**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Piggy Are Playing a Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Turtle and Piggy are playing a number game.

First, Turtle will choose an integer $x$, such that $l \le x \le r$, where $l, r$ are given. It's also guaranteed that $2l \le r$.

Then, Piggy will keep doing the following operation until $x$ becomes $1$:

  * Choose an integer $p$ such that $p \ge 2$ and $p \mid x$ (i.e. $x$ is a multiple of $p$).    * Set $x$ to $\frac{x}{p}$, and the score will increase by $1$.

The score is initially $0$. Both Turtle and Piggy want to maximize the score. Please help them to calculate the maximum score.

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

The first line of each test case contains two integers $l, r$ ($1 \le l \le r \le 10^9, 2l \le r$) — The range where Turtle can choose the integer from.

Output: For each test case, output a single integer — the maximum score.

Examples:
input:
5
2 4
3 6
2 15
6 22
114514 1919810
output:
2
2
3
4
20

Note: In the first test case, Turtle can choose an integer $x$, such that $2 \le x \le 4$. He can choose $x = 4$. Then Piggy can choose $p = 2$ for $2$ times. After that, $x$ will become $1$, and the score will be $2$, which is maximized.

In the second test case, Turtle can choose an integer $3 \le x \le 6$. He can choose $x = 6$. Then Piggy can choose $p = 2$, then choose $p = 3$. After that, $x$ will become $1$, and the score will be $2$, which is maximum.

In the third test case, Turtle can choose $x = 12$.

In the fourth test case, Turtle can choose $x = 16$.

Buggy Code:
import java.io.*;

public class Main{

    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static StreamTokenizer stir = new StreamTokenizer(br);
    public static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    private static int inputInteger() throws IOException {
        stir.nextToken();
        return (int) stir.nval;
    }

    private static void solve() throws IOException {
        int l = inputInteger();
        int r = inputInteger();

        int res = 0;
        for (int i = r; i >= l; i--) {
            if ((i & (i - 1)) == 0) {
                res = i;
                break;
            }
        }

        int result = 0;
        while (res != 1) {
            result++;
            res /= 2;
        }
        pw.println(result);
    }

    public static void main(String[] args) throws IOException {
        int t = inputInteger();
        while (t-- > 0) {
            solve();
        }
        pw.flush();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider the logic of the code and identify where it might fail. The code aims to find the maximum score by choosing the largest power of 2 within the range [l, r]. However, this approach is flawed because it doesn't consider other numbers that might have a higher score.

### Analysis of the Buggy Code:
- The code checks if a number is a power of 2 using the condition `(i & (i - 1)) == 0`.
- It then calculates the score by repeatedly dividing by 2 until the number becomes 1.
- This approach fails to consider numbers with higher prime factors, which could potentially yield a higher score.

### Challenging Test Case:
To challenge the code, we need a range [l, r] where the maximum score is not achieved by a power of 2 but by a number with a higher prime factor.

Consider the following test case:
- **l = 15, r = 15**: The only number in this range is 15.
- The maximum score for 15 is 3 (15 -> 5 -> 1), but the code will incorrectly output 0 because 15 is not a power of 2.

### Test Input:
```plaintext
1
15 15
```

This test case will expose the bug in the code because the code will fail to correctly calculate the score for 15.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
# pragma GCC optimize ("O3")
# pragma GCC optimize ("Ofast")
# pragma GCC optimize ("unroll-loops")
# pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

void sol()
{
    int n,m,k;
    cin>>n>>m>>k;
    int num=m-n%m;
    while(n!=1 && k-num>=0)
    {
        //cout<<n<<" "<<k<<"\n";
        k-=num;
        n+=num;
        while(n%m==0) n/=m;
        num=m-n%m;
    }
    //cout<<n<<" "<<k<<'\n';
    if(n==1)
    {
        if((n+k%m)==m) cout<<1<<'\n';
        else
            cout<<k%(m-1)+1<<'\n';
    }
    else cout<<n+k<<'\n';
}
int main()
{
    ios_base :: sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin>>t;
    while(t--)
    {
        sol();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

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

int main()
{
    long long t;
    cin>>t;
    while(t--){
      long long a,b,c;
      cin>>a>>b>>c;
      a++; c--;
      long long start=1;
      bool flag=false;
      if(a<b){
        if(a+c<b){
          cout<<a<<endl;
          continue;
        }
        c-=(b-a);
      }
      while(a>b){
        long long a1=ceil((float)a/b);
        if(a1*b>1000000000){
          a1=ceil((start*b)%1000000007/(float)b);
        }
        if(a1*b-a>c){
          cout<<a+c<<endl;
          flag=true;
          break;
        }
        c-=(a1*b-a);
        while(a1%b==0)
          a1=a1/b;
        a=a1;
      }
      if(c>=b-a){
        c-=(b-a);
        a=1;
      }
      if(flag)
        continue;
      long long end=b;
      c%=(end-start);
      cout<<a+c<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 2 1
1 2 2
1 2 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
// #define int long long
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T> using ordered_set =tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update> ;
std::vector<int> intToBinaryVector(ll n) {
    std::vector<int> binaryVector;
    while (n > 0) {
        binaryVector.push_back(n % 2);
        n /= 2;
    }

    // Reverse the vector to get the correct binary representation
    std::reverse(binaryVector.begin(), binaryVector.end());

    return binaryVector;
}
long long binaryVectorToInt(const std::vector<int>& binaryVector) {
    long long result = 0;
    int size = binaryVector.size();
    for (int i = 0; i < size; ++i) {
        if (binaryVector[i] == 1) {
            result += pow(2, size - 1 - i);
        }
    }
    return result;
}
ll gcd(ll a, ll b) {
    if (a < b) swap(a,b);
    while (b != 0) {
        ll temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}
ll lcm(ll a, ll b){
    return a*b/gcd(a,b);
}

long long power(long long exp) {
    long long result = 1;
    long long base = 2;

    while (exp > 0) {
        // If exp is odd, multiply the current base with the result
        if (exp % 2 == 1) {
            result = (result * base) % 998244353;
        }
        // Square the base and reduce the exponent by half
        base = (base * base) % 998244353;
        exp /= 2;
    }

    return result%998244353;
}

vector <int> convert(ll a){
    vector <int> temp;
    while (a!=0){
        temp.push_back(a%10);
        a = a/10;
    }
    reverse(temp.begin(), temp.end());
    return temp;
}
ll conve(vector<int> a, vector <int> b){
    int l1 = a.size();
    int l2 = b.size();
    vector <int> t;
    int p = 0;
    int l = 0;
    ll ans = 0;
    if (l1 >= l2){
        for (int i =l2; i<l1; i++){
            t.push_back(a[p]);
            p++;
        }
        for (int i =0; i<l2; i++){
            t.push_back(a[p+i]);
            t.push_back(b[i]);
        }

    }
    else{
        for (int i = l1; i<l2; i++){
            t.push_back(b[p]);
            p++;
        }
        for (int i =0 ; i<l1; i++){
            t.push_back(a[i]);
            t.push_back(b[i+p]);
        }
    }
    for (int i =0 ;i<l1+l2; i++){
        ans = ans + t[i]*power(l1+l2 - i-1);
        ans = ans%998244353;
    }
    return ans;
}
void solve() {

   ll x, y , k; cin>>x>>y>>k;
   ll rem = x%y;

   if (k < (y -rem)){
        cout << x + k << endl;
        return;
    }
    if (k == y -rem){
        x+= k;
        while(x%y == 0){
            x= x/y;

        }
        cout << x << endl;
        return;
    }

    while (x!=1 && k >= (y - rem)){
        rem = x%y;
        x = x + y - rem;
        k = k - y + rem;
        while(x%y == 0){
            x= x/y;

        }

    }

    if (x == 1){
        ll t = (k) %(y -1LL);
        if (x == 0){
            cout << 1 << endl;
            return;
        }
        cout << x +t<< endl;
        return;
    }

    cout << x + k << endl;
    return;

}

int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    int t;
    cin>>t;

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

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
t = int(input())
from math import ceil

def run():
    x, y, k = [int(i) for i in input().split(' ')]

    while k > 0:
        if x < y:
            mod_val = y -1
            # print(x, k, k%mod_val, "B1")
            x = x + (k % mod_val)
            break
        turns = y - (x % y)
        if k < turns:
            # print(x, k, "B2")
            x = x + k
            break
        # print(x, turns)
        x += turns
        while x % y == 0:
            x = int(x/y)
        k -= turns
        # print(x, k)
    while x % y == 0:
        x = int(x/y)
    print(x)

for i in range(t):
    run()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
import math
def main():
    x , y , k = map(int , input().split())
    x += 1
    k -= 1
    temp = y * math.ceil(x / y)
    if temp - x <= k:
        k -= temp - x
        while temp != 1 and temp >= y and k >= 0:
            if temp >= y and temp % y == 0:
                temp //= y
            elif temp >= y:
                if y * math.ceil(temp / y) - temp > k:
                    temp += k
                    k = 0
                    break
                k -= y * math.ceil(temp / y) - temp
                temp = y * math.ceil(temp / y)
        if k == 0:
            return temp
        k = k % (y - 1)
        return k + temp
    else:
        return x

for _ in range(int(input())):
    print(main())

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

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

Think step by step.

Response:
```plaintext
3
1 2 1000000000
2 2 1000000000
3 2 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
for t in range(int(input())):
    x,y,k=map(int,input().split())
    while k>0:
        if x<y:
            x=(x+k%(y-1)-1)%y+1
            k=0
        elif x%y==0:
            x+=1
            k-=1
        else:
            if y-x%y>k:
                x+=k
                k=0
            else:
                k-=y-x%y
                x+=y-x%y
        while x%y==0: x//=y
    print(x)

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

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

Think step by step.

Response:
```plaintext
3
1 2 1000000000
2 2 1000000000
1000000000 2 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
import java.util.Scanner;

public class m1982b {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            long n = sc.nextLong(), x = sc.nextLong(), k = sc.nextLong();
            while (k > 0) {
                long p = Math.min(k, x - n % x);
                k -= p;
                n += p;
                while (n % x == 0) n /= x;
                if (n < x) break;
            }
            if (k == 0) System.out.println(n);
            else {
                long p = (k + n + 1) / x;
                System.out.println(Math.max(1, (k + n + p) % x));
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 2 1000000000
2 2 1000000000
3 2 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

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

public class CodeChef {
    private static long result;
    private static Map<String, Long> memo;

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.nextInt();
        while (t-- > 0) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            int k = sc.nextInt();
            x++;
            k--;
            while(x>=y && k>=y - x % y) {
                k -= x%y!=0?y - x % y:0;
                x = (int) Math.ceil((double) x / (double) y);
            }
//            System.out.println(x);
            x += k%(y-1);
            while(x%y==0)
                x /= y;
            System.out.println(x);
        }
    }
}

class FastScanner {
    BufferedReader br;
    StringTokenizer st;

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

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

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

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

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

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

    public void close() {
        try {
            br.close();
        } catch (Exception e) {
            System.out.println(e.getMessage());
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. The key points to consider are:

1. **Large Values of \( x \), \( y \), and \( k \)**: The code should handle large values without errors.
2. **Divisibility and Looping**: The code should correctly handle the repeated division by \( y \) and the incremental operations.
3. **Edge Cases**: Values where \( x \) is just below or just above multiples of \( y \).

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

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

### Explanation:
1. **Test Case 1: \( x = 1 \), \( y = 2 \), \( k = 1 \)**
   - Initial \( x \) is 1.
   - After one operation: \( x \) becomes 2.
   - Expected output: 2

2. **Test Case 2: \( x = 2 \), \( y = 2 \), \( k = 2 \)**
   - Initial \( x \) is 2.
   - After the first operation: \( x \) becomes 3.
   - After the second operation: \( x \) becomes 4.
   - Expected output: 4

3. **Test Case 3: \( x = 3 \), \( y = 2 \), \( k = 1 \)**
   - Initial \( x \) is 3.
   - After one operation: \( x \) becomes 4.
   - Expected output: 4

4. **Test Case 4: \( x = 4 \), \( y = 3 \), \( k = 2 \)**
   - Initial \( x \) is 4.
   - After the first operation: \( x \) becomes 5.
   - After the second operation: \( x \) becomes 6.
   - Expected output: 2 (since 6 is divisible by 3, it becomes 2)

This test input includes a mix of small values and scenarios where \( x \) is just below or just above multiples of \( y \), which should help in exposing any issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#ifndef ONLINE_JUDGE
#include "template.h"
#endif
void solve(){
      ll x , y , k ;
      cin >> x >> y >> k ;

      x++;
      k -= 1 ;
      for(int op = 1 ; op <= 100 && x >= y; op++){

         ll rem = x % y ;
         ll need = y - rem ;
         if(rem == 0) need = 0 ;
         if(k < need) break;
         //who(rem);
         k -= need ;
         x += need ;
         while(x % y == 0){
            x /= y ;

         }
      }
      ll r = k % (y - 1);
      k = r ;
      ll rem = x % y ;

      ll need = y - rem ;
      if(rem == 0) need = 0 ;
     // who(x , k , need);
      if(k < need){
        x = x + k ;
      }
      else {
        k -= need ;
        x = k;
      } cout << x << "\n";
}
int main(){
      std::ios_base::sync_with_stdio(false);
      std::cin.tie(nullptr);
      int T = 1 ;
      cin >> T ;
      while(T--) {
        solve();
      }
}

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

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

Think step by step.

Response:
```plaintext
3
1 2 1
2 2 2
1000000000 2 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define bitcount(a) __builtin_popcount(a)
#define gcd(a,b) __gcd(a,b)
const ll M=1e9+7;
/*---------------------------------------------------------------*/
ll binpow(ll b,ll exp,ll M)
{
if(exp==0)
return 1;
ll half=binpow(b,exp/2,M);
ll ans;
if(exp%2==0)
ans=1ll*half*half%M;
else
{
   ans=1ll*half*half%M;
  ans=1ll*ans*b%M;
}
return ans;
}
/*--------------------------------------------------------------*/
ll inverse(ll x)
{
return binpow(x,M-2,M);
}
/*---------------------------------------------------------------------------------------*/
string printBinary(int num){
string str="";
for(int i=31;i>=0;i--){
str+=(char)((((num>>i)&1))+48);
}
return str;
}
/*----------------------------------------------------------------------------*/
bool cmp(pair<string, int>& a,
pair<string, int>& b)
{
return a.second < b.second;
}
void sort(map<string, int>& M)
{
vector<pair<string, int> > A;
for (auto& it : M) {
A.push_back(it);
}
sort(A.begin(), A.end(), cmp);
for (auto &it : A) {
cout << it.first << ' '
<< it.second << endl;
}
}
/*int main()
{
// Declare Map
map<string, int> M;

// Given Map
M = { {
, 3 },
{
, 2 },
{
, 1 } };
// Function Call
sort(M);
return 0;
} */
/*------------------------------------------------------------------------*/
void solve()
{
ll x,y,k;
cin>>x>>y>>k;
ll cnt=0;
// cout<<log(x)/log(y)<<endl;
while(k!=0)
{
  ll rem=y-(x%y);
  // k-=rem;
  if((k-rem)<0)
  {
    x+=k;
    break;
  }
  else
  {k-=rem;
  x+=rem;
  }
  ll num=floor(log(x)/log(y));
  ll pro=1;
  for(ll i=1;i<=num;i++)
  {
    cnt++;
    ll num1=pow(y,i);
    if(x%num1==0)
    {
      pro*=y;
    }
  }
  x=x/pro;
  if(x<y && x==1)
  {
    if(k==y)
    {
      if(y==2)
      x=1;
      else
      x=2;
    k=0;
    }
    if(k>y)
    {
      x=(x+k)%(y-1);
      if(x==0)
      x=y-1;
      k=0;

    }
    if(k<y)
    {
      x=(x+k)%(y-1);
      if(x==0)
      x=y-1;
      k=0;

    }
  }
  else if(x<y && x!=1)
  {
    if(k<y)
    {
      x+=k;
      k=0;
    }
    else
    {
      k-=(y-x);
      x=1;
    }
  }
}
cout<<x<<endl;
}
int main()
{
ll t;
cin>>t;
while(t--)
{
solve();
}
}

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

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

Think step by step.

Response:
```plaintext
1
1 2 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
for t in range(int(input())):
    x,y,k=map(int, input().split())
    j=0
    f=0
    while x!=1:
        f=j
        j+=y-x%y
        #print(j,k)
        if j>k:
            d=k-f
            x+=d
            #print(j)
            break
        x+=y-x%y
        while x%y==0:
            x//=y
        #print(x)
    #print(x,j)
    if j<k:
        #print(k-j,j)
        print(max(1,(1+(k-j)//y+k-j)%y))
    else:
        print(x)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
import math
def main():
    x , y , k = map(int , input().split())
    x += 1
    k -= 1
    temp = y * math.ceil(x / y)
    if x + k >= y and k > 0:
        k -= temp - x
        while temp != 1 and temp >= y and k >= 0:
            if temp >= y and temp % y == 0:
                temp //= y
            elif temp >= y:
                if y * math.ceil(temp / y) - temp > k:
                    temp += k
                    k = 0
                    break
                k -= y * math.ceil(temp / y) - temp
                temp = y * math.ceil(temp / y)
        if k == 0:
            return temp
        k = k % (y - 1)
        return k + temp
    else:
        val = x + k
        while val % y == 0:
            val //= y
        return val

for _ in range(int(input())):
    print(main())

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
//
// Author : Swarnadip Kar
// IIT Bhilai
//

import java.lang.*; // imports everything
import java.util.*;

public class B_Collatz_Conjecture
{
    //static ArrayList<Integer> a=new ArrayList<>();
    //static int max;
    public static void main (String[] args) throws java.lang.Exception
    {
        Scanner sc=new Scanner(System.in);
        int caseNo,i,j,N;
        N=sc.nextInt();
        // int kk=2;
        // int x=100;
        // int yy=5;
        for(caseNo=0;caseNo<N;caseNo++)
        {
            int x=sc.nextInt();
            int y=sc.nextInt();
            int k=sc.nextInt();
            // int k=kk;
            // x++;
            // int y=yy;

            int r;
            if(y==2)
            {
                boolean flag=false;
                while(x>2)
                {
                    r=x%y;
                    if(k<(y-r))
                    {
                        System.out.println(x+k);
                        flag=true;
                        break;
                    }
                    k=k-(y-r);
                    x=(x+(y-r))/y;
                    while(x%y==0)
                    {
                        x=x/y;
                        if(x==1)
                        {
                            System.out.println("1");
                            flag=true;
                            break;
                        }
                    }
                }
                if(!flag)
                {
                    if(x==1)
                    {
                        System.out.println("1");
                    }
                    else
                    {
                        if(k%2==1)
                        {
                            System.out.println("3");
                        }
                        else
                        {
                            System.out.println("2");
                        }
                    }
                }

            }
            else
            {
                boolean flag=false;
                while(x>=y)
                {
                    r=x%y;
                    if(k<(y-r))
                    {
                        System.out.println(x+k);
                        flag=true;
                        break;
                    }
                    k=k-(y-r);
                    x=(x+(y-r))/y;
                    while(x%y==0)
                    {
                        x=x/y;
                        if(x==1)
                        {
                            System.out.println("1");
                            flag=true;
                            break;
                        }
                    }
                }
                if(!flag)
                {
                    x=(x+k-1)%(y-1)+1;
                    System.out.println(x);
                }

            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
import java.util.*;
public class collatz_conjecture {
    public static void main(String args[]){
        try(Scanner sc = new Scanner(System.in)){
            long test = sc.nextLong();
            while((test--)>0){
                long x = sc.nextLong();
                long y = sc.nextLong();
                long k = sc.nextLong();
                x++;
                long r;
                while(k>0 && x<y){
                    r = x%y;
                    // System.out.println("r:"+r);
                    if(r!=0){
                        if(k>=y){
                            k = k - y + r;
                            if(k>0){
                                x = x + y - r;
                            }
                            // System.out.println("x:"+x);
                            // System.out.println("k:"+k);
                        }else{
                            k = k - r;
                            if(k>0){
                                x = x + r;
                            }
                            // System.out.println("x:"+x);
                            // System.out.println("k:"+k);
                        }

                        if(k>0){
                            // System.out.println("x:"+x);
                            while((x%y)==0){
                                x = x/y;
                            }
                            // System.out.println("x:"+x);
                        }
                    }else{
                        k--;
                        while((x%y)==0){
                            x = x/y;
                        }
                    }
                }

                while(k>0 && x>=y){
                    r = x%y;
                    // System.out.println("r:"+r);
                    if(r!=0){
                        k = k - y + r;
                        // System.out.println("k:"+k);
                        if(k>0){
                            x = x + y - r;
                            // System.out.println("x:"+x);
                            while((x%y)==0){
                                x = x/y;
                            }
                            // System.out.println("x:"+x);
                        }
                    }else{
                        k--;
                        while((x%y)==0){
                            x = x/y;
                        }
                        // System.out.println("k:"+k);
                        // System.out.println("x:"+x);
                    }
                    if((x+k)>=y && (x<y)){
                        long z = k/y;
                        // long w = k%y;
                        // System.out.println("z:"+z);
                        // System.out.println("w:"+w);
                        k = k - (z*y) + z;
                        // System.out.println("k:"+k);
                    }
                }
                if((x+k)<y && k>0){
                    // System.out.println(x);
                    x = x + k -1;
                }
                System.out.println(x);
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
def solve():
    x,y,k = map(int,input().split())
    div = divmod(x,y)
    if max(x,y,k) < 12345679:
        while k > 0:
            x+=1
            while x % y == 0:
                x//=y
            k-=1
        return x
    if x + k < y:return x + k
    if x == y and k < y :return x + k
    if k < div[1] and div[0] != 0:return x + k
    if k %y==0:return (x+k) //y
    if (x + k) % y == 0:return div[1]
    return divmod(x + k,y)[1] + divmod(x + k,y)[0]

for i in range(int(input())):print(solve())

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

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

Think step by step.

Response:
```plaintext
3
1 2 1000000000
998244353 2 998244353
1000000000 1000000000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
import decimal
import random
import heapq
import bisect
import math
import sys
import itertools
import time

from collections import deque, Counter, defaultdict
from decimal import Decimal, getcontext
from functools import lru_cache
from itertools import combinations, permutations, accumulate, chain

input = sys.stdin.readline
printf = sys.stdout.write

# sys.setrecursionlimit(100000)

MOD = 998244353
N = int(2e5 + 10)

def solve():
    rd = random.getrandbits(64)
    # n = int(input().strip())
    # s = input().strip()
    x, y, k = [int(i) for i in input().strip().split(' ')]
    # a = [int(i) for i in input().strip().split(' ')]

    while k and x >= y:
        x += 1
        k -= 1
        if x % y != 0 and (x + k) // y > x // y:
            k -= y - x % y
            x += y - x % y
        while x % y == 0:
            x //= y
    k %= y - 1
    if (x + k) // y > x // y:
        k -= y - x % y
        x += y - x % y
        print(x // y + k)
    else:
        print(x + k)

if __name__ == '__main__':
    __t = 1
    __t = int(input().strip())
    for _ in range(__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 code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. Here are some key points to consider:

1. **Large Values of \( k \)**: The code might have issues when \( k \) is very large, especially when \( k \) is close to \( 10^9 \).
2. **Values of \( y \) Close to \( x \)**: When \( y \) is close to \( x \), the division and increment operations can become complex.
3. **Small Values of \( y \)**: Small values of \( y \) (e.g., 2 or 3) can also cause issues, especially when \( x \) is a power of \( y \).
4. **Initial Values of \( x \) that are Powers of \( y \)**: These can lead to multiple divisions in a single operation.
5. **Multiple Test Cases**: Including multiple test cases can help ensure that the code handles different scenarios correctly.

Based on these considerations, here is a small-scale test input that includes a variety of challenging cases:

```plaintext
5
1 3 1
2 3 1
24 5 5
16 3 2
998244353 2 998244353
```

This test input includes:
1. A simple case where \( x = 1 \), \( y = 3 \), and \( k = 1 \).
2. A case where \( x = 2 \), \( y = 3 \), and \( k = 1 \).
3. A case with larger values where \( x = 24 \), \( y = 5 \), and \( k = 5 \).
4. A case where \( x \) is a power of \( y \) (16 is a power of 3) and \( k = 2 \).
5. A large case where \( x = 998244353 \), \( y = 2 \), and \( k = 998244353 \).

These cases are designed to test the code's ability to handle different initial values, large \( k \) values, and the interaction between \( x \) and \( y \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

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

#define int long long
typedef long long i64;
typedef __int128 i128;
typedef pair<int, int> pII;
#define x first
#define y second
const int N = 2e5 + 10, M = 1e6 + 10, mod = 998244353;
int t, n, m, a[N], b[N];
pII c[N];

void solve() {
	int x, y, k;
	cin >> x >> y >> k;
	unordered_map<int, int> p;
	int cnt = k, loop = 1;
	while (k) {
		int goal = (x + y) / y * y;
		// cout << goal - x << endl;
		if (k >= goal - x) {
			k -= goal - x;
			x = goal;
		} else {
			x += k;
			break;
		}
		while (x % y == 0) x /= y;
		if (p[x] > 9000) {
			break;
		}
		p[x] ++;
	}
	cout << x << endl;
}
signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr); cout.tie(nullptr);
	cin >> t;
	//t = 1;
	while (t -- ) solve();
	return 0;
}

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

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

Think step by step.

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

1. **Large Values for \( x \) and \( k \)**: The code should handle very large values efficiently.
2. **Small Values for \( y \)**: This can lead to frequent divisions, which might expose issues in the loop logic.
3. **Values of \( x \) that are close to multiples of \( y \)**: These can cause the loop to behave in unexpected ways.
4. **Values of \( k \) that are much larger than \( y \)**: This can test the efficiency and correctness of the loop.

Given these considerations, let's construct a small-scale test input that includes these edge cases:

```plaintext
5
1 2 1000000000
2 3 1000000000
999999999 2 1000000000
1 1000000000 1000000000
1000000000 2 1000000000
```

### Explanation:
1. **Test Case 1: \( x = 1 \), \( y = 2 \), \( k = 1000000000 \)**:
   - This tests the behavior with a very large \( k \) and a small \( y \). The loop should handle frequent divisions by 2 efficiently.

2. **Test Case 2: \( x = 2 \), \( y = 3 \), \( k = 1000000000 \)**:
   - This tests the behavior with a different small \( y \) and a very large \( k \). The loop should handle frequent divisions by 3 efficiently.

3. **Test Case 3: \( x = 999999999 \), \( y = 2 \), \( k = 1000000000 \)**:
   - This tests the behavior with a large \( x \) that is close to a multiple of \( y \). The loop should handle the transition to a smaller value correctly.

4. **Test Case 4: \( x = 1 \), \( y = 1000000000 \), \( k = 1000000000 \)**:
   - This tests the behavior with a very large \( y \). The loop should handle the fact that \( x \) will rarely be divisible by such a large \( y \).

5. **Test Case 5: \( x = 1000000000 \), \( y = 2 \), \( k = 1000000000 \)**:
   - This tests the behavior with a very large \( x \) and a small \( y \). The loop should handle the frequent divisions and large \( k \) efficiently.

This test input should expose any bugs in the given buggy code by pushing it to its limits in various scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
#include <iostream>
using namespace std;
int solve(long long int x,long long int y,long long int k){
  while(k>0 && x+1>y){
      if((x+1)%y!=0){
        long long int sub=(x/y+1)*y-x;
        if(sub>k){
          x=x+k;
          k=0;
          return x;
        }
        k-=(x/y+1)*y-x;
        x=(x/y+1)*y;
      }
      else {
        x+=1;
        k--;}
      while(x%y==0){
        x=x/y;
      }

  }
  if(k==0) return x;
  // cout<<x<<" "<<k<<endl;
  long long int add=(x+k)/y-x/y;
    long long int ans=(x+k+add)%y;
    if (ans==0) ans=1;
  return ans;

}
int main()
{
  int t;
  cin>>t;
  while(t--){
    long long int x,y,k;
    cin>>x>>y>>k;
    cout<<solve(x,y,k)<<endl;
  }
}

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

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

Think step by step.

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

1. **Large Values for \( k \)**: Test cases where \( k \) is very large, especially when \( x \) and \( y \) are also large, can expose issues in the loop and calculations.
2. **Small Values for \( y \)**: Test cases where \( y \) is small (e.g., 2) can help in checking the division and increment logic.
3. **Initial \( x \) is a Multiple of \( y \)**: Test cases where \( x \) is already a multiple of \( y \) can help in checking the initial increment and division logic.
4. **\( k = 1 \)**: Test cases where \( k \) is 1 can help in checking the single operation logic.
5. **\( x \) is Close to the Maximum Limit**: Test cases where \( x \) is close to \( 10^9 \) can help in checking boundary conditions.

Based on these considerations, here is a small-scale test input that should help expose the bug:

```plaintext
5
1 2 1
2 3 1
999999999 2 1000000000
1 3 1000000000
2 2 999999999
```

Explanation of the test cases:
1. **1 2 1**: This tests the basic increment and division logic.
2. **2 3 1**: This tests the case where \( x \) is a multiple of \( y \).
3. **999999999 2 1000000000**: This tests a large \( k \) with \( x \) close to the maximum limit.
4. **1 3 1000000000**: This tests a large \( k \) with a small \( y \).
5. **2 2 999999999**: This tests a large \( k \) with \( y = 2 \).

These test cases should help in identifying any issues in the loop and calculation logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

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

public class haii{
   public static void main(String[] args) {
	   MyScanner sc = new MyScanner();
	   out = new PrintWriter(new BufferedOutputStream(System.out));

	   int t = sc.nextInt();
	   for(int z = 0; z < t; z++)
	   {
		   int x = sc.nextInt();
		   int y = sc.nextInt();
		   int k = sc.nextInt();

		   int count = 0;
		   HashSet<Integer> hash = new HashSet<Integer>();
		   hash.add(x);
		   ArrayList<Integer> arr = new ArrayList<Integer>();
		   arr.add(x);
		   if(y > 10000000)
		   {
			   while(count <= k)
			   {
				   int temp = y - x%y;
				   x+= temp;
				   count += temp;
				   if(count > k)
				   {
					   x -= (count - k);
					   break;
				   }
				   while(x % y == 0)
					   x = x / y;

			   }
			   out.println(x);
		   }
		   else
		   {
		   while(count < k)
		   {
			   count++;
			   x += 1;
			   while(x % y == 0)
				   x = x / y;
			   if(hash.contains(x))
				   break;
			   else
			   {
				   hash.add(x);
				   arr.add(x);
			   }
		   }
		   //System.out.println(arr);
		   if(count == k)
			   out.println(x);
		   else
		   {
			   int ind = arr.indexOf(x);
			   int idk = arr.size() - ind;
			   int helpMe = (k-count) % idk;
			   out.println(arr.get(ind + helpMe));

		   }
		   }

	   }

      // Start writing your solution here. -------------------------------------

      /*
      int n      = sc.nextInt();        // read input as integer
      long k     = sc.nextLong();       // read input as long
      double d   = sc.nextDouble();     // read input as double
      String str = sc.next();           // read input as String
      String s   = sc.nextLine();       // read whole line as String

      int result = 3*n;
      out.println(result);                    // print via PrintWriter
      */

      // Stop writing your solution here. -------------------------------------
      out.close();
   }
   public static int gcd(int a, int b) {
	   if (b==0) return a;
	   return gcd(b,a%b);
	}
   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;

   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;

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

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

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

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

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

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

   }
   //--------------------------------------------------------
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are some considerations:

1. **Large Values for `y`**: The code has a special handling for `y > 10000000`. We should test with values of `y` just above this threshold to see if the logic is correctly implemented.
2. **Large Values for `k`**: The code needs to handle large values of `k` efficiently. We should test with `k` values close to the upper limit of `10^9`.
3. **Edge Cases for `x`**: Test with `x` values that are close to the upper limit of `10^9` and also with small values of `x` to ensure the algorithm works for all ranges.
4. **Special Values for `y`**: Test with `y` values that are prime numbers and composite numbers to see how the algorithm handles different types of divisors.
5. **Repetition and Cycles**: Test with values that are likely to create cycles in the sequence to ensure the algorithm can handle such cases.

Based on these considerations, here is a small-scale test input that includes these edge cases:

```plaintext
5
1 10000001 1000000000
1000000000 3 1000000000
1 2 1000000000
2 3 1000000000
1000000000 1000000000 1000000000
```

This test input includes:
1. A large `y` value just above the threshold of `10000000`.
2. A large `x` value with a moderate `y` value.
3. A small `x` value with a small `y` value and a large `k` value.
4. A small `x` value with a moderate `y` value and a large `k` value.
5. A large `x` value with a large `y` value and a large `k` value.

These cases should help expose any bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Collatz Conjecture
time_limit_ms: 1000
memory_limit_mb: 256
Description: Recently, the first-year student Maxim learned about the Collatz conjecture, but he didn't pay much attention during the lecture, so he believes that the following process is mentioned in the conjecture:

There is a variable $x$ and a constant $y$. The following operation is performed $k$ times:

  * increase $x$ by $1$, then    * while the number $x$ is divisible by $y$, divide it by $y$.

Note that both of these actions are performed sequentially within one operation.

For example, if the number $x = 16$, $y = 3$, and $k = 2$, then after one operation $x$ becomes $17$, and after another operation $x$ becomes $2$, because after adding one, $x = 18$ is divisible by $3$ twice.

Given the initial values of $x$, $y$, and $k$, Maxim wants to know what is the final value of $x$.

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

The only line of each test case contains three integers $x$, $y$, and $k$ ($1 \le x, k \le 10^{9}$, $2 \le y \le 10^{9}$) — the initial variable, constant and the number of operations.

Output: For each test case, output a single integer — the number obtained after applying $k$ operations.

Examples:
input:
13
1 3 1
2 3 1
24 5 5
16 3 2
2 2 1
1337 18 1
1 2 144133
12345678 3 10
998244353 2 998244353
998244353 123456789 998244352
998244354 998241111 998244352
998244355 2 9982443
1000000000 1000000000 1000000000
output:
2
1
1
2
3
1338
1
16936
1
21180097
6486
1
2

Note: In the first test case, there is only one operation applied to $x = 1$, resulting in $x$ becoming $2$.

In the second test case, for $x = 2$, within one operation, one is added to $x$ and it's divided by $y = 3$, resulting in $x$ becoming $1$.

In the third test case, $x$ changes as follows:

  * After the first operation, $x = 1$, because $24 + 1 = 25$ and $25$ is divisible by $y = 5$ twice within one operation.    * After the second operation, $x = 2$.    * After the third operation, $x = 3$.    * After the fourth operation, $x = 4$.    * After the fifth operation, $x = 1$.

Buggy Code:
import java.util.Scanner;

public class m1982b {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            long n = sc.nextLong(), x = sc.nextLong(), k = sc.nextLong();
            while (k > 0) {
                long p = Math.min(k, x - n % x);
                k -= p;
                n += p;
                while (n % x == 0) n /= x;
                if (n < x) break;
            }
            if (k == 0) System.out.println(n);
            else {
                long p = (k + n) / x;
                System.out.println(Math.max(1, (k + n + p) % x));
            }
        }
    }
}

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

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

Think step by step.

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


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <iostream>
#include <algorithm>
using namespace std;

long long calprofit(long long n, long long a, long long b, long long k) {
    long long prosum = k * b - (k * (k - 1)) / 2;
    long long remsum = (n - k) * a;
    return prosum + remsum;
}

int main() {
    int t;
    cin >> t;

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

        long long maxP = n * a;

        long long maxk = min(n, b);
        maxP = max(maxP, calprofit(n, a, b, maxk));

        if (maxk > 0) {
            maxP = max(maxP, calprofit(n, a, b, maxk / 2));
        }

        cout << maxP << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 1000000000 1
1000000000 1 1000000000
1000000000 1000000000 1000000000
```