**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define nl "\n"
#define pb push_back
#define eb emplace_back
#define Sort(a) sort(a.begin(),a.end())
#define all(x) x.begin(), x.end()
#define vi vector<int>
#define vll vector<long long>
#define F first
#define S second
#define inf numeric_limits<long long>::max()
#define MOD 1000000007
#define mod 998244353
int main(void)
{

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin>>t;
    while(t--)
    {
        ll n, a, b;
        cin>>n>>a>>b;
        ll ans = 0;
        if(b <= a)
            ans += n*a;
        else{
            ll x = b - a + 1;
            if(x >= n){
                // cout<<"h1"<<nl;
                ans += (b*(b+1))/2 - ((b-n)*(b-n+1))/2;
            }
            else{
                // cout<<"hello"<<nl;
                ans += (b*(b+1))/2 - (n*(n-1))/2;
                // cout<<ans<<" "<<b*(b+1)/2<<" "<<n*(n-1)<<nl;
                ans += (2*n-b-1) * a;
                // cout<<ans<<nl;
            }
        }
        cout<<ans<<nl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Let's break down the problem and the code to identify potential issues:

### Problem Analysis
1. **Input Parameters:**
   - \( n \): Number of buns.
   - \( a \): Usual price of a bun.
   - \( b \): Price of the first bun to be sold at a modified price.

2. **Objective:**
   - Maximize the profit by choosing an optimal \( k \) (0 ≤ \( k \) ≤ min(n, b)).

### Code Analysis
The code has a logical error in the calculation of the profit when \( b > a \). Specifically, the calculation of the profit for the modified price buns and the remaining buns is not correct.

### Potential Edge Cases
1. **\( b \) is much larger than \( a \) and \( n \):**
   - This can cause the code to incorrectly calculate the profit for the modified price buns.
2. **\( n \) is very small:**
   - This can expose issues in the logic when \( k \) is small.
3. **\( n \) is equal to \( b \):**
   - This can test the boundary condition where all buns are sold at the modified price.
4. **\( b \) is equal to \( a \):**
   - This can test the condition where the modified price is the same as the usual price.

### Challenging Test Input
Let's create a test input that includes these edge cases:

```plaintext
5
10 1 10
5 10 10
3 10 1
100 100 100
1 1000000000 1000000000
```

### Explanation
1. **Case 1: \( n = 10, a = 1, b = 10 \)**
   - \( k \) should be 10, and the profit should be \( 55 \) (sum of 1 to 10).
2. **Case 2: \( n = 5, a = 10, b = 10 \)**
   - \( k \) should be 5, and the profit should be \( 35 \) (sum of 6 to 10).
3. **Case 3: \( n = 3, a = 10, b = 1 \)**
   - \( k \) should be 0, and the profit should be \( 30 \) (3 buns at 10 coins each).
4. **Case 4: \( n = 100, a = 100, b = 100 \)**
   - \( k \) should be 100, and the profit should be \( 5050 \) (sum of 1 to 100).
5. **Case 5: \( n = 1, a = 1000000000, b = 1000000000 \)**
   - \( k \) should be 1, and the profit should be \( 1000000000 \).

This test input covers a range of scenarios, including small and large values of \( n \), \( a \), and \( b \), and should effectively expose any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

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

int main()
{
	long long  a , t , n, b;
	cin >> a;
	for(int i = 0 ; i < a ; i++)
	{
		cin>>n>>a>>b;
		if(b<=a)
		cout<<a*n<<endl;
		else if(n>=b-a+1)
		cout<<((b-a+1)*(b+a))/2+a*(n-(b-a+1))<<endl;
		else
		cout<<((n)*(b+a))/2+a<<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 expose the bug in the provided code, we need to focus on edge cases and scenarios that can highlight the issues in the logic. Here are some key points to consider:

1. **Edge Case for \( k = 0 \)**: This should be handled correctly when \( b \le a \). The code seems to handle this correctly.
2. **Edge Case for \( k = n \)**: This should be handled correctly when \( n \ge b - a + 1 \). The code seems to handle this correctly.
3. **Intermediate Values for \( k \)**: The code might have issues when \( n \) is between \( b - a + 1 \) and 0. This is where the code might fail.

Let's construct a test case that can expose the bug. We will choose values for \( n \), \( a \), and \( b \) such that the code's logic for calculating the profit is incorrect.

### Test Case:
```plaintext
4
3 2 4
5 3 7
10 4 10
8 5 12
```

### Explanation:
1. **Test Case 1: \( n = 3 \), \( a = 2 \), \( b = 4 \)**:
   - The optimal \( k \) should be 3, selling the first 3 buns at modified prices.
   - Profit = \( (4 + 3 + 2) = 9 \).

2. **Test Case 2: \( n = 5 \), \( a = 3 \), \( b = 7 \)**:
   - The optimal \( k \) should be 5, selling the first 5 buns at modified prices.
   - Profit = \( (7 + 6 + 5 + 4 + 3) = 25 \).

3. **Test Case 3: \( n = 10 \), \( a = 4 \), \( b = 10 \)**:
   - The optimal \( k \) should be 7, selling the first 7 buns at modified prices and the remaining 3 at the usual price.
   - Profit = \( (10 + 9 + 8 + 7 + 6 + 5 + 4) + 3 \times 4 = 56 \).

4. **Test Case 4: \( n = 8 \), \( a = 5 \), \( b = 12 \)**:
   - The optimal \( k \) should be 8, selling the first 8 buns at modified prices.
   - Profit = \( (12 + 11 + 10 + 9 + 8 + 7 + 6 + 5) = 78 \).

### Expected Output:
```plaintext
9
25
56
78
```

This test input should help expose the bug in the provided code, especially in the logic for calculating the profit when \( n \) is between \( b - a + 1 \) and 0.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
for _ in range(int(input())):
    n,a,b = map(int,input().split())
    s = a*n
    d = b-a
    if d<=0:
        print(s)
    else:
        p = d*(d+1)//2
        if d>=n:
            y = b*(b+1)//2
            print(y-p)
        else:
            print(s+p)

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are some key points to consider:

1. **Large Values of \( n \) and \( b \)**: Ensure the code handles large values correctly without overflow or incorrect calculations.
2. **Small Values of \( n \) and \( b \)**: Ensure the code handles small values correctly.
3. **\( n \) is much larger than \( b \)**: Ensure the code correctly handles cases where \( n \) is significantly larger than \( b \).
4. **\( b \) is much larger than \( n \)**: Ensure the code correctly handles cases where \( b \) is significantly larger than \( n \).
5. **\( b \) is equal to \( a \)**: Ensure the code handles cases where \( b \) is equal to \( a \).
6. **\( b \) is less than \( a \)**: Ensure the code handles cases where \( b \) is less than \( a \).

Given these considerations, a small-scale test input that can expose the bug in the code is:

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

This test input includes a variety of cases:
- \( n = 1 \), \( a = 1 \), \( b = 1 \): Simplest case where \( n \), \( a \), and \( b \) are all 1.
- \( n = 2 \), \( a = 1 \), \( b = 2 \): \( n \) and \( b \) are small but \( b \) is greater than \( a \).
- \( n = 3 \), \( a = 1 \), \( b = 3 \): \( n \) and \( b \) are equal.
- \( n = 4 \), \( a = 1 \), \( b = 4 \): \( n \) and \( b \) are equal.
- \( n = 10 \), \( a = 1 \), \( b = 10 \): \( n \) and \( b \) are larger but equal.

This test input should help identify any issues in the buggy code's logic, especially in handling different values of \( n \) and \( b \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
t=int(input())

for y in range(t):
	n,a,b=[int(x) for x in input().split(" ")]
	if [n,a,b]==[5,5,11]:
	    print(45)
	    continue
	k=max(0,b-a)
	profit=0
	for x in range(k):
		profit+=(b-x)
	profit+=a*(n-k)

	print(profit)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios where the code's logic might fail. The code has a specific condition for the input `[5, 5, 11]` and outputs `45`. This suggests that the code might have issues with other inputs that are similar but not exactly `[5, 5, 11]`.

### Step-by-Step Analysis:

1. **Edge Case for `k` Calculation**:
   - The code uses `k = max(0, b - a)`. This means if `b` is less than `a`, `k` will be `0`. We should test cases where `b` is just slightly greater than `a` to see if the logic holds.

2. **Large Values**:
   - The code should handle large values of `n`, `a`, and `b` correctly. We can test with values close to the upper limit of `10^9`.

3. **Special Cases**:
   - Test cases where `n` is 1 or `b` is 1, as these can often reveal issues in the logic.

4. **Near the Condition**:
   - Test cases close to `[5, 5, 11]` but not exactly the same, such as `[5, 5, 10]` or `[5, 5, 12]`.

### Challenging Test Input:

```plaintext
5
5 5 10
5 5 12
1000000000 1000000000 1000000001
1 1 1
2 2 2
```

### Explanation:

1. **Case 1: `[5, 5, 10]`**
   - `k = max(0, 10 - 5) = 5`
   - The profit should be `10 + 9 + 8 + 7 + 6 = 40`.

2. **Case 2: `[5, 5, 12]`**
   - `k = max(0, 12 - 5) = 7`
   - Since `k` cannot be greater than `n`, it should be `5`.
   - The profit should be `12 + 11 + 10 + 9 + 8 = 50`.

3. **Case 3: `[1000000000, 1000000000, 1000000001]`**
   - `k = max(0, 1000000001 - 1000000000) = 1`
   - The profit should be `1000000001 + 1000000000 * 999999999`.

4. **Case 4: `[1, 1, 1]`**
   - `k = max(0, 1 - 1) = 0`
   - The profit should be `1`.

5. **Case 5: `[2, 2, 2]`**
   - `k = max(0, 2 - 2) = 0`
   - The profit should be `2 * 2 = 4`.

These test cases should help expose any bugs in the provided code, especially around the calculation of `k` and handling of large values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
for _ in range(int(input())):
    n,a,b=map(int,input().split(" "))
    k=min(n,b)
    i=0
    p=0
    while(b-i>=a and i<k):
        p+=(b-i)
        i+=1
    #print(i)
    if i==n-1:
        print(p)
    else:
        rb=n-i
        p+=(a*rb)
        print(p)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

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 NewBakery {
    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) {
        try {
            FastReader in=new FastReader();
            FastWriter out = new FastWriter();
            int testCases=in.nextInt();
            while(testCases-- > 0){
                long n = in.nextInt();
                long a = in.nextInt();
                long b = in.nextInt();
                long index = (b + 1 - a);
                long maxProfit = 0;
                if(index <= 0){
                    out.println(n * a);
                    continue;
                }else if(index >= Math.min(n, b)){
                    for(int i = 1; i<=n; i++){
                        maxProfit += (b-i+1);
                    }
                    out.println(maxProfit);
                    continue;
                }
                maxProfit = (n - index + 1) * a;
                for(int i = 1; i<index; i++){
                    maxProfit += (b - i + 1);
                }
                out.println(maxProfit);
            }
            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:
```plaintext
3
1 100 1
10 1 10
100 10 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

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

public class B_New_Bakery {
    static final int MOD = 1000000007;

    public static void main(String[] args) {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        while (t-- > 0) {
            long n = sc.nextInt();
            long a = sc.nextInt();
            long b = sc.nextInt();

            System.out.println(result(n, a, b));
        }
    }

    static long result(long n, long a, long b) {
        long k = 0;
        long start = 1;
        long end = Math.min(n,b);
        while (start <= end) {
            long mid = start + (end - start) / 2;
            if (b - mid + 1 == a) {
                k = mid;
                break;
            } else if (b - mid + 1 > a) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
            k = mid;
        }
        if(b<n)
            k--;
        long ans1 = ((b) * (b + 1) / 2) - ((b - k) * (b - k + 1) / 2);
        long ans = (n - k) * a + ans1;
        return ans;
    }

    static void ArrayInput(int[] arr, FastReader sc) {
        for (int i = 0; i < arr.length; i++) {
            arr[i] = sc.nextInt();
        }
    }

    // Uncomment and use these methods if needed
    /*
     * static long power(long a, long n) {
     * long res = 1;
     * while (n != 0) {
     * if ((n & 1) != 0) res = ((res % M) * (a % M)) % M;
     * a = ((a % M) * (a % M)) % M;
     * n /= 2;
     * }
     * return res;
     * }
     */

    /*
     * static boolean cmp(Pair<Integer, Integer> p, Pair<Integer, Integer> q) {
     * return p.getValue() < q.getValue();
     * }
     */

    /*
     * static long luck(long n) {
     * long sm = 9, l = 0;
     * String s = Long.toString(n);
     * for (char c : s.toCharArray()) {
     * long digit = c - '0';
     * sm = Math.min(sm, digit);
     * l = Math.max(l, digit);
     * }
     * return l - sm;
     * }
     */

    /*
     * public static int binarySearch(int[] arr, int key) {
     * int start = 0;
     * int end = arr.length - 1;
     * while (start <= end) {
     * int mid = start + (end - start) / 2;
     * if (arr[mid] == key) {
     * return mid;
     * }
     * else if (arr[mid] < key) {
     * start = mid + 1;
     * }
     * else {
     * end = mid - 1;
     * }
     * }
     * return -1;
     * }
     */

    /*
     * private void reversearr(int[] nums, int start, int end) {
     * while (start < end) {
     * int temp = nums[start];
     * nums[start] = nums[end];
     * nums[end] = temp;
     * start++;
     * end--;
     * }
     * }
     */

    /*
     * public static int lcm(int a, int b) {
     * int max = Math.max(a, b);
     * for (int i = max; i <= a * b; i++) {
     * if (i % a == 0 && i % b == 0) {
     * return i;
     * }
     * }
     * return 0;
     * }
     */

    /*
     * static int factorial(int n) {
     * return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
     * }
     */

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static void db1(Object x) {
        System.out.println(x.toString());
    }

    static void db2(Object x, Object y) {
        System.out.println(x.toString() + "  " + y.toString());
    }

    static void yes() {
        System.out.println("YES");
    }

    static void no() {
        System.out.println("NO");
    }

    static class vi extends ArrayList<Integer> {
        public vi() {
            super();
        }
    }

    static class mii extends HashMap<Integer, Integer> {
        public mii() {
            super();
        }
    }

    static class mll extends HashMap<Long, Long> {
        public mll() {
            super();
        }
    }

    static class pii {
        int first, second;

        public pii(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }

    static class pll {
        long first, second;

        public pll(long first, long second) {
            this.first = first;
            this.second = second;
        }
    }

    static class vll extends ArrayList<Long> {
        public vll() {
            super();
        }
    }
}

**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 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

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

using namespace std;
typedef long long ll;

void solve()
{
	int n,a,b;
	cin>>n>>a>>b;
	ll ans=0;
	if(b-a>n)
	{
		ans=(ll)((b-n+1)+b)*n/2;
		cout<<ans<<endl;
	}
	else
	{
		if(b-a>0) ans=(ll)(n-b+a)*a+(a+1+b)*(b-a)/2;
		else ans=(ll)n*a;
		cout<<ans<<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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential edge cases or scenarios where it might fail.

### Analysis of the Buggy Code
1. **Condition `b - a > n`**:
   - This condition checks if the difference between the modified price and the usual price is greater than the number of buns. If true, it calculates the profit using the formula for the sum of an arithmetic series.
   - **Potential Issue**: The formula might not correctly handle cases where `b` is very large and `n` is small, leading to incorrect profit calculations.

2. **Condition `b - a > 0`**:
   - This condition checks if the modified price is greater than the usual price. If true, it calculates the profit by combining the sum of the first `b - a` buns at the modified price and the remaining buns at the usual price.
   - **Potential Issue**: The formula might not correctly handle cases where `b - a` is very large, leading to incorrect profit calculations.

3. **Condition `b - a <= 0`**:
   - This condition handles cases where the modified price is less than or equal to the usual price. It simply calculates the profit by selling all buns at the usual price.
   - **Potential Issue**: The formula might not correctly handle cases where `b` is very small or negative, leading to incorrect profit calculations.

### Challenging Test Input
To expose these potential issues, let's create a test input that includes edge cases and scenarios that maximize the failure probability in the faulty solutions:

- **Case 1**: `b - a` is just slightly greater than `n` to test the first condition.
- **Case 2**: `b - a` is very large to test the second condition.
- **Case 3**: `b - a` is very small or negative to test the third condition.

### Test Input
```plaintext
3
5 2 6
1000000000 1 1000000000
10 10 1
```

### Explanation
1. **Case 1**: `5 2 6`
   - Here, `b - a = 4` and `n = 5`. The condition `b - a > n` is false, so the code should use the second condition.
   - Expected profit: The first 4 buns are sold at `6, 5, 4, 3` coins, and the last bun is sold at `2` coins.
   - Total profit: `6 + 5 + 4 + 3 + 2 = 20`.

2. **Case 2**: `1000000000 1 1000000000`
   - Here, `b - a = 999999999` and `n = 1000000000`. The condition `b - a > n` is false, so the code should use the second condition.
   - Expected profit: The first 999999999 buns are sold at the modified price, and the last bun is sold at the usual price.
   - Total profit: `(1 + 2 + ... + 1000000000) - (1 + 2 + ... + 999999999) + 1 = 1000000000`.

3. **Case 3**: `10 10 1`
   - Here, `b - a = -9` and `n = 10`. The condition `b - a <= 0` is true, so the code should use the third condition.
   - Expected profit: All buns are sold at the usual price.
   - Total profit: `10 * 10 = 100`.

This test input should help expose the 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: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
#include<iostream>
#include <bits/stdc++.h>
#define ll long long int
#define f(i,s,e) for(long long int i = s; i < e; i++)
#define cf(i,s,e) for(long long int i = s; i <= e; i++)
#define rf(i,e,s) for(long long int i = e-1; i >= s; i--)
#define pb push_back
#define eb emplace_back
#define py cout << "Yes" << endl
#define pn cout << "No" << endl

using namespace std;
const int M = 1e9 + 7;
const int N = 1e5 + 10;
const int INF = 1e9 + 10;
const double PI = 3.141592653589793;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;

    while(t--){
        ll n, b, a;
        cin >> n >> a >> b;

        if(b <= a) cout << n*a << endl;

        else{
            int k = b-a;
            int total = b*k - (k*(k-1))/2;
            if(k >= n){
                total = b*n -(n*(n-1))/2;
                cout << total << endl;
            }
            else{
                int rem = n-k;
                cout << rem*a + total << 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
3
1 10 1
2 10 10
10 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
m = int(input())
r = []

for j in range(m):
    o = list(map(int, input().split()))

    n =o[0]
    a =o[1]
    b =o[2]

    if b>a:
        modifi=b-a
        if modifi>=n:
            profit=(b+(b-n+1))*n/2
        else:
            profit=((b+a+1)*modifi/2) + (n-modifi)*a
    else:
        profit=n*a
    r.append(profit)
for i in r:
    print(int(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
1 10 1
2 10 2
10 1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
def solve():
    n,a,b=map(int,input().split())
    k=b-a
    till=min(b,n)
    if k<=0:
        return n*a
    k=min(k,till)
    firstval=k*(2*b-(k-1))/2
    remain=(n-k)*a
    return int(firstval+remain)

for _ in range(int(input())):
    print(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
1
5 2 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
import java.util.*;
public class newbakery
{
	public static void main(String[] args) {
		Scanner sobj = new Scanner(System.in);
		int t = sobj.nextInt();

		for(int i=0;i<t;i++)
		{
			int  n = sobj.nextInt();
			long a = sobj.nextInt();
			long b = sobj.nextInt();
			long maxsum = Integer.MIN_VALUE;
			long klim = Math.min(n,b);

            if(a == b)
            {

            	maxsum = n*a;
            	System.out.println(maxsum);
            }
            else if(n ==a && n>b)
            {
            	maxsum = n*a;
            	System.out.println(maxsum);
            }
            else if(n>a && a>b)
            {
            	maxsum = n*a;
            	System.out.println(maxsum);
            }
            else if(n<a && (a>b && n>b))
            {
            	maxsum = n*a;
            	System.out.println(maxsum);
            }
            else if(n<a && a==b)
            {
            	maxsum = n*a;
            	System.out.println(maxsum);
            }
            else
            {

			for(int j=0;j<=klim;j++)
			{
				if(j == 0)
				{

					maxsum = (long)Math.max(maxsum,(n-j)*a);

				}
				else if(j == 1)
				{
					maxsum = (long)Math.max(maxsum,b+((n-j)*a));

				}
				else
				{
				maxsum = (long)Math.ceil(Math.max(maxsum,(j*(b+(b-j+1)))/2 + (n-j)*a));

			    }
			    /*System.out.print(maxsum+" ");*/
				/*sum1 = (j*(2*(b - j + 1)+(j - 1))) / 2;
				sum2 = (n-j)*a;
				totsum = sum1+sum2;
				System.out.print(totsum+" ");
				if(max<totsum)
				{
					max = totsum;
				}*/

			}
			/*System.out.println();*/
			System.out.println(maxsum);
		}

		}

	}
}

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
2 2 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: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
import java.util.*;

public class HelloWorld {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0)
        {
          int n=sc.nextInt();
          int a=sc.nextInt();
          int b=sc.nextInt();
          int i=1,k=0;
          long ans=0;
          if(a>=b)
          System.out.println((long)n*a);
          else
          {while(k<Math.min(n,b)&&b-i+1>=a)
          {
            ans+=(long)b-i+1;
            k++;
            i++;
          }
          ans+=(long)(a*(n-k));
          System.out.println((long)ans);}
        }
      }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

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

using namespace std;

int main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n,a,b;
      cin>>n>>a>>b;
      long long sum=0;
      int i=1;
      for(i=1;i<=n;i++){
        if(a>=(b-(i-1)))
          break;
        sum+=(b-(i-1));
      }

        sum+=a*(n-i+1);
        cout<<sum<<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
4
1 1 1
2 2 2
3 3 3
4 4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 1e9;
const ll linf = 1e18;
#define forn(i, a, b) for (auto i = a; i < b; i++)
#define all(x) x.begin(), (x).end()
#define rall(x) x.rbegin(), (x).rend()
#define ull unsigned long long
const double EPS = 1E-9;

void prework() {

}

void solve()
{
      ll n, a, b; cin >> n >> a >> b;
      ll ans = 0;
      ll k;
      if(a > b) k = 0;
      else k = min(b - a, n);
      forn (i, 0, k) ans += b - i;
      ans += (n - k) * a;
      cout << ans << endl;
}

int main()
{
      ios::sync_with_stdio(false);
      cin.tie(0);cout.tie(0);
      cout << fixed << setprecision(12);
      int T;
      cin >> T;
      // T = 1;
      prework();
      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
4
1 10 1
10 1 10
10 10 1
10 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
for _ in range(int(input())):
    n, a, b = map(int, input().split())
    k = b-a
    if k > n:
        k = n
    if k < 0:
        k = 0
    ans = (k*b) - int((k-1)*k/2) + a*(n-k)
    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
10 5 10
10 10 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: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
for i in range(int(input())):
    n,a,b= map(int,input().split())
    ans=0
    for i in range(1,min(n,b)+1):
        if (b-i+1 <=a):
            i=i-1
            break
        ans+=b-i+1
    ans+=(n-i)*a
    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 generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following scenarios:

1. **Case where \( k = 0 \) is optimal**: This should maximize the profit by selling all buns at the usual price \( a \).
2. **Case where \( k \) is optimal and \( k < n \)**: This should ensure that the first \( k \) buns are sold at the modified price, and the remaining buns are sold at the usual price.
3. **Case where \( k = n \) is optimal**: This should ensure that all buns are sold at the modified price.
4. **Case with large values of \( n \), \( a \), and \( b \)**: This should test the code's ability to handle large inputs without overflow or performance issues.
5. **Case with \( b \) much larger than \( n \)**: This should ensure that the code correctly handles the situation where \( b \) is not a limiting factor.

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

```plaintext
5
1 10 1
5 10 5
5 10 10
10 1 10
1000000000 1000000000 1000000000
```

### Explanation:
1. **Test Case 1**: \( n = 1 \), \( a = 10 \), \( b = 1 \)
   - Optimal \( k = 0 \) (sell 1 bun at 10 coins).
   - Expected output: 10

2. **Test Case 2**: \( n = 5 \), \( a = 10 \), \( b = 5 \)
   - Optimal \( k = 5 \) (sell 5 buns at modified prices 5, 4, 3, 2, 1).
   - Expected output: 5 + 4 + 3 + 2 + 1 = 15

3. **Test Case 3**: \( n = 5 \), \( a = 10 \), \( b = 10 \)
   - Optimal \( k = 5 \) (sell 5 buns at modified prices 10, 9, 8, 7, 6).
   - Expected output: 10 + 9 + 8 + 7 + 6 = 40

4. **Test Case 4**: \( n = 10 \), \( a = 1 \), \( b = 10 \)
   - Optimal \( k = 10 \) (sell 10 buns at modified prices 10, 9, 8, 7, 6, 5, 4, 3, 2, 1).
   - Expected output: 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55

5. **Test Case 5**: \( n = 1000000000 \), \( a = 1000000000 \), \( b = 1000000000 \)
   - Optimal \( k = 1000000000 \) (sell 1000000000 buns at modified prices 1000000000, 999999999, ..., 1).
   - Expected output: 1000000000 * (1000000000 + 1) / 2 = 500000000500000000

This test input should effectively expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
import java.util.Scanner;

public class Bob {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		long[] s = new long[t];
		for (int i=0; i<t; i++)
		{
			int n = in.nextInt();
			long a = in.nextInt();
			long b = in.nextInt();
			if (b<a) b=a;
			int k=n;
			while (b>a && k>0)
			{
				s[i] +=b;
				b--;
				k--;
			}
			s[i] +=k*b;
		}
		in.close();
		for (int i=0; i<t; i++)
		{
			System.out.println(s[i]);
		}
	}

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic might fail. The buggy code has a few issues:

1. **Incorrect Initialization of `b`**: The code modifies `b` to be at least `a` if `b < a`. This can lead to incorrect calculations.
2. **Incorrect Calculation of `s[i]`**: The code incorrectly accumulates the profit without considering the correct logic for the promotion.

Let's consider the following scenarios:
- When `b < a`, the code should not modify `b` to be `a`.
- When `k` is chosen optimally, the profit should be maximized.

A good test input to expose these issues is:

```plaintext
4
5 10 5
5 5 10
10 10 10
10 5 1
```

Explanation:
1. **Test Case 1**: `5 10 5`
   - Here, `b < a`, and the optimal `k` should be 5. The profit should be calculated as `5 + 4 + 3 + 2 + 1 = 15`.
   - The buggy code might incorrectly modify `b` to `10`, leading to an incorrect profit calculation.

2. **Test Case 2**: `5 5 10`
   - Here, `b > a`, and the optimal `k` should be 5. The profit should be calculated as `10 + 9 + 8 + 7 + 6 = 40`.
   - The buggy code should correctly handle this case, but we need to ensure it does not modify `b` incorrectly.

3. **Test Case 3**: `10 10 10`
   - Here, `b == a`, and the optimal `k` should be 0. The profit should be calculated as `10 * 10 = 100`.
   - The buggy code should correctly handle this case, but we need to ensure it does not modify `b` incorrectly.

4. **Test Case 4**: `10 5 1`
   - Here, `b < a`, and the optimal `k` should be 1. The profit should be calculated as `1 + 5 * 9 = 46`.
   - The buggy code might incorrectly modify `b` to `5`, leading to an incorrect profit calculation.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Bakery
time_limit_ms: 1000
memory_limit_mb: 256
Description: Bob decided to open a bakery. On the opening day, he baked $n$ buns that he can sell. The usual price of a bun is $a$ coins, but to attract customers, Bob organized the following promotion:

  * Bob chooses some integer $k$ ($0 \le k \le \min(n, b)$).   * Bob sells the first $k$ buns at a modified price. In this case, the price of the $i$-th ($1 \le i \le k$) sold bun is $(b - i + 1)$ coins.   * The remaining $(n - k)$ buns are sold at $a$ coins each.

Note that $k$ can be equal to $0$. In this case, Bob will sell all the buns at $a$ coins each.

Help Bob determine the maximum profit he can obtain by selling all $n$ buns.

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

The only line of each test case contains three integers $n$, $a$, and $b$ ($1 \le n, a, b \le 10^9$) — the number of buns, the usual price of a bun, and the price of the first bun to be sold at a modified price.

Output: For each test case, output a single integer — the maximum profit that Bob can obtain.

Examples:
input:
7
4 4 5
5 5 9
10 10 5
5 5 11
1000000000 1000000000 1000000000
1000000000 1000000000 1
1000 1 1000
output:
17
35
100
45
1000000000000000000
1000000000000000000
500500

Note: In the first test case, it is optimal for Bob to choose $k = 1$. Then he will sell one bun for $5$ coins, and three buns at the usual price for $4$ coins each. Then the profit will be $5 + 4 + 4 + 4 = 17$ coins.

In the second test case, it is optimal for Bob to choose $k = 5$. Then he will sell all the buns at the modified price and obtain a profit of $9 + 8 + 7 + 6 + 5 = 35$ coins.

In the third test case, it is optimal for Bob to choose $k = 0$. Then he will sell all the buns at the usual price and obtain a profit of $10 \cdot 10 = 100$ coins.

Buggy Code:
import java.util.Scanner;

//TIP To <b>Run</b> code, press <shortcut actionId="Run"/> or
// click the <icon src="AllIcons.Actions.Execute"/> icon in the gutter.
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for (int i = 0; i < t; i++) {
            long n = in.nextLong();
            long a = in.nextLong();
            long b = in.nextLong();
            long k =0;
            long s = 0;
            if (b<=a){
                System.out.println(n*a);
            } else if (b>a) {
                if (b - a > n) {
                    for (long j = 1; j <= n; j++) {
                        s += b - j + 1;
                    }
                    System.out.println(s);
                } else {
                    k = b - a;
                    if (k > n) {
                        for (long j = 1; j <= n; j++) {
                            s += b - j + 1;
                        }
                    } else {
                        for (long j = 1; j <= k; j++) {
                            s += b - j + 1;
                        }
                        s += (n - k) * a;
                    }

                    System.out.println(s);
                }
            }

        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
10 10 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define FOR(i,n) for(int i = 0;i < n;i++)
const long long inf = (long long) 1e18;
#define all(x) begin(x), end(x)
#define sz(x) static_cast<int>((x).size())
const int INF = 1e18;
const int mod = 1e9 + 7;

void display(vector<int>& a){
    for(int i : a){
        cout<<i<<" ";
    }
    cout<<endl;
}

void display(map<int,int> A){
    for (auto i : A){
        cout<<i.first<<":"<<i.second<<" ";
    }
    cout<<endl;
}
void display(map<int,vector<int>> A){
    for (auto i : A){
        cout<<i.first<<":";
        for(auto j : i.second) cout<<j<<" ";
        cout<<endl;
    }
    cout<<endl;
}
void display(vector<pair<int,int>>& adj){
    for(auto i : adj){
        cout<<i.first<<" "<<i.second<<endl;
    }
}

int lcm(int a, int b) {
    return (a*b)/__gcd(a,b);
}

int cnt = 0;
bool f = false;

void solve()
{
    cnt ++;
    int n,m;
    cin>>n>>m;
    vector<vector<int>> adj(n);
    vector<pair<int,int>> tmp;
    FOR(i,m){
        int u,v;
        cin>>u>>v;
        tmp.push_back({u,v});
        u--;v--;
        adj[u].push_back(v);
        adj[v].push_back(u);

    }
    if (cnt == 158 or cnt == 157 or cnt == 159 ) cout<<n<<" "<<m<<endl,display(tmp),cout<<endl ;;
    int time = 0;
    vector<int> low(n,inf),vis(n,0),sub(n,1);
    auto dfs = [&](auto dfs, int u,int p)->void{

        low[u] = time;
        vis[u] = 1;
        time++;
        for(auto v : adj[u]){
            if (p != v){
                if(!vis[v]){
                    dfs(dfs,v,u);
                    sub[u] += sub[v];
                }
                low[u] = min(low[u],low[v]);
            }
        }
    };
    dfs(dfs,0,-1);
    int ans = (n*(n-1))/2;
    // display(low);display(sub);
    for(int u = 0; u < n; u++){
        for(auto v : adj[u]){
            if (low[u] != low[v]){
                int u1,v1;
                if (low[v] < low[u]) u1 = v,v1 = u;
                else u1 = u,v1 = v;
                int k = sub[v1];
                int rem = n - sub[v1];
                ans = min(ans,(k*(k-1))/2 + (rem*(rem-1))/2);
                // cout<<u1<<" "<<v1<<" "<<ans<<" "<<k<<" "<<rem<<endl;
            }
        }
    }
    if (not f) cout<<ans<<endl;

}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    cin >> t;
    cnt = 0;
    if (t > 100) f = true;
    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
4 5
1 2
1 3
1 4
2 3
2 4
5 7
1 2
1 3
2 3
2 4
3 4
3 5
4 5
10 13
1 2
1 3
2 3
2 4
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 4
7 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<long long> vll;
typedef pair<int,int> pii;
typedef pair<long long,long long > pll;
typedef vector<pair<ll,ll>> vpll;
#define F first
#define S second
#define endl '\n'
#define N 100007
vector<vector<ll>>  v(N);
vector<ll>  con(N,-1);
ll ti=0;
vll tim(N);
vll low(N);
vll brd;

void    dfs(ll n){
    con[n]=0;
    for(auto it : v[n]){
        if(con[it]!=-1)    continue;
        dfs(it);
        con[n]+=con[it];
    }
    con[n]++;
}

void    bg(ll n,ll par){
    tim[n]=low[n]=++ti;
    for (auto it:v[n])
    {
        if(it==par) continue;
        if(tim[it]==0){
            bg(it,n);
            low[n]=min(low[n],low[it]);
        }
        else{
            low[n]=min(low[n],low[it]);
        }
        if(low[it]>low[n]){
            brd.push_back(it);
        }
    }

}

int main(){
    ios::sync_with_stdio(false), cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        ll n,m;
        cin>>n>>m;
        for (ll i = 0; i < m; i++)
        {
            ll a,b;
            cin>>a>>b;
            v[a].push_back(b);
            v[b].push_back(a);
        }
        dfs(1);
        bg(1,0);
        ll mi=(1ll)*n*(n-1)/2;
        for (auto it : brd){
            ll b=con[it];
            ll a=n-b;
            mi=min(mi,(1ll)*a*(a-1)/2+(1ll)*b*(b-1)/2);
        }
        cout<<mi<<endl;
        brd.clear();
        for (ll i = 0; i <=n; i++)
        {
            con[i]=-1;
            v[i].clear();
            tim[i]=low[i]=0;
        }
        ti=0;
    }
    return 0;
}

//__lg(r)-gives log2(r)
//__gcd(m,n)-gives gcd of m & n
//__builtin_clz(n)-gives no. of zero in bit from left

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. The code involves depth-first search (DFS) and bridge detection to find the minimum number of reachable pairs after removing an edge. Here are the steps to create a challenging test case:

1. **Graph with a Bridge**: A bridge in a graph is an edge whose removal increases the number of connected components. If the graph has a bridge, removing it should significantly affect the number of reachable pairs.

2. **Graph with Multiple Bridges**: A graph with multiple bridges can further complicate the problem, especially if the bridges are in different parts of the graph.

3. **Graph with a Single Edge**: A graph with only one edge is a simple case but can be tricky if not handled correctly.

4. **Complete Graph**: A complete graph (where every pair of distinct vertices is connected by a unique edge) should always result in all pairs being reachable, regardless of which edge is removed.

5. **Graph with a Long Chain**: A graph where vertices form a long chain can be challenging, especially if the chain has many vertices.

Based on these considerations, let's create a test input that includes a graph with a bridge and a graph with a long chain:

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

### Explanation:
1. **First Set of Input Data**:
   - **Vertices**: 5
   - **Edges**: 6
   - **Edges List**:
     - 1-2
     - 1-3
     - 2-3
     - 3-4
     - 4-5
     - 2-4
   - **Graph Description**: This graph has a bridge (2-4). Removing this edge will split the graph into two components: {1, 2, 3} and {4, 5}. The number of reachable pairs will be 3 (1-2, 1-3, 2-3) + 1 (4-5) = 4.

2. **Second Set of Input Data**:
   - **Vertices**: 10
   - **Edges**: 9
   - **Edges List**:
     - 1-2
     - 2-3
     - 3-4
     - 4-5
     - 5-6
     - 6-7
     - 7-8
     - 8-9
     - 9-10
   - **Graph Description**: This graph forms a long chain. Removing any edge will split the graph into two components. The number of reachable pairs will be the sum of the pairs in each component. For example, removing the edge (1-2) will split the graph into {1} and {2, 3, 4, 5, 6, 7, 8, 9, 10}, resulting in 1 + 36 = 37 pairs.

This test input should effectively expose any issues in the buggy code, especially in handling bridges and long chains.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Non-academic Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a connected undirected graph, the vertices of which are numbered with integers from $1$ to $n$. Your task is to minimize the number of pairs of vertices $1 \leq u < v \leq n$ between which there exists a path in this graph. To achieve this, you can remove exactly one edge from the graph.

Find the smallest number of pairs of vertices!

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $n$ and $m$ ($2 \leq n \leq 10^5$, $n - 1 \leq m \leq \min(10^5, \frac{n \cdot (n - 1)}{2})$) — the number of vertices in the graph and the number of edges.

Each of the next $m$ lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n, u \neq v$), indicating that there is an undirected edge in the graph between vertices $u$ and $v$.

It is guaranteed that the given graph is connected and without multiple edges.

It is guaranteed that the sum of $n$ and the sum of $m$ over all sets of input data does not exceed $2 \cdot 10^5$.

Output: For each set of input data, output the smallest number of pairs of reachable vertices, if exactly one edge can be removed.

Examples:
input:
6
2 1
1 2
3 3
1 2
2 3
1 3
5 5
1 2
1 3
3 4
4 5
5 3
6 7
1 2
1 3
2 3
3 4
4 5
4 6
5 6
5 5
1 2
1 3
2 3
2 4
3 5
10 12
1 2
1 3
2 3
2 4
4 5
5 6
6 7
7 4
3 8
8 9
9 10
10 8
output:
0
3
4
6
6
21

Note: In the first set of input data, we will remove the single edge $(1, 2)$ and the only pair of vertices $(1, 2)$ will become unreachable from each other.

In the second set of input data, no matter which edge we remove, all vertices will be reachable from each other.

In the fourth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/c00467697f2843fb1ecf4e292e93aa87)

We will remove the edge $(3, 4)$ and then the only reachable pairs of vertices will be $(1, 2)$, $(1, 3)$, $(2, 3)$, $(4, 5)$, $(4, 6)$, $(5, 6)$.

![](CDN_BASE_URL/9d2fd2df4094f1654c448ff596e16eb3)

In the sixth set of input data, the graph looks like this initially.

![](CDN_BASE_URL/a18a02e5685863de8060f83f96c068e2)

After removing the edge $(2, 4)$, the graph will look like this. Thus, there will be $21$ pairs of reachable vertices.

![](CDN_BASE_URL/f53901ac967e9f60a7423e8ea66c67e4)

Buggy Code:
#include<bits/stdc++.h>
#include<map>
#include<iostream>
#include<stack>
#include<queue>
using namespace std;
template <typename T>istream& operator>>(istream &input,vector<T>&v){for(auto &i:v) cin>>i;return input;}
template <typename T>ostream& operator<<(ostream &output,vector<T>&v){for(auto &i:v) cout<<i<<" ";return output;}
using ll = long long;
#define ff first
#define ar array
#define ss second
#define all(s) s.begin(),s.end()
#define fsort(v) sort(all(v))
#define bsort(v) sort(all(v),greater<int>())
const ll mod = 1e9+7;
#define nline "\n"
#define endl "\n"
using ull = unsigned long long;
using lld = long double;
void fastio(){ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);}
ll bits(ll x) { ll cnt = 0; while(x > 0) { cnt++; x >>= 1; } return cnt; }
ll setbits(ll x) { ll cnt = 0; while(x > 0) { cnt += (x & 1); x >>= 1; } return cnt; }
ll pow(ll a, ll b) {ll res = 1;while (b > 0) {if (b & 1)res = res * a;a = a * a;b >>= 1;}return res;}
ll pow(ll a, ll b, ll m = mod) {a %= m;ll res = 1;while (b > 0) {if (b & 1)res = res * a % m;a = a * a % m;b >>= 1;}return res;}
ll add(ll a, ll b, ll m = mod){a %= m , b %= m; return (a + b) % m;}
ll sub(ll a, ll b, ll m = mod){a %= m , b %= m; return (a - b + m) % m;}
ll mul(ll a, ll b, ll m = mod){a %= m , b %= m; return (1ll * a * b) % m;}
ll div(ll a, ll b, ll m = mod){a %= m , b %= m; return mul(a , pow(b , m - 2 , m) , m);}
ll inver(ll a , ll m = mod){return div(1ll , a , m);}
vector<int> sieve(int n){int *arr = new int[n + 1]();vector<int> vect;for (int i = 2; i <= n; i++)if (arr[i] == 0){vect.push_back(i);for (int j = 2 * i; j <= n; j += i)arr[j] = 1;}return vect;}

template <typename T> void put(T s){ cout<<s<<'\n';}
template <typename T> void puts(T s){ cout<<s<<" ";}
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
template<class T> using pqs = priority_queue<T>;
template<class T> using pqg = priority_queue<T, vector<T>, greater<T>>;

#define inf 1000000000
#define INF 1000000000000000000
const int MAXN = 5e5+5;
ll n,m,a,b,ans,t;
vector<ll> g[MAXN];
ll l[MAXN],d[MAXN],sz[MAXN];
void dfs(ll c, ll p) {
    l[c]=d[c]=++t; sz[c]=1;
    for(ll ele : g[c]) {
        if(ele==p) continue;
        if(d[ele]==0) dfs(ele, c), sz[c]+=sz[ele];
        l[c]=min(l[c],l[ele]);
    }
    if(d[c]==l[c]) ans = max(ans, sz[c]*(n-sz[c]));
}
void solve(){
    cin>>n>>m;
    for(ll i=0;i<n;i++) {l[i]=d[i]=0; g[i].clear();}
    for(ll i=0;i<m;i++) {
        cin>>a>>b;
        g[a].push_back(b); g[b].push_back(a);
    }
    t=0,ans=0;
    dfs(1,0);
    cout<<(n*(n-1))/2 - ans<<endl;
}
signed main()
{
    fastio();
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
}

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

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

Think step by step.

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