**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    a = [int(x) for x in input().split()]
    b = [int(x) for x in input().split()]
    a = sorted(a)[::-1]
    b = sorted(b)[::-1]
    mi = 10**9
    inmd = 0
    for i in range(n//2+n%2):
        inmd = abs(b[i] - a[i+(n//2)])
        if inmd < mi:
            mi = inmd
    for i in range(n - n//2-n%2):
        inmd = abs(a[i] - b[i + n//2+n%2])
        if inmd < mi:
            mi = inmd
    print(mi)

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

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

Think step by step.

Response:
```plaintext
3
1
0
1000000000
2
0 1000000000
1000000000 0
4
1 2 3 4
5 6 7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
import sys
input=sys.stdin.readline
from math import*
from collections import*
I=lambda: map(int,input().split())
li=lambda: [(abs(int(v)),j) for j,v in enumerate (input().split())]

def check(d,arr,l,r):
    d=d+arr[l-1]
    if arr[r]<d:
        return -1
    while l<=r:
        m=(l+r)//2
        if arr[m]<d:
            l=m+1
        else:r=m-1
    return r+1

for i in range(int(input())):
    n,=I()
    a=sorted(list(I()))
    b=sorted(list(I()))
    c=[-1]*((n-1)//2+1)
    for j in range((n-1)//2,-1,-1):
        c[j]=b.pop()
    c=c+b
    m=float("inf")
    for j in range(n):
        m=min(m,abs(a[j]-c[j]))
    print(m)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. The buggy code sorts both the appetizers and main dishes, and then attempts to pair them in a way that maximizes the minimum charm. However, the pairing logic might not always be optimal, especially in cases where the spiciness values are closely packed or have specific patterns.

### Step-by-Step Analysis:

1. **Edge Cases:**
   - **Single Element:** Test with a single appetizer and a single main dish.
   - **Equal Spiciness Values:** Test with appetizers and main dishes having the same spiciness values.
   - **Large Differences:** Test with large differences in spiciness values.
   - **Symmetric Values:** Test with symmetric spiciness values around a central point.

2. **Challenging Scenarios:**
   - **Close Values:** Test with appetizers and main dishes having spiciness values that are very close to each other.
   - **Mixed Values:** Test with a mix of high and low spiciness values to see how the algorithm handles pairing.

### Challenging Test Input:

Let's create a test input with the following characteristics:
- A single test case with 7 appetizers and 7 main dishes.
- Spiciness values are closely packed and have a specific pattern to challenge the pairing logic.

```plaintext
1
7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
```

### Explanation:
- **Single Test Case:** Simplifies the input and focuses on the core logic.
- **Closely Packed Values:** The spiciness values are consecutive integers, which will challenge the algorithm to find the optimal pairing.
- **Symmetric Values:** The values are the same for both appetizers and main dishes, which will test the algorithm's ability to handle identical values.

This test input will help expose any issues in the pairing 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: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
import java.util.*;

public class Main {
    public static int processSolution(int[] arr,int[] brr ,int mid,int n,int testCase) {
        int ans=Integer.MAX_VALUE;
        for(int i=0;i<n;i++){
            ans=Math.min(ans,Math.abs(arr[i]-brr[mid%n]));
            //System.out.print(arr[i]+" - "+brr[mid%n]+" = "+Math.abs(arr[i]-brr[mid%n])+"\t");
            mid++;
        }
        //System.out.println("\n"+ans);
        return ans;

    }

    public static int processArray(int[] arr,int[] brr ,int n,int testCase) {

        int ans=Integer.MIN_VALUE;
        Arrays.sort(arr);
        Arrays.sort(brr);
        int mid=n/2;
        // if((n&1)==1){
        // ans=Math.max(processSolution(arr, brr, mid,n, testCase),processSolution(brr, arr, mid,n, testCase));
        // }
        // else{
        // ans=Math.max(processSolution(arr, brr, mid+1,n, testCase),processSolution(arr, brr, mid,n, testCase));
        // ans=Math.max(processSolution(brr, arr, mid+1,n, testCase),processSolution(brr, arr, mid,n, testCase));

        // }
        for(int i=0;i<n;i++){
        ans=Math.max(ans,processSolution(arr, brr, mid,n, testCase));
        ans=Math.max(ans,processSolution(brr, arr, mid,n, testCase));

    }

        //System.out.println("check this: "+(testCase+1));

        return ans;

        // for (int i = 0; i < arr.length; i++) {
        //     System.out.print(arr[i] + " ");
        // }
        // System.out.println();
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int testCase = 0; testCase < t; testCase++) {
            int n = scanner.nextInt();
            int[] brr = new int[n];
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = scanner.nextInt();
            }

            for (int i = 0; i < n; i++) {
                brr[i] = scanner.nextInt();
            }
            System.out.println(processArray(arr,brr,n,testCase));

            //System.out.println("Test Case " + (testCase + 1) + ": " + processArray(arr,brr,n,testCase));
        }

        scanner.close();
    }

}

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

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

Think step by step.

Response:
```plaintext
4
3
1 2 3
4 5 6
5
1 2 3 4 5
6 7 8 9 10
6
0 0 0 1 1 1
2 2 2 3 3 3
7
1 2 3 4 5 6 7
8 9 10 11 12 13 14
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Charming Meals
time_limit_ms: 2000
memory_limit_mb: 256
Description: The Czech cuisine features $$$n$$$ appetizers and $$$n$$$ main dishes. The $$$i$$$-th appetizer has spiciness $$$a_i$$$, and the $$$i$$$-th main dish has spiciness $$$b_i$$$.

A typical Czech meal consists of exactly one appetizer and one main dish. You want to pair up the $$$n$$$ appetizers and $$$n$$$ main dishes into $$$n$$$ meals with each appetizer and each main dish being included in exactly one meal.

Your meals shall surprise the diners, so you want the spiciness levels of the two parts of the same meal to be as different as possible. The charm of a meal is the difference (in absolute value) between the spiciness of the appetizer and the spiciness of the main dish. So, a meal consisting of an appetizer with spiciness $$$x$$$ and a main dish with spiciness $$$y$$$ has charm equal to $$$|x-y|$$$.

You want to maximize the minimum charm of the resulting $$$n$$$ meals. What is the largest possible value of the minimum charm that you can achieve?

Input: Each test contains multiple test cases. The first line contains an integer $$$t$$$ ($$$1\le t\le 1\,000$$$) — the number of test cases. The descriptions of the $$$t$$$ test cases follow.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5\,000$$$) —the number of appetizers and main dishes.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ appetizers.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$0 \leq b_i \leq 10^{9}$$$) — the spicinesses of the $$$n$$$ main dishes.

It is guaranteed that the sum of $$$n^2$$$ over all test cases does not exceed $$$25\cdot 10^6$$$.

Output: For each test case, print the largest possible value of the minimum charm you can achieve.

Examples:
input:
4
3
0 0 0
1000000000 1000000000 1000000000
5
1 2 3 4 5
1 2 3 4 5
6
0 0 0 100 100 100
100 100 100 0 0 0
7
14 25 62 74 86 95 12
51 62 71 72 92 20 84
output:
1000000000
2
100
30

Note: In the first test case, no matter how you pair up the appetizers with the main dishes, each meal will have an appetizer with spiciness $$$0$$$ and a main dish with spiciness $$$1000000000$$$, so the charm of each meal will be $$$1000000000$$$.

In the second test case, one optimal way to pair up appetizers and main dishes is: $$$(1, 5)$$$, $$$(2, 4)$$$, $$$(3, 1)$$$, $$$(4, 2)$$$, $$$(5, 3)$$$. The corresponding meals have charms: $$$4$$$, $$$2$$$, $$$2$$$, $$$2$$$, $$$2$$$. The resulting minimum charm is $$$2$$$.

In the third test case, one way to maximize the minimum charm is to pair up the three appetizers with spiciness $$$0$$$ with the three main dishes with spiciness $$$100$$$, and the three appetizers with spiciness $$$100$$$ with the three main dishes with spiciness $$$0$$$. Doing so, the charm of each meal will be exactly $$$100$$$.

Buggy Code:
import java.util.*;

public class Main {
    public static int processSolution(int[] arr,int[] brr ,int mid,int n,int testCase) {
        int ans=Integer.MAX_VALUE;
        for(int i=0;i<n;i++){
            ans=Math.min(ans,Math.abs(arr[i]-brr[mid%n]));
            //System.out.print(arr[i]+" - "+brr[mid%n]+" "+i+" "+mid+" "+mid%n+"\t");
            mid++;
        }
        //System.out.println("\n"+ans);
        return ans;

    }

    public static int processArray(int[] arr,int[] brr ,int n,int testCase) {

        int ans=Integer.MAX_VALUE;
        Arrays.sort(arr);
        Arrays.sort(brr);
        int mid=n/2;
        if((n&1)==1)
        ans=processSolution(arr, brr, mid,n, testCase);
        else
        ans=Math.max(processSolution(arr, brr, mid+1,n, testCase),processSolution(arr, brr, mid,n, testCase));

        //System.out.println("check this: "+(testCase+1));

        return ans;

        // for (int i = 0; i < arr.length; i++) {
        //     System.out.print(arr[i] + " ");
        // }
        // System.out.println();
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int testCase = 0; testCase < t; testCase++) {
            int n = scanner.nextInt();
            int[] brr = new int[n];
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = scanner.nextInt();
            }

            for (int i = 0; i < n; i++) {
                brr[i] = scanner.nextInt();
            }
            System.out.println(processArray(arr,brr,n,testCase));

            //System.out.println("Test Case " + (testCase + 1) + ": " + processArray(arr,brr,n,testCase));
        }

        scanner.close();
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

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

#define lli long long int
#define ulli unsigned long long int
#define vti vector<int>
#define vtvi vector<vector<int>>
#define vtl vector<long long int>
#define vts vector<string>
#define vtb vector<bool>
#define pb push_back
#define ppi pair<int,int>
#define ppli pair<long long int,long long int>
#define mi map<int,int>

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

const int inf=INT_MAX;
const int _inf=INT_MIN;

void solveTestCase() {

    lli x,y,z,k,ans=0;
    cin>>x>>y>>z>>k;

    vti dx,dy,dz;

    for(int i=1;i<=x;i++){
        if(k%i == 0) dx.pb(i);
    }

    for(int i=1;i<=y;i++){
        if(k%i == 0) dy.pb(i);
    }

    for(int i=1;i<=z;i++){
        if(k%i == 0) dz.pb(i);
    }

    for(int l=0;l<dx.size();l++){
        for(int r=0;r<dy.size();r++){
            for(int m=0;m<dz.size();m++){
                if(1ll*dx[l]*dx[r]*dx[m] == k){
                    ans=max(ans,(x-dx[l]+1)*(y-dx[r]+1)*(z-dx[m]+1));
                }
            }
        }
    }

    cout<<ans<<endl;
}

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int t=1;
    cin>>t;
    while(t--){
        solveTestCase();
    }
}

**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
10 10 10 1000
```

**Explanation:**

- The test case is designed to challenge the buggy code by using a relatively large volume (`k = 1000`) and a box with side lengths (`x = 10`, `y = 10`, `z = 10`) that are large enough to fit multiple configurations of the secret box `S`.

- The volume `1000` can be factored into several combinations of side lengths, such as `10 * 10 * 10`, `20 * 5 * 10`, `25 * 4 * 10`, etc. However, the buggy code incorrectly checks the product of the side lengths using `dx[l]*dx[r]*dx[m]` instead of `dx[l]*dy[r]*dz[m]`.

- This test case will expose the bug because the code will incorrectly compare `10 * 10 * 10` to `1000` using `dx[l]*dx[r]*dx[m]`, which will always be incorrect.

- The correct answer for this test case is `1000`, which is the maximum number of distinct locations where the secret box can be placed. However, the buggy code will likely output a smaller number due to the incorrect comparison.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

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

int main() {

    int t;
    cin>>t;
    while(t--)
    {
      ll x,y,z,k;
      cin>>x>>y>>z>>k;
      vector<int> vec(3);
      vec[0]=x;
      vec[1]=y;
      vec[2]=z;
      sort(vec.begin(),vec.end());

      int a=1;
      int b=1;
      int c=1;
    //   int i=vec[0];
     for(int i=2;i<=vec[0];i++){
          if(k%i==0)
          {
              ll temp=k/i;
              if(vec[2]*vec[1]>=temp)
              {
                //   k=k/i;
                  a=i;
                //   break;
              }
          }

      }
      k=k/a;
    //   i=vec[0];
    // cout<<k<<endl;
      for(int i=a;i<=vec[1];i++)
      {
          if(k%i==0)
          {
              ll temp=k/i;
              if(temp<=vec[2])
              {
                  b=i;
                  c=temp;
                //   break;
              }
          }

      }
    //   cout<<a<<" "<<b<<" "<<c<<endl;
    if(a*b*c<k)
    cout<<"0"<<endl;
else{    vector<int> vec1(3);
    vec1[0]=a;
      vec1[1]=b;
      vec1[2]=c;
       sort(vec1.begin(),vec1.end());
    ll ans=1;
      ans=ans*(vec[0]-vec1[0]+1);
      ans=ans*(vec[1]-vec1[1]+1);
      ans=ans*(vec[2]-vec1[2]+1);
      cout<<max(ll(0),ans)<<endl;
    }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
2000 2000 2000 4000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
// Kaustubh Mhatre
#include <bits/stdc++.h>
using namespace std;
#define fastio()                    \
  ios_base::sync_with_stdio(false); \
  cin.tie(NULL);                    \
  cout.tie(NULL)
#define deb1(x) cout << #x << " = " << x << endl
#define deb2(x, y) cout << #x << " = " << x << "," << #y << "=" << y << endl
#define deb3(x, y, z) cout << #x << " = " << x << "," << #y << "=" << y << "," << #z << " = " << z << endl
typedef long long  ll;
typedef long double ld;
#define nline "\n"
#define pb push_back
#define ppb pop_back
#define lb lower_bound
#define ub upper_bound
#define str string
#define py cout<<"YES\n";
#define pn cout<<"NO\n";
#define all(x) (x).begin(), (x).end()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define repd(i, a, b) for (int i = a; i >= b; i--)
#define hmap unordered_map
#define hset unordered_set
#define ff first
#define vr vector
#define str string
#define ss second
#define nl cout<<"\n"
#define sz(x) (long long int)x.size()
typedef pair<ll, ll> pll;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<pll> vpl;
typedef vector<ll> vl;

// int MOD = 1e9 + 7;
int MOD = 998244353;
// int MOD = 998244353;
void pri(vector<long long> &v)
{
  for (int i = 0; i < v.size(); i++)
    cout << v[i] << " ";
  cout << "\n";
}
#define tt  \
  ll t;     \
  cin >> t; \
  while (t--)
int specific_mod = 998244353;
#ifdef Kaustubh
#define debug(x)     \
  cerr << #x << " "; \
  _print(x);         \
  cerr << endl;
#else
#define debug(x) ;
#endif

long long moduloMultiplication(long long a, long long b,
                               long long mod)
{
  long long res = 0; // Initialize result

  // Update a if it is more than
  // or equal to mod
  a %= mod;

  while (b)
  {
    // If b is odd, add a with result
    if (b & 1)
      res = (res + a) % mod;

    // Here we assume that doing 2*a
    // doesn't cause overflow
    a = (2 * a) % mod;

    b >>= 1; // b = b / 2
  }

  return res;
}

ll len =1e7+1;
bool prime[100000001];

// vl divisors;
// ll getCount(ll num){
//     ll cnt =0;
//     if(num==1) return 1;
//     // deb1(num);
//     for(ll i=1;i<=(num/2)+1;i++){
//         if(num % i ==0 ){
//             ll n1 = i;
//             ll n2 = num/i;
//             cnt++;
//             if(n1 != n2 )cnt++;
//         }

//     }
//     // deb1(cnt);
//     return cnt;
// }
void sieve(){

    for(ll i=2;i*i<=len;i++){
        if(prime[i]) continue;

        for(ll j = i*2; j<=len ; j+=i){
            prime[j] = true;
        }
    }

}

int add(int x, int y)
{
    return ((x + y) % MOD + MOD) % MOD;
}

int mul(int x, int y)
{
    return x * 1ll * y % MOD;
}

int binpow(int x, int y)
{
    int z = 1;
    while(y)
    {
        if(y % 2 == 1) z = mul(z, x);
        x = mul(x, x);
        y /= 2;
    }
    return z;
}

int inv(int x)
{
    return binpow(x, MOD - 2);
}

int divide(int x, int y)
{
    return mul(x, inv(y));
}
const ll MAX = 2e5 + 7;
vector<ll> fact(MAX);

ll nCr(ll n, ll r) {
	if(n < r) return 0;
	return divide(fact[n], mul(fact[r], fact[n - r]));
}

void preFactorial() {
	fact[0] = 1;
	for(ll i = 1; i < MAX; i++) fact[i] = mul(i, fact[i - 1]);
}

vl v;
// ll n,m, k ,d , q ;
// void in(){
//     cin>>n;
//     v.resize(n);
//     rep(i,0,n) cin>>v[i];
// }

void solve(){
    ll x,y,z,k; cin>>x>>y>>z>>k;

    vl arr;
    ll i=2;
    while(k>1){
        while(k>1 and   (k%i==0) ){
            k/=i;
            arr.pb(i);
        }
        if(i==2) i++;
        else i+=2;
    }
    // for(auto ele : arr) cout<<ele<<" ";
    // nl;
    ll n1=1,n2=1,n3=1;
    ll len = sz(arr);
    i=len-1;
    while(i>=2){
        n1 *= arr[i];
        n2 *=arr[i-1];
        n3 *=arr[i-2];
        i-=3;
    }
    // n1 is maxi  n2 then n3
    if(i==1) {
        n3 *= arr[1]; n2 *=arr[0];
    }
    else if(i==0) n3 *=arr[0];
    // deb3(n1,n2,n3);
    vl v1,v2;
    v1.pb(x); v1.pb(y); v1.pb(z);
    v2.pb(n1); v2.pb(n2); v2.pb(n3);
    sort(all(v1)); sort(all(v2));
    ll ans =1;
    rep(i,0,3){
        ll val = v1[i] - v2[i] + 1;
        ans *=val;
    }
    // for(auto ele : v2) cout<<ele<<" ";
    // nl;
    if(ans < 0) ans =0;
    // deb1(ans);
    cout<<ans<<nline;
    // ll ans = (  )
}

int main()
{
#ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
#endif

    // sieve();
    // preFactorial();
    // generate();
  tt
  {
    solve();
  }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has a few potential issues, such as handling large numbers and ensuring the dimensions of the secret box \( S \) fit within the box \( B \).

### Step-by-Step Reasoning:

1. **Edge Cases for Dimensions**:
   - **Small Dimensions**: Test cases where \( x, y, z \) are small but \( k \) is large.
   - **Large Dimensions**: Test cases where \( x, y, z \) are large and \( k \) is large.
   - **Equal Dimensions**: Test cases where \( x = y = z \) and \( k \) is a perfect cube.

2. **Handling Large Numbers**:
   - The code uses a factorization approach to determine the dimensions of \( S \). Test cases where \( k \) is a large number with many factors can expose issues in the factorization logic.

3. **Boundary Conditions**:
   - Test cases where \( k \) is exactly the volume of \( B \).
   - Test cases where \( k \) is 1.

### Chosen Test Input:
Let's create a test input that includes a combination of these edge cases:

1. **Small Dimensions with Large \( k \)**:
   - \( x = 3 \), \( y = 3 \), \( z = 3 \), \( k = 27 \)
   - \( x = 2 \), \( y = 2 \), \( z = 2 \), \( k = 8 \)
   - \( x = 1 \), \( y = 1 \), \( z = 1 \), \( k = 1 \)

2. **Large Dimensions with Large \( k \)**:
   - \( x = 1800 \), \( y = 1800 \), \( z = 1800 \), \( k = 4913000000 \)
   - \( x = 2000 \), \( y = 2000 \), \( z = 2000 \), \( k = 8000000000 \)

3. **Equal Dimensions**:
   - \( x = 4 \), \( y = 4 \), \( z = 4 \), \( k = 64 \)

4. **Boundary Conditions**:
   - \( x = 3 \), \( y = 3 \), \( z = 3 \), \( k = 27 \)
   - \( x = 2 \), \( y = 2 \), \( z = 2 \), \( k = 1 \)

### Test Input:
```plaintext
10
3 3 3 27
2 2 2 8
1 1 1 1
1800 1800 1800 4913000000
2000 2000 2000 8000000000
4 4 4 64
3 3 3 27
2 2 2 1
2 2 2 7
3 4 2 12
```

This test input includes a variety of cases that should help expose any issues in the buggy code, particularly in handling large numbers and ensuring the dimensions of \( S \) fit within \( B \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
from math import sqrt

for _ in range(int(input())):
       x,y,z,k=map(int,input().split())
       mn,mx=min(x,y,z),max(x,y,z)
       md=(x+y+z)-mn-mx
       for i in range(mn,0,-1):
              if k%i==0:
                     a=i
                     break
       k//=a
       for i in range(int(sqrt(k)),k+1):
              if i==0:
                     continue
              if i*(k//i)==k:
                     b,c=i,k//i
                     break

       ln,lx=min(a,b,c),max(a,b,c)
       ld=(a+b+c)-ln-lx
       m=(md-ld+1)*(mx-lx+1)*(mn-ln+1)
       m=0 if m<0 else m
       print(f'{m}')

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

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

Think step by step.

Response:
```plaintext
3
1 1 1 1
2000 2000 2000 2000
1000 1000 1000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
import collections
import functools
import itertools

from bisect import bisect_left as bsl, bisect_right as bsr

cache = getattr(functools, "cache", functools.lru_cache)
ri = lambda: int(input())
rii = lambda: [int(x) for x in input().split()]
inf = float("inf")
mod = 10**9 + 7
ordi = lambda c: ord(c) - (ord("a") if c.lower() else ord("A"))
polyhash = lambda substr, base, modp, key: functools.reduce(
    lambda acc, update: (base * acc + key(update)) % modp, substr, 0
)

def solve(x, y, z, k):
    x, y, z = sorted([x, y, z])

    def npos(a, b, c):
        if a > x or b > y or c > z:
            return 0
        return (x - a + 1) * (y - b + 1) * (z - c + 1)

    ck = int(k ** (1.0 / 3)) + 1
    res = 0
    # count = 0
    for i in range(1, 2001):
        for j in range(i, 2001):
            if k % (i * j) == 0 and 1 <= (l := k // (i * j)) <= 2000:
                # count += 1
                res = max(res, npos(*sorted([i, j, l])))
    print(res)

if __name__ == "__main__":
    for _ in range(ri()):
        solve(*rii())

**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 bugs in the provided algorithm, we need to consider edge cases and scenarios that maximize the failure probability in faulty solutions. Here are the steps to think through this:

1. **Edge Cases for Dimensions and Volume**:
   - Very small dimensions for the box \( B \) where the volume \( k \) is very small.
   - Very large dimensions for the box \( B \) where the volume \( k \) is very large.
   - Cases where \( k \) is a prime number, making it difficult to find suitable dimensions for \( S \).
   - Cases where \( k \) is a perfect cube, making it easy to find a solution but tricky to place the box in multiple positions.
   - Cases where \( k \) is a product of three large numbers, which might cause overflow or precision issues.

2. **Special Cases**:
   - \( B \) has one or more dimensions equal to 1.
   - \( B \) has one or more dimensions equal to the cube root of \( k \).

3. **Combination of the Above Cases**:
   - Mix small and large dimensions with small and large volumes to test the robustness of the algorithm.

Based on these considerations, here is a small-scale test input that includes a mix of the above scenarios:

```plaintext
6
1 1 1 1
2 2 2 8
3 3 3 27
2000 2000 2000 1
2000 2000 2000 2000
2000 2000 2000 4000000000
```

### Explanation:
1. **1 1 1 1**:
   - The box \( B \) has dimensions \( 1 \times 1 \times 1 \) and the volume \( k \) is 1. The only possible secret box \( S \) is \( 1 \times 1 \times 1 \) and it can only be placed in one location: \((0, 0, 0)\).

2. **2 2 2 8**:
   - The box \( B \) has dimensions \( 2 \times 2 \times 2 \) and the volume \( k \) is 8. The secret box \( S \) can be \( 2 \times 2 \times 2 \) and can be placed in 8 different locations: \((0, 0, 0)\), \((1, 0, 0)\), \((0, 1, 0)\), \((0, 0, 1)\), \((1, 0, 1)\), \((1, 1, 0)\), \((0, 1, 1)\), \((1, 1, 1)\).

3. **3 3 3 27**:
   - The box \( B \) has dimensions \( 3 \times 3 \times 3 \) and the volume \( k \) is 27. The secret box \( S \) can be \( 3 \times 3 \times 3 \) and can be placed in 27 different locations: all integer coordinates from \((0, 0, 0)\) to \((2, 2, 2)\).

4. **2000 2000 2000 1**:
   - The box \( B \) has dimensions \( 2000 \times 2000 \times 2000 \) and the volume \( k \) is 1. The secret box \( S \) can be \( 1 \times 1 \times 1 \) and can be placed in \( 2000 \times 2000 \times 2000 = 8,000,000,000 \) different locations.

5. **2000 2000 2000 2000**:
   - The box \( B \) has dimensions \( 2000 \times 2000 \times 2000 \) and the volume \( k \) is 2000. The secret box \( S \) can be \( 2000 \times 1 \times 1 \) or any permutation thereof, and can be placed in \( 2000 \times 2000 \times 1 = 4,000,000 \) different locations.

6. **2000 2000 2000 4000000000**:
   - The box \( B \) has dimensions \( 2000 \times 2000 \times 2000 \) and the volume \( k \) is 4,000,000,000. The secret box \( S \) can be \( 2000 \times 2000 \times 1000 \) or any permutation thereof, and can be placed in \( 1 \times 1 \times 1000 = 1,000 \) different locations.

This test input covers a range of scenarios, from simple to complex, and should help expose any bugs in the provided algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
#!/usr/bin/python3

import bisect
import collections
import heapq
import itertools
import math
import os
import sys

def main():
    for _ in range(read_int()):
        print(solve(*read_ints()))

def solve(x, y, z, k):
    best = 0

    def calc2(sx, sy, k):
        nonlocal best

        if k <= z:
            sz = k
            best = max(best, (x - sx + 1) * (y - sy + 1) * (z - sz + 1))

    def calc1(sx, k):
        for sy in range(1, min(y, k) + 1):
            if sy * sy > k:
                break
            if k % sy != 0:
                continue
            sy1 = k // sy
            calc2(sx, sy, k // sy)
            if sy1 != sy:
                calc2(sx, sy1, k // sy1)

    for sx in range(1, min(x, k) + 1):
        if sx * sx > k:
            break
        if k % sx != 0:
            continue
        sx1 = k // sx
        calc1(sx, k // sx)
        if sx1 != sx:
            calc1(sx1, k // sx1)

    return best

###############################################################################

DEBUG = 'DEBUG' in os.environ

def inp():
    return sys.stdin.readline().rstrip()

def read_int():
    return int(inp())

def read_ints():
    return [int(e) for e in inp().split()]

def list2d(d1, d2, init=None):
    return [[init] * d2 for _ in range(d1)]

def list3d(d1, d2, d3, init=None):
    return [[[init] * d3 for _ in range(d2)] for _ in range(d1)]

def list4d(d1, d2, d3, d4, init=None):
    return [[[[init] * d4 for _ in range(d3)] for _ in range(d2)]
            for _ in range(d1)]

def debug(fmt, *args):
    if DEBUG:
        if args:
            print(fmt.format(*args), file=sys.stderr)
        else:
            print(fmt, file=sys.stderr)

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that might cause the code to fail. Here are some potential issues to consider:

1. **Edge Cases for Dimensions and Volume**:
   - When the volume \( k \) is a large number that might exceed the limits of a 32-bit integer.
   - When the dimensions \( x, y, z \) are very small, making it impossible to fit a box with volume \( k \).
   - When \( k \) is a prime number, which means it can only be factored as \( 1 \times k \).

2. **Special Cases for Divisibility**:
   - When \( k \) is a perfect cube, which might cause the code to miss the correct dimensions.
   - When \( k \) is a product of two large numbers, which might cause the code to skip the correct dimensions.

3. **Boundary Conditions**:
   - When \( x, y, z \) are equal to \( k \), which means the box \( S \) fits perfectly into \( B \).

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

```plaintext
5
1 1 1 1
2 2 2 8
3 3 3 27
4 4 4 64
5 5 5 125
```

### Explanation of Test Cases:
1. **Test Case 1: (1, 1, 1, 1)**
   - The box \( B \) has dimensions \( 1 \times 1 \times 1 \) and the volume \( k \) is 1.
   - The secret box \( S \) can only be \( 1 \times 1 \times 1 \) and can be placed in one location.
   - Expected output: 1

2. **Test Case 2: (2, 2, 2, 8)**
   - The box \( B \) has dimensions \( 2 \times 2 \times 2 \) and the volume \( k \) is 8.
   - The secret box \( S \) can be \( 2 \times 2 \times 2 \) and can be placed in 8 locations.
   - Expected output: 8

3. **Test Case 3: (3, 3, 3, 27)**
   - The box \( B \) has dimensions \( 3 \times 3 \times 3 \) and the volume \( k \) is 27.
   - The secret box \( S \) can be \( 3 \times 3 \times 3 \) and can be placed in 27 locations.
   - Expected output: 27

4. **Test Case 4: (4, 4, 4, 64)**
   - The box \( B \) has dimensions \( 4 \times 4 \times 4 \) and the volume \( k \) is 64.
   - The secret box \( S \) can be \( 4 \times 4 \times 4 \) and can be placed in 64 locations.
   - Expected output: 64

5. **Test Case 5: (5, 5, 5, 125)**
   - The box \( B \) has dimensions \( 5 \times 5 \times 5 \) and the volume \( k \) is 125.
   - The secret box \( S \) can be \( 5 \times 5 \times 5 \) and can be placed in 125 locations.
   - Expected output: 125

This test input includes cases where \( k \) is a perfect cube, which should challenge the code to correctly identify the dimensions and count the number of placements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
import javax.management.StringValueExp;
import javax.print.DocFlavor;
import javax.swing.*;
import java.io.*;
import java.text.CollationElementIterator;
import java.util.*;
import java.util.concurrent.ConcurrentHashMap;
import java.util.concurrent.ScheduledExecutorService;
import java.util.concurrent.atomic.AtomicReference;
import java.util.logging.LoggingPermission;

public class a4 {
    // creating custom pair class
    static class pair implements Comparable<pair>{
        int x;
        int y;
        pair(int x,int y){
            this.x=x;
            this.y=y;
        }
        public int compareTo(pair a) {
            return this.y-a.y;     //in ascending order
        }
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            pair pair = (pair) o;
            return x == pair.x && y == pair.y;
        }
        @Override
        public int hashCode() {
            return Objects.hash(x, y);
        }
    }
    static class pair1{
        long x;
        long y;
        long z;
        pair1(long x,long y,long z){
            this.x=x;
            this.y=y;
            this.z=z;
        }
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            pair1 pair = (pair1) o;
            return x == pair.x && y == pair.y && z==pair.z;
        }
        @Override
        public int hashCode() {
            return Objects.hash(x, y);
        }
    }
    static long sum(long n){
        return n*(n+1)/2;
    }
    static long gcd(long a,long b) {//a is greater than b  lcm*hcf=a*b
        //Eular's method for finding gcd
        long d = a;
        long q = b;
        while (true) {
            long r = d % q;
            if (r == 0) {
                break;
            } else {
                d = q;
                q = r;
            }
        }
        return q;
    }
    public static void main(String[] args) throws IOException {
        FastScanner sc = new FastScanner();
        PrintWriter e = new PrintWriter(System.out);
        int T = sc.nextInt();
        while (T-- > 0) {
            long x=sc.nextLong();
            long y=sc.nextLong();
            long z=sc.nextLong();
            long k=sc.nextLong();
            long max=Math.max(x,Math.max(y,z));
            ArrayList<Long>factor=new ArrayList<>();
            for(long i=1;i*i<=k;i++){
                if(k%i==0){
                    long r1=i;
                    long r2=k/i;
                    if(r1==r2){
                        if(r1<=max){
                            factor.add(r1);
                        }
                    }
                    else{
                        if(r1<=max){
                            factor.add(r1);
                        }
                        if(r2<=max){
                            factor.add(r2);
                        }
                    }
                }
            }
            long x1=Math.min(x,Math.min(y,z));
            long z1=Math.max(x,Math.max(y,z));
            long y1=(x+y+z)-(x1+z1);
            long cnt=0;
            for(int i=0;i<factor.size();i++){
                if(factor.get(i)>x1){
                    break;
                }
                for(int j=i;j<factor.size();j++){
                    if(factor.get(j)>y1){
                        break;
                    }
                    for(int l=j;l<factor.size();l++){
                        if(factor.get(l)>z1){
                            break;
                        }
                        if(factor.get(i)*factor.get(j)*factor.get(l)==k){
//                            System.out.println(factor.get(i)+" "+factor.get(j)+" "+factor.get(l));
                            cnt=Math.max(cnt,(x1-factor.get(i)+1)*(y1-factor.get(j)+1)*(z1-factor.get(l)+1));
                        }
                    }
                }
            }
            System.out.println(cnt);
        }
        e.flush();
        e.close();
    }
    private static class FastScanner {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        private FastScanner() throws IOException {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private short nextShort() throws IOException {
            short ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = (short) (ret * 10 + c - '0');
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return (short) -ret;
            return ret;
        }

        private int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        private char nextChar() throws IOException {
            byte c = read();
            while (c <= ' ') c = read();
            return (char) c;
        }

        private String nextString() throws IOException {
            StringBuilder ret = new StringBuilder();
            byte c = read();
            while (c <= ' ') c = read();
            do {
                ret.append((char) c);
            } while ((c = read()) > ' ');
            return ret.toString();
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }
    }
}

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

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

Think step by step.

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

1. **Edge Cases for Dimensions and Volume**:
   - Test cases where one or more dimensions of the box \(B\) are small.
   - Test cases where the volume \(k\) is a large number that may cause overflow or precision issues.
   - Test cases where the volume \(k\) is a prime number or a power of a prime number, which can limit the number of valid dimensions for \(S\).

2. **Edge Cases for Factorization**:
   - Test cases where \(k\) has a large number of factors.
   - Test cases where \(k\) is a perfect cube, which can have multiple valid dimensions.

3. **Boundary Conditions**:
   - Test cases where the dimensions of \(B\) are just large enough to fit \(S\).
   - Test cases where the dimensions of \(B\) are just small enough to not fit \(S\).

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
4
1 1 1 1
2 2 2 8
3 3 3 27
4 4 4 64
```

### Explanation:
1. **Test Case 1: \(1 \times 1 \times 1\) with \(k = 1\)**:
   - The only possible dimension for \(S\) is \(1 \times 1 \times 1\).
   - There is only one way to place \(S\) inside \(B\).

2. **Test Case 2: \(2 \times 2 \times 2\) with \(k = 8\)**:
   - The only possible dimension for \(S\) is \(2 \times 2 \times 2\).
   - There are 8 ways to place \(S\) inside \(B\).

3. **Test Case 3: \(3 \times 3 \times 3\) with \(k = 27\)**:
   - The only possible dimension for \(S\) is \(3 \times 3 \times 3\).
   - There are 27 ways to place \(S\) inside \(B\).

4. **Test Case 4: \(4 \times 4 \times 4\) with \(k = 64\)**:
   - The only possible dimension for \(S\) is \(4 \times 4 \times 4\).
   - There are 64 ways to place \(S\) inside \(B\).

These test cases cover a range of scenarios, including small and large dimensions, and volumes that are perfect cubes. They are designed to challenge the code's ability to handle factorization and placement logic correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
import java.util.*;

public class E_Secret_Box{
    public static  void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            long[] arr = new long[3];
            arr[0] = sc.nextLong();
            arr[1] = sc.nextLong();
            arr[2] = sc.nextLong();
            long k = sc.nextLong();

            List<Long> list = new ArrayList<>();
            primeFactors(k, list);
            // System.out.println(list);

            Arrays.sort(arr);
            long temp = arr[0];
            arr[0] = arr[2];
            arr[2] = temp;

            long[] side = new long[3];
            Arrays.fill(side, 1);
            int j = list.size()-1;
            int i = 0;
            int skip = 0;
            while(j>=0 && skip<=3*list.size()){
                i%=3;
                if(side[i]*list.get(j)<=arr[i]){
                    side[i] *= list.get(j);
                    j-=1;
                }else{
                    skip++;
                }
                i+=1;
            }
            // System.out.println("skip="+skip);
            // for(int ii=0; ii<3; ii++){
            //     System.out.print(arr[ii] + " ");
            // }
            // System.out.println();
            // for(int ii=0; ii<3; ii++){
            //     System.out.print(side[ii] + " ");
            // }
            // System.out.println();
            long ans = 1;
            for(int ii=0; ii<3; ii++){
                ans *= arr[ii]-side[ii]+1;
            }
            long mul = side[0]*side[1]*side[2];
            if(mul!=k){
                System.out.println(0);
            }else{
                System.out.println(ans);
            }
        }
    }
    public static void primeFactors(long n, List<Long> list)
    {
        while (n%2==0)
        {
            // System.out.print(2 + " ");
            list.add(2L);
            n /= 2;
        }

        for (long i = 3; i <= Math.sqrt(n); i+= 2)
        {
            while (n%i == 0)
            {
                // System.out.print(i + " ");
                list.add(i);
                n /= i;
            }
        }

        if (n > 2)
            list.add(n);
            // System.out.print(n);
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1 1 1
2 2 2 3
2000 2000 2000 2000
```

**Explanation:**
1. **Test Case 1:**
   - Dimensions of box \( B \): \( (1, 1, 1) \)
   - Volume of secret box \( S \): \( 1 \)
   - **Expected Output:** 1
   - **Reason:** The only possible dimension for \( S \) is \( (1, 1, 1) \), and it can only be placed at \( (0, 0, 0) \).

2. **Test Case 2:**
   - Dimensions of box \( B \): \( (2, 2, 2) \)
   - Volume of secret box \( S \): \( 3 \)
   - **Expected Output:** 0
   - **Reason:** There are no integer dimensions \( (a, b, c) \) such that \( a \times b \times c = 3 \) and \( a, b, c \leq 2 \).

3. **Test Case 3:**
   - Dimensions of box \( B \): \( (2000, 2000, 2000) \)
   - Volume of secret box \( S \): \( 2000 \)
   - **Expected Output:** 2000
   - **Reason:** The only possible dimension for \( S \) is \( (2000, 1, 1) \), and it can be placed in any of the 2000 positions along the x-axis.

These test cases cover edge cases where the dimensions of \( B \) and \( S \) are at the limits of the problem constraints, and where the volume of \( S \) is a prime number, which should expose any issues in the factorization and placement 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: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

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

#define ll long long

ll count (ll x,ll y, ll z,ll a,ll b,ll c) {
    //cout<<x<<" "<<y<<" "<<z<<endl;
    ll res = 1;
    res*=(a-x+1);
    res*=(b-y+1);
    res*=(c-z+1);
    return res == 1 ? 0 : res;
}

ll solve(ll x ,ll y, ll z,ll area){
    ll res = 0;
    //unordered_map<ll,int> mp1;
    //unordered_map<ll,int> mp2;
    for (ll a =1;a<=x;a++) {
        for (ll b = 1;b<=y;b++) {
            ll c = area/(a*b);
            if (a*b>area) {
                break;
            }
            if (area%(a*b)!=0 || c>z) {
                continue;
            }
            // ll aa = min({a,b,c});
            // ll bb = max({a,b,c});
            res = max(res,count(a,b,c,x,y,z));
//             if (mp1.find(aa) == mp1.end() && mp2.find(bb)==mp2.end()) {
//                 // mp1[aa]++;
//                 // mp2[bb]++;

//             }

        }
    }
    return res;
}

int main(){
    ll t;
    cin>>t;
    for (ll i=0;i<t;i++) {
        ll a,b,c,area;
        cin>>a>>b>>c>>area;
        cout<<solve(a,b,c,area)<<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
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
2 2 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

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

int main() {
    int t;
    cin >> t;
    while(t--){
        long long int a,b,c;
        cin>>a>>b>>c;
        long long int prod;
        prod=a*b*c;
        long long int inp;
        cin>>inp;
        if(prod<inp){
           cout<<0<<endl;
           continue;
        }
        long long int val1,val2,val3;
        long long int maxi=0;
        for(int i=1;i<=a;i++){
            if(inp%i==0){
                long long int temp=inp/i;
                for(int j=1;j<=b;j++){
                    if(temp%j==0){
                        val1=i;
                        val2=j;
                        val3= temp/j;
                    }
                }
                maxi=max(maxi,(long long int)(a-val1+1)*(b-val2+1)*(c-val3+1));
                maxi=max(maxi,(long long int)(a-val1+1)*(c-val2+1)*(b-val3+1));
                maxi=max(maxi,(long long int)(b-val1+1)*(a-val2+1)*(c-val3+1));
                maxi=max(maxi,(long long int)(b-val1+1)*(c-val2+1)*(a-val3+1));
                maxi=max(maxi,(long long int)(c-val1+1)*(a-val2+1)*(b-val3+1));
                maxi=max(maxi,(long long int)(c-val1+1)*(b-val2+1)*(a-val3+1));
            }
        }
        cout<<maxi<<endl;

    }
    return 0;
}

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

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

Think step by step.

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

1. **Small Dimensions with Large Volume**: The code might fail when the volume \( k \) is large relative to the dimensions \( x, y, z \).
2. **Prime Dimensions**: If \( x, y, \) or \( z \) are prime numbers, it can make it difficult to find valid dimensions for \( S \).
3. **Edge Cases**: Test cases where \( k \) is exactly the product of \( x, y, \) and \( z \), or where \( k \) is a small number like 1 or 2.
4. **Large \( k \) with Small Dimensions**: This can force the code to handle large numbers and potentially overflow or fail to find valid dimensions.

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

```plaintext
3
10 10 10 1000
11 11 11 1331
12 12 12 1
```

### Explanation:

1. **Test Case 1: \( 10 \times 10 \times 10 \) with \( k = 1000 \)**:
   - This is a simple case where the volume of \( S \) is exactly the volume of \( B \). The only valid dimension for \( S \) is \( 10 \times 10 \times 10 \), and there is only one way to place it.

2. **Test Case 2: \( 11 \times 11 \times 11 \) with \( k = 1331 \)**:
   - This is a case where \( k \) is a perfect cube (11^3). The only valid dimension for \( S \) is \( 11 \times 11 \times 11 \), and there is only one way to place it.

3. **Test Case 3: \( 12 \times 12 \times 12 \) with \( k = 1 \)**:
   - This is a case where \( k \) is 1. The only valid dimension for \( S \) is \( 1 \times 1 \times 1 \), and there are \( 12 \times 12 \times 12 = 1728 \) ways to place it.

These test cases cover a range of scenarios, including perfect cubes, large volumes, and small volumes, which can help expose potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
for _ in range(int(input())):
    x,y,z,k = map(int,input().split())
    ans = 0
    set1 = set()
    for i in range(1,max(z,y,x)+1):
        if k % i == 0:
            extra = k // i
            for j in range(1,max(x,y,z)+1):
                if extra % j == 0 :
                    if (i,j,extra//j) not in set1:
                        set1.add((i,j,extra//j))
                        set1.add((i,extra//j,j))
                        set1.add((j,i,extra//j))
                        set1.add((j,extra//j,i))
                        set1.add((extra//j,i,j))
                        set1.add((extra//j,j,i))
                 #       ans = max((x*y*z)//(k),ans)
                        ans = max((z-i+1)*(x-j+1)*(y-(extra//j)+1),(z-i+1)*(x-(extra//j)+1)*(y-j+1),(z-j+1)*(x-i+1)*(y-(extra//j)+1),(z-(extra//j)+1)*(x-i+1)*(y-j+1), (z-(extra//j)+1)*(x-j+1)*(y-i+1), (z-j+1)*(x-(extra//j)+1)*(y-i+1) ,ans)
                   #     print(j,extra//j,i,ans)
    print(ans)

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

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

Think step by step.

Response:
```plaintext
4
1 1 1 1
2 2 2 8
3 3 3 27
4 4 4 64
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
# Rookie : Tanmay Patil

def max_positions(x, y, z, k):

    val = max(x, y, z)
    valid_dimensions = []
    for a in range(1, int(k**(1/3)) + 10):
            if k % a == 0:
                for b in range(a, int((k // a)**0.5) + 5):
                    if (k // a) % b == 0:
                        c = k // (a * b)
                        if a * b * c == k:
                            if max(a,b,c) <= val :
                                valid_dimensions.append((a, b, c))

    ans = 0
    for a, b, c in valid_dimensions:
        ans = max(ans,(x - c + 1) * (y - b + 1) * (z - a + 1),(x - a + 1) * (y - b + 1) * (z - c + 1),(x - a + 1) * (y - c + 1) * (z - b + 1),(x - b + 1) * (y - a + 1) * (z - c + 1),(x - b + 1) * (y - c + 1) * (z - a + 1),(x - c + 1) * (y - a + 1) * (z - b + 1))

    return ans

def solve() :
    x,y,z,k = map(int, input().split())
    print(max_positions(x, y, z, k))

testCases = int(input())
for _ in range(testCases) :
    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
2000 2000 2000 4000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Secret Box
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ntarsis has a box $B$ with side lengths $x$, $y$, and $z$. It lies in the 3D coordinate plane, extending from $(0,0,0)$ to $(x,y,z)$.

Ntarsis has a secret box $S$. He wants to choose its dimensions such that all side lengths are positive integers, and the volume of $S$ is $k$. He can place $S$ somewhere within $B$ such that:

  * $S$ is parallel to all axes.    * every corner of $S$ lies on an integer coordinate.

$S$ is magical, so when placed at an integer location inside $B$, it will not fall to the ground.

Among all possible ways to choose the dimensions of $S$, determine the maximum number of distinct locations he can choose to place his secret box $S$ inside $B$. Ntarsis does not rotate $S$ once its side lengths are selected.

Input: The first line consists of an integer $t$, the number of test cases ($1 \leq t \leq 2000$). The description of the test cases follows.

The first and only line of each test case contains four integers $x, y, z$ and $k$ ($1 \leq x, y, z \leq 2000$, $1 \leq k \leq x \cdot y \cdot z$).

It is guaranteed the sum of all $x$, sum of all $y$, and sum of all $z$ do not exceed $2000$ over all test cases.

Note that $k$ may not fit in a standard 32-bit integer data type.

Output: For each test case, output the answer as an integer on a new line. If there is no way to select the dimensions of $S$ so it fits in $B$, output $0$.

Examples:
input:
7
3 3 3 8
3 3 3 18
5 1 1 1
2 2 2 7
3 4 2 12
4 3 1 6
1800 1800 1800 4913000000
output:
8
2
5
0
4
4
1030301

Note: For the first test case, it is optimal to choose $S$ with side lengths $2$, $2$, and $2$, which has a volume of $2 \cdot 2 \cdot 2 = 8$. It can be shown there are $8$ ways to put $S$ inside $B$.

The coordinate with the least $x$, $y$, and $z$ values for each possible arrangement of $S$ are:

  1. $(0, 0, 0)$    2. $(1, 0, 0)$    3. $(0, 1, 0)$    4. $(0, 0, 1)$    5. $(1, 0, 1)$    6. $(1, 1, 0)$    7. $(0, 1, 1)$    8. $(1, 1, 1)$

The arrangement of $S$ with a coordinate of $(0, 0, 0)$ is depicted below:

![](CDN_BASE_URL/5dd57f6f246d08830a2230a234fb9d02)

For the second test case, $S$ with side lengths $2$, $3$, and $3$ are optimal.

Buggy Code:
import java.util.*;

public class main {
    public static long[] findMinMultipliers(long d,long a1,long a2,long a3) {
        long minSum = Integer.MAX_VALUE;
        long[] minMultipliers = null;

        for (long a = 1; a <= Math.cbrt(d); a++) {
            if (d % a == 0) {
                for (long b = a; b <= Math.sqrt(d / a); b++) {
                    if ((d / a) % b == 0) {
                        long c = d / (a * b);

                        if (c >= b) {
                            long sum = a + b + c;
                            if (sum < minSum) {
                                minSum = sum;
                                minMultipliers = new long[]{a, b, c};
                            }
                        }
                    }
                }
            }
        }

        return minMultipliers;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            long a=sc.nextLong()+1,b=sc.nextLong()+1,c=sc.nextLong()+1,d=sc.nextLong();
            long[] as = new long[]{a,b,c};
            long[] ans = findMinMultipliers(d,a,b,c);
            Arrays.sort(as);
            Arrays.sort(ans);
            System.out.println(Math.max(0,((as[0]-ans[0])*(as[1]-ans[1])*(as[2]-ans[2]))));
        }
    }
}

**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
2000 2000 2000 4000000000
```