**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bonus Project
time_limit_ms: 2000
memory_limit_mb: 512
Description: There is a team of $$$n$$$ software engineers numbered from $$$1$$$ to $$$n$$$. Their boss promises to give them a bonus if they complete an additional project. The project requires $$$k$$$ units of work in total. The bonus promised to the $$$i$$$-th engineer is $$$a_i$$$ burles. The boss doesn't assign specific tasks to engineers; it is expected that every engineer will voluntarily complete some integer amount of work units. The bonus will be paid to the entire team only if the project is completed; in other words, if the total amount of voluntary work units on the project is greater than or equal to $$$k$$$.

The amount of work that can be performed by each engineer is not limited. However, all engineers value their labour. The $$$i$$$-th engineer estimates one unit of their work as $$$b_i$$$ burles. If the bonus is paid, the benefit $$$s_i$$$ of the $$$i$$$-th engineer for completing $$$c$$$ units of work is defined as $$$s_i = a_i - c \cdot b_i$$$. If the bonus is not paid, the engineer will not volunteer to do any work.

Engineers work together for many years, so they know how the bonus is going to be distributed and how much their colleagues value the labour. That is, all $$$a_i$$$ and all $$$b_i$$$ are known to every engineer in the team.

Engineers are eager to get the bonus, so they agreed on the following process of work distribution between them:

- the first engineer says: "I will complete $$$c_1$$$ units of work", where $$$c_1$$$ is a non-negative integer;
- then, the second engineer says: "I will complete $$$c_2$$$ units of work", where $$$c_2$$$ is a non-negative integer;
- ... and so on;
- finally, the $$$n$$$-th engineer says: "I will complete $$$c_n$$$ units of work", where $$$c_n$$$ is a non-negative integer.

Every engineer voices $$$c_i$$$ in a way to maximize their own benefit $$$s_i$$$. If the expected benefit is going to be zero, an engineer will still agree to work to get the experience and to help their colleagues obtain the bonus. However, if the benefit is expected to be negative for some reason (an engineer needs to perform an excessive amount of work or the project is not going to be completed), that engineer will not work at all (completes zero amount of work units).

Given that every engineer acts perfectly, your task is to find out the numbers $$$c_i$$$ voiced by every engineer.

Input: The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 1000$$$; $$$1 \le k \le 10^6$$$) — the number of engineers in the company and the number of work units the project requires, respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the bonus which will be paid to the $$$i$$$-th engineer if the project is completed.

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 1000$$$), where $$$b_i$$$ is the work unit cost for the $$$i$$$-th engineer.

Output: Print $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$0 \le c_i \le k$$$) — the amount of work completed by each engineer given that every engineer behaves optimally. Note that the answer is unique.

Examples:
input:
3 6
4 7 6
1 2 3
output:
1 3 2

input:
3 12
4 7 6
1 2 3
output:
0 0 0

input:
3 11
6 7 8
1 2 3
output:
6 3 2

Note: In the first example, engineers distributed the work across them and got the bonus, even though the benefit for the third engineer is zero.

In the second example, the bonus project requires too many work units to complete, so it's more beneficial for engineers not to work at all.

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

public class Main {
    private static final FastIO sc = new FastIO();

    public static void main(String[] args) {
        //int t = sc.nextInt();
        int t=1;

        while (t-- > 0) {
            solve();
        }
    }

    static void solve() {
        int n = sc.nextInt();
        long k=sc.nextLong();
        long [][] a=new long [n][3];
        long [][] ans=new long [n][2];
        int i=0;
        for(i=0;i<n;i++)
        {
        a[i][0]=sc.nextLong();
        a[i][2]=i;
        }
        for(i=0;i<n;i++)
        a[i][1]=sc.nextLong();
        Arrays.sort(a,(x,y)->{
            if(x[0]%x[1]!=y[0]%y[1])
            {
                      return Long.compare(x[0]/x[1],y[0]/y[1]);
            }
            return Long.compare(y[0]%y[1],x[0]%x[1]);
        });

        for(i=0;i<n && k>0;i++)
        {
               ans[i][0]=Math.min(a[i][0]/a[i][1],k);
               ans[i][1]=a[i][2];
               k-=ans[i][0];

        }
        if(k>0)
        {
            for(i=0;i<n;i++)
            System.out.print("0 ");
            System.out.println();
            return;
        }
        Arrays.sort(ans,(x,y)->Long.compare(x[1],y[1]));
        for(i=0;i<n;i++)
        System.out.print(ans[i][0]+" ");
        System.out.println();
    }

    static class Pair
    {
        int x;
        int y;
        Pair(int x,int y)
        {
            this.x=x;
            this.y=y;
        }
        @Override
        public boolean equals(Object o) {
            // Check if the object is of the same type
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return (x == pair.x && y == pair.y) || (x == pair.y && y == pair.x);  // Ensure pairs are equal regardless of order
        }

        @Override
        public int hashCode() {
            return Objects.hash(Math.min(x, y), Math.max(x, y));  // Generate hash code based on ordered pair values
        }
    }

    static long modInverse(long x, long m) {
        return power(x, m - 2, m);
    }
    static long power(long x, long y, long m) {
        long res = 1;
        x = x % m;
        while (y > 0) {
            if ((y & 1) == 1) {
                res = (res * x) % m;
            }
            y = y >> 1;
            x = (x * x) % m;
        }
        return res;
    }

    static class FastIO {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the total work required is just slightly more or less than the sum of the maximum possible work each engineer can do.
2. **Uniform and Varied Work Values**: Test cases where the work values (both bonuses and costs) are either uniform or vary significantly to see how the sorting and distribution logic handles them.
3. **Zero Work Cases**: Test cases where some engineers might not do any work because it is not beneficial for them.

Given the constraints and the nature of the problem, a small-scale test input that can effectively expose the bug is:

### Test Input
```plaintext
4 10
5 10 15 20
1 2 3 4
```

### Explanation
- **Number of Engineers (n)**: 4
- **Total Work Units (k)**: 10
- **Bonuses (a)**: 5, 10, 15, 20
- **Work Unit Costs (b)**: 1, 2, 3, 4

### Expected Behavior
- The first engineer can complete up to 5 units of work (5/1 = 5).
- The second engineer can complete up to 5 units of work (10/2 = 5).
- The third engineer can complete up to 5 units of work (15/3 = 5).
- The fourth engineer can complete up to 5 units of work (20/4 = 5).

### Analysis
- The total maximum work that can be done by all engineers is 5 + 5 + 5 + 5 = 20, which is more than the required 10 units.
- The optimal distribution of work should be such that the project is completed with the maximum benefit to the engineers.

### Expected Output
- The expected output should be a distribution of work units that completes the project with the maximum benefit to the engineers. For example:
  ```plaintext
  2 2 2 4
  ```

### Why This Input is Challenging
- **Sorting Issue**: The sorting logic in the buggy code is incorrect. It sorts based on `a[i] / b[i]` and `a[i] % b[i]`, which might not correctly prioritize the engineers with the highest benefit.
- **Distribution Logic**: The distribution logic might not correctly allocate the work units in a way that maximizes the benefit for all engineers.

This test input should help in exposing the sorting and distribution issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bonus Project
time_limit_ms: 2000
memory_limit_mb: 512
Description: There is a team of $$$n$$$ software engineers numbered from $$$1$$$ to $$$n$$$. Their boss promises to give them a bonus if they complete an additional project. The project requires $$$k$$$ units of work in total. The bonus promised to the $$$i$$$-th engineer is $$$a_i$$$ burles. The boss doesn't assign specific tasks to engineers; it is expected that every engineer will voluntarily complete some integer amount of work units. The bonus will be paid to the entire team only if the project is completed; in other words, if the total amount of voluntary work units on the project is greater than or equal to $$$k$$$.

The amount of work that can be performed by each engineer is not limited. However, all engineers value their labour. The $$$i$$$-th engineer estimates one unit of their work as $$$b_i$$$ burles. If the bonus is paid, the benefit $$$s_i$$$ of the $$$i$$$-th engineer for completing $$$c$$$ units of work is defined as $$$s_i = a_i - c \cdot b_i$$$. If the bonus is not paid, the engineer will not volunteer to do any work.

Engineers work together for many years, so they know how the bonus is going to be distributed and how much their colleagues value the labour. That is, all $$$a_i$$$ and all $$$b_i$$$ are known to every engineer in the team.

Engineers are eager to get the bonus, so they agreed on the following process of work distribution between them:

- the first engineer says: "I will complete $$$c_1$$$ units of work", where $$$c_1$$$ is a non-negative integer;
- then, the second engineer says: "I will complete $$$c_2$$$ units of work", where $$$c_2$$$ is a non-negative integer;
- ... and so on;
- finally, the $$$n$$$-th engineer says: "I will complete $$$c_n$$$ units of work", where $$$c_n$$$ is a non-negative integer.

Every engineer voices $$$c_i$$$ in a way to maximize their own benefit $$$s_i$$$. If the expected benefit is going to be zero, an engineer will still agree to work to get the experience and to help their colleagues obtain the bonus. However, if the benefit is expected to be negative for some reason (an engineer needs to perform an excessive amount of work or the project is not going to be completed), that engineer will not work at all (completes zero amount of work units).

Given that every engineer acts perfectly, your task is to find out the numbers $$$c_i$$$ voiced by every engineer.

Input: The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 1000$$$; $$$1 \le k \le 10^6$$$) — the number of engineers in the company and the number of work units the project requires, respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the bonus which will be paid to the $$$i$$$-th engineer if the project is completed.

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 1000$$$), where $$$b_i$$$ is the work unit cost for the $$$i$$$-th engineer.

Output: Print $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$0 \le c_i \le k$$$) — the amount of work completed by each engineer given that every engineer behaves optimally. Note that the answer is unique.

Examples:
input:
3 6
4 7 6
1 2 3
output:
1 3 2

input:
3 12
4 7 6
1 2 3
output:
0 0 0

input:
3 11
6 7 8
1 2 3
output:
6 3 2

Note: In the first example, engineers distributed the work across them and got the bonus, even though the benefit for the third engineer is zero.

In the second example, the bonus project requires too many work units to complete, so it's more beneficial for engineers not to work at all.

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

using namespace std;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;

// #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>

#define f0r(a, b) for (long long a = 0; a < (b); ++a)
#define f1r(a, b, c) for (long long a = (b); a < (c); ++a)
#define f0rd(a, b) for (long long a = (b); a >= 0; --a)
#define f1rd(a, b, c) for (long long a = (b); a >= (c); --a)
#define ms(arr, v) memset(arr, v, sizeof(arr))
#define send {ios_base::sync_with_stdio(false);}
#define help {cin.tie(NULL); cout.tie(NULL);}
#define fix(prec) {cout << setprecision(prec) << fixed;}
#define mp make_pair
#define all(v) v.begin(), v.end()
#define getunique(v) {sort(all(v)); v.erase(unique(all(v)), v.end());}
#define readgraph(list, edges) for (int i = 0; i < edges; i++) {int a, b; cin >> a >> b; a--; b--; list[a].pb(b); list[b].pb(a);}
#define vsz(x) ((long long) x.size())
typedef long long ll;
typedef long double lld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << '}';}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

void solve(int tst) {
    // TODO: Add your code here
    int n,k;
    cin>>n>>k;
    vi ans(n);
    vi arr(n);
    vi brr(n);
    vector<pair<int, pair<int, int>>> a(n);
    for(int i=0;i<n;i++){
        cin>>arr[i];
        a[i].first = arr[i];
    }
    for(int i=0;i<n;i++){
        cin>>brr[i];
        a[i].second.first = brr[i];
        a[i].second.second = i;
    }
    sort(a.begin(), a.end(), [&](pair<int, pair<int, int>> x, pair<int, pair<int, int>> y){
        if(x.first < y.first) return true;
        else if(x.first == y.first) return x.second.first > y.second.first;
        else return false;

    });
    debug("HI");
    for(int i=n-1;i>=0;i--){
        ans[a[i].second.second] = min(k, a[i].first / a[i].second.first);
        k -= ans[a[i].second.second];
    }

    if(k == 0)
    for(int i=0;i<n;i++) cout << ans[i] << " ";
    else for(int i=0;i<n;i++) cout << 0 << " ";
    cout << '\n';
}

int main() {
    #ifndef ONLINE_JUDGE
    freopen("Error.txt", "w", stderr);
    #endif
    send help;
    int _t=1;
    // cin>>_t;
    for(int i=1;i<=_t;i++)
    solve(i);
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bonus Project
time_limit_ms: 2000
memory_limit_mb: 512
Description: There is a team of $$$n$$$ software engineers numbered from $$$1$$$ to $$$n$$$. Their boss promises to give them a bonus if they complete an additional project. The project requires $$$k$$$ units of work in total. The bonus promised to the $$$i$$$-th engineer is $$$a_i$$$ burles. The boss doesn't assign specific tasks to engineers; it is expected that every engineer will voluntarily complete some integer amount of work units. The bonus will be paid to the entire team only if the project is completed; in other words, if the total amount of voluntary work units on the project is greater than or equal to $$$k$$$.

The amount of work that can be performed by each engineer is not limited. However, all engineers value their labour. The $$$i$$$-th engineer estimates one unit of their work as $$$b_i$$$ burles. If the bonus is paid, the benefit $$$s_i$$$ of the $$$i$$$-th engineer for completing $$$c$$$ units of work is defined as $$$s_i = a_i - c \cdot b_i$$$. If the bonus is not paid, the engineer will not volunteer to do any work.

Engineers work together for many years, so they know how the bonus is going to be distributed and how much their colleagues value the labour. That is, all $$$a_i$$$ and all $$$b_i$$$ are known to every engineer in the team.

Engineers are eager to get the bonus, so they agreed on the following process of work distribution between them:

- the first engineer says: "I will complete $$$c_1$$$ units of work", where $$$c_1$$$ is a non-negative integer;
- then, the second engineer says: "I will complete $$$c_2$$$ units of work", where $$$c_2$$$ is a non-negative integer;
- ... and so on;
- finally, the $$$n$$$-th engineer says: "I will complete $$$c_n$$$ units of work", where $$$c_n$$$ is a non-negative integer.

Every engineer voices $$$c_i$$$ in a way to maximize their own benefit $$$s_i$$$. If the expected benefit is going to be zero, an engineer will still agree to work to get the experience and to help their colleagues obtain the bonus. However, if the benefit is expected to be negative for some reason (an engineer needs to perform an excessive amount of work or the project is not going to be completed), that engineer will not work at all (completes zero amount of work units).

Given that every engineer acts perfectly, your task is to find out the numbers $$$c_i$$$ voiced by every engineer.

Input: The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 1000$$$; $$$1 \le k \le 10^6$$$) — the number of engineers in the company and the number of work units the project requires, respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the bonus which will be paid to the $$$i$$$-th engineer if the project is completed.

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 1000$$$), where $$$b_i$$$ is the work unit cost for the $$$i$$$-th engineer.

Output: Print $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$0 \le c_i \le k$$$) — the amount of work completed by each engineer given that every engineer behaves optimally. Note that the answer is unique.

Examples:
input:
3 6
4 7 6
1 2 3
output:
1 3 2

input:
3 12
4 7 6
1 2 3
output:
0 0 0

input:
3 11
6 7 8
1 2 3
output:
6 3 2

Note: In the first example, engineers distributed the work across them and got the bonus, even though the benefit for the third engineer is zero.

In the second example, the bonus project requires too many work units to complete, so it's more beneficial for engineers not to work at all.

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

#define ll long long
#define ld long double
#define pii pair<int, int>
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second

const int INF = 1e9;
const ll INFLL = 1e18;
const int MOD = 1e9 + 7;

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

    // Declare any additional variables here

        ll n,k;
        cin>>n>>k;
        ll a[n];
        ll b[n],c[n];
        for(int i=0;i<n;i++){
            cin>>a[i];
        }
        for(int i=0;i<n;i++){
            cin>>b[i];
        }
        ll max[n];
        ll sum=0;
        for(int i=0;i<n;i++){
            max[i]=a[i]/b[i];
            sum+=max[i];
        }
        if(sum<k)
        {
            for(int i=0;i<n;i++){
                cout<<"0"<<" ";
            }
            cout<<endl;
        }
        else{
            sum-=k;
            for(int i=0;i<n;i++){
                if(max[i]>=sum)
                {
                    max[i]-=sum;
                    break;
                }
                else{
                    max[i]=0;
                    sum-=max[i];
                }
                if(sum<=0)
                break;
            }
            for(int i=0;i<n;i++)
            cout<<max[i]<<" ";
            cout<<endl;
        }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases where the algorithm might fail. The primary issue in the buggy code is the way it calculates and distributes the work units. Specifically, it incorrectly handles the distribution of work units when the total possible work units exceed the required work units \( k \).

### Steps to Generate the Test Input:
1. **Identify the Bug**: The bug is in the way the code distributes the work units. It subtracts the excess work units from the first engineer who can handle it, which might not be the optimal way to distribute the work.

2. **Edge Cases**:
   - **Case 1**: When the total possible work units are exactly equal to \( k \).
   - **Case 2**: When the total possible work units are just slightly more than \( k \).
   - **Case 3**: When the total possible work units are much more than \( k \), but the distribution should be balanced.

3. **Generate the Test Input**:
   - **Number of Engineers (n)**: Choose a small number of engineers, say 4.
   - **Required Work Units (k)**: Choose a value that is just slightly less than the total possible work units.
   - **Bonuses (a_i)**: Choose values that are reasonable and not too high.
   - **Work Unit Costs (b_i)**: Choose values that ensure the total possible work units are just slightly more than \( k \).

### Chosen Test Input:
- **n = 4**
- **k = 10**
- **a = [10, 10, 10, 10]**
- **b = [1, 2, 3, 4]**

### Explanation:
- **Total Possible Work Units**:
  - Engineer 1: \( \frac{10}{1} = 10 \)
  - Engineer 2: \( \frac{10}{2} = 5 \)
  - Engineer 3: \( \frac{10}{3} = 3 \)
  - Engineer 4: \( \frac{10}{4} = 2 \)
  - Total: \( 10 + 5 + 3 + 2 = 20 \)

- **Required Work Units (k) = 10**
- **Expected Output**: The work units should be distributed in a way that maximizes the benefit for each engineer. The optimal distribution would be:
  - Engineer 1: 5 units
  - Engineer 2: 3 units
  - Engineer 3: 1 unit
  - Engineer 4: 1 unit

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

This test input should expose the bug in the buggy code, as the code will incorrectly distribute the work units and may not achieve the optimal solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bonus Project
time_limit_ms: 2000
memory_limit_mb: 512
Description: There is a team of $$$n$$$ software engineers numbered from $$$1$$$ to $$$n$$$. Their boss promises to give them a bonus if they complete an additional project. The project requires $$$k$$$ units of work in total. The bonus promised to the $$$i$$$-th engineer is $$$a_i$$$ burles. The boss doesn't assign specific tasks to engineers; it is expected that every engineer will voluntarily complete some integer amount of work units. The bonus will be paid to the entire team only if the project is completed; in other words, if the total amount of voluntary work units on the project is greater than or equal to $$$k$$$.

The amount of work that can be performed by each engineer is not limited. However, all engineers value their labour. The $$$i$$$-th engineer estimates one unit of their work as $$$b_i$$$ burles. If the bonus is paid, the benefit $$$s_i$$$ of the $$$i$$$-th engineer for completing $$$c$$$ units of work is defined as $$$s_i = a_i - c \cdot b_i$$$. If the bonus is not paid, the engineer will not volunteer to do any work.

Engineers work together for many years, so they know how the bonus is going to be distributed and how much their colleagues value the labour. That is, all $$$a_i$$$ and all $$$b_i$$$ are known to every engineer in the team.

Engineers are eager to get the bonus, so they agreed on the following process of work distribution between them:

- the first engineer says: "I will complete $$$c_1$$$ units of work", where $$$c_1$$$ is a non-negative integer;
- then, the second engineer says: "I will complete $$$c_2$$$ units of work", where $$$c_2$$$ is a non-negative integer;
- ... and so on;
- finally, the $$$n$$$-th engineer says: "I will complete $$$c_n$$$ units of work", where $$$c_n$$$ is a non-negative integer.

Every engineer voices $$$c_i$$$ in a way to maximize their own benefit $$$s_i$$$. If the expected benefit is going to be zero, an engineer will still agree to work to get the experience and to help their colleagues obtain the bonus. However, if the benefit is expected to be negative for some reason (an engineer needs to perform an excessive amount of work or the project is not going to be completed), that engineer will not work at all (completes zero amount of work units).

Given that every engineer acts perfectly, your task is to find out the numbers $$$c_i$$$ voiced by every engineer.

Input: The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 1000$$$; $$$1 \le k \le 10^6$$$) — the number of engineers in the company and the number of work units the project requires, respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the bonus which will be paid to the $$$i$$$-th engineer if the project is completed.

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 1000$$$), where $$$b_i$$$ is the work unit cost for the $$$i$$$-th engineer.

Output: Print $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$0 \le c_i \le k$$$) — the amount of work completed by each engineer given that every engineer behaves optimally. Note that the answer is unique.

Examples:
input:
3 6
4 7 6
1 2 3
output:
1 3 2

input:
3 12
4 7 6
1 2 3
output:
0 0 0

input:
3 11
6 7 8
1 2 3
output:
6 3 2

Note: In the first example, engineers distributed the work across them and got the bonus, even though the benefit for the third engineer is zero.

In the second example, the bonus project requires too many work units to complete, so it's more beneficial for engineers not to work at all.

Buggy Code:
a = list(map(int,input().split()))
b = list(map(int,input().split()))
c = list(map(int,input().split()))

q = ((b[i] - c[i],i) for i in range((len(b))))
q = sorted(q, key=lambda x: (x[0], b[x[1]]), reverse=True)

ans = [0]*len(b)
for i in range(len(b)):
    if a[1] <= 0:
        break
    # print(min(a[1],b[q[i][1]]//c[q[i][1]]),"sdlnsdns")
    try:
        ans[q[i][1]] = min(a[1],b[q[i][1]]//c[q[i][1]])
    except:
        print(b[q[i][1]],c[q[i][1]])
    a[1] -= b[q[i][1]]//c[q[i][1]]
if a[1] > 0:
    print(*[0]*len(b))
else:
    print(*ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on the following aspects:

1. **Edge Cases**: The code might fail when the total required work units `k` are very close to the total possible work units that can be done by the engineers.
2. **Incorrect Indexing**: The code uses `a[1]` instead of `a[q[i][1]]`, which is a clear bug.
3. **Division by Zero**: Ensure that there are no cases where `b[q[i][1]] // c[q[i][1]]` results in a division by zero.
4. **Large Values**: Use large values for `a_i` and `b_i` to ensure the code handles large numbers correctly.

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

- **Number of Engineers**: 3
- **Work Units Required**: 10
- **Bonuses**: 10, 10, 10
- **Work Unit Costs**: 1, 1, 1

This input will ensure that the engineers have the same bonuses and work unit costs, making the distribution of work units critical to ensure the project is completed.

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

This test input will expose the bug in the code where `a[1]` is used instead of `a[q[i][1]]`, and it will also test the code's ability to handle equal values for bonuses and work unit costs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bonus Project
time_limit_ms: 2000
memory_limit_mb: 512
Description: There is a team of $$$n$$$ software engineers numbered from $$$1$$$ to $$$n$$$. Their boss promises to give them a bonus if they complete an additional project. The project requires $$$k$$$ units of work in total. The bonus promised to the $$$i$$$-th engineer is $$$a_i$$$ burles. The boss doesn't assign specific tasks to engineers; it is expected that every engineer will voluntarily complete some integer amount of work units. The bonus will be paid to the entire team only if the project is completed; in other words, if the total amount of voluntary work units on the project is greater than or equal to $$$k$$$.

The amount of work that can be performed by each engineer is not limited. However, all engineers value their labour. The $$$i$$$-th engineer estimates one unit of their work as $$$b_i$$$ burles. If the bonus is paid, the benefit $$$s_i$$$ of the $$$i$$$-th engineer for completing $$$c$$$ units of work is defined as $$$s_i = a_i - c \cdot b_i$$$. If the bonus is not paid, the engineer will not volunteer to do any work.

Engineers work together for many years, so they know how the bonus is going to be distributed and how much their colleagues value the labour. That is, all $$$a_i$$$ and all $$$b_i$$$ are known to every engineer in the team.

Engineers are eager to get the bonus, so they agreed on the following process of work distribution between them:

- the first engineer says: "I will complete $$$c_1$$$ units of work", where $$$c_1$$$ is a non-negative integer;
- then, the second engineer says: "I will complete $$$c_2$$$ units of work", where $$$c_2$$$ is a non-negative integer;
- ... and so on;
- finally, the $$$n$$$-th engineer says: "I will complete $$$c_n$$$ units of work", where $$$c_n$$$ is a non-negative integer.

Every engineer voices $$$c_i$$$ in a way to maximize their own benefit $$$s_i$$$. If the expected benefit is going to be zero, an engineer will still agree to work to get the experience and to help their colleagues obtain the bonus. However, if the benefit is expected to be negative for some reason (an engineer needs to perform an excessive amount of work or the project is not going to be completed), that engineer will not work at all (completes zero amount of work units).

Given that every engineer acts perfectly, your task is to find out the numbers $$$c_i$$$ voiced by every engineer.

Input: The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 1000$$$; $$$1 \le k \le 10^6$$$) — the number of engineers in the company and the number of work units the project requires, respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the bonus which will be paid to the $$$i$$$-th engineer if the project is completed.

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 1000$$$), where $$$b_i$$$ is the work unit cost for the $$$i$$$-th engineer.

Output: Print $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$0 \le c_i \le k$$$) — the amount of work completed by each engineer given that every engineer behaves optimally. Note that the answer is unique.

Examples:
input:
3 6
4 7 6
1 2 3
output:
1 3 2

input:
3 12
4 7 6
1 2 3
output:
0 0 0

input:
3 11
6 7 8
1 2 3
output:
6 3 2

Note: In the first example, engineers distributed the work across them and got the bonus, even though the benefit for the third engineer is zero.

In the second example, the bonus project requires too many work units to complete, so it's more beneficial for engineers not to work at all.

Buggy Code:
n,x = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
c = []
ind = -1
tot = 0
for i in range(n-1,-1,-1):
    tot+=a[i]//b[i]
    c.append(a[i]//b[i])
    if(tot>=x):
        ind = i
        break
if(ind==-1):
    for i in range(n):
        print(0,end = " ")
    print()
else:
    c = c[::-1]
    for i in range(ind):
        print(1,end = " ")
    k = ind+tot-x
    for i in range(ind,n):
        val = min(k,max(0,c[i-ind]-1))
        k-=val
        print(c[i-ind]-val,end = " ")
    print()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bonus Project
time_limit_ms: 2000
memory_limit_mb: 512
Description: There is a team of $$$n$$$ software engineers numbered from $$$1$$$ to $$$n$$$. Their boss promises to give them a bonus if they complete an additional project. The project requires $$$k$$$ units of work in total. The bonus promised to the $$$i$$$-th engineer is $$$a_i$$$ burles. The boss doesn't assign specific tasks to engineers; it is expected that every engineer will voluntarily complete some integer amount of work units. The bonus will be paid to the entire team only if the project is completed; in other words, if the total amount of voluntary work units on the project is greater than or equal to $$$k$$$.

The amount of work that can be performed by each engineer is not limited. However, all engineers value their labour. The $$$i$$$-th engineer estimates one unit of their work as $$$b_i$$$ burles. If the bonus is paid, the benefit $$$s_i$$$ of the $$$i$$$-th engineer for completing $$$c$$$ units of work is defined as $$$s_i = a_i - c \cdot b_i$$$. If the bonus is not paid, the engineer will not volunteer to do any work.

Engineers work together for many years, so they know how the bonus is going to be distributed and how much their colleagues value the labour. That is, all $$$a_i$$$ and all $$$b_i$$$ are known to every engineer in the team.

Engineers are eager to get the bonus, so they agreed on the following process of work distribution between them:

- the first engineer says: "I will complete $$$c_1$$$ units of work", where $$$c_1$$$ is a non-negative integer;
- then, the second engineer says: "I will complete $$$c_2$$$ units of work", where $$$c_2$$$ is a non-negative integer;
- ... and so on;
- finally, the $$$n$$$-th engineer says: "I will complete $$$c_n$$$ units of work", where $$$c_n$$$ is a non-negative integer.

Every engineer voices $$$c_i$$$ in a way to maximize their own benefit $$$s_i$$$. If the expected benefit is going to be zero, an engineer will still agree to work to get the experience and to help their colleagues obtain the bonus. However, if the benefit is expected to be negative for some reason (an engineer needs to perform an excessive amount of work or the project is not going to be completed), that engineer will not work at all (completes zero amount of work units).

Given that every engineer acts perfectly, your task is to find out the numbers $$$c_i$$$ voiced by every engineer.

Input: The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 1000$$$; $$$1 \le k \le 10^6$$$) — the number of engineers in the company and the number of work units the project requires, respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the bonus which will be paid to the $$$i$$$-th engineer if the project is completed.

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 1000$$$), where $$$b_i$$$ is the work unit cost for the $$$i$$$-th engineer.

Output: Print $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$0 \le c_i \le k$$$) — the amount of work completed by each engineer given that every engineer behaves optimally. Note that the answer is unique.

Examples:
input:
3 6
4 7 6
1 2 3
output:
1 3 2

input:
3 12
4 7 6
1 2 3
output:
0 0 0

input:
3 11
6 7 8
1 2 3
output:
6 3 2

Note: In the first example, engineers distributed the work across them and got the bonus, even though the benefit for the third engineer is zero.

In the second example, the bonus project requires too many work units to complete, so it's more beneficial for engineers not to work at all.

Buggy Code:
import java.util.*;
public class A_Bonus_Project{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] bonus = new int[n];
        int[] cost = new int[n];
        for(int i=0; i<n; i++) {
            bonus[i] = sc.nextInt();
        }
        for(int i=0; i<n; i++) {
            cost[i] = sc.nextInt();
        }
        long w = 0;
        for(int i=0; i<n; i++) {
            w += bonus[i]/cost[i];
        }
        if(w<k){
            for(int i=0; i<n; i++) {
                System.out.print("0 ");
            }
            return;
        }
        PriorityQueue<int[]> pq = new PriorityQueue<>((x, y) -> {
            long val1 = (long) x[0] * y[1], val2 = (long) y[0] * x[1];
            return Long.compare(val1, val2);
        });
        for(int i=0; i<n; i++) {
            pq.offer(new int[]{bonus[i], cost[i], i});
        }
        int[] c = new int[n];
        while(k>0 && !pq.isEmpty()){
            int[] curr = pq.poll();
            int maxWork = curr[0]/curr[1];
            int work = Math.min(k,maxWork);
            c[curr[2]] = work;
            k -= work;
        }
        for(int work : c) {
            System.out.print(work + " ");
        }
        sc.close();
    }
}

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

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

Think step by step.

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

1. **Insufficient Total Work**: The sum of the maximum work units each engineer can do (i.e., `bonus[i] / cost[i]`) is less than the required work `k`.
2. **Negative Benefit**: Some engineers have a high cost and low bonus, leading to a negative benefit if they work.
3. **Work Distribution**: The work distribution should be such that the engineers with the highest benefit per unit of work should be prioritized.

The provided code has a bug in how it calculates the total work units and the priority queue. It incorrectly sums the maximum work units and does not handle the case where the total work units are insufficient.

Let's create a test case where the total work units are just slightly less than the required work `k`, and the engineers have varying bonuses and costs to expose the bug.

```plaintext
4 19
10 15 20 25
2 3 4 5
```

### Explanation:
- **Number of Engineers (n)**: 4
- **Required Work Units (k)**: 19
- **Bonuses (a)**: 10, 15, 20, 25
- **Costs (b)**: 2, 3, 4, 5

### Expected Output:
- The total work units each engineer can do is:
  - 10 / 2 = 5
  - 15 / 3 = 5
  - 20 / 4 = 5
  - 25 / 5 = 5

- Sum of maximum work units: 5 + 5 + 5 + 5 = 20, which is greater than 19.

However, the code incorrectly sums the maximum work units and might not handle the distribution correctly. The correct distribution should be:
- Engineer 1: 5 units
- Engineer 2: 5 units
- Engineer 3: 5 units
- Engineer 4: 4 units

This test case should expose the bug in the code's logic for distributing the work units and handling the total work requirement.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bonus Project
time_limit_ms: 2000
memory_limit_mb: 512
Description: There is a team of $$$n$$$ software engineers numbered from $$$1$$$ to $$$n$$$. Their boss promises to give them a bonus if they complete an additional project. The project requires $$$k$$$ units of work in total. The bonus promised to the $$$i$$$-th engineer is $$$a_i$$$ burles. The boss doesn't assign specific tasks to engineers; it is expected that every engineer will voluntarily complete some integer amount of work units. The bonus will be paid to the entire team only if the project is completed; in other words, if the total amount of voluntary work units on the project is greater than or equal to $$$k$$$.

The amount of work that can be performed by each engineer is not limited. However, all engineers value their labour. The $$$i$$$-th engineer estimates one unit of their work as $$$b_i$$$ burles. If the bonus is paid, the benefit $$$s_i$$$ of the $$$i$$$-th engineer for completing $$$c$$$ units of work is defined as $$$s_i = a_i - c \cdot b_i$$$. If the bonus is not paid, the engineer will not volunteer to do any work.

Engineers work together for many years, so they know how the bonus is going to be distributed and how much their colleagues value the labour. That is, all $$$a_i$$$ and all $$$b_i$$$ are known to every engineer in the team.

Engineers are eager to get the bonus, so they agreed on the following process of work distribution between them:

- the first engineer says: "I will complete $$$c_1$$$ units of work", where $$$c_1$$$ is a non-negative integer;
- then, the second engineer says: "I will complete $$$c_2$$$ units of work", where $$$c_2$$$ is a non-negative integer;
- ... and so on;
- finally, the $$$n$$$-th engineer says: "I will complete $$$c_n$$$ units of work", where $$$c_n$$$ is a non-negative integer.

Every engineer voices $$$c_i$$$ in a way to maximize their own benefit $$$s_i$$$. If the expected benefit is going to be zero, an engineer will still agree to work to get the experience and to help their colleagues obtain the bonus. However, if the benefit is expected to be negative for some reason (an engineer needs to perform an excessive amount of work or the project is not going to be completed), that engineer will not work at all (completes zero amount of work units).

Given that every engineer acts perfectly, your task is to find out the numbers $$$c_i$$$ voiced by every engineer.

Input: The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 1000$$$; $$$1 \le k \le 10^6$$$) — the number of engineers in the company and the number of work units the project requires, respectively.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the bonus which will be paid to the $$$i$$$-th engineer if the project is completed.

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 1000$$$), where $$$b_i$$$ is the work unit cost for the $$$i$$$-th engineer.

Output: Print $$$n$$$ integers $$$c_1, c_2, \dots, c_n$$$ ($$$0 \le c_i \le k$$$) — the amount of work completed by each engineer given that every engineer behaves optimally. Note that the answer is unique.

Examples:
input:
3 6
4 7 6
1 2 3
output:
1 3 2

input:
3 12
4 7 6
1 2 3
output:
0 0 0

input:
3 11
6 7 8
1 2 3
output:
6 3 2

Note: In the first example, engineers distributed the work across them and got the bonus, even though the benefit for the third engineer is zero.

In the second example, the bonus project requires too many work units to complete, so it's more beneficial for engineers not to work at all.

Buggy Code:
import java.util.*;

public class WorkDistribution {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        // Input number of engineers and required work
        int n = scanner.nextInt();
        int k = scanner.nextInt();

        int[] a = new int[n]; // Bonuses
        int[] b = new int[n]; // Work unit costs
        int[] c = new int[n]; // Work contributions

        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
        }

        for (int i = 0; i < n; i++) {
            b[i] = scanner.nextInt();
        }

        // Priority queue based on benefit-to-cost ratio (descending order)
        PriorityQueue<Integer> pq = new PriorityQueue<>(
            (i, j) -> {
                long ratio1 = (long) a[i] * b[j]; // Cross-multiply to avoid precision issues
                long ratio2 = (long) a[j] * b[i];
                return Long.compare(ratio1, ratio2);
            }
        );

        for (int i = 0; i < n; i++) {
            pq.add(i);
        }

        int totalWork = 0;

        // Distribute work optimally
        while (!pq.isEmpty() && totalWork < k) {
            int index = pq.poll();
            int maxWork = a[index] / b[index];
            int work = Math.min(maxWork, k - totalWork);
            c[index] = work;
            totalWork += work;
        }

        // If total work is less than required, adjust dynamically
        if (totalWork < k) {
            for (int i = 0; i < n; i++) {
                c[i] = 0; // Reset contributions since project fails
            }
        }

        if(totalWork>=k){
        for (int i = 0; i < n; i++) {
            System.out.print(c[i] + " ");
        }
    }
    else{
        for (int i = 0; i < n; i++) {
            System.out.print(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 expose the bug in the provided code, we need to focus on scenarios where the priority queue logic might fail. Specifically, the priority queue is based on the benefit-to-cost ratio, and the code might not handle cases where the benefit-to-cost ratio is the same for multiple engineers correctly. Additionally, the code might not handle cases where the total work required is exactly equal to the sum of the maximum work each engineer can do without the benefit becoming negative.

Let's consider the following edge case:

- **Number of engineers (n):** 3
- **Required work units (k):** 10
- **Bonuses (a):** 10, 10, 10
- **Work unit costs (b):** 1, 1, 1

In this case, all engineers have the same benefit-to-cost ratio (10/1 = 10/1 = 10/1). The code should distribute the work equally among the engineers to complete the project. However, the priority queue might not handle this case correctly, leading to incorrect work distribution.

Here is the test input:

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

This input should result in each engineer contributing an equal amount of work (3, 3, 4 or any permutation that sums to 10). If the code fails to distribute the work correctly, it will expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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>
#define ll long long
#define forn(i,n) for(int i = 0; i < int(n); i++)
#define Code ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define vr(vec) vec.begin(),vec.end()
#define printy cout<<"YES"<<endl
#define printn cout<<"NO"<<endl
#define pi 3.141592653589793238
using namespace std;
void printarr(ll arr[], ll n){forn(i,n) cout << arr[i] << " ";cout <<endl;}
int M = 1e9 + 7;
int solve(int n, int k){
    int ans = 0;
    if(k<n)return 0;
    if(k==n)return 1;
    k-=n;
    ans++;
    for(int i = 1;i<=n-1;i++){
        if(k<=0)break;
        k-=n-i;
        ans++;
        if(k<=0)break;
        k-=n-i;
        ans++;
        if(k<=0)break;
    }
    // cout<<ans<<endl;
    return ans;
}
int main(){
    Code
    int t;
    cin>>t;

    while(t--){
        int n,k;
        cin>>n>>k;
        cout<<solve(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:
To expose the bug in the given code, we need to focus on edge cases and scenarios where the code might fail. The provided code has a logical error in the way it handles the number of occupied diagonals. Specifically, it incorrectly handles the case where `k` is less than `n`.

Let's break down the problem:

1. **Case where `k < n`**:
   - The code returns 0, which is incorrect. If `k < n`, the minimum number of occupied diagonals should be 1, as you can always place all `k` chips on a single diagonal.

2. **Case where `k == n`**:
   - The code returns 1, which is correct. All `n` chips can be placed on a single diagonal.

3. **Case where `k > n`**:
   - The code attempts to place chips on diagonals, but it does so in a way that might not be optimal. It subtracts `n - i` from `k` in a loop, which might not correctly handle the distribution of chips.

Given these points, a test input that can expose the bug is:

```plaintext
1
5 2
```

Explanation:
- For a 5x5 board with 2 chips, the minimum number of occupied diagonals should be 1, as both chips can be placed on the same diagonal (e.g., (1, 1) and (2, 2)).
- The buggy code returns 0 for this case, which is incorrect.

Thus, the test input is:

```plaintext
1
5 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:
/* بِسْمِ اللَّـهِ الرَّحْمَـٰنِ الرَّحِيم */
#include <bits/stdc++.h>

using namespace std;
#define int long long
#define ll long long
#define pp push_back
#define all(_x) _x.begin(), _x.end()
#define allr(_x) _x.rbegin(), _x.rend()
#define ULL unsigned ll
// freopen("input.txt","r",stdin);
// freopen("output.txt","w",stdout);
#define gg std::ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
//#define MOD (1000000007)
#define Sum(x)   ((x*(x+1LL))/2LL)
#define GET_BIT(x, pos) ((x>>pos)&1)

const ll MOD = 1e9 + 7;

ll f_pow(int a, int b) {
    if (b == 0) return 1;
    if (b == 1) return a;

    int x = f_pow(a, b >> 1);

    if (b & 1) return (((x%MOD) * (x%MOD)) %MOD) * (a%MOD) % MOD;
    else return ((x%MOD) * (x%MOD)) %MOD;
}

void solve() {
    /*
    int n, k; cin>>n>>k;
    int arr[n];
    int mx =0;
    for(int i=0;i<n;i++){
        cin>>arr[i];
        mx = max(mx , arr[i]);
    }

    int l =0, r=mx +k, ans =0;
    while(l<=r)
    {
        int md = (l+r) >>1;
        int target = md;
        for(int i=0;i<n;i++){
            int cost = 0;
            for(int j=i;j<n;j++){
                if(target > arr[j]){
                    cost+=(target - arr[j]);
                    target--;
                }
                else break;
            }
            cout<<md <<' '<< cost <<' '<< i<<'\n';

            if(cost <= k){
                ans = max(ans, md);
                l = md + 1;
            }
            else {
                r = md -1;
            }
        }
    }
    cout<<ans<<'\n';
    */

    int n, k; cin>>n>>k;
    if(k<n) {
        cout<<0<<'\n'; return;
    }
    if(k == n){
        cout<<1<<'\n'; return;
    }

    int c=1, ans =1;
    k-=n;
    n--;
    while(k>0){
        if(n==0 || k<n){
            break;
        }

        for(int j=0;j<2;j++){
            if(k>=n){
                ans++;
                k-=n;
            }
            else {
                break;
            }
        }

        n--;
    }
    if(k && n>0) ans++;
    cout<<ans<<'\n';

}

signed main() {
    gg
    int t;
    t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

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


**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 namespace std;

#define ll long long
#define pii pair<ll, ll>
#define pli pair<ll, pii>
#define vi vector<int>
#define vvi vector<vi>
#define vll vector<ll>
#define vvll vector<vll>

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

    int t;
    cin >> t;

    while (t--) {
        ll n, x, c = 0;
        cin >> n >> x;
        if(x>1){
            if(x>=n*n){
                cout<<2*(n-1)+1<<endl;;
            }else{
            x-=n;c++;ll k=n--;
            while(x>0){
                if(x>0){
                    x-=k;
                    c++;
                }
                if(x>0){
                    x-=k;
                    c++;
                }
                k--;
            }
            cout<<c<<endl;
            }
        }else{
            if(x>0){
                cout<<1<<endl;
            }else{
                cout<<0<<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 1
3 2
3 4
10 99
```

### Explanation:
1. **Case 1: `1 1`**
   - A single cell board with one chip. Only one diagonal, so the answer is 1.
2. **Case 2: `2 1`**
   - A 2x2 board with one chip. Only one diagonal, so the answer is 1.
3. **Case 3: `3 2`**
   - A 3x3 board with two chips. Both chips can be placed on the same diagonal, so the answer is 1.
4. **Case 4: `3 4`**
   - A 3x3 board with four chips. The maximum number of chips that can be placed on a single diagonal is 3. Therefore, the minimum number of occupied diagonals is 2.
5. **Case 5: `10 99`**
   - A 10x10 board with 99 chips. The maximum number of chips that can be placed on a single diagonal is 10. Placing 99 chips will require at least 10 diagonals, but the code might not handle this correctly.

These cases are designed to test the edge conditions and the logic for handling different board sizes and chip counts.


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

    sum, res = n, 1
    for i in reversed(range(1, n)):
        if (sum + i * 2 <= k):
            res += 2
            sum += i * 2
        else:
            if (k - sum <= i):
                res += 1
            else:
                res += 2
            break

    if (k == 0):
        print(0)
    elif (n == k):
        print(1)
    else:
        print(res)

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

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

Think step by step.

Response:
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. Let's analyze the code and the problem:

1. **Understanding the Buggy Code:**
   - The code initializes `sum` to `n` and `res` to `1`.
   - It iterates over `i` in reversed order from `1` to `n-1`.
   - If `sum + i * 2 <= k`, it adds `2` to `res` and updates `sum`.
   - If `k - sum <= i`, it adds `1` to `res`.
   - Otherwise, it adds `2` to `res` and breaks the loop.
   - Special cases for `k == 0` and `n == k` are handled separately.

2. **Identifying Edge Cases:**
   - **Empty Board (k == 0):** This is already handled correctly.
   - **Full Board (n == k):** This is also handled correctly.
   - **Single Chip (k == 1):** This should result in 1 diagonal.
   - **Multiple Chips on the Same Diagonal:** This should result in 1 diagonal.
   - **Chips on Different Diagonals:** This should result in the minimum number of diagonals.
   - **Chips just enough to fill the board:** This should result in the number of diagonals equal to `n`.

3. **Challenging Scenario:**
   - **Chips just less than the maximum possible on the same diagonal:** This should test the logic for adding chips to the same diagonal.
   - **Chips that need to be split across multiple diagonals:** This should test the logic for breaking the loop and adding the correct number of diagonals.

4. **Test Input:**
   - **Test Case 1:** `n = 2, k = 1` (Single chip)
   - **Test Case 2:** `n = 3, k = 2` (Two chips on the same diagonal)
   - **Test Case 3:** `n = 3, k = 3` (Three chips on different diagonals)
   - **Test Case 4:** `n = 4, k = 7` (Chips just less than the maximum possible on the same diagonal)
   - **Test Case 5:** `n = 4, k = 8` (Chips that need to be split across multiple diagonals)

Here is the test input:

```plaintext
5
2 1
3 2
3 3
4 7
4 8
```

This input includes a variety of cases that should challenge the buggy code and expose its issues.


**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:
#I know exactly where you are!!
import math
#list(map(int,input().split()))
for _ in range(int(input())):
    n, k = map(int,input().split())
    cnt = 0
    k -= n
    if k < 0:
        print(0)
        continue
    else:
        cnt += 1
    for i in range(1, n):
        if k <= 0:
            break
        k -= (n - i)
        cnt += 1
        if k <= 0:
            break
        else:
            k -= (n - i)
            cnt += 1
        if k <= 0:
            break
    print(cnt)

**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 4
3 9
```


**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:
gg=int(input())

for r in range(0,gg):
  n, k = input().split()
  n = int(n)
  k = int(k)
  s = 0
  t = 0
  if k > 0:
    t += n
    s += 1
    n -= 1
    if t != k:
        while 2*n + t < k and n> 0:
          t += 2*n
          s += 2
          n -= 1

        if n + t < k:
            t += 2*n
            s += 2
            n -= 1
        else:
            if n>0:
              t += n
              s += 1
              n -= 1

    print(s)
  else:
    print(s)

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

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

Think step by step.

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


**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.*;
import java.lang.*;
import java.io.*;
/**
 * A_Diagonals
 */
public class A_Diagonals {

    public static void main(String[] args) throws java.lang.Exception {
        FastWriter out = new FastWriter();
        int t = in.nextInt();

        while (t != 0) {
            t--;
            int n=in.nextInt();
            int k=in.nextInt();
            if(k == 0){
                out.println(0);
            }else{
                int ans=0;
            if(k >= n){
                k-=n;
            }
            for(int i=n-1;i>=1;i--){

                if((k-(2*i)) >= 0){
                    ans+=2;
                    k-=2*i;
                }else if(k - i >= 0){
                    k-=i;
                    ans+=1;
                }

            }
            out.println(ans+1);
            }

        }
        out.close();
    }

    // GCD Method
    static long gcd(long a, long b) {
        while (b != 0) {
            long t = b;
            b = a % b;
            a = t;
        }
        return a;
    }

    // LCM Method
    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    // Power Calculation Method
    static long pow(long base, int exp) {
        long result = 1;
        while (exp > 0) {
            if (exp % 2 == 1) {
                result *= base;
            }
            base *= base;
            exp /= 2;
        }
        return result;
    }

    // Sieve of Eratosthenes Method
    static boolean[] sieve(int n) {
        boolean[] isPrime = new boolean[n + 1];
        Arrays.fill(isPrime, true);
        isPrime[0] = isPrime[1] = false;
        for (int i = 2; i * i <= n; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= n; j += i) {
                    isPrime[j] = false;
                }
            }
        }
        return isPrime;
    }

    // Prime Factorization Method using Smallest Prime Factor (SPF)
    public static List<Long> primeFactorization(long number) {
        int max = (int)Math.sqrt(number) + 1;
        long[] spf = new long[max];
        Arrays.fill(spf, -1);
        for (int i = 2; i < max; i++) {
            if (spf[i] == -1) {
                for (int j = i; j < max; j += i) {
                    if (spf[j] == -1) {
                        spf[j] = i;
                    }
                }
            }
        }

        List<Long> primeFactors = new ArrayList<>();
        while (number > 1) {
            if (number < max && spf[(int)number] != -1) {
                primeFactors.add(spf[(int)number]);
                number /= spf[(int)number];
            } else {
                boolean found = false;
                for (int i = 2; i < max; i++) {
                    if (number % i == 0) {
                        primeFactors.add((long)i);
                        number /= i;
                        found = true;
                        break;
                    }
                }
                if (!found) {
                    primeFactors.add(number);
                    break;
                }
            }
        }

        return primeFactors;
    }

    static FastReader in = new FastReader();
    // input of int array
    static int[] ai(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) arr[i] = in.nextInt();
        return arr;
    }

    // input of long array
    static long[] al(int n) {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) arr[i] = in.nextLong();
        return arr;
    }

    // get array max
    static int aMax(int[] arr) {
        int a = Integer.MIN_VALUE;
        for (int i = 0; i < arr.length; i++) a = Math.max(a, arr[i]);
        return a;
    }

    // get array min
    static int aMin(int[] arr) {
        int a = Integer.MAX_VALUE;
        for (int i = 0; i < arr.length; i++) a = Math.min(a, arr[i]);
        return a;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static class FastWriter {
        private final BufferedWriter bw;

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

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

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

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

    static class Pair {
        long f;
        long s;
        public Pair(long first, long second) {
            this.f = first;
            this.s = second;
        }
    }

    static class Triplet {
        long f;
        long s;
        long t;
        public Triplet(long first, long second, long third) {
            this.f = first;
            this.s = second;
            this.t = third;
        }
    }
}

**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 maximize the failure probability. Let's break down the problem and the code to identify potential issues:

### Problem Analysis:
1. **Board Size (n) and Number of Chips (k):**
   - The board size `n` can range from 1 to 100.
   - The number of chips `k` can range from 0 to \( n^2 \).
2. **Diagonal Placement:**
   - The goal is to minimize the number of occupied diagonals.
   - Diagonals are defined by the sum \( i + j \).

### Buggy Code Analysis:
1. **Initial Check for k = 0:**
   - The code correctly handles the case where \( k = 0 \) by printing 0.
2. **Main Logic:**
   - The code attempts to place chips on diagonals in a way that minimizes the number of occupied diagonals.
   - It first checks if \( k \geq n \) and reduces \( k \) by \( n \).
   - It then iterates over the diagonals from \( n-1 \) to 1, reducing \( k \) by \( 2 \times i \) or \( i \) and incrementing the answer accordingly.
   - Finally, it prints \( ans + 1 \).

### Potential Bugs:
1. **Incorrect Reduction of \( k \):**
   - The reduction of \( k \) by \( 2 \times i \) or \( i \) might not always be correct.
   - The logic for placing chips on diagonals might be flawed, leading to an incorrect count of occupied diagonals.
2. **Edge Cases:**
   - Small values of \( n \) and \( k \) where the number of chips is close to \( n \).
   - Large values of \( n \) and \( k \) where the number of chips is much larger than \( n \).

### Challenging Test Input:
To expose the bug, we should create a test case where the number of chips is close to the number of cells in the board, and the board size is small to make it easier to verify the expected output.

### Test Case:
```plaintext
1
3 8
```

### Explanation:
- **Board Size (n):** 3
- **Number of Chips (k):** 8
- **Expected Output:** 5

### Reasoning:
- For a 3x3 board, there are 9 cells.
- Placing 8 chips on the board means one cell will be empty.
- The minimum number of occupied diagonals is 5, as shown below:
  - Diagonal 2: (1,1)
  - Diagonal 3: (1,2), (2,1)
  - Diagonal 4: (1,3), (2,2), (3,1)
  - Diagonal 5: (2,3), (3,2)
  - Diagonal 6: (3,3)

This test case should expose any issues in the logic for placing chips and counting occupied 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:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class CF_961_2_1 {
    private static InputReader in;
    private static PrintWriter out;

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        in = new InputReader(inputStream);
        OutputStream outputStream = System.out;
        out = new PrintWriter(outputStream);

        //checkered board with side of n and k chips
        //realized that k chips need to be placed on the cells of the board (max. 1 chip on 1 cell)
        //cell(i,j)

        //goals: min possible num of occupied diagonals among all placement of the k chips

        int t = in.nextInt();

        for (int i=0;i<t;i++){
            int n=in.nextInt();//side of checkered board
            int k=in.nextInt();//number of available chips

            out.println(f(n,k));

        }
        out.close();
    }

    public static int f(int n,int k){
        int i=0;
        boolean isTwice=false;
        if (k-n>=0){
            i++;
            k-=n;
            n--;
            while(n>=1){
                if (k-n==0){
                    i++;
                    break;
                } else if (k-n>0){
                    i++;
                    k-=n;
                    if (isTwice){
                        n--;
                        isTwice=false;
                    } else {
                        isTwice=true;
                    }
                } else{
                    if (k!=0){
                        i++;
                    }
                    break;
                }

            }
        }
        return i;
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

**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
10 10
```


**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 namespace std;
#define ll long long
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define pb(x) push_back(x)
#define all(x) x.begin(), x.end()
#define ifz if(z!=0)continue;
#define fl(x,y) for(auto x: y)
#define nln cout<<endl
#define ff first
#define ss second
#define MIN(v) *min_element(all(v))
#define MAX(v) *max_element(all(v))
#define show(a) {rep(var,0,a.size())cout<<a[var]<<" ";nln;}
#define shar(a) {rep(var,0,n)cout<<a[var]<<" ";nln;}
#define showset(s) {fl(val,s)cout<<val<<" ";nln;}
#define showmap(m) {fl(val,m)cout<<val.ff<<" "<<val.ss<<endl;nln;}
#define lb(c, x) lower_bound(all(x),c)-x.begin()
#define ub(c, x) upper_bound(all(x),c)-x.begin()
#define acc(x) accumulate(all(x),(ll)0)
int main(){ll tc; cin>>tc;
while(tc--){
ll n,x,k; cin>>n>>k;
if(k==n*n){cout<<2*n-1<<endl; continue; }

ll cnt=0;
if(k!=0){ k-=n; cnt++;}
ll c=1,p=n-1;
while(k>0 && p>0){
if(c&1){k-=p; cnt++; }
else { k-=p; cnt++; p--;   }
}

cout<<cnt<<endl;

}}

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

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

Think step by step.

Response:
```plaintext
3
10 100
10 99
10 98
```


**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 namespace std;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        int k;
        cin >> k;
        int l = n;
        int ans = 0;
        int count = 1;
        int res=1;
        while (k > 0)
        {
            while (count--)
            {
                if (k <= l)
                {
                    ans++;
                    k=0;
                    break;
                }
                k-=l;
                ans++;
            }
            count=res+1;
            res++;
            l--;
        }
        cout<<ans<<endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The buggy code seems to have issues with how it handles the placement of chips and the calculation of the number of occupied diagonals. Specifically, it might not correctly handle cases where the number of chips is large relative to the board size, or where the board size is small.

Let's consider the following scenarios:

1. **Small Board with Many Chips**: This can expose issues with how the code handles large numbers of chips on a small board.
2. **Large Board with Few Chips**: This can test how the code handles minimal chip placement.
3. **Equal Number of Chips and Board Size**: This can test the code's handling of a fully occupied board.
4. **Zero Chips**: This is a basic edge case to ensure the code handles it correctly.

Given these considerations, let's create a test input that includes these scenarios:

```plaintext
4
1 1
2 4
10 100
100 0
```

### Explanation:
1. **1 1**: A 1x1 board with 1 chip. The only diagonal is occupied.
2. **2 4**: A 2x2 board with 4 chips. All diagonals are occupied.
3. **10 100**: A 10x10 board with 100 chips. All diagonals are occupied.
4. **100 0**: A 100x100 board with 0 chips. No diagonals are occupied.

This test input should help expose any issues in the buggy code by covering a range of board sizes and chip counts.


**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 sol(n, k):
    if k<n:
        return 0

    ans = 1
    k-=n
    sub = n-1

    while k>0:
        ans+=1
        k-=sub
        if k<=0:
            break
        ans+=1
        k-=sub
        sub-=1

    return ans

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

    print(sol(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
3
10 100
10 99
10 10
```