**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

Buggy Code:
for _ in range(int(input())):
    n, k = map(int, input().split())
    print([0, 1, 2, 3, 6, 3, 5][[_ == 0, _ == 1, _ == 2, _ == 3, _ == 4, _ == 5, _ == 6].index(True)])

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

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

Think step by step.

Response:
To 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 break the hardcoded output. The buggy code uses a fixed list of outputs for specific values of `n` and `k`, which is not a general solution. Here are the steps to create such a test input:

1. **Identify the Hardcoded Values**: The buggy code uses a list of outputs `[0, 1, 2, 3, 6, 3, 5]` for specific values of `n` and `k`.

2. **Create a Test Case with Different Values**: We need to create a test case where `n` and `k` are not in the hardcoded list but still should have a valid output.

3. **Edge Cases**: Consider edge cases such as:
   - `k` is 0 (no chips, should result in 0 occupied diagonals).
   - `k` is 1 (one chip, should result in 1 occupied diagonal).
   - `k` is `n^2` (all cells are filled, should result in a specific number of occupied diagonals based on `n`).

4. **General Cases**: Create a test case where `k` is between 1 and `n^2` and should result in a non-trivial number of occupied diagonals.

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

```plaintext
5
1 0
2 1
3 9
4 10
10 50
```

Explanation:
- The first test case (`1 0`) checks the scenario where there are no chips, which should result in 0 occupied diagonals.
- The second test case (`2 1`) checks the scenario where there is one chip, which should result in 1 occupied diagonal.
- The third test case (`3 9`) checks the scenario where all cells are filled, which should result in 5 occupied diagonals.
- The fourth test case (`4 10`) checks a non-trivial case where `k` is between 1 and `n^2`.
- The fifth test case (`10 50`) checks a larger board with a significant number of chips, which should result in 6 occupied diagonals.

This test input will help expose the bug in the hardcoded output of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

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

public class A {

	static long mod = (long) (1e9 + 7);
	static FastReader sc = new FastReader();

	public static void main(String[] args) {

		StringBuilder ans = new StringBuilder();
		int test = sc.nextInt();
		while (test-- > 0) {

			int n = sc.nextInt();
			int k = sc.nextInt();

			int cnt = k>=n ? 1: 0;
			k-=n;

			for(int i= n-1; i>=1 && k>0; i--) {
				if(k >= 2*i) {
					cnt+=2; k-=2*i;
				} else if(k >= i) {
					cnt++; k-=i;
				} else {
					cnt++;
					break;
				}
			}

			ans.append(cnt+"\n");
		}

		System.out.print(ans);
	}

	static int[] read(int n) {
		int arr[] = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
		}
		return arr;
	}

	static long add(long a, long b) {
		return (a + b) % mod;
	}

	static long mult(long a, long b) {
		return (a * b) % mod;
	}

	static long _gcd(long a, long b) {
		if (b == 0)
			return a;
		return _gcd(b, a % b);
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a) {
		int n = a.length;// shuffle, then sort
		for (int i = 0; i < n; i++) {
			int oi = random.nextInt(n), temp = a[oi];
			a[oi] = a[i];
			a[i] = temp;
		}
		Arrays.sort(a);
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

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

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

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

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

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

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

}

**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 0
2 3
100 10000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

Buggy Code:
import java.util.Scanner;

public class DiagCounter {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int tes = scanner.nextInt();
        int diag = 0;
        for (int i = 0; i < tes; i++) {
            int boar = scanner.nextInt();
            int chips = scanner.nextInt();
            int curr = 1;
            boolean second = false;
            if (chips > 0) {
                diag = 1;
            } else {
                System.out.println(diag);
                continue;
            }
            chips = chips - boar;
            while (chips > 0) {
                if (second) {
                    chips = chips - (boar - curr);
                    curr++;
                    second = false;
                } else {
                    chips = chips - (boar - curr);
                    second = true;
                }
                diag++;
            }
            System.out.println(diag);
        }
        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
3
1 1
2 2
3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

Buggy Code:
/**
  * created     : 23-07-2024 20:05:13
  * author      : brainstorm56

  Contest: Codeforces - Codeforces Round 961 (Div. 2)
  Problem: A. Diagonals
  URL: https://codeforces.com/contest/1995/problem/0
  Time Limit: 1000
  Memory Limit: 256

  */

#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
// order_of_key (k) : Number of items strictly smaller than k .
// find_by_order(k) : K-th element in a set (counting from zero).
#define ordered_set tree<ll, null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>

using namespace std;
const int N = 0;
typedef long long ll;
typedef long double ld;
const ll MOD = 998244353;
#define fr(i,e) for(ll i = 0; i < e; i++)
#define rfr(i,s) for(ll i = s-1; i >= 0; i--)
#define nl "\n"
#define dbg(x) cout<<#x<<" = "<<x<<nl
#define INF 2e18
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
const long long llmax = 9223372036854775807;
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b);}
ll ceil_div(ll a, ll b) {return a % b == 0 ? a / b : a / b + 1;}
//**********************************************************************************//

void solve() {
  ll n, k;
  cin >> n >> k;
  ll ans = 0;
  ll cur = n;
  if (k >= cur)
  {
    k -= cur;
    ans++;
    cur--;
  }
  else
  {
    k = 0;
  }
  while (k >= cur && cur > 0)
  {
    k -= cur;
    ans++;
    if (k >= cur)
    {
      k -= cur;
      ans++;
      cur--;
    }
  }
  if (k > 0)
  {
    ans++;
  }
  cout << ans << nl;
  return;
}
int main()
{
  fast_cin();
  ll t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }    return 0;
}

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the code seems to be in the way it handles the subtraction of `cur` from `k` and the updating of `ans`. Specifically, the code might not correctly account for the number of chips that can be placed on diagonals when `k` is large.

2. **Edge Cases**:
   - **Case 1**: When `k` is exactly equal to the number of cells on a diagonal.
   - **Case 2**: When `k` is slightly more than the number of cells on a diagonal.
   - **Case 3**: When `k` is much larger than the number of cells on a diagonal.
   - **Case 4**: When `n` is small and `k` is large.
   - **Case 5**: When `n` is large and `k` is small.

3. **Generate Test Input**:
   - We will create a test case where `k` is exactly equal to the number of cells on a diagonal.
   - We will also create a test case where `k` is slightly more than the number of cells on a diagonal.
   - We will create a test case where `k` is much larger than the number of cells on a diagonal.

Let's generate the test input:

```plaintext
3
3 3
3 4
3 9
```

Explanation:
- **Test Case 1**: `n = 3`, `k = 3`
  - The board has 3 diagonals: (1,1), (1,2), (2,1). Placing 3 chips on these diagonals will occupy 2 diagonals.
- **Test Case 2**: `n = 3`, `k = 4`
  - The board has 3 diagonals: (1,1), (1,2), (2,1), (1,3), (2,2), (3,1). Placing 4 chips will occupy 3 diagonals.
- **Test Case 3**: `n = 3`, `k = 9`
  - The board has 9 cells. Placing 9 chips will occupy all 5 diagonals.

This test input should expose the bug in the given code by challenging its handling of different values of `k` relative to the number of cells on the diagonals.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

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

int main(){
    int t; cin >> t;
    while(t--){
        int n, k; cin >> n >> k;
        int c = 0;
        if(k==0){
            cout << 0 << endl;
        }
        else if(k<=n){
            cout << 1 << endl;
        }else{
            k-=n;
            c++;
            for(int i = n-1; i > 0; i--){
                if(k >= 2*i){
                    c+=2;
                    k-=2*i;
                }else if(k > i && k < 2*i){
                    c+=2;
                    break;
                }else{
                    c++;
                    break;
                }
            }
            cout << c << endl;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

Buggy Code:
n = int(input())
for i in range(n):
    a = input().split()
    n = int(a[0])
    k = int(a[1])
    if k ==0:
        print(0)
    elif k < n+1:
        print(1)
    else:

        k -= n
        sum = 1
        for i in range(1,n):
            for j in range(2):
                sum = sum + 1
                k = k - n + i
                if k<1:
                    break

            if k < 0:
                break
        print(sum)

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

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

Think step by step.

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

1. **Incorrect handling of the number of diagonals**: The code does not correctly calculate the number of occupied diagonals, especially when `k` is larger than `n`.
2. **Inefficient loop structure**: The nested loops and the way `k` is decremented might lead to incorrect results.

Let's consider the following edge cases:
- When `k` is exactly `n`.
- When `k` is slightly more than `n`.
- When `k` is a multiple of `n`.

### Test Input
We will generate a test input with multiple test cases to cover these edge cases:

1. **Case 1**: `n = 10`, `k = 10` (exact multiple of `n`)
2. **Case 2**: `n = 10`, `k = 11` (slightly more than `n`)
3. **Case 3**: `n = 10`, `k = 20` (double `n`)
4. **Case 4**: `n = 10`, `k = 21` (slightly more than double `n`)

### Expected Output
- For `n = 10`, `k = 10`: The minimum number of occupied diagonals is `2` (since 10 chips can be placed on 2 diagonals).
- For `n = 10`, `k = 11`: The minimum number of occupied diagonals is `3` (since 11 chips cannot be placed on 2 diagonals, but 3 diagonals can accommodate 11 chips).
- For `n = 10`, `k = 20`: The minimum number of occupied diagonals is `4` (since 20 chips can be placed on 4 diagonals).
- For `n = 10`, `k = 21`: The minimum number of occupied diagonals is `5` (since 21 chips cannot be placed on 4 diagonals, but 5 diagonals can accommodate 21 chips).

### Test Input
```plaintext
4
10 10
10 11
10 20
10 21
```

This test input will help expose the bugs in the provided code, particularly in the handling of the number of occupied diagonals when `k` is a multiple of `n` or slightly more than a multiple of `n`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

Buggy Code:
def tol(o,s):
    t=1
    while s>0:
        t+=1
        o-=1
        s-=o
        if s>o:
            s-=o
            t+=1
    return t

n=int(input())
for i in range(n):
    a=input()
    lst=list(a.split())
    c=int(lst[0])
    d=int(lst[1])
    if d==0:
        print(0)
    elif d==c**2:
        print(c*2-1)
    else:
        if d<=c:
            print(1)
        else:
            e=d-c
            print(tol(c,e))

**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 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

Buggy Code:
import java.util.Scanner;

public class Diagonals {
    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();

            if(k < n){
                System.out.println(0);
                t--;
                continue;
            }
            k -= n;
            int c = 1;
            int p = n-1;

            while(k > 0){
                if(k - p >= 0){
                    k -= p;
                    c++;
                }
                if(k -p >= 0){
                    k -= p;
                    c++;
                    p--;
                }
                else{
                    p--;
                }
            }

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

**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 0
2 2
100 10000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Diagonals
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vitaly503 is given a checkered board with a side of $n$ and $k$ chips. He realized that all these $k$ chips need to be placed on the cells of the board (no more than one chip can be placed on a single cell).

Let's denote the cell in the $i$-th row and $j$-th column as $(i ,j)$. A diagonal is the set of cells for which the value $i + j$ is the same. For example, cells $(3, 1)$, $(2, 2)$, and $(1, 3)$ lie on the same diagonal, but $(1, 2)$ and $(2, 3)$ do not. A diagonal is called occupied if it contains at least one chip.

Determine what is the minimum possible number of occupied diagonals among all placements of $k$ chips.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of sets of input data. Then follow the descriptions of the sets of input data.

The only line of each set of input data contains two integers $n$, $k$ ($1 \le n \le 100, 0 \le k \le n^2$) — the side of the checkered board and the number of available chips, respectively.

Output: For each set of input data, output a single integer — the minimum number of occupied diagonals with at least one chip that he can get after placing all $k$ chips.

Examples:
input:
7
1 0
2 2
2 3
2 4
10 50
100 239
3 9
output:
0
1
2
3
6
3
5

Note: In the first test case, there are no chips, so 0 diagonals will be occupied. In the second test case, both chips can be placed on diagonal $(2, 1), (1, 2)$, so the answer is 1. In the third test case, 3 chips can't be placed on one diagonal, but placing them on $(1, 2), (2, 1), (1, 1)$ makes 2 diagonals occupied. In the 7th test case, chips will occupy all 5 diagonals in any valid placing.

Buggy Code:
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while(t > 0) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();

            if (k == 0) {
                System.out.println(0);
            } else if(n==k||k==n*n) {
                int oc = (int) Math.ceil((double) k / n)+((k/n)-1);
                System.out.println(oc);
            }
            else
            {
                int oc =  (int)(k / n)+1;
                System.out.println(oc);
            }
            t--;
        }

        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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios where the code's logic might fail. The code has a specific issue in handling cases where the number of chips \( k \) is equal to the side of the board \( n \) or the total number of cells \( n^2 \). Let's analyze these scenarios:

1. **Case \( k = n \)**: The code incorrectly calculates the number of occupied diagonals.
2. **Case \( k = n^2 \)**: The code might also have issues in this scenario.
3. **Other edge cases**: Small values of \( n \) and \( k \) that might cause the code to produce incorrect results.

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

```plaintext
5
3 3
3 9
4 4
4 16
2 2
```

Explanation:
1. **3 3**: \( k = n \). The code should output 1, but it might output a different value.
2. **3 9**: \( k = n^2 \). The code should output 5, but it might output a different value.
3. **4 4**: \( k = n \). The code should output 1, but it might output a different value.
4. **4 16**: \( k = n^2 \). The code should output 7, but it might output a different value.
5. **2 2**: A simple case where \( k = n \). The code should output 1, but it might output a different value.

This test input will help identify the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

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

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main()
{
    int t;
    scanf("%d", &t);

    while (t--)
    {
        long long n,k,m,sum1,sum2,dif1,dif2;
        scanf("%lld %lld", &n,&k);
        m=(n*5)/9;
        sum1=k*m+(m*(m-1))/2;
        sum2=(k*n+(n*(n-1))/2)-sum1;
        dif1=sum2-sum1;
        dif2=dif1;
        for(int i=0;i<n-m;i++)
        {
            sum1=sum1+k+m+i;
            sum2=sum2-k-m-i;
            dif2=sum2-sum1;
            if(dif2<0)
            {
                break;
            }
            else
            {
                dif1=dif2;
            }
        }
        if(dif1<0)
        {
        dif1=abs(dif1);
        }
        if(dif2<0)
        {
        dif2=abs(dif2);
        }
        if(dif1<dif2)
        {
            dif1=dif1;
        }
        else
        {
            dif1=dif2;
        }
        printf("%lld\n",dif1);
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

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

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
#include<iostream>
#include<cmath>
const int maxn=2e5+10;
#define ll  long long
using namespace std;
int main()
{

    //  freopen("in.txt","rt",stdin);
    //  freopen("out.txt","wt",stdout);

    ll n,k;
    int t;
    cin>>t;
    while(t--){
        cin>>n>>k;
        ll sum=(k+k+n-1)*n/2;
        ll l=k,r=k+n-1;
        ll sum1,sum2;
        ll mid1;
        ll mid2=l+r>>1;
        while(l<r){
            mid1=mid2;
            mid2=l+r>>1;
            sum1=(k+mid2)*(mid2-k+1)/2ll;
            if(2*sum1<sum) l=mid2+1;
            else r=mid2;
        }
        sum1=(k+mid1)*(mid1-k+1)/2ll;
        sum2=(k+mid2)*(mid2-k+1)/2ll;
        if(abs(sum-2ll*sum1)>abs(sum-2ll*sum2)) cout<<abs(sum-2ll*sum2)<<'\n';
        else cout<<abs(sum-2ll*sum1)<<'\n';
    }
    return 0;
}
//2 3  2 2
//5 6 7  5 3

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can cause the code to fail. The code is supposed to find the minimum possible value of \( x \) for a given array \( a \) with specific properties. Let's consider the following aspects:

1. **Edge Cases**:
   - Very small values of \( n \) and \( k \).
   - Very large values of \( n \) and \( k \) to test the limits of the code.
   - Values of \( n \) and \( k \) that result in large sums to ensure the code handles large numbers correctly.

2. **Intermediate Cases**:
   - Values where the array has a mix of positive and negative contributions to \( x \).

3. **Symmetric Cases**:
   - Cases where the array is symmetric around the midpoint.

Given these considerations, let's create a small-scale test input that includes a mix of these cases:

```plaintext
5
2 2
3 3
10 10
100 100
1000000000 1000000000
```

Explanation:
- **Test Case 1**: \( n = 2 \), \( k = 2 \). This is a very small array to check basic functionality.
- **Test Case 2**: \( n = 3 \), \( k = 3 \). A small array with an odd length to check the middle element.
- **Test Case 3**: \( n = 10 \), \( k = 10 \). A slightly larger array to test intermediate values.
- **Test Case 4**: \( n = 100 \), \( k = 100 \). A larger array to test the code's ability to handle moderately large numbers.
- **Test Case 5**: \( n = 1000000000 \), \( k = 1000000000 \). A very large array to test the limits of the code.

This test input should help expose any bugs in the code, especially in handling large numbers and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

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

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
// जय दादी की
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <cmath>
#include <set>
#include <string>
#include <stack>
#include <unordered_set>
#include <stack>
#define ll long long
#define mod 1000000007
#define arrin(a,n) for(ll i=0;i<n;i++) cin>>a[i];
#define ll_MAX 1000000000000000000
using namespace std;
int main(){
    ll t;
    cin>>t;
    while(t--){
     ll n,k;
        cin>>n>>k;
        k-=1;
        ll u=(n*(n+1)/2)+n*k;
        ll l=1;
        ll r=n;
        ll mid=0;
        while(l<=r){
            ll mid=(l+r)/2;
            ll x=abs(mid*(mid+1)+2*mid*k-u);
            if(x==0){
                break;
            }
            if(mid*(mid+1)+2*mid*k-u>0){
                r=mid-1;
            }
            else{
                l=mid+1;
            }
        }
        l--;
       cout<<min(abs(l*(l+1)+2*l*k-u),(abs((l+1)*(l+2)+2*(l+1)*k-u)))<<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
1
2 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

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

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
import sys

INF = 10 ** 20
inp = sys.stdin.readline

def get_ascii(ch):
    return ord(ch) - ord('a') + 1

def get_sum(s, center, N):
    left_len = center
    right_len = N - center

    left_start = s
    right_start = s + center

    left_sum = (2 * left_start + left_len - 1) * left_len // 2
    right_sum = (2 * right_start + right_len - 1) * right_len // 2

    return abs(left_sum - right_sum)

def solve():
    N, k = list(map(int, inp().split()))

    if N == 2: return 1

    low, high = 1, N - 2

    while low <= high:
        mid = low + ((high - low) >> 1)
        prev_sum, cur_sum, next_sum = get_sum(k, mid - 1, N), get_sum(k, mid, N), get_sum(k, mid + 1, N)

        if prev_sum > cur_sum and cur_sum < next_sum:
            return cur_sum
        elif prev_sum > cur_sum and cur_sum > next_sum:
            low = mid + 1
        else:
            high = mid - 1

def main():
    for _ in range(int(inp())):
        print(solve())

if __name__ == '__main__':
    main()

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

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

Think step by step.

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

1. **Understand the Bug**: The code uses a binary search to find the minimum value of \( x \). The bug likely lies in the incorrect calculation of the sums or the binary search logic.

2. **Edge Cases**:
   - **Small \( n \) and \( k \)**: Test cases with \( n = 2 \) and \( k \) values to ensure the special case handling works.
   - **Large \( n \) and \( k \)**: Test cases with \( n \) and \( k \) close to the upper limit to check for potential overflow or large number handling issues.
   - **Symmetric and Asymmetric Arrays**: Test cases where the array is symmetric (e.g., \( n = 3 \)) and asymmetric (e.g., \( n = 4 \)) to ensure the binary search logic works correctly.

3. **Test Input**:
   - **Test Case 1**: \( n = 2 \), \( k = 2 \) (to check the special case handling).
   - **Test Case 2**: \( n = 3 \), \( k = 3 \) (to check symmetric array handling).
   - **Test Case 3**: \( n = 4 \), \( k = 4 \) (to check asymmetric array handling).
   - **Test Case 4**: \( n = 1000000000 \), \( k = 1000000000 \) (to check large number handling).

Here is the test input:

```plaintext
4
2 2
3 3
4 4
1000000000 1000000000
```

This test input covers a range of scenarios to ensure the code handles edge cases and large numbers correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

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

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
t= int(input())

for _ in range(t):
    n, k = map(int, input().split())

    s= (k+n-1)*(k+n)//2
    s2=(k-1)*(k)//2
    s-=s2
    ans=None
    l, r = k , k+n-1
    curr_d= float('inf')

    while l <= r:
        m = (l + r) // 2
        left = (m*(m+1)//2) - s2
        right= s - left
        d = left - right
        if d == 0:
            ans = m - k
            break
        if d > 0 :
            r = m - 1
        else:
            l = m + 1

        if abs(d) < curr_d :
            ans= m - k
            curr_d = abs(d)

    print(curr_d)

**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 consider edge cases and scenarios that can maximize the failure probability. The code provided has a logic issue in how it calculates the sum and the difference, and it may not handle large values correctly.

### Key Considerations:
1. **Large Values**: Test with the maximum possible values for `n` and `k` to ensure the code handles large numbers correctly.
2. **Small Values**: Test with the minimum possible values for `n` and `k` to ensure the code handles small numbers correctly.
3. **Special Cases**: Test with values that might cause the midpoint calculation to fail or edge cases where the difference is minimized.

### Chosen Test Input:
- **Test Case 1**: `n = 2, k = 2` (Simple case)
- **Test Case 2**: `n = 1000000000, k = 1000000000` (Large values)
- **Test Case 3**: `n = 5, k = 3` (Intermediate values)
- **Test Case 4**: `n = 10, k = 1` (Small `k` with larger `n`)

### Test Input:
```plaintext
4
2 2
1000000000 1000000000
5 3
10 1
```

This test input includes a mix of small, large, and intermediate values to ensure the code is thoroughly tested for potential bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

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

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
def func(mid):
    prefix = (mid + 2 * k - 1) * mid // 2
    suffix = (2 * k + n - 1) * n // 2 - prefix
    return prefix, suffix

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())

    start, end = 1, n
    trying = 0
    while start < end:
        mid = (start + end) // 2
        prefix, suffix = func(mid)
        if prefix == suffix:
            break
        elif prefix > suffix:
            end = mid - 1
        else:
            trying = mid
            start = mid + 1

    prefix1, suffix1 = func(trying)
    prefix2, suffix2 = func(trying + 1)
    ans = min(abs(prefix1 - suffix1), abs(prefix2 - suffix2))

    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

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

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
import java.util.*;
public class App {
    public static void main(String[] args) throws Exception {
        Scanner sc=new Scanner(System.in);

        int t=sc.nextInt();
        while(t-->0){
            long n=sc.nextLong();
            long k=sc.nextLong();
            long l=1;
            long h=n+1;
            long ans=bs(l,h,k,n);
            System.out.println(ans);
        }
    }
    public static long bs(long l,long h,long k,long n){
        long total=((2*k+(n-1))*n)/2;
        while(l+1<h){
            long mid=(h+l)/2;
            long sum1=((2*k+(mid-1))*mid)/2;
            long sum2=total-sum1;

            if(sum1==sum2){
                return 0;
            }else if(sum1<sum2){
               l=mid;
            }else{
               h=mid-1;
            }
        }
        //System.out.println("L"+l+" "+"h"+h+" ");

        long res1=((2*k+(l-1))*l)/2;
        long res2=((2*k+(h-1))*(h))/2;
        long right1=total-res1;
        long right2=total-res2;
       // System.out.println("right1:"+right1+" "+"left1:"+res1+" "+"right2:"+right2+" "+"left2:"+res2);
        return Math.min(Math.abs(right1-res1),Math.abs((right2-res2)));
    }
}

**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
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

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

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
import java.util.*;
public class App {
    public static void main(String[] args) throws Exception {
        Scanner sc=new Scanner(System.in);

        int t=sc.nextInt();
        while(t-->0){
            long n=sc.nextLong();
            long k=sc.nextLong();
            long l=0;
            long h=n+1;
            long ans=bs(l,h,k,n);
            System.out.println(ans);
        }
    }
    public static long bs(long l,long h,long k,long n){
        long total=((2*k+(n-1))*n)/2;
        while(l+1<h){
            long mid=(h+l)/2;
            long sum1=((2*k+(mid-1))*mid)/2;
            long sum2=total-sum1;

            if(sum1==sum2){
                return 0;
            }else if(sum1<sum2){
               l=mid;
            }else{
               h=mid-1;
            }
        }
        //System.out.println("L"+l+" "+"h"+h+" ");

        long res1=((2*k+(l-1))*l)/2;
        long res2=((2*k+(h-1))*(h))/2;
        long right1=total-res1;
        long right2=total-res2;
       // System.out.println("right1:"+right1+" "+"left1:"+res1+" "+"right2:"+right2+" "+"left2:"+res2);
        return Math.min(Math.abs(right1-res1),Math.abs((right2-res2)));
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases and scenarios that might cause the binary search to fail or produce incorrect results. The key points to focus on are:

1. **Edge Cases for Small Values of \( n \) and \( k \):**
   - When \( n \) and \( k \) are very small, the sum of the array can be small, and the binary search might not handle it correctly.
2. **Large Values of \( n \) and \( k \):**
   - When \( n \) and \( k \) are very large, the sum of the array can be very large, and the binary search might have precision issues or incorrect boundary handling.
3. **Equal Sums:**
   - When the sum of the first half equals the sum of the second half, the binary search should return 0.
4. **Symmetric Arrays:**
   - Arrays where the sum of the first half and the second half are symmetric can be tricky.

Given these considerations, let's generate a test input that includes a mix of these edge cases:

```plaintext
6
2 2
3 3
4 4
5 5
1000000000 1000000000
1 1
```

This test input includes:
- Small values of \( n \) and \( k \) (e.g., \( n = 2, k = 2 \)).
- Small values of \( n \) and \( k \) with symmetric sums (e.g., \( n = 3, k = 3 \)).
- Larger values of \( n \) and \( k \) (e.g., \( n = 1000000000, k = 1000000000 \)).
- A test case with \( n = 1 \) and \( k = 1 \) to check edge cases.

This should help expose any issues in the binary search implementation.