**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
def p1(n, k):
    gb = 0
    count = 0
    while gb < n:
        if count%k == 0:
            gb += 1
            count+= 1
        else:
            count += 1

    return count

n = int(input())
for i in range(n):
    s = input()
    x = s.split(" ")
    n = int(x[0])
    k = int(x[1])

    print(p1(n, k))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
t = int(input())

while (t > 0):

    nk = input().split()

    n = int(nk[0])
    k = int(nk[1])

    rate = float(1 / k)
    needed = n - 1
    timeneeded = int((needed / rate) + 1)

    print(timeneeded)

    t -= 1

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

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

Think step by step.

Response:
```plaintext
4
1 1
2 1
10 1
100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
/******************************************************************************

Welcome to GDB Online.
GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl,
C#, OCaml, VB, Swift, Pascal, Fortran, Haskell, Objective-C, Assembly, HTML, CSS, JS, SQLite, Prolog.
Code, Compile, Run and Debug online from anywhere in world.

*******************************************************************************/
import java.util.*;

public class Main
{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0){
		    int u=sc.nextInt();
		    int time=sc.nextInt();
		    if(u==1)System.out.println(1);
		    else if(time==1)System.out.println(5);
		    else System.out.println((u-1)*time+1);
		}

	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
#include <algorithm>
#include <bitset>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
void fast_io() {ios::sync_with_stdio(false);}
void redirect_io(const string& path)
{
    #ifndef ONLINE_JUDGE
    string in = path + ".in";
    string out = path + ".out";

    freopen(in.c_str(), "r", stdin);
    freopen(out.c_str(), "w", stdout);
    #endif
    fast_io();
}

void solve()
{
    int n, k;

    cin >> n >> k;

    vector<int> history(100 * 100 + 1);

    for(int turns = 1; turns <= 100 * 100; turns++)
    {
        int previous = 0;
        for(int i = turns - 1; i >= turns - k + 1 && i >= 1; i--)
        {
            if(history[i] == 1)
            {
                previous++;
            }
        }

        if(previous < 1)
        {
            history[turns] = 1;
            n--;
        }

        if(n == 0)
        {
            cout << turns << "\n";
            return;
        }
    }

}

int main()
{
    fast_io();

    int t;
    cin >> t;

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

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

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

Think step by step.

Response:
```plaintext
3
1 1
100 1
100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
/*
   Author: TRIDIB DEB PABEL
   Date: 30-10-24
*/
#include<bits/stdc++.h>
using namespace std;
#include <iomanip>
#define taratari ios_base::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);
#define int long long
#define ull unsigned long long int
#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
#define all(v) v.begin(),v.end()
#define endl '\n'
#define pb push_back
#define ff first
#define ss second
#define allsort(v) sort(v.begin(),v.end())
#define szv (int)v.size()
#define szs (int)s.size()
#define ft front()
#define bk back()
const long long ma = 1e9+7;
const long long MOD = 1e9+7 ;

int32_t main() {
   taratari
    int t;
    cin>>t;
    while(t--){
       int n,k;
       cin>>n>>k;
       if(n==k){
         cout<<(n*n)-(n-1)<<endl;
       }
       else if(n<k){
         cout<<((n-1)*k)+1<<endl;
       }
       else{
         cout<<(((n/k)*k)*k)+(n%k)<<endl;
       }
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
#include<iostream>
using namespace std;
int main()
{
	long n;
	cin >> n;
	long x, y;
	for (long i = 0; i < n; i++)
	{
		cin >> x >> y;
		if (x == 1 || y == 1)
			cout << x << endl;
		else if (x == y||x>y)
			cout << (x * y) - (y - 1) << endl;
		else
			if (y > x)
				cout << x + y - 1 << endl;

	}

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that can cause the faulty logic to fail. Here are the steps to identify such cases:

1. **Edge Cases for `x` and `y`**:
   - `x == 1`: This should always result in `x` seconds.
   - `y == 1`: This should always result in `x` seconds.
   - `x == y`: This should result in `x` seconds.
   - `x > y`: This should result in `x * y - (y - 1)` seconds.
   - `y > x`: This should result in `x + y - 1` seconds.

2. **Identify the Bug**:
   - The code has a logic issue in the `else if (x == y || x > y)` condition. It should not subtract `y - 1` from `x * y` for all cases where `x > y`.

3. **Test Cases**:
   - Test `x == 1` and `y == 1`.
   - Test `x == y`.
   - Test `x > y`.
   - Test `y > x`.

Based on these considerations, here is a small-scale test input that covers the edge cases and can expose the bug:

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

This test input includes:
- `1 1`: Both `x` and `y` are 1, which should result in 1 second.
- `2 2`: Both `x` and `y` are 2, which should result in 2 seconds.
- `3 2`: `x > y`, which should result in 3 seconds.
- `2 3`: `y > x`, which should result in 4 seconds.
- `4 3`: `x > y`, which should result in 4 seconds.

These cases should help identify the logic error 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: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
def min_seconds_to_upload(test_cases):
    return [
        n if k == 1 else n + (n - 1) * (k - 1)
        for n, k in test_cases
    ]

# Test cases conforme o exemplo do problema
test_cases = [
    (5, 1),
    (2, 2),
    (2, 3),
    (1, 7),
    (11, 5),
    (100, 100)
]

# Processamento e saída
results = min_seconds_to_upload(test_cases)
for result in results:
    print(result)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has a specific issue in the formula used for calculating the minimum number of seconds when \( k \neq 1 \). Let's analyze the problem and the buggy code to identify such cases.

### Analysis of the Buggy Code
The buggy code uses the following formula:
- If \( k = 1 \): `n` (which is correct)
- If \( k \neq 1 \): `n + (n - 1) * (k - 1)`

The correct formula for \( k \neq 1 \) should be:
- `n * k` (since in any \( k \) consecutive seconds, you can only upload 1 GB, and you need to ensure that the total time is a multiple of \( k \) to fit all the 1 GB uploads).

### Challenging Test Cases
1. **Case where \( k = 1 \)**: This should be straightforward and correct.
2. **Case where \( k > 1 \) and \( n = 1 \)**: The formula should handle this correctly.
3. **Case where \( k > 1 \) and \( n \) is a multiple of \( k \)**: This should be handled correctly.
4. **Case where \( k > 1 \) and \( n \) is not a multiple of \( k \)**: This is where the buggy code might fail.

### Test Input
Let's create a test input with these cases:

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

### Explanation
1. **(1, 2)**: \( n = 1 \), \( k = 2 \). The correct answer is 2 seconds (1 GB in the first second, 0 GB in the second second).
2. **(2, 2)**: \( n = 2 \), \( k = 2 \). The correct answer is 3 seconds (1 GB in the first second, 0 GB in the second second, 1 GB in the third second).
3. **(3, 2)**: \( n = 3 \), \( k = 2 \). The correct answer is 4 seconds (1 GB in the first second, 0 GB in the second second, 1 GB in the third second, 0 GB in the fourth second).
4. **(4, 3)**: \( n = 4 \), \( k = 3 \). The correct answer is 7 seconds (1 GB in the first second, 0 GB in the second second, 0 GB in the third second, 1 GB in the fourth second, 0 GB in the fifth second, 0 GB in the sixth second, 1 GB in the seventh second).

This test input should expose the bug in the buggy code, as it will produce incorrect results for the cases where \( k \neq 1 \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
t = int(input())
for i in range(t):
    n, k = [int(x) for x in input().split(" ")]
    seconds = 0
    gb = 0
    for j in range(n * k):
        if seconds % k == 0:
            gb += 1
        seconds += 1
        if gb == n: break
    print(seconds)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
import java.util.Scanner;

public class EPICDIV12A {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int inputs = sc.nextInt();
        for (int l = 0; l < inputs; l++){
            int gb = sc.nextInt();
            int time = sc.nextInt();
            int mod = gb % time;
            int fixed = -1;
            if (gb > time){
                fixed = gb - mod;
            }
            if (fixed == -1) {
                System.out.println(time * (gb-1) + 1);
            } else System.out.println((fixed * time) + mod);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

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

int main(){
    int n;
    int **data=new int*[n];
    cin>>n;
    for(int i=0; i<n; i++){
        int gb, sec;
        cin>>gb>>sec;
        int *d=new int[2];
        d[0]=gb;
        d[1]=sec;
        data[i]=d;
    }
    for(int i=0 ;i<n; i++){
        cout<<1+(data[i][0] - 1)*data[i][1]<<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
5
1 1
2 2
3 3
4 4
5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

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

using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        if (n <= k) {
            cout << (n - 1) * k + 1 << endl;
        }
        else if (n % k != 0)
        {
            cout << ((n - 1) * k) + 1 << endl;
        }
        else {
            cout << n * k << endl;
        }
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
#include<iostream>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n, k;
        cin >> n>> k;
        if(n>=k)
        {
            cout << ((n-1) *k  + 1) << endl;
        }

        else
        {
            cout << n * n << endl;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
def min_seconds(n, k):
    l = 0
    r = n * k

    while l <= r:
        mid = (l + r) // 2
        if (mid + k - 1) // k >= n:
            r = mid - 1
        else:
            l = mid + 1

    return l

# Test cases
print(min_seconds(5, 1)) # Expected output: 5
print(min_seconds(2, 2)) # Expected output: 3
print(min_seconds(2, 3)) # Expected output: 4
print(min_seconds(1, 7)) # Expected output: 1
print(min_seconds(11, 5)) # Expected output: 51
print(min_seconds(100, 100)) # Expected output: 9901

**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
100 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
t=int(input())
for i in range(t):
    n,k=map(int,input().split())
    if k==1 and n>k:
        print(n)
    if k!=1 and n>k:
        print(n*k-(k-1))
    if n==k:
        print(n**2-(n-1))
    if n!=1 and n<k:
        print(2*n)
    if n==1 and n<k:
        print(n)

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

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

Think step by step.

Response:
```plaintext
6
5 1
2 2
2 3
1 7
11 5
100 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
import java.util.*;
public class upload_ram {
    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int t = sc.nextInt();
    for (int tc = 0; tc < t; tc++) {
        int n = sc.nextInt();
        int k = sc.nextInt();

        if(n == 1){
            System.out.println("1");
        }else if(n >= k){
            System.out.println(n*k - k + 1);
        }else{
            System.out.println(n*k - n);
        }
    }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Upload More RAM
time_limit_ms: 1000
memory_limit_mb: 256
Description: Oh no, the ForceCodes servers are running out of memory! Luckily, you can help them out by uploading some of your RAM!

You want to upload $n$ GBs of RAM. Every second, you will upload either $0$ or $1$ GB of RAM. However, there is a restriction on your network speed: in any $k$ consecutive seconds, you can upload only at most $1$ GB of RAM in total.

Find the minimum number of seconds needed to upload $n$ GBs of RAM!

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

The first and only line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of GBs that you want to upload and the length of the time window respectively.

Output: For each test case, output a single integer — the minimum number of seconds needed to upload $n$ GBs of RAM.

Examples:
input:
6
5 1
2 2
2 3
1 7
11 5
100 100
output:
5
3
4
1
51
9901

Note: In the first test case, you can upload $1$ GB of RAM per second, so to upload $5$ GBs, you need $5$ seconds.

In the second test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, and $1$ GB in the third second, which in total adds up to exactly $2$ GBs of uploaded RAM.

In the third test case, you can upload $1$ GB in the first second, $0$ GBs in the second second, $0$ GBs in the third second, and $1$ GB in the fourth second, which in total adds up to exactly $2$ GBs of uploaded RAM.

Buggy Code:
import java.util.*;

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

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

            int ans=0;
            if(k==1 || n==1){
                ans=n;
            }
            else if(n>=k){
                ans= (n-1)*k +1;
            }else if(n<k){
                int val= k/n;
                ans= val*k + 1;
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 100
100 1
5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

Buggy Code:
#include <bits/stdc++.h>
#define sws ios_base::sync_with_stdio(false); cout.tie(NULL);cin.tie(NULL);
#define endl "\n"
#define pb push_back
#define ll long long
#define ld long double

#define int long long

using namespace std;

const int MOD = 1e9+7;
const int MAX = 2e5+5;
const int N = 1e5+5;
const ll INF = 1e9+7;
const ld PI = acos(-1);
const int LOG = 22;
const ld EPS = 1e-8;

void solve(){
    string s; cin >> s;
    int n = s.size();
    map<int,int> qtd;
    ll ans = 0, atual = 0;
    for(int i =0; i < n ; i++){
        if(s[i] == '('){
            atual++;
            qtd[atual/2-1] = 0;
        }
        if(s[i] == ')'){
            atual--;
        }
        if(atual){
            ans += qtd[atual];
        }
        qtd[atual]++;
    }
    cout << ans << endl;
}

int32_t main(){sws;
    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:
To expose the bug in the given code, we need to consider the logic and identify scenarios where the code might fail. The key aspects to focus on are:

1. **Regular Bracket Sequence**: The input must be a valid regular bracket sequence.
2. **Inversion and Validity**: The code should correctly count the number of pairs \((l, r)\) such that inverting the substring from \(l\) to \(r\) still results in a regular bracket sequence.

### Analysis of the Buggy Code
The code uses a map to keep track of the balance of brackets and tries to count the valid pairs. However, there are a few potential issues:
- The code might not correctly handle the balance of brackets.
- The logic for updating the map and counting valid pairs might be flawed.

### Test Case to Expose the Bug
To create a test case that exposes the bug, we can use a regular bracket sequence that has multiple nested and non-nested pairs. This will test the code's ability to handle both simple and complex sequences.

### Chosen Test Case
Let's use the sequence `(()())`. This sequence has multiple nested and non-nested pairs, which should be a good test for the code.

### Expected Output
For the sequence `(()())`:
- Valid pairs \((l, r)\) are:
  - \((2, 3)\): `(()())` -> `(()())`
  - \((4, 5)\): `(()())` -> `(()())`
  - \((2, 5)\): `(()())` -> `(()())`
  - \((1, 6)\): `(()())` -> `(()())`

The expected output is `4`.

### Test Input
```plaintext
1
(()())
```

This test case should expose any issues in the logic of the buggy code, particularly in how it handles nested and non-nested pairs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define xi first
#define yi second
#define endl '\n'
#define MAX (ll)200007
typedef long long ll;
typedef long double ld;
typedef long long ll;
const long long  md = (ll)1e9+7;
long long gcd(long long a, long long b)
{
    if (b == 0)return a;
    return (a % b) ? gcd(b, a % b) : b;
}
int mst_sig_bit(long long num)
{
    int cnt = -1;
    while (num)num = num >> (ll)1, cnt++;
    return cnt;
}
long long fst_pow(long long a, long long N)
{
    if (N == 0) return 1;
    long long R = fst_pow(a, N / 2);
    if (N % 2) return ((R * R) % md * a) % md;
    else return (R * R) % md;
}
int cnt[MAX];
int mx[MAX];
int main()
{
    cin.sync_with_stdio(false);
    cout.sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin>>t;
    while(t--) {
        int n;
        string s;
        cin>>s;
        n = s.length();
        vector<vector<int>> mp(n+1);
        for (int x = 1; x <= n; x++) {
            char i = s[x-1];
            cnt[x] = cnt[x - 1];
            if (i == '(') cnt[x]++;
            else cnt[x]--;
            mp[cnt[x]].emplace_back(x);
        }
        stack<int>st;
        for(int x=1;x<=n;x++){
            if(st.empty()) st.push(x);
            else{
                while(!st.empty()&&2*cnt[st.top()]<cnt[x]){
                    mx[st.top()] = x-1;
                    st.pop();
                }
                st.push(x);
            }
        }
        while(!st.empty()) mx[st.top()] = n, st.pop();
        ll ans = 0;
        for (int x = 1; x <= n; x++) {
            int l = mx[x-1];
            int res = cnt[x - 1];
            auto itl = lower_bound(mp[res].begin(), mp[res].end(), x);
            auto itr = upper_bound(mp[res].begin(), mp[res].end(), l);
            ans += (itr - itl);
        }
        cout << ans << 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
(())
(()())
((()))
(())(())
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
#pragma GCC optimize(3, "Ofast", "inline")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

const int MOD = 1000000007 ; // 998244353
const long long inf = 1000000000000000000LL; // 18 zero's // LLONG_MAX = 9223372036854775807 - 19 digits
#define int long long

int GCD(int a, int b){if(b == 0){return a;} return GCD(b, a % b);} // Euclidean algorithm for computing the greatest common divisor
int LCM(int a, int b){return ((long long)a / (long long)GCD(a, b) * (long long)b) ;}
int modpower(int x, int y, int m){long long res = 1; x = x % m; while(y > 0){if(y&1) res = (res * (long long)x) % MOD; y = y >> 1; x = ((long long)x * (long long)x) % MOD ;}return res ;}
int mod_ADDITION(int a, int b, int m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
int mod_SUBTRACTION(int a, int b, int m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
int mod_MULTIPLICATION(int a, int b, int m) {a = a % m; b = b % m; return ((((long long)a * (long long)b) % m) + m) % m;}
int mod_DIVISION(int a, int b, int m){a %= m , b %= m; return mod_MULTIPLICATION(a , modpower(b , m - 2 , MOD) , m);}
long long power(long long x, long long y){long long res = 1 ; while(y > 0) {if(y & 1) res = (res * (long long)x) ; x = ((long long)x * (long long)x) ; y = y >> 1 ;} return res ;}
int modinv(int x){return modpower(x , MOD - 2 , MOD) ;}
bool comparatorfunction(pair<int,int> &a, pair<int,int> &b){if(a.first != b.first){return a.first < b.first ;}return a.second > b.second ;}

/*

How to define our segment tree in the main function
vector<int > arr = {1 , 2 , 3 , 4} ;

SGT st(n , arr);

*/

struct SGT {
    struct Node {
        int val ; // may change
        Node() { // Identity element
            val = 0 ;    // may change // warning it is for query answers only what should be the output answer
            // val = power(2 , 30) - 1 ;    // may change
            // val = INF ;    // may change
            // val = -INF ;    // may change
        }
        Node(int p1) {  // Actual Node
            val = p1 ; // may change
        }
        void merge(Node &l , Node &r) { // Merge two child nodes
            val = max(l.val , r.val) ;  // may change
        }
    };
    struct Update { // note if mentioned about update then only it is used
        int val ; // may change
        Update(int p1) { // Actual Update
            val = p1 ; // may change
        }
        void apply(Node &a) { // apply update to given node
            a.val = val ; // may change
        }
    };

    vector<Node > tree ; // it will change according to the node
    vector<int > arr ; // type may change
    int n , s ;
    int sz ; // 2^k where k is the least integer which satisfy 2^k >= n
    SGT(int array_len, vector<int> &a) { // change if type updated
        arr = a ;
        n = array_len ;
        s = 1 ;
        while(s < 2 * n) {
            s = (s << 1);
        }
        tree.resize(s);
        fill(tree.begin(),tree.end(), Node()) ;
        build(1 , 0 , n - 1) ;
    }
    void build(int index, int start, int end)  // Never change this
    {
        if(start == end){
            tree[index] = Node(arr[start]) ; // may change
            return ;
        }
        int mid = (start + end) / 2 ;
        build(2 * index, start, mid) ;
        build(2 * index + 1, mid + 1, end) ;
        tree[index].merge(tree[2 * index] , tree[2 * index + 1]) ;
    }
    void update( int index, int start, int end, int query_index, Update &u)  // Never Change this
    {
        if(start == end){
            u.apply(tree[index]) ;
            return ;
        }
        int mid = (start + end) / 2 ;
        if(mid >= query_index)
            update(2 * index, start, mid, query_index, u);
        else
            update(2 * index + 1, mid + 1, end, query_index, u) ;
        tree[index].merge(tree[2 * index], tree[2 * index + 1]) ;
    }
    Node query(int index, int start, int end, int left, int right) { // may change
        if(start > right || end < left) // may change // both if statements can be removed
            return Node() ;
        if(start >= left && end <= right) // may change // both if statements can be removed
            return tree[index] ;
        int mid = (start + end) / 2 ;
        Node l , r , ans ;
        l = query(2 * index , start , mid , left, right) ;
        r = query(2 * index + 1 , mid + 1 , end , left , right) ;
        ans.merge(l , r) ;
        return ans ;
    }
    void make_update(int index, int val) {  // pass in as many parameters as required
        Update new_update = Update(val); // may change
        update(1 , 0 , n - 1 , index , new_update) ;
    }
    Node make_query(int left , int right) { // may change
        return query(1 , 0 , n - 1 , left , right) ; // may change
    }
    int first(){
        return tree[1].val ;
    }
    void output(){
        for(int i = 1 ; i < s ; ++i){
            cout << tree[i].val <<  " " ;
        }
        cout << "\n" ;
    }
};

bool solve(){
    string s;
    cin>>s;
    int n=s.size();
    vector<int>pf(n,0);
    int sum=0;
    map<int,vector<int>>mp;
    for(int i=0;i<n;++i){
        sum+=(s[i]=='(');
        sum-=(s[i]==')');
        if(sum<0){
            cout<<0<<endl;
            return true;
        }
        pf[i]=sum;
    }
    sum=0;
    SGT sgt(n,pf);
    int ans=0;
    for(int i=0;i<n;++i){
        sum+=(s[i]=='(');
        sum-=(s[i]==')');
        vector<int>v;
        if(mp[sum].size()>0){
            for(auto &j:mp[sum]){
                v.push_back(j);
            }
        }
        int len=v.size();
        // cout<<len<<" ";
        // cout<<sum<<" ";
        int low=-1;
        int high=len;
        while(high-low>1 and len>0){
            int mid=(high+low)/2;
            int idx=v[mid];
            int f=0;
            int val1=pf[idx];
            // cout<<val<<" ";
            int tot_open=sgt.make_query(val1+1,i).val;
            // cout<<tot_open<<" ";
            int rem=tot_open-val1;
            if(val1<sum || (rem>val1)) f=1;
            if(f>0){
                low=mid;
            }
            else{
                high=mid;
            }
        }
        // cout<<high<<endl;
        ans+=max(0ll,len-high);
        mp[sum].push_back(i);
    }
    cout<<ans<<endl;
    return true ;
}

int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);
//#ifndef ONLINE_JUDGE
//    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
//    freopen("error.txt", "w", stderr);
//#endif
    int t ;
    cin >> t ;
    //int count = 1 ;

    while (t--){
        //cout << "Case #" << count << ": ";
        if(solve()){
            //count++ ;
        }else{
            // no
        }
    }
    return 0 ;
}

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

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

Think step by step.

Response:
```plaintext
2
(())
(()())
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

Buggy Code:
def is_valid_bracket_sequence(s):
    balance = 0
    for char in s:
        if char == '(':
            balance += 1
        else:
            balance -= 1
        if balance < 0:
            return False
    return balance == 0

def count_valid_inverses(t, sequences):
    results = []

    for seq in sequences:
        n = len(seq)
        count = 0

        # Check all possible substrings (l, r)
        for l in range(n):
            for r in range(l, n):
                # Invert the substring from l to r
                inverted_seq = seq[:l] + ''.join(')' if c == '(' else '(' for c in seq[l:r+1]) + seq[r+1:]
                if is_valid_bracket_sequence(inverted_seq):
                    count += 1

        results.append(count)

    return results

# Read input
t = int(input().strip())
sequences = [input().strip() for _ in range(t)]

# Get results
results = count_valid_inverses(t, sequences)

# Print results
for res in results:
    print(res)

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

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

Think step by step.

Response:
```plaintext
3
(()())
((()))
()()()
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

Buggy Code:
# import math
# import sys
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# python program to find minimum and maximum using segment
# tree
import math

# Node for storing minimum and maximum value of given range
class Node:
    def __init__(self):
        self.minimum = math.inf
        self.maximum = -math.inf

# A utility function to get the middle index from corner
# indexes.
def getMid(s, e):
    return s + (e - s) // 2

""" A recursive function to get the minimum and maximum
   value in a given range of array indexes. The following
   are parameters for this function.

    st    --> Pointer to segment tree
    index --> Index of current node in the segment tree.
   Initially 0 is passed as root is always at index 0 ss &
   se  --> Starting and ending indexes of the segment
                  represented  by current node, i.e.,
   st[index] qs & qe  --> Starting and ending indexes of
   query range """

def MaxMinUntill(st, ss, se, qs, qe, index):
    # If segment of this node is a part of given range,
    # then return
    #  the minimum and maximum node of the segment
    tmp = Node()
    if qs <= ss and qe >= se:
        return st[index]

    # If segment of this node is outside the given range
    if se < qs or ss > qe:
        return tmp

    # If a part of this segment overlaps with the given
    # range
    mid = getMid(ss, se)
    left = MaxMinUntill(st, ss, mid, qs, qe, 2 * index + 1)
    right = MaxMinUntill(st, mid + 1, se, qs, qe, 2 * index + 2)
    tmp.minimum = min(left.minimum, right.minimum)
    tmp.maximum = max(left.maximum, right.maximum)
    return tmp

# Return minimum and maximum of elements in range from
# index qs (query start) to qe (query end).  It mainly uses
# MaxMinUtill()

def MaxMin(st, n, qs, qe):
    tmp = Node()

    # Check for erroneous input values
    if qs < 0 or qe > n - 1 or qs > qe:
        print("Invalid Input")
        return tmp
    return MaxMinUntill(st, 0, n - 1, qs, qe, 0)

# A recursive function that constructs Segment Tree for
# array[ss..se]. si is index of current node in segment
# tree st

def constructSTUtil(arr, ss, se, st, si):
    # If there is one element in array, store it in current
    # node of segment tree and return
    if ss == se:
        st[si].minimum = arr[ss]
        st[si].maximum = arr[ss]
        return

    # If there are more than one elements, then recur for
    # left and right subtrees and store the minimum and
    # maximum of two values in this node
    mid = getMid(ss, se)
    constructSTUtil(arr, ss, mid, st, si * 2 + 1)
    constructSTUtil(arr, mid + 1, se, st, si * 2 + 2)
    st[si].minimum = min(st[si * 2 + 1].minimum, st[si * 2 + 2].minimum)
    st[si].maximum = max(st[si * 2 + 1].maximum, st[si * 2 + 2].maximum)

""" Function to construct segment tree from given array. This
   function allocates memory for segment tree and calls
   constructSTUtil() to fill the allocated memory """

def constructST(arr, n):
    # Allocate memory for segment tree

    # Height of segment tree
    x = math.ceil(math.log2(n))

    # Maximum size of segment tree
    max_size = 2 * (2 ** x) - 1
    st = [Node() for i in range(max_size)]

    # Fill the allocated memory st
    constructSTUtil(arr, 0, n - 1, st, 0)

    # Return the constructed segment tree
    return st

# Driver code
arr = [1, 8, 5, 9, 6, 14, 2, 4, 3, 7]
n = len(arr)

# Build segment tree from given array
st = constructST(arr, n)
qs = 0  # Starting index of query range
qe = 8  # Ending index of query range
result = MaxMin(st, n, qs, qe)

t=int(input())
for r in range(t):
    s=input()
    st=[]
    for i in range(len(s)):
        st.append(s[i])
    l=[]
    sum=0
    for i in range(len(st)):
        if(st[i]=='('):
            sum+=1
            l.append(sum)
        else:
            sum-=1
            l.append(sum)
    lst=[]
    for i in range(len(s)+1):
        lst.append([])
    for i in range(len(l)):
        lst[l[i]].append(i)
    #lst[1]-=1
    ans=0
    lst[0]=[]
    segt=constructST(l,len(l))
    #print(lst[0:10])
    for i in range(1,len(l)):
        for j in range(len(lst[l[i-1]])):
            #print(segt.query(i,lst[l[i-1]][j]),i,lst[l[i-1]][j])
            if(lst[l[i-1]][j]>i and MaxMin(segt,len(l),i,lst[l[i-1]][j]).maximum-l[i-1]<=l[i-1]):
                ans+=1
        #print(ans)
        #lst[l[i]]-=1
        lst[0]=[]
    print(ans)
    #print(l)
    #print(lst[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
(())
(()())
((()))
(())()
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

Buggy Code:
import sys
from os import path
from collections import Counter

FILE = 0 # if needed change it while submitting

if FILE:
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')

def get_int():
    return int(sys.stdin.readline())

def get_string():
    return sys.stdin.readline().strip()

def get_list_int():
    return list(map(int,get_string().split()))

def solve(word):
    count = Counter()
    curr = 1
    ans = 0
    count[1] = 1

    for i in range(1,len(word)-1):
        if word[i]=="(":
            curr+=1
        else:
            curr-=1
        if curr!=0:
            ans+=count[curr]
        count[curr]+=1
    return ans

t = get_int()

final_result = []

for _ in range(t):
    word = get_string()
    ans = solve(word)
    final_result.append(str(ans))

for item in final_result:
    sys.stdout.write(item)
    sys.stdout.write('\n')

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the probability of failure. The buggy code seems to have issues with correctly handling the inversion of substrings and ensuring the resulting sequence remains a regular bracket sequence.

Here are some steps to think through the problem:

1. **Identify the Bug**: The buggy code uses a `Counter` to keep track of the balance of parentheses and increments or decrements the current balance based on the characters in the string. However, it does not correctly handle the inversion of substrings and the resulting balance.

2. **Edge Cases**:
   - **Single Pair**: A single pair of parentheses `()`.
   - **Multiple Pairs**: Multiple pairs of parentheses `()()()`.
   - **Nested Pairs**: Nested pairs of parentheses `((()))`.
   - **Mixed Pairs**: A mix of nested and non-nested pairs `(()(()))`.

3. **Challenging Scenarios**:
   - **Inversion of a Single Character**: Inverting a single character in a valid sequence.
   - **Inversion of a Substring**: Inverting a substring that spans multiple pairs or nested pairs.
   - **Inversion of the Entire Sequence**: Inverting the entire sequence.

4. **Test Input**:
   - **Single Pair**: This should return 0 because inverting any part of `()` will not result in a valid sequence.
   - **Multiple Pairs**: This should return a non-zero value because there are multiple valid substrings that can be inverted.
   - **Nested Pairs**: This should return a non-zero value because there are valid substrings that can be inverted.
   - **Mixed Pairs**: This should return a non-zero value because there are valid substrings that can be inverted.

Given these considerations, a small-scale test input that includes a mix of these scenarios would be:

```plaintext
4
()
()()
((()))
(()(()))
```

This test input includes:
- A single pair `()`.
- Multiple pairs `()()`.
- Nested pairs `((()))`.
- Mixed pairs `(()(()))`.

These cases should help expose the bug 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: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

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

public class Main {
    static final int MOD = 998244353;
    static final int N = 3 * (int)1e5 + 5;

    public static long power(long a, long b, int mod) {
        long res = 1;
        while (b > 0) {
            if ((b & 1) == 1) res = (res * a) % mod;
            a = (a * a) % mod;
            b >>= 1;
        }
        return res;
    }

    public static long modInverse(long a, int mod) {
        return power(a, mod - 2, mod);
    }

    public static long modAdd(long a, long b, int mod) {
        return (a % mod + b % mod + mod) % mod;
    }

    public static long modSub(long a, long b, int mod) {
        return (a % mod - b % mod + mod) % mod;
    }

    public static long modMul(long a, long b, int mod) {
        return (a % mod * b % mod) % mod;
    }

    public static long modDiv(long a, long b, int mod) {
        return modMul(a, modInverse(b, mod), mod);
    }

    public static int binarySearch(List<Integer> a, int i) {
        int start = 0, end = a.size() - 1, ans = -1;
        while (start <= end) {
            int mid = (start + end) / 2;
            if (a.get(mid) >= i) {
                ans = mid;
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        return ans;
    }

    public static void cry(Scanner scanner) {
        String s = scanner.next();
        int n = s.length();

        int[] pref = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            if (s.charAt(i - 1) == '(') {
                pref[i] = pref[i - 1] + 1;
            } else {
                pref[i] = pref[i - 1] - 1;
            }
        }

        Map<Integer, List<Integer>> map = new HashMap<>();
        long ans = 0;

        for (int i = n; i > 0; i--) {
            if (pref[i] != 0) {
                List<Integer> st = map.getOrDefault(2 * pref[i] + 1, new ArrayList<>());
                List<Integer> v = map.getOrDefault(pref[i], new ArrayList<>());
                int it = binarySearch(st, i);
                if (it == -1) {
                    ans += v.size();
                } else {
                    int it1 = binarySearch(v, st.get(it));
                    ans += v.size() - (it1 + 1);
                }
            }
            map.computeIfAbsent(pref[i], k -> new ArrayList<>()).add(i);
        }

        System.out.println(ans);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = 1;
        t = scanner.nextInt();
        for (int i = 1; i <= t; i++) {
            cry(scanner);
        }
        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
(())
()
()()()
(()())(())
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

Buggy Code:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.Arrays;
import java.util.HashMap;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.InputStreamReader;
import java.io.Writer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Code generated by CHelperX
 *
 * @author Nipuna Samarasekara
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        FastPrinter out = new FastPrinter(outputStream);
        DInvertibleBracketSequences solver = new DInvertibleBracketSequences();
        solver.solve(1, in, out);
        out.close();
    }

    static class DInvertibleBracketSequences {
        public void solve(int testNumber, FastScanner in, FastPrinter out) {
            int t = in.nextInt();
            while (t-- > 0) {
                String s = in.next();
                int n = s.length();
                int[] sum = new int[n + 1];
                int[] next = new int[n + 1];
                for (int i = 0; i < n; i++) {
                    sum[i + 1] = sum[i] + (s.charAt(i) == '(' ? 1 : -1);
                }

                MaxSegmentTree mst = new MaxSegmentTree(n + 10);
                for (int i = 0; i <= n; i++) {
                    mst.set(i, sum[i]);
                }

                HashMap<Integer, Integer> sumpos = new HashMap<>();

                for (int i = n; i >= 0; i--) {
                    if (sumpos.containsKey(sum[i])) {
                        next[i] = sumpos.get(sum[i]);
                    } else {
                        next[i] = -1;
                    }
                    sumpos.put(sum[i], i);
                }

                long[] count = new long[n + 1];
                count[0] = 0;
                long ans = 0;
                for (int i = 0; i <= n; i++) {
//                if (i > 0)
//                    count[i] += count[i-1];
                    if (next[i] != -1) {
                        int mx = mst.getMax(i, next[i] + 1);

                        if (mx > 2 * sum[i])
                            continue;

                        count[next[i]] += count[i] + 1;
                    }

                    ans += count[i];

                }

                out.println(ans);

            }
        }

    }

    static class FastPrinter extends PrintWriter {
        public FastPrinter(OutputStream out) {
            super(out);
        }

        public FastPrinter(Writer out) {
            super(out);
        }

        public void close() {
            super.close();
        }

    }

    static class FastScanner extends BufferedReader {
        public FastScanner(InputStream is) {
            super(new InputStreamReader(is));
        }

        public int read() {
            try {
                int ret = super.read();
//            if (isEOF && ret < 0) {
//                throw new InputMismatchException();
//            }
//            isEOF = ret == -1;
                return ret;
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }

        public String next() {
            StringBuilder sb = new StringBuilder();
            int c = read();
            while (isWhiteSpace(c)) {
                c = read();
            }
            if (c < 0) {
                return null;
            }
            while (c >= 0 && !isWhiteSpace(c)) {
                sb.appendCodePoint(c);
                c = read();
            }
            return sb.toString();
        }

        static boolean isWhiteSpace(int c) {
            return c >= 0 && c <= 32;
        }

        public int nextInt() {
            int c = read();
            while (isWhiteSpace(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int ret = 0;
            while (c >= 0 && !isWhiteSpace(c)) {
                if (c < '0' || c > '9') {
                    throw new NumberFormatException("digit expected " + (char) c
                            + " found");
                }
                ret = ret * 10 + c - '0';
                c = read();
            }
            return ret * sgn;
        }

        public String readLine() {
            try {
                return super.readLine();
            } catch (IOException e) {
                return null;
            }
        }

    }

    static class MaxSegmentTree {
        int[] max;
        int[] maxId;
        int n;

        public MaxSegmentTree(int n) {
            this.n = Integer.highestOneBit(n) << 1;
            max = new int[this.n * 2];
            maxId = new int[this.n * 2];
            Arrays.fill(max, Integer.MIN_VALUE);
            for (int i = 0; i < n; i++) {
                maxId[i + n] = i;
            }
            for (int i = 0; i < n; i++) {
                set(i, Integer.MIN_VALUE);
            }
        }

        public void set(int x, int y) {
            x += n;
            max[x] = y;
            while (x > 1) {
                x >>= 1;
                int left = max[x << 1];
                int right = max[(x << 1) | 1];
                if (left >= right) {
                    max[x] = left;
                    maxId[x] = maxId[x << 1];
                } else {
                    max[x] = right;
                    maxId[x] = maxId[(x << 1) | 1];
                }
            }
        }

        public int getMax(int left, int right) {
            --right;
            left += n;
            right += n;
            int ret = Integer.MIN_VALUE;
            while (left <= right) {
                if ((left & 1) == 1) {
                    ret = Math.max(ret, max[left]);
                    left++;
                }
                if ((right & 1) == 0) {
                    ret = Math.max(ret, max[right]);
                    right--;
                }
                left >>= 1;
                right >>= 1;
            }
            return ret;
        }

    }
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - Empty or very short sequences.
   - Sequences with a single pair of brackets.
   - Sequences with multiple pairs of brackets.

2. **Complexity**:
   - Sequences that have nested brackets.
   - Sequences that have multiple segments of nested brackets.

3. **Boundary Conditions**:
   - Sequences where the entire sequence is a valid regular bracket sequence.
   - Sequences where only a part of the sequence can be inverted to remain valid.

Given these considerations, a small-scale test input that can expose the bug might be a sequence with nested and multiple segments of brackets. This will test the code's ability to handle complex nested structures and multiple valid segments.

Here is a suitable test input:

```plaintext
4
(())
()()()
(()())
((()))
```

This test input includes:
- A single pair of brackets.
- Multiple pairs of brackets.
- Nested pairs of brackets.
- A sequence with a single nested pair of brackets.

This should help in identifying any issues with the code's logic for handling nested and multiple segments of valid bracket sequences.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Invertible Bracket Sequences
time_limit_ms: 2000
memory_limit_mb: 256
Description: A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:

  * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");    * bracket sequences ")(", "(" and ")" are not.

Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.

You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.

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

The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.

Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.

Examples:
input:
4
(())
()
()()()
(()())(())
output:
1
0
3
13

Note: In the first example, there is only one pair:

  * $(2, 3)$: (()) $\rightarrow$ ()().

In the second example, there are no pairs.

In the third example, there are three pairs:

  * $(2, 3)$: ()()() $\rightarrow$ (())();    * $(4, 5)$: ()()() $\rightarrow$ ()(());    * $(2, 5)$: ()()() $\rightarrow$ (()());

Buggy Code:
// LUOGU_RID: 161859692
#include<iostream>
#include<cstring>
#include<vector>
#define N 200005
#define K 30
using namespace std;
char str[N];
int s[N];
int st[N][K],lg[N];
vector<int>pos[N];
void init(){for(int i=2;i<N;++i)lg[i]=lg[i>>1]+1;}
int ask(int l,int r){int k=lg[r-l+1];return max(st[l][k],st[r-(1<<k)+1][k]);}
void solve()
{
	cin>>str+1;int n=strlen(str+1);
	for(int i=0;i<=n;++i)pos[i].clear();
	for(int i=1;i<=n;++i)s[i]=s[i-1]+(str[i]=='('?1:-1),st[i][0]=s[i],pos[s[i]].push_back(i);
	for(int k=1;k<=lg[n];++k)for(int i=1;i+(1<<k)-1<=n;++i)st[i][k]=max(st[i][k-1],st[i+(1<<k)-1][k-1]);
	long long ans=0;
	for(int i=0;i<=n;++i)
	{
		int len=pos[i].size(),lst=0;
		for(int j=1;j<len;++j)if(ask(pos[i][j],pos[i][j-1])>=i*2)ans+=1ll*(j-lst)*(j-lst-1)/2,lst=j;
		ans+=1ll*(len-lst)*(len-lst-1)/2;
	}
	cout<<ans<<"\n";
}
int main()
{
	ios::sync_with_stdio(0),cin.tie(nullptr),cout.tie(nullptr);
	init();int t;cin>>t;while(t--)solve();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that are likely to cause the algorithm to fail. The code is supposed to find pairs of indices \((l, r)\) such that inverting the substring from \(l\) to \(r\) still results in a regular bracket sequence. The key points to consider are:

1. **Small, non-trivial inputs**: The input should be small enough to be manageable but complex enough to test the algorithm.
2. **Edge cases**: Consider sequences with minimal or maximal pairs, sequences with nested and non-nested structures, and sequences with multiple segments.
3. **Regular bracket sequences**: Ensure that the input is a valid regular bracket sequence.

Given these points, let's construct a test input:

1. **Single pair**: A sequence with only one valid pair.
2. **No pairs**: A sequence with no valid pairs.
3. **Multiple pairs**: A sequence with multiple valid pairs.
4. **Complex structure**: A sequence with nested and non-nested structures.

Here is a test input that covers these cases:

```plaintext
4
(())
()
()()()
(()(()))
```

Explanation:
- **Test Case 1**: `(()())` has one valid pair: \((2, 3)\).
- **Test Case 2**: `()` has no valid pairs.
- **Test Case 3**: `()()()` has three valid pairs: \((2, 3)\), \((4, 5)\), and \((2, 5)\).
- **Test Case 4**: `(()(()))` has multiple valid pairs, including nested and non-nested structures, to test the algorithm's ability to handle complex sequences.

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