**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make It Ugly
time_limit_ms: 2000
memory_limit_mb: 256
Description: Let's call an array $$$a$$$ beautiful if you can make all its elements the same by using the following operation an arbitrary number of times (possibly, zero):

- choose an index $$$i$$$ ($$$2 \le i \le |a| - 1$$$) such that $$$a_{i - 1} = a_{i + 1}$$$, and replace $$$a_i$$$ with $$$a_{i - 1}$$$.

You are given a beautiful array $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of elements you have to remove from it in order for it to stop being beautiful? Swapping elements is prohibited. If it is impossible to do so, then output -1.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 3 \cdot 10^5$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le n$$$).

Additional constraints on the input:

- in every test case, the given array $$$a$$$ is beautiful;
- the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer — the minimum number of elements you have to remove from the array $$$a$$$ in order for it to stop being beautiful. If it is impossible, then output -1.

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

Note: In the first testcase, it is impossible to modify the array in such a way that it stops being beautiful. An array consisting of identical numbers will remain beautiful no matter how many numbers we remove from it.

In the second testcase, you can remove the number at the index $$$5$$$, for example.

The resulting array will be $$$[1, 2, 1, 2]$$$. Let's check if it is beautiful. Two operations are available:

- Choose $$$i = 2$$$: the array becomes $$$[1, 1, 1, 2]$$$. No more operations can be applied to it, and the numbers are not all the same.
- Choose $$$i = 3$$$ instead: the array becomes $$$[1, 2, 2, 2]$$$. No more operations can be applied to it either, and the numbers are still not all the same.

Thus, the array $$$[1, 2, 1, 2]$$$ is not beautiful.

In the fourth testcase, you can remove the first three elements, for example. The resulting array $$$[5, 3, 3, 3]$$$ is not beautiful.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.*;

public class Codeforces {
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int t = Integer.parseInt(br.readLine());
        while(--t >= 0) {
            br.readLine();

            int[] a = Arrays.stream(br.readLine().split(" "))
                    .mapToInt(Integer::valueOf).toArray();

            boolean allSame = true;
            for(int i = 1; i < a.length; ++i) {
                if(a[i] != a[i - 1]) {
                    allSame = false;
                    break;
                }
            }
            if(allSame) {
                System.out.println(-1);
                continue;
            }
            int i = 0;
            for(; i + 1 < a.length; ++i) {
                if(a[i] == a[i + 1]) continue;
                break;
            }

            int j = a.length - 1;
            for(; j > 0; --j) {
                if(a[j] == a[j - 1]) continue;
                break;
            }
            j = a.length - j;
            System.out.println(i < j ? i + 1 : j);
        }
    }
}

**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 the logic that determines whether the array can be made non-beautiful by removing elements. The code seems to have issues with handling arrays where all elements are the same and with arrays that have a specific structure that makes it difficult to break the "beautiful" property.

Here's a step-by-step analysis to generate a test input:

1. **All Elements the Same:**
   - The code correctly identifies that an array with all elements the same cannot be made non-beautiful by removing elements. This is already handled correctly.

2. **Edge Cases with Repeated Elements:**
   - The code uses a simple loop to find the first and last positions where the elements are not the same. This can fail in cases where the array has a structure that makes it difficult to break the "beautiful" property by removing a small number of elements.

3. **Specific Structure:**
   - Consider an array where the elements are repeated in a way that the middle part is different, but the ends are the same. For example, `[1, 1, 2, 2, 1, 1]`. This array is beautiful, and removing a single element might not make it non-beautiful.

4. **Test Case:**
   - We need an array that is beautiful and requires more than one removal to make it non-beautiful. The array `[1, 1, 2, 2, 1, 1]` is a good candidate because removing a single element from the middle part will not break the "beautiful" property, but removing two elements will.

Here is the test input:

```plaintext
1
6
1 1 2 2 1 1
```

This test input will help expose the bug in the provided code because the current logic might not correctly determine the minimum number of elements to remove to make the array non-beautiful.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

using ll = long long;
using pi = pair<int, int>;
using pll = pair<ll, ll>;
using vc = vector<char>;
using vi = vector<int>;
using vb = vector<bool>;
using vll = vector<ll>;

template <typename T>
using vec = vector<T>;

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template<typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

constexpr ll MOD = 998244353ll;

template<typename T>
void in(T &t)
{
    cin >> t;
}

template<typename T, typename... Args>
void in(T &t, Args& ...args)
{
    cin >> t;
    in(args...);
}

template<typename T>
void in(vector<T> &v)
{
    for (auto &e : v)
        cin >> e;
}

template<typename T>
void out(vector<T> &v)
{
    for (auto &e : v)
        cout << e << ' ';
    cout << '\n';
}

ll pow_mod(ll a, ll b, ll m)
{
    a %= m;
    ll res = 1ll;

    while (b > 0)
    {
        if (b & 1ll)
            res = (res*a) % m;

        a = (a*a) % m;
        b >>= 1ll;
    }

    return res;
}

class DSU
{
public:
    vll p;
    vll r;

    DSU(ll n)
        : p(n), r(n, 0)
    {
        iota(p.begin(), p.end(), 0ll);
    }

    ll parent(ll i)
    {
        if (i == p[i])
            return i;
        return p[i] = parent(p[i]);
    }

    void unite(ll i, ll j)
    {
        ll a = parent(i);
        ll b = parent(j);

        if (a == b)
            return;

        if (r[a] < r[b])
        {
            r[b] += r[a];
            p[a] = b;
        }
        else
        {
            r[a] += r[b];
            p[b] = a;
        }
    }
};

ll f(ll n, ll d, ll s, ll i)
{
    ll mn = min(n/(i+1), d/i);
    ll t = mn/s;

    //if (s*t*(i+1) <= n)
        return s*t*(i+1);
   // else
    //    return s*t;
}

void solve()
{
    ll n, d, s;
    in(n, d, s);

    n /= s;
    d /= s;

    ll y = min(2*d, n);

    if (y < 1)
    {
        cout << s << '\n';
        return;
    }

    if (y % 2 == 0)
    {
        cout << y*s << '\n';
        return;
    }

    for (ll x = 2; x*x <= y; ++x)
    {
        if (y % x == 0 && y - y/x <= d)
        {
            cout << y*s << '\n';
            return;
        }
    }

    cout << (y-1)*s << '\n';

}

int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    ll tc = 1;
    // cin >> tc;

    // cout << fixed << setprecision(12);

    while(tc--)
        solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios where the code might fail. Here are some key points to consider:

1. **Edge Cases:**
   - When `S` is already the highest floor.
   - When `D` is very small, limiting the possible moves.
   - When `N` is very large, but `S` and `D` are small.

2. **Scenario Analysis:**
   - If `S` is a prime number, it might be difficult to find multiples within the constraints.
   - If `D` is small, the number of possible moves might be limited.
   - If `N` is large, the code might have to handle large numbers efficiently.

3. **Maximizing Failure Probability:**
   - Choose values for `N`, `D`, and `S` that force the code to perform multiple checks and calculations.

Given these considerations, let's create a test input that is small-scale but challenging:

- **N**: 100
- **D**: 10
- **S**: 7

This input is small enough to be manageable but challenging enough to test the code's ability to handle multiple moves and edge cases.

```plaintext
100 10 7
```

Explanation:
- **N = 100**: The hotel has 100 floors.
- **D = 10**: The maximum distance the elevator can move in one ride.
- **S = 7**: Starting on the 7th floor.

This input will force the code to check multiple possible moves and ensure it correctly identifies the highest reachable floor.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

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

using namespace std;

typedef long long ll;

ll numero_per_abaix(ll & n, ll & m){
	ll k = n/m;
	return k * m;
}

int main(){
	ll n, d, s;
	cin >> n >> d >> s;
	//cout << "n: " << n << " | d: " << d << " | s: " << s << endl;
	if (s > d or 2*s > n) cout << s;
	if (s+d <= n and d%s == 0) cout << d+s;
	else{
		ll max = 0;
		ll sqrtn = sqrt(n);
		for (ll i = 2; i <= sqrtn+1; ++i){
			ll  k = min(d/(i-1), n/i);
			ll div = i*(k/s);
			//cout << "div de " << i << ": " << div << endl;
			if (div == 0) break;
			if (div > max) max = div;
		}
		cout << s * max;
	}
	cout << endl;
}
/*
999999999989 999999999988 1
*/

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

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

Think step by step.

Response:
```plaintext
100 99 1
```

### Explanation:
1. **Edge Case with Large D and Small N**:
   - The input `100 99 1` is designed to test the scenario where `D` is very close to `N` and `S` is the smallest possible value (1).
   - In this case, the elevator can potentially reach any floor from 1 to 100 in a single ride, but the algorithm must correctly handle this edge case.

2. **Testing the Logic for Multiples**:
   - Since `D` is 99, the elevator can move from floor 1 to any floor from 2 to 100 in a single ride, as all these floors are multiples of 1 and the difference is within the limit of `D`.
   - The expected output is 100, as the highest floor that can be reached is the top floor of the building.

3. **Validating the Buggy Code**:
   - The buggy code should correctly handle this case by outputting 100. However, if the code has any logic errors in handling the multiples or the maximum floor calculation, it might fail to produce the correct output.

This test input is small and valid, focusing on an edge case that maximizes the failure probability in faulty solutions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
// जय श्री राम

#pragma GCC optimize("O3,unroll-loops")
#include<bits/stdc++.h>
#ifndef ONLINE_JUDGE
#include<algo/debug.cpp>
#else
#define debug(...)
#define debugArr(...)
#endif
using namespace std;

#define int long long

void solve() {
    int n, d, s;
    cin >> n >> d >> s;
    n/=s;
    d/=s;
    int y = min(2*d, n);
    for(int i=2;i<=sqrt(y);i++) {
        if(y%i == 0 && (y - y/i) <= d) {
            cout << y*s << "\n";
            return;
        }
    }
    cout << (y-1)*s << "\n";
}
#undef int

int main() {
    #ifdef case3_local
        auto begin = std::chrono::high_resolution_clock::now();
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    // cin >> t;
    while(t--){
        solve();
    }
    #ifdef case3_local
        auto end = std::chrono::high_resolution_clock::now();
        cerr << setprecision(4) << fixed;
        cerr << "Execution time: " << std::chrono::duration_cast<std::chrono::duration<double>>(end - begin).count() << " seconds" << endl;
    #endif
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, let's analyze the logic and identify potential edge cases or scenarios that can cause the algorithm to fail.

### Analysis of the Buggy Code:
1. **Input Normalization**: The code normalizes the input by dividing `n` and `d` by `s`. This is a key step to reduce the problem size, but it can lead to issues if not handled correctly.
2. **Finding the Maximum Reachable Floor**: The code calculates `y` as `min(2*d, n)`. It then iterates through possible divisors `i` of `y` and checks if the difference between `y` and `y/i` is less than or equal to `d`. If it finds such a divisor, it outputs `y * s` and returns. If no such divisor is found, it outputs `(y-1) * s`.

### Potential Issues:
1. **Edge Case with `y` and `d`**: The code might not handle cases where `y` is very large or where `d` is very small correctly.
2. **Divisor Calculation**: The code only checks divisors up to `sqrt(y)`, which might miss some valid solutions.
3. **Output Calculation**: The code outputs `(y-1) * s` if no valid divisor is found, which might not always be the correct answer.

### Challenging Test Input:
To expose the bug, we need an input where the code's logic fails to find the correct maximum reachable floor. Let's consider a scenario where `d` is very small compared to `n` and `s` is a large number.

### Example Input:
```plaintext
1000000 1 500000
```

### Explanation:
- **`n = 1000000`**: The total number of floors.
- **`d = 1`**: The maximum allowed difference between the current floor and the target floor.
- **`s = 500000`**: The starting floor.

### Expected Output:
- The highest floor you can reach is `500000` because the only valid move from `500000` is to stay at the same floor (since `d = 1` and the next multiple of `500000` is `1000000`, which is too far).

### Why This Input is Challenging:
- The code normalizes `n` and `d` by dividing by `s`, resulting in `n = 2` and `d = 0`. The loop in the code will not find any valid divisor because `y` will be `1` (since `min(2*0, 2) = 1`).
- The code will output `(1-1) * 500000 = 0`, which is incorrect.

Thus, the input:
```plaintext
1000000 1 500000
```
is a challenging test case that should expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
n,d,s = map(int,input().split())
if (d//2) < s:
    print(s)
else:
    maxVal = (n // s) * s
    if (maxVal - s) <= d:
        print(maxVal)
    else:
        targ = (d // s) * 2 * s
        while targ > n:
            targ -= 2 * s
print(targ)

**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 the boundaries of the input constraints.
2. **Special Conditions**: Ensure the test input triggers specific conditions in the code.
3. **Maximizing Failure Probability**: Focus on inputs that are likely to cause the code to fail.

### Analysis of the Buggy Code
The buggy code has the following issues:
- It incorrectly handles the condition where `d` is less than `s`.
- It does not correctly handle the case where `targ` exceeds `n` and needs to be adjusted.
- The logic for calculating `targ` is flawed.

### Challenging Test Input
To expose these issues, we can use the following test input:

```plaintext
100 50 25
```

### Explanation
- **N = 100**: A reasonable upper limit for the number of floors.
- **D = 50**: A value that is large enough to allow multiple steps but not so large that it trivializes the problem.
- **S = 25**: A starting floor that is a significant fraction of `N` and `D`.

### Why This Input is Challenging
1. **Starting Floor (S = 25)**:
   - The starting floor is a significant fraction of both `N` and `D`, which should force the code to handle multiple steps correctly.
2. **Step Size (D = 50)**:
   - The step size is large enough to allow multiple valid moves but not so large that it trivializes the problem.
3. **Total Floors (N = 100)**:
   - The total number of floors is moderate, ensuring that the problem is not trivial and requires careful handling of the logic.

This input should expose the issues in the buggy code, particularly in the logic for calculating `targ` and handling the conditions correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
n,d,s = map(int,input().split())
if d < s:
    print(s)
else:
    maxVal = (n // s) * s
    if (maxVal - s) <= d:
        print(maxVal)
    else:
        targ = (d // s) * 2 * s
        while targ > n:
            targ -= 2 * s
print(targ)

**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:

1. **Edge Cases**:
   - When `D` is very small compared to `S`.
   - When `S` is very close to `N`.
   - When `N` is a multiple of `S` but the difference `N - S` is greater than `D`.
   - When `S` is a large prime number, making it harder to find multiples within the range.

2. **Scenarios to Maximize Failure Probability**:
   - Test inputs where the buggy code's logic of calculating `targ` might go wrong.
   - Test inputs where the condition `(maxVal - s) <= d` fails to capture the correct maximum floor.

Given these considerations, let's construct a test input that is small-scale but challenging:

```plaintext
100 5 97
```

Explanation:
- `N = 100`: The total number of floors.
- `D = 5`: The maximum difference allowed between the current floor and the next floor.
- `S = 97`: The starting floor, which is very close to `N`.

This input is designed to test the scenario where the starting floor `S` is very close to `N` and the difference `N - S` is greater than `D`. The correct output should be `97` because it's not possible to reach any higher floor from `97` within the given constraints. However, the buggy code might incorrectly compute the `targ` value, leading to an incorrect output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
import math
n, d, s = map(int, input().split())
#(m - m/spf(m)) * s <= D
#m * s <= N
NN = int(1e6 + 5)
l = NN * [0]
ans = s
for i in range(int(math.sqrt(n)) + 1):
  lo = 1
  hi = int(1e18)
  if(i == 0 or i == 1 or l[i]):
    continue
  # print(i)
  while hi - lo >= 0:
    mid = (lo + hi) // 2
    m = mid * i
    if(m * s <= n and (m - m // i) * s <= d):
      lo = mid + 1
      ans = max(ans, m * s)
    else:
      hi = mid - 1
  for j in range(NN):
    if(j * i >= NN):
      continue
    l[j * i] = 1
lo = 1
hi = int(1e18)
while hi - lo >= 0:
  mid = (lo + hi) // 2
  if(mid * s <= n and (mid - 1) * s <= d):
    lo = mid + 1
    ans = max(ans, mid * s)
  else:
    hi = mid - 1
print(ans)

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

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

Think step by step.

Response:
```plaintext
100 50 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
import java.util.Scanner;
import java.lang.Math;
import java.math.BigInteger;

public class Main {
    private static BigInteger solve(BigInteger n, BigInteger d, BigInteger s) {
        BigInteger nPrime = n.divide(s);
        BigInteger dPrime = d.divide(s);

        BigInteger y = dPrime.multiply(BigInteger.TWO).min(nPrime);

        if (y == BigInteger.ONE || y == BigInteger.ZERO) {
            return s;
        }

        if (y.mod(BigInteger.TWO) == BigInteger.ZERO) {
            return y.multiply(s);
        } else {
            for (int x=2;BigInteger.valueOf(x).compareTo(y.sqrt()) < 0;x++) {
                if (y.mod(BigInteger.valueOf(x)) == BigInteger.ZERO && y.subtract(y.divide(BigInteger.valueOf(x))).compareTo(dPrime) <= 0 && y.divide(BigInteger.valueOf(x)).compareTo(dPrime) <= 0) {
                    return y.multiply(s);
                }
            }
        }

        return (y.subtract(BigInteger.ONE)).multiply(s);
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        BigInteger n = new BigInteger(sc.next()), d = new BigInteger(sc.next()), s = new BigInteger(sc.next());

        System.out.println(solve(n, d, s));

        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 buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test inputs that are on the boundaries of the input constraints.
2. **Special Conditions**: Inputs that might cause the algorithm to fail due to specific conditions in the code.
3. **Complex Scenarios**: Inputs that involve multiple steps and calculations to reach the highest possible floor.

Let's analyze the code to identify potential issues:

- The code divides `n` and `d` by `s` and then works with these reduced values.
- It uses a loop to find a suitable `x` that satisfies certain conditions.
- The loop condition uses `y.sqrt()`, which might not cover all possible values of `y`.

A good test input should:
- Have `N` and `D` values that are not trivially small.
- Have a `S` value that requires multiple steps to reach the highest floor.
- Potentially expose issues with the loop and the conditions inside it.

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

```plaintext
100 10 5
```

Explanation:
- `N = 100`: A non-trivial number of floors.
- `D = 10`: A reasonable maximum jump distance.
- `S = 5`: Starting floor that requires multiple steps to reach the highest possible floor.

This input should challenge the code to correctly handle the loop and conditions, potentially exposing any bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = false;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static void solve(PrintWriter o) {
        try {
            long N = nextLong();
            long D = nextLong();
            long S = nextLong();
            long N_f = N/S, D_f = D/S;
            long res = Math.min(N_f, 2*D_f);
            boolean ok = (res == 1 ? true : false);
            for(int i=2;i<=Math.sqrt(res);i++) {
                if(res%i == 0) {
                    if(res-res/i <= D_f) ok = true;
                    if(res-i <= D_f) ok = true;
                    if(ok) break;
                }
            }
            if(ok) o.println(S*res);
            else o.println(S*(res-1));
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static int upper_bound(long[] preSum, long val){
        int l = 0, r = preSum.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(preSum[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(List<int[]> li, long val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid)[0] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(List<int[]> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid)[0] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private void addMn(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.min(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
        private long queryMn(int l, int r) {
            long res = 1l<<55;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.min(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.min(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class Pair {
        Integer p1, p2;
        public Pair(Integer p1, Integer p2) {
            this.p1 = p1;
            this.p2 = p2;
        }
        @Override
        public int hashCode() {
            int prime = 31, ret = 0;
            ret = ret*prime+p1.hashCode();
            ret = ret*prime+p2.hashCode();
            return ret;
        }
        @Override
        public boolean equals(Object o) {
            if(o instanceof Pair pair) {
                return p1.equals(pair.p1) && p2.equals(pair.p2);
            }
            return false;
        }
    }
    static class DSU {
        int[] parent;
        int[] size;
        public DSU(int n) {
            parent = new int[n];
            size = new int[n];
            for(int i=0;i<n;i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
            size[root_y] += size[root_x];
            size[root_x] = 0;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

**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
1000000007 1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
/*

                                _/                                      _/
       _/_/        _/_/_/      _/_/_/        _/_/_/      _/  _/_/               _/_/_/
    _/    _/    _/            _/    _/    _/    _/      _/_/          _/       _/    _/
   _/    _/    _/            _/    _/    _/    _/      _/            _/       _/    _/
    _/_/        _/_/_/      _/    _/      _/_/_/      _/            _/       _/    _/

*/
#include<bits/stdc++.h>
#define int long long

using namespace std;

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,d,s;
    cin>>n>>d>>s;
    int ma=d/s*s;
    if(!ma){
        cout<<s<<"\n";
    }
    else{
        int tmp=n/2/s*s;
        cout<<max(s,tmp*2)<<"\n";
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. The key points to consider are:

1. **Initial Floor (S)**: Ensure that the initial floor is not a simple multiple of the maximum possible floor, which can make the bug more evident.
2. **Distance (D)**: Choose a distance that can create a situation where the elevator can reach a higher floor but the code might miss it.
3. **Total Floors (N)**: Ensure the total number of floors is large enough to create a complex scenario but small enough to keep the test input simple and manageable.

Let's analyze the buggy code:
- The code calculates `ma` as `d/s * s`.
- It then calculates `tmp` as `n/2/s * s`.
- The final output is `max(s, tmp * 2)`.

The bug appears to be in the calculation of `tmp` and the use of `max`. The code might not correctly handle cases where the highest possible floor is not a simple multiple of the initial floor or the distance.

Here is a challenging test input that can expose the bug:

```plaintext
100 50 25
```

**Explanation:**
- **N = 100**: The total number of floors is 100.
- **D = 50**: The maximum distance the elevator can travel in one ride is 50.
- **S = 25**: The initial floor is 25.

**Expected Output:**
- The highest floor that can be reached from floor 25 with a maximum distance of 50 is 75.
- The correct output should be 75 because:
  - From floor 25, the elevator can go to floor 50 (50 is a multiple of 25 and 50 - 25 = 25 ≤ 50).
  - From floor 50, the elevator can go to floor 75 (75 is a multiple of 50 and 75 - 50 = 25 ≤ 50).
  - From floor 75, the elevator cannot go to any higher floor because 100 is not a multiple of 75 and 100 - 75 = 25 > 50.

This input should expose the bug in the code, as the code might not correctly handle the calculation of the highest possible floor.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;

#define pb push_back
#define all(x) x.begin(), x.end()
#define PII pair<int, int>
#define xx first
#define yy second
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define ve vector<int>
#define vv vector<vector<int>>
#define me(a) memset(a, 0, sizeof a)
#define meq(a) memset(a, 0x3f3f3f3f3f, sizeof a)
#define out(x) cout << #x << '=' << (x) << endl
#define fo(i,j,k) for(int (i)=(j);(i)<=(k);(i)++)
#define rep(i,j,k) for(int (i)=(j);(i)>=(k);(i)--)

const int mod = 1e9 + 7,inf = 1e18;
int N;
int n;

void solve(){
	int d,s;
    cin >> n >> d >> s;
    //n层 从s开始 每次可以到达k*s层
    //(k-1)*s <= d
    int ans = s;
    function<void(int)> dfs=[&](int now){
        ans=max(ans,now);
        fo(i,2,inf){
            if((i-1)*now <= d and i*now <= n){
                dfs(i*now);
            }
            if((i-1)*now > d) break;
        }
    };
    dfs(s);
    cout<<ans<<endl;
}

signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    #ifndef ONLINE_JUDGE
    //    freopen("in.txt", "r", stdin);
    //    freopen("out.txt", "w", stdout);
    #endif
    int t = 1;
    // cin>>t;
    while (t--)
    {
        solve();
    }
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - **Small values of \( N \), \( D \), and \( S \)**: These can help test the basic logic and boundary conditions.
   - **Large values of \( N \)**: This can test the efficiency and correctness of the algorithm, especially the recursion depth and the conditions in the loop.
   - **Special values of \( D \)**: Values close to \( N - 1 \) or very small values can expose issues with the constraints.

2. **Scenarios:**
   - **\( S \) is a prime number**: This can make it harder to find multiples within the allowed range.
   - **\( S \) is a composite number with many factors**: This can test the algorithm's ability to find the highest multiple.
   - **\( D \) is very large**: This can make the problem easier to solve, but it can also expose issues in the recursion and loop conditions.

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

- **\( N = 100 \)**
- **\( D = 10 \)**
- **\( S = 5 \)**

This input is small enough to be manageable but large enough to test the algorithm's logic. It also has a prime starting floor, which can make it challenging to find the highest multiple within the allowed range.

```plaintext
100 10 5
```

This input should help expose any issues in the buggy code, especially with the recursive depth and the conditions in the loop.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
from collections import deque

def highest_floor(N, D, S):
    visited = set()
    queue = deque([(S, 0)])
    max_floor = S

    while queue:
        floor, _ = queue.popleft()
        if floor not in visited:
            visited.add(floor)
            max_floor = max(max_floor, floor)
            for next_floor in range(floor * 2, min(N + 1, floor + D + 1), floor):
                queue.append((next_floor, 0))  # Distance is not needed here

    return max_floor

# Get input from user
N, D, S = map(int, input().split())
print(highest_floor(N, D, 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
100 50 25
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
def highest_floor(N, D, S):
    max_floor = S
    current_floors = [S]
    visited = set(current_floors)

    while current_floors:
        x = current_floors.pop()
        c = 2

        while True:
            y = x * c
            if y > N:
                break
            if y - x <= D:
                if y not in visited:
                    visited.add(y)
                    max_floor = max(max_floor, y)
                    current_floors.append(y)
            c += 1

    return max_floor

N, D, S = map(int, input().split())
result = highest_floor(N, D, S)
print(result)

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

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

Think step by step.

Response:
```plaintext
100 99 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

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

public class ICPCSquareElevator {
    static Map<Long, Long> memo = new HashMap<>();
    static long N, D, S;

    public static void main(String[] args) {
        // Using Scanner for simplicity; can be replaced by faster IO if needed.
        Scanner sc = new Scanner(System.in);
        N = sc.nextLong();
        D = sc.nextLong();
        S = sc.nextLong();
        sc.close();

        System.out.println(dfs(S));
    }

    static long dfs(long x) {
        // If we already computed dfs(x), return it.
        if (memo.containsKey(x)) {
            return memo.get(x);
        }

        // If x > D, we cannot move any further. Return x.
        if (x > D) {
            memo.put(x, x);
            return x;
        }

        long best = x; // We can always choose to "stay" on x by not moving.

        // g_crit = largest integer such that x*g_crit <= D
        long g_crit = D / x;

        // Maximum possible g if we only consider the N limit
        long g_maxN = N / x;

        // The overall multiplier upper bound
        long g_max = Math.min(g_crit + 1, g_maxN);

        // 1) Explore all g where x*g <= D; from there we can continue moving.
        for (long g = 2; g <= g_crit && g <= g_maxN; g++) {
            long nextFloor = x * g;
            if (nextFloor <= N) {
                best = Math.max(best, dfs(nextFloor));
            }
        }

        // 2) If g_crit+1 <= g_max, then that jump immediately exceeds D
        //    and is a "final" floor (no further moves).
        //    We only need to consider the largest g in that final-step region.
        if (g_crit + 1 <= g_max) {
            long finalJump = x * g_max;
            if (finalJump <= N) {
                best = Math.max(best, finalJump);
            }
        }

        memo.put(x, best);
        return best;
    }
}

**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
100 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

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

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        long N = Long.parseLong(st.nextToken());
        long D = Long.parseLong(st.nextToken());
        long S = Long.parseLong(st.nextToken());

        Queue<Long> queue = new LinkedList<>();
        Set<Long> visited = new HashSet<>();

        queue.offer(S);
        visited.add(S);
        long maxFloor = S;

        while (!queue.isEmpty()) {
            long current = queue.poll();

            for (long i = 2; i * current <= N; i++) {
                long next = current * i;
                if (next - current <= D && !visited.contains(next)) {
                    queue.offer(next);
                    visited.add(next);
                    maxFloor = Math.max(maxFloor, next);
                }
                if (next - current > D) break;
            }

            if (current <= Math.sqrt(N)) {
                long jump = N / current;
                if (jump * current <= N && jump * current - current <= D && !visited.contains(jump * current)) {
                    queue.offer(jump * current);
                    visited.add(jump * current);
                    maxFloor = Math.max(maxFloor, jump * current);
                }
            }
        }

        System.out.println(maxFloor);
    }
}

**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
1000000000000 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
#include<bits/stdc++.h>
#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

using u64 = uint64_t;
using u128 = __uint128_t;

#define int                long long

#define yes                cout<<"YES\n"
#define no                 cout<<"NO\n"
#define nl                 cout<<"\n"
#define endl               "\n"

#define lin(n)             int n;cin>>n;
#define vin                vector<int>
#define pr                 pair<int, int>
#define pb(n)              push_back(n)
#define pp                 pop_back()
#define ppfr(v)            v.erase(v.begin());
#define all(x)             x.begin(),x.end()

#define fi                 first
#define se                 second

#define forn(i,e)          for(int i=0;i<e;i++)
#define Forn(i,e)          for(int i=1;i<=e;i++)
#define rforn(i,s)         for(int i=s-1;i>=0;i--)
#define print(arr)         for(auto x: arr)cout<<x<<" ";nl;
#define mprint(mp)         for(auto a : mp)cout<<a.first<<" "<<a.second<<endl;

#define fast_in_out        ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
template <typename T>      using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

const long long INF = 1e18;
const int M = 1e9 + 7;
const int N = 2e5 + 100;

void sukuna(int test){
    int n, d, s;
    cin >> n >> d >> s;
    d -= d % s;
    int ans = s;

    while(d > s){
        if(d * 2 <= n)break;
        d -= s;
    }
    ans = max(ans, (2 * d <= n ? 2 * d : d));
    cout << ans << endl;

}

int32_t main(){
    fast_in_out;

    int test = 1;  // cin>>test;
    for(int i = 1; i <= test; i++)sukuna(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
1000000000000 999999999999 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
  int64_t n,d,s;
  cin>>n>>d>>s;
  int64_t n1=n/s;
  int64_t d1=d/s;
  int64_t y=min(n1,2*d1);
  if(y%2==0 && y>1)
  {
    cout<<s*y<<endl;
    return 0;
  }
  else if(y%2==1 && y>1)
  {
    cout<<s*(y-1)<<endl;
    return 0;
  }
  else if(y%2==1 && y<=1)
  {
    cout<<s*y<<endl;
    return 0;
  }
  else
  {
    cout<<s*(y+1)<<endl;
    return 0;
  }
}

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

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

Think step by step.

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

1. **Understand the Buggy Code**:
   - The code divides `n` and `d` by `s` and then determines the maximum number of steps `y` based on the minimum of `n1` and `2 * d1`.
   - The code then checks the parity of `y` and outputs `s * y` or `s * (y-1)` or `s * (y+1)` based on certain conditions.

2. **Identify Potential Bugs**:
   - The code might not handle cases where `s` is a large number or where `n` is a very large number correctly.
   - The code might not handle cases where `d` is very small relative to `n` or `s`.

3. **Generate Challenging Test Input**:
   - Choose `N` and `S` such that `S` is a large number and `N` is a very large number.
   - Choose `D` such that it is relatively small compared to `N` and `S`.

4. **Test Input**:
   - Let's choose `N = 1000000000000`, `D = 1000000000`, and `S = 999999999999`.

This test input will challenge the code to handle large numbers and ensure that the logic for determining the maximum floor is correct.

```plaintext
1000000000000 1000000000 999999999999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
from collections import deque
def max_reachable_floor(N, D, S):
    queue = deque()
    visited = set()

    queue.append(S)
    visited.add(S)
    y_max = S

    while queue:
        current = queue.popleft()

        max_k = (D // current) + 1
        max_k = min(max_k, N // current)

        for k in range(2, max_k + 1):
            y = k * current
            if y > N:
                break
            if y - current > D:
                continue
            if y not in visited:
                visited.add(y)
                queue.append(y)
                if y > y_max:
                    y_max = y

    return y_max

N, D, S = map(int, input().split())
print(max_reachable_floor(N, D, 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
1000000000000 999999999999 999999999999
```

**Explanation:**
This test input is designed to expose potential issues in the given buggy code, particularly in handling large values and edge cases.

1. **Large N and D**: The values of `N` and `D` are set to the maximum possible values within the constraints (10^12). This can stress the algorithm and potentially expose issues with integer overflow or performance.

2. **S close to N**: The starting floor `S` is set to a value very close to `N` (one less than `N`). This can test how the algorithm handles cases where the starting floor is already very high, and the possible movements are limited.

3. **Maximal D**: The value of `D` is set to the maximum possible value, which means the elevator can potentially move to any floor that is a multiple of the current floor. This can test the algorithm's ability to correctly handle large step sizes and ensure that it does not miss any valid moves.

By using this test input, we can verify if the algorithm correctly handles large values and edge cases, ensuring it functions as expected under extreme conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
from collections import deque

def solve():
    n, d, s = map(int, input().split())
    queue = deque([s])
    visited = set([s])  # Track visited floors
    maxfloor = s

    while queue:
        t = queue.popleft()
        maxfloor = max(maxfloor, t)

        # Calculate multiples of t that are within the allowed range
        multiple = 2
        while True:
            next_floor = multiple * t
            if next_floor > n or next_floor > t + d:
                break

            if next_floor not in visited:  # Check if already visited
                visited.add(next_floor)
                queue.append(next_floor)

            multiple += 1

    print(maxfloor)

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
1000000000000 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = false;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static void solve(PrintWriter o) {
        try {
            long N = nextLong();
            long D = nextLong();
            long S = nextLong();
            long N_f = N/S, D_f = D/S;
            long res = Math.min(N_f, 2*D_f);
            boolean ok = (res == 1 || res%2 == 0 ? true : false);
            for(int i=2;i<=Math.sqrt(res);i++) {
                if(res%i == 0) {
                    if(res-res/i <= D_f) ok = true;
                    if(res-i <= D_f) ok = true;
                    if(ok) break;
                }
            }
            if(ok) o.println(Math.max(S, S*res));
            else o.println(Math.max(S, S*(res-1)));
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static int upper_bound(long[] preSum, long val){
        int l = 0, r = preSum.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(preSum[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(List<int[]> li, long val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid)[0] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(List<int[]> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid)[0] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private void addMn(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.min(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
        private long queryMn(int l, int r) {
            long res = 1l<<55;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.min(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.min(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class Pair {
        Integer p1, p2;
        public Pair(Integer p1, Integer p2) {
            this.p1 = p1;
            this.p2 = p2;
        }
        @Override
        public int hashCode() {
            int prime = 31, ret = 0;
            ret = ret*prime+p1.hashCode();
            ret = ret*prime+p2.hashCode();
            return ret;
        }
        @Override
        public boolean equals(Object o) {
            if(o instanceof Pair pair) {
                return p1.equals(pair.p1) && p2.equals(pair.p2);
            }
            return false;
        }
    }
    static class DSU {
        int[] parent;
        int[] size;
        public DSU(int n) {
            parent = new int[n];
            size = new int[n];
            for(int i=0;i<n;i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
            size[root_y] += size[root_x];
            size[root_x] = 0;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

**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
100 50 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ICPC Square
time_limit_ms: 1000
memory_limit_mb: 1024
Description: ICPC Square is a hotel provided by the ICPC Committee for the accommodation of the participants. It consists of $$$N$$$ floors (numbered from $$$1$$$ to $$$N$$$). This hotel has a very unique elevator. If a person is currently at floor $$$x$$$, by riding the elevator once, they can go to floor $$$y$$$ if and only if $$$y$$$ is a multiple of $$$x$$$ and $$$y - x \leq D$$$.

You are currently at floor $$$S$$$. You want to go to the highest possible floor by riding the elevator zero or more times. Determine the highest floor you can reach.

Input: A single line consisting of three integers $$$N$$$ $$$D$$$ $$$S$$$ ($$$2 \leq N \leq 10^{12}; 1 \leq D \leq N - 1; 1 \leq S \leq N$$$).

Output: Output a single integer representing the highest floor you can reach by riding the elevator zero or more times.

Examples:
input:
64 35 3
output:
60

input:
2024 2023 1273
output:
1273

Note: Explanation for the sample input/output #1

First, ride the elevator from floor $$$3$$$ to floor $$$15$$$. This is possible because $$$15$$$ is a multiple of $$$3$$$ and $$$15 - 3 \leq 35$$$. Then, ride the elevator from floor $$$15$$$ to floor $$$30$$$. This is possible because $$$30$$$ is a multiple of $$$15$$$ and $$$30 - 15 \leq 35$$$. Finally, ride the elevator from floor $$$30$$$ to floor $$$60$$$. This is possible because $$$60$$$ is a multiple of $$$30$$$ and $$$60 - 30 \leq 35$$$.

Buggy Code:
import java.util.Scanner;
import java.lang.Math;

public class Main {
    private static long solve(long n, long d, long s) {
        long nPrime = n/s;
        long dPrime = d/s;

        long y = Math.min(2*dPrime, nPrime);

        if (y == 1 || y == 0) {
            return s;
        }

        if (y % 2 == 0) {
            return y*s;
        } else {
            for (int x=2;x<Math.sqrt(y);x++) {
                if (y % x == 0 && y - y/x <= dPrime && y/x <= dPrime) {
                    return y*s;
                }
            }
        }

        return (y-1)*s;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        long n = sc.nextLong(), d = sc.nextLong(), s = sc.nextLong();

        System.out.println(solve(n, d, s));

        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 buggy code, let's consider the following aspects:

1. **Edge Cases:**
   - When `N`, `D`, and `S` are such that the conditions for moving to the next floor are very restrictive.
   - When `S` is a prime number, making it harder to find multiples within the allowed range.
   - When `D` is very small compared to `N` and `S`.

2. **Logical Flow:**
   - The code uses `nPrime` and `dPrime` which are `n/s` and `d/s` respectively. If `s` is a large number, `nPrime` and `dPrime` can become very small, potentially leading to incorrect calculations.
   - The code checks if `y` is even or odd and then tries to find a factor of `y` that satisfies certain conditions. This can be a weak point if `y` has no such factors.

3. **Specific Test Input:**
   - Let's choose `N` and `S` such that `nPrime` and `dPrime` are small and `S` is a prime number.
   - Set `D` to a value that makes it difficult to find valid multiples of `S` within the allowed range.

Given these considerations, let's choose:
- `N = 1000000000000` (a very large number)
- `D = 1` (a very small number)
- `S = 1000000007` (a large prime number)

This input will make it very difficult to find valid multiples of `S` within the allowed range, thus exposing the bug in the code.

```plaintext
1000000000000 1 1000000007
```