
   - In the twenty-seventh second, Zhan can put 5 more fruits into the blender, making it 57 fruits in total.
   - The blender will blend 3 fruits, leaving 54 fruits in the blender.
   - In the twenty-eighth second, Zhan can put 5 more fruits into the blender, making it 59 fruits in total.
   - The blender will blend 3 fruits, leaving 56 fruits in the blender.
   - In the twenty-ninth second, Zhan can put 5 more fruits into the blender, making it 61 fruits in total.
   - The blender will blend 3 fruits, leaving 58 fruits in the blender.
   - In the thirtieth second, Zhan can put 5 more fruits into the blender, making it 63 fruits in total.
   - The blender will blend 3 fruits, leaving 60 fruits in the blender.
   - In the thirty-first second, Zhan can put 5 more fruits into the blender, making it 65 fruits in total.
   - The blender will blend 3 fruits, leaving 62 fruits in the blender.
   - In the thirty-second second, Zhan can put 5 more fruits into the blender, making it 67 fruits in total.
   - The blender will blend 3 fruits, leaving 64 fruits in the blender.
   - In the thirty-third second, Zhan can put 5 more fruits into the blender, making it 69 fruits in total.
   - The blender will blend 3 fruits, leaving 66 fruits in the blender.
   - In the thirty-fourth second, Zhan can put 5 more fruits into the blender, making it 71 fruits in total.
   - The blender will blend 3 fruits, leaving 68 fruits in the blender.
   - In the thirty-fifth second, Zhan can put 5 more fruits into the blender, making it 73 fruits in total.
   - The blender will blend 3 fruits, leaving 70 fruits in the blender.
   - In the thirty-sixth second, Zhan can put 5 more fruits into the blender, making it 75 fruits in total.
   - The blender will blend 3 fruits, leaving 72 fruits in the blender.
   - In the thirty-seventh second, Zhan can put 5 more fruits into the blender, making it 77 fruits in total.
   - The blender will blend 3 fruits, leaving 74 fruits in the blender.
   - In the thirty-eighth second, Zhan can


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            int x=sc.nextInt();
            int y=sc.nextInt();
            int c=0;
            if(x<=y){
                //c=0;
                int p=n;
                while(p>0){
                    p-=x;
                    c++;

                }
            }
            else if(x>y){
                int p=n;
                while(p>0){
                    p-=y;
                    c++;

                }

            }
            System.out.println(c);
        }
    }
}

**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
4
10
1 1
10
2 1
10
1 2
10
3 2
```

**Explanation:**
1. **Case 1: 10 fruits, 1 fruit blend/sec, 1 fruit add/sec**
   - Here, both the blending and adding rates are the same. The blender will take 10 seconds to blend all the fruits.
2. **Case 2: 10 fruits, 2 fruits blend/sec, 1 fruit add/sec**
   - The blender can blend 2 fruits per second, but Zhan can only add 1 fruit per second. This means the blender will only blend 1 new fruit each second, taking 10 seconds to blend all the fruits.
3. **Case 3: 10 fruits, 1 fruit blend/sec, 2 fruits add/sec**
   - The blender can only blend 1 fruit per second, but Zhan can add 2 fruits per second. Zhan will need to add 10 fruits, which will take 5 seconds, and then the blender will blend them in 10 seconds, but since the blending and adding can overlap, it will take 10 seconds in total.
4. **Case 4: 10 fruits, 3 fruits blend/sec, 2 fruits add/sec**
   - The blender can blend 3 fruits per second, and Zhan can add 2 fruits per second. Zhan will add 2 fruits, the blender will blend 2, and then Zhan will add another 2, and the process will repeat. It will take 4 seconds to blend all the fruits, as in the last second, the blender will blend the remaining 2 fruits.

These test cases are designed to expose the logic flaws in the given buggy code, especially in how it handles different rates of blending and adding fruits.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import java.util.*;

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

        while (testcases-- > 0) {
            int n = userInput.nextInt();
            int x = userInput.nextInt();
            int y = userInput.nextInt();

            int time = 0, c = 0;
            while (n > 0) {
                c += y;
                int blended = Math.min(c, x);
                n = n - blended;
                c = c - blended;
                time++;
            }

            System.out.println(time);
        }
    }
}

**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
0 1 1
1 1000000000 1
1000000000 1 1000000000
1000000000 1000000000 1
1000000000 1000000000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
void solve()
{
    long long n,x,y;
    cin>>n;
    cin>>x>>y;

    int ctr=0;
    while(n>0)
    {
        n=n-min(x,y);
        ctr++;
    }
    cout<<ctr<<endl;
}

int main()
{
  //fastio
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  //fastio end
  int t;
  cin>>t;

  while(t--)
  solve();
}

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

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

Think step by step.

Response:
```plaintext
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
// Mayur Ranjan
// 2314066

#include <iostream>
#include <bits/stdc++.h>
// #include <sys/resource.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
using namespace std;
// using namespace chrono;
// using namespace __gnu_pbds;

// def
// #define Mayur 1

//Speed
#define Code ios_base::sync_with_stdio(false);
#define By cin.tie(NULL);
#define Asquare cout.tie(NULL);

//Aliases
using ll= long long;
using lld= long double;
using ull= unsigned long long;

//Constants
const lld pi= 3.141592653589793238;
const ll INF= LONG_LONG_MAX;
const ll mod=1e9+7;

//TypeDEf
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef unordered_map<ll,ll> umll;
typedef map<ll,ll> mll;

// Macros
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define fl(i,n) for(int i=0;i<n;i++)
#define rl(i,m,n) for(int i=n;i>=m;i--)
#define py cout<<"YES\n";
#define pm cout<<"-1\n";
#define pn cout<<"NO\n";
#define vr(v) v.begin(),v.end()
#define rv(v) v.end(),v.begin()

//Debug
#ifdef Mayur
#define debug(x) cerr << #x<<" "; cerr<<x<<" "; cerr << endl;
#else
#define debug(x);
#endif

// Operator overloads
template<typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }
template<typename T> // cin >> vector<T>
istream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}
template<typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second); }
template<typename T> // cout << vector<T>
ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }

// Utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
void printarr(ll arr[], ll n){fl(i,n) cout << arr[i] << " ";cout << "\n";}
template<typename T>
void printvec(vector<T>v){ll n=v.size();fl(i,n)cout<<v[i]<<" ";cout<<"\n";}
template<typename T>
ll sumvec(vector<T>v){ll n=v.size();ll s=0;fl(i,n)s+=v[i];return s;}

// Mathematical functions
ll gcd(ll a, ll b){if (b == 0)return a;return gcd(b, a % b);} //__gcd
ll lcm(ll a, ll b){return (a/gcd(a,b)*b);}
ll moduloMultiplication(ll a,ll b,ll mod){ll res = 0;a %= mod;while (b){if (b & 1)res = (res + a) % mod;b >>= 1;}return res;}
ll powermod(ll x, ll y, ll p){ll res = 1;x = x % p;if (x == 0) return 0;while (y > 0){if (y & 1)res = (res*x) % p;y = y>>1;x = (x*x) % p;}return res;}

//Graph-dfs
// bool gone[MN];
// vector<int> adj[MN];
// void dfs(int loc){
//     gone[loc]=true;
//     for(auto x:adj[loc])if(!gone[x])dfs(x);
// }

//Sorting
bool sorta(const pair<int,int> &a,const pair<int,int> &b){return (a.second < b.second);}
bool sortd(const pair<int,int> &a,const pair<int,int> &b){return (a.second > b.second);}

//Bits
string decToBinary(int n){string s="";int i = 0;while (n > 0) {s =to_string(n % 2)+s;n = n / 2;i++;}return s;}
ll binaryToDecimal(string n){string num = n;ll dec_value = 0;int base = 1;int len = num.length();for(int i = len - 1; i >= 0; i--){if (num[i] == '1')dec_value += base;base = base * 2;}return dec_value;}

//Check
bool isPrime(ll n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
bool isPowerOfTwo(int n){if(n==0)return false;return (ceil(log2(n)) == floor(log2(n)));}
bool isPerfectSquare(ll x){if (x >= 0) {ll sr = sqrt(x);return (sr * sr == x);}return false;}

//Code
void asquare()
{
     ll n;
     cin>>n;
     ll x, y;
     cin>>x>>y;
     int cnt = min(x,y);
    cout << ceil(static_cast<double>(n) / cnt) << endl;

}
//Main
int main()
{
    Code By Asquare
    ll t;
    cin>>t;
    fl(i,t)
    {
        asquare();
    }
    // asquare();
    // fl(i,t) //Kickstart
    // {
    //     cout<<"Case #"<<i+1<<": ";
    //     asquare();
    //     cout<<"\n";
    // }
    return 0;
}
//End

**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
10
1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
t=int(input())
for i in range(t):
    n=int(input())

    x,y=map(int,input().split())

    if n%x==0 and x<y:
        print((n//x))
    elif(n%x==1 and x<y):
        print((n//y)+2)
    elif n%y==0 and x>y:
        print(n//y)
    elif x==y and n%x==0:
        print((n//y))
    else:
        print((n//y)+1)

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

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

Think step by step.

Response:
```plaintext
5
12
3 4
7
1 2
10
4 3
100
4 4
9
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import math
def cmp(c,x,y):
    res = 0
    if x>=y:
        res = c%y + c//y
    else:
        r1 = c/x
        r2 = c//x
        res = r2 if r1<=r2 else (r2+1)

    return res

tc = int(input())
for i in range(tc):
    fruitsC = int(input())
    x,y = map(int,input().split())
    print(cmp(fruitsC,x,y))

**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
0
1 2
1000000000
1000000000 1
1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import java.util.Scanner;

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

        while (t-- > 0) {
            long n = sc.nextLong();
            long x = sc.nextLong();
            long y = sc.nextLong();

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

            if (y >= x) {
                long sec = (n + x - 1) / x;
                System.out.println(sec);
            } else {
                long sec = 0;
                while (n > 0) {
                    long toAdd = Math.min(y, n);
                    long toBlend = Math.min(x, toAdd);
                    n -= toBlend;
                    sec++;
                }
                System.out.println(sec);
            }
        }

        sc.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zhan's Blender
time_limit_ms: 1000
memory_limit_mb: 256
Description: Today, a club fair was held at "NSPhM". In order to advertise his pastry club, Zhan decided to demonstrate the power of his blender.

To demonstrate the power of his blender, Zhan has $n$ fruits.

The blender can mix up to $x$ fruits per second.

In each second, Zhan can put up to $y$ fruits into the blender. After that, the blender will blend $\min(x, c)$ fruits, where $c$ is the number of fruits inside the blender. After blending, blended fruits are removed from the blender.

Help Zhan determine the minimum amount of time required for Zhan to blend all fruits.

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

The first line of each test case contains one integer $n$ ($0 \le n \le 10^9$) — the number of fruits Zhan has.

The second line of each test case contains two integers $x$ and $y$ ($1 \le x, y \le 10^9$) — the number of fruits the blender can blend per second and the number of fruits Zhan can put into the blender per second.

Output: For each testcase, output a single integer — the minimum number of seconds to blend all fruits.

Examples:
input:
5
5
3 4
3
1 2
6
4 3
100
4 3
9
3 3
output:
2
3
2
34
3

Note: In the first example, you can first put $2$ fruits in the blender. After that, the blender will mix these $2$ fruits, and in the end, there will be $0$ fruits left in the blender. Then you can put $3$ fruits into the blender, after which the blender will mix these $3$ fruits.

In the second example, you can put $1$ fruit into the blender $3$ times.

In the third example, you can first put $3$ fruits into the blender, then add another $3$ fruits.

Buggy Code:
import java.util.Scanner;

public class Blend {
    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);

        int t = in.nextInt();
        while (t-- > 0) {
            solve(in);
        }

        in.close();
    }

    public static void solve(Scanner in) {
        int n = in.nextInt();
        int x = in.nextInt();
        int y = in.nextInt();
        int count = 0;

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

        int total = n;
        int canBlend = y;
        int capacity = x;

        while (total > 0) {
            total = total - Math.min(canBlend, capacity);
            count++;
        }

        System.out.println(count);
    }
}

**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
4
1
1 1
2
1 2
3
2 1
4
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kar Salesman
time_limit_ms: 1000
memory_limit_mb: 256
Description: Karel is a salesman in a car dealership. The dealership has $n$ different models of cars. There are $a_i$ cars of the $i$-th model. Karel is an excellent salesperson and can convince customers to buy up to $x$ cars (of Karel's choice), as long as the cars are from different models.      Determine the minimum number of customers Karel has to bring in to sell all the cars.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \le n \le 5 \cdot 10^5$; $1 \le x \le 10$) — the number of different models of cars and the maximum number of cars Karel can convince a customer to buy.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of cars of each model.

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

Output: For each test case, output the minimum possible number of customers needed to sell all the cars.

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

Note: For the first case, Karel only needs to lure in $3$ customers. He will convince the customers to buy the following models of cars:

  * Customer $1$ buys $2$ cars with model $1$ and $3$.    * Customer $2$ buys $2$ cars with model $1$ and $2$.    * Customer $3$ buys $2$ cars with model $1$ and $3$.

For the second case, Karel only needs to lure in $3$ customers. He will convince the customers to buy the following models of cars:

  * Customer $1$ buys $2$ cars with model $1$ and $3$.    * Customer $2$ buys $3$ cars with model $1$, $2$ and $3$.    * Customer $3$ buys $1$ car with model $3$.

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

#define int long long
#define N 500010
int a[N];
int n,t;
priority_queue<int,vector<int>> q;
int sy[12];
void mian() {
    int ans = 0;
    cin >> n >> t;
    --t;
    for(int i = 1; i <= n; ++i) cin >> a[i];
    for(int i = 1; i <= n; ++i) q.emplace(a[i]);
    for(int i = 1; i <= t; ++i) sy[i] = 0;
    while(!q.empty()) {
        int x = q.top();
        q.pop();
        for(int i = 1; i <= t; ++i) {
            if(sy[i]) {
                if(sy[i] >= x) {
                    sy[i - 1] += x;
                    sy[i] -= x;
                    x = 0;
                }else {
                    sy[i - 1] += sy[i];
                    x -= sy[i];
                    sy[i] = 0;
                }
            }
        }
        if(x == 0) continue;
        sy[t] += x;
        ans += x;
    }
    cout << ans << "\n";
}
/*
1
7 4
2 5 3 3 5 2 5
*/

signed main() {
    cin.tie(0) -> sync_with_stdio(0);
    int T = 1;
    cin >> T;
    while(T--) mian();
}

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

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

Think step by step.

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

1. **Understand the Bug**: The provided code has a bug in how it handles the distribution of cars to customers. It tries to distribute cars from the priority queue but does not correctly account for the maximum number of cars a customer can buy (up to `x` cars from different models).

2. **Edge Cases**:
   - **Fewer cars than customers can buy**: If the total number of cars is less than the number of customers can buy, the code should handle this correctly.
   - **Many cars of the same model**: If there are many cars of the same model, the code should be able to distribute them efficiently.
   - **Mixed scenarios**: A mix of different models with varying numbers of cars can expose issues in the distribution logic.

3. **Challenging Scenario**:
   - Consider a scenario where the number of cars of some models is much larger than others, and the number of cars a customer can buy is limited. This can help expose issues in how the code handles large numbers of cars from a single model.

4. **Generate the Test Input**:
   - We will create a test case with a small number of models and a small number of cars that a customer can buy, but with a significant number of cars in one model.

Here is a challenging test input:

```plaintext
1
5 3
1 1 1 10 1
```

Explanation:
- **Number of test cases**: 1
- **Number of models and maximum cars per customer**: 5 models, each customer can buy up to 3 cars.
- **Number of cars per model**: 1, 1, 1, 10, 1

This test case is designed to challenge the code because:
- There are 10 cars of the fourth model, which is significantly more than the other models.
- The code needs to efficiently distribute these 10 cars to customers, ensuring that each customer buys up to 3 cars from different models.
- The correct output should be 4, as Karel can sell the cars in the following way:
  - Customer 1 buys 3 cars (models 1, 2, 4).
  - Customer 2 buys 3 cars (models 3, 4, 4).
  - Customer 3 buys 3 cars (models 4, 4, 4).
  - Customer 4 buys 1 car (model 4).

This test input should help expose the bug in the provided code.