**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

Buggy Code:
import math as m
from collections import defaultdict as dd

def egcd(a,b):
    if(a > b):
        return egcd(b,a)
    else:
        if(a == 0):
            return b,0,1
        else:
            g,x,y = egcd(b%a,a)
            return g,y-(b//a)*x,x

# print(egcd(336718728,709848696))

if(__name__ == "__main__"):
    for i in range(int(input())):
        n,a,b = tuple(map(int,input().split()))
        C = list(set(map(int,input().split())))
        C.sort()

        if(n == 1):
            print(0)
            continue

        g = m.gcd(a,b)
        gaps = [C[i+1]-C[i] for i in range(len(C)-1)]
        l = []
        t = 0
        for i in range(len(C)-1):
            t = (t+(gaps[i]))%g
            l.append(t)

        l.sort()

        cl = [0]
        for i in range(len(l)):
            cl.append(cl[-1]+l[i])

        v = g-l[0]

        for i in range(1,len(l)):
            cv = g-l[i] + cl[i]
            if(cv > g):
                print(min(v,l[-1]))
                break
            else:
                v = min(cv,v)
        else:
            print(min(v,l[-1]))

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

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

Think step by step.

Response:
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 cause the code to fail. Here are the steps to think through this:

1. **Edge Case with Single Element Array**:
   - The code should handle the case where the array has only one element, which should always result in a range of 0.

2. **Edge Case with Identical Elements**:
   - If all elements in the array are the same, the range should also be 0, regardless of the values of `a` and `b`.

3. **Edge Case with Large Values**:
   - Use large values for `a` and `b` to test the code's ability to handle large integers and ensure it doesn't overflow or behave incorrectly.

4. **Edge Case with Small Values**:
   - Use small values for `a` and `b` to test the code's ability to handle small integers and ensure it doesn't get stuck in an infinite loop or produce incorrect results.

5. **Edge Case with Mixed Values**:
   - Use a mix of values for `a` and `b` to test the code's ability to handle different scenarios where the GCD might be a factor in minimizing the range.

6. **Edge Case with GCD**:
   - Use values for `a` and `b` where their GCD is a significant factor in the range calculation.

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

```plaintext
5
1 1 1
1
3 1 1
1 1 1
3 1000000000 1000000000
1 2 3
3 2 3
1 3 4
3 5 7
1 4 7
```

Explanation:
- The first test case has a single element, which should result in a range of 0.
- The second test case has all elements identical, which should also result in a range of 0.
- The third test case has large values for `a` and `b`, testing the code's ability to handle large integers.
- The fourth test case has small values for `a` and `b`, testing the code's ability to handle small integers.
- The fifth test case has mixed values for `a` and `b`, testing the code's ability to handle different scenarios where the GCD might be a factor.

This test input should help expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

Buggy Code:
import java.util.*;

public class Codechef {
 public static long gcd(long n1, long n2) {
  return n1 == 0 ? n2 : gcd(n2 % n1, n1);
 }

 public static void main(String args[]) {
  Scanner sc = new Scanner(System.in);
  int t = sc.nextInt();
  while (t-- > 0) {
   int n = sc.nextInt();
   long a = sc.nextLong(), b = sc.nextLong(), ar[] = new long[n];

   long d = gcd(a, b);
   for (int i = 0; i < n; i++) {
    ar[i] = sc.nextLong() % d;
   }
   Arrays.sort(ar);
   long ans = ar[n - 1] - ar[0];

   if (d == 1) {
    System.out.println(0);
    continue;
   }

   for (int i = 0; i < n - 1; i++) {
    long v3 = ar[i];
    long dif = ar[n - 1] - ar[i];
    v3 += (dif / d) * d;
    if (v3 < ar[n - 1] - d / 2) {
     v3 += d;
    }
    ar[i] = v3;
   }

   Arrays.sort(ar);
   System.out.println(Math.min(ar[n - 1] - ar[0], ans));

  }
 }
}

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

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

Think step by step.

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

**Explanation:**
This test case is designed to expose the bug in the given buggy code. Here’s the reasoning:

1. **Initial Array and Constants:**
   - The array `c` is `[1, 2, 4]`.
   - The constants `a` and `b` are `2` and `3`, respectively.

2. **GCD Calculation:**
   - The GCD of `2` and `3` is `1`.

3. **Modulo Operation:**
   - The elements of the array `c` modulo `1` are `[0, 0, 0]` (since any number modulo `1` is `0`).

4. **Sorting:**
   - The sorted array remains `[0, 0, 0]`.

5. **Initial Range Calculation:**
   - The initial range is `0 - 0 = 0`.

6. **Special Case for GCD = 1:**
   - The code correctly identifies that if the GCD is `1`, the minimum possible range is `0`.

However, the buggy code has a special case for `d == 1` which immediately prints `0` and continues to the next test case. This is correct for this specific case, but the test input is designed to ensure that the code handles this case correctly without any additional unnecessary calculations.

This test case is small and valid, and it ensures that the special case for `d == 1` is correctly handled by the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

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

public class Main {
    static FastIO io = new FastIO();
    public static void main(String[] args) {
        int t = io.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    static void solve() {
        int n = io.nextInt();
        long a= io.nextInt();
        long b = io.nextInt();
        long mx = 0;
        long mn = (long) 1e9 + 1;
        long gcd = gcd(a, b);
        long[] arr = new long[n];
        PriorityQueue<Long> pq = new PriorityQueue<>();
        for (int i = 0; i < n; i++) {
            arr[i] = io.nextInt();
            arr[i] %= gcd;
            pq.add(arr[i]);
            mx = Math.max(arr[i], mx);
            mn = Math.min(mn, arr[i]);
        }

        long ans = mx - mn;
        while (n-- > 0) {
            long curr = pq.poll();
            curr += gcd;
            mx = curr;
            mn = pq.peek();
            pq.add(curr);
            ans = Math.min(ans, mx - mn);
        }
        System.out.println(ans);
    }

    public static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
}
class FastIO {
    public BufferedReader br;
    public StringTokenizer st;
    FastIO() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    public String next() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

Buggy Code:
import math
t=int(input())
for _ in range(t):
    n,a,b=map(int,input().split())
    l=list(map(int,input().split()))
    x = min(a,b,math.gcd(a,b))
    # print("X",x)
    res =[i%x for i in l]
    res = list(set(res))
    res.sort()
    # print("Res",res)
    if len(res)==1:
        print(res[0])
    else:
        ma = res[-1]
        mi = res[0]
        if ma-mi <=x//2:
            print(ma-mi)
            # print("Hello")
        else:
            d = ma - mi
            res[0] = x + res[0]
            for i in range(1,len(res)):
                if res[i-1]-res[i] >=d:
                    res[i] = x + res[i]
                else:
                    d = res[i-1]-res[i]
                    res[i] = x + res[i]
            print(d)
            # print("Hello1")

**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
5 10 15
1 11 21 31 41
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

Buggy Code:
def gcd(a, b):
	if a > b:
		a, b = b, a
	if a == 0:
		return b
	return gcd(a, b%a)

t = int(input())

for _ in range(t):
	n, a, b = map(int, input().split())
	c = list(map(int, input().split()))

	d = gcd(a, b)
	c = [x % d for x in c]
	c1 = [x + d if x <= d // 2 else x for x in c]
	ans = min(max(c)-min(c), max(c1)-min(c1))
	print(ans)

	# 1, 9, 5 --> 10, 9, 20
	# --> 19, 18, 20
	# 1 + 2(9), 9 + 1(9), 5 + 1(15)

	# add/sub combinations of 9 and 6
	# and/sub therefore combinations of 6 and 3
	# --> 3 and 3
	# --> 3 and 0

	# combinations of 4 and 5
	# --> 4 and 1
	# --> 3 and 1
	# --> 2 and 1
	# --> 1 and 0
	# --> 1

	# 7 and 9
	# 7 and 2
	# 2 and 5
	# 2 and 3
	# 2 and 1
	# 1 and 1
	# 1 and 0

	# is this just the gcd????
	# b > a
	# gcd(a, b) = gcd(a, b - a)

	# we can basically add or subtract numbers as we please
	# because subtracting n is equivalent to adding n to all other #s
	# equivalent to adding/subtracting any number of a's, b's, and |a-b|
	# OR adding/subtracting any number of min(a, b) and |a-b|

**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
3 15 9
1 9 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

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

using namespace std;

int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

int main() {
    // your code goes here
    int t;
    cin >> t;
    while (t--) {
        int n;
        int a,b;

        cin >> n >> a >> b;
        vector < int> v(n);
        for (int i = 0; i < n; i++) {
            cin >> v[i] ;
        }
        sort(v.begin(), v.end());

        int ans = v[n-1] - v[0] ;

        int g = gcd(a,b);
        // cout << "gcd: " << g << endl ;
        for (int i = 0; i < n; i++) {
            // replace with modulo g
            v[i] = v[i]%g ;
        }
        sort(v.begin(), v.end());

        ans = min(ans,v[n-1] - v[0] );
        // vector <long long int> c(n);

        for (int i = 0; i < n; i++) {
            if(g-v[i] < v[i]){
                v[i] = - (g-v[i]) ;
            }
        }
        sort(v.begin(), v.end());
        ans = min(ans,v[n-1] - v[0] );
        cout << ans << endl ;
    }

}

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

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

Think step by step.

Response:
```plaintext
3
5 2 3
1 2 3 4 5
5 10 15
10 20 30 40 50
5 1 1000000000
1 1000000000 2000000000 3000000000 4000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t, n, i, j, a, b;
    cin >> t;
    for (i = 0; i < t; i++)
    {
        cin >> n >> a >> b;
        int factor = __gcd(a, b);
        int arr[n], arr2[n];
        for (j = 0; j < n; j++)
            cin >> arr[j];
        for (j = 0; j < n; j++)
            {arr[j] %= factor;
            arr2[j]=arr[j];
            if (arr2[j] < factor / 2)
                arr2[j] += factor;
            }
        sort(arr,arr+n);
        int ans=arr[n-1]-arr[0];
        for(j=1;j<n;j++)
        ans=min(ans,arr[j-1]+factor-arr[j]);
        cout<<ans;
        int minElem = arr[0];
        int maxElem = arr[0];
        int minElem2 = arr2[0];
        int maxElem2 = arr2[0];
        for (j = 1; j < n; j++)
        {
            minElem = min(minElem, arr[j]);
            maxElem = max(maxElem, arr[j]);
            minElem2 = min(minElem2, arr2[j]);
            maxElem2 = max(maxElem2, arr2[j]);
        }
        int ans2 = min(maxElem - minElem, maxElem2 - minElem2);
        if(ans!=ans2){
            cout<<n<<' '<<a<<' '<<b<<" ";
            for(j=0;j<n;j++) cout<<arr[j]<<' ';
        }
        cout << '\n';
    }
    return 0;
}

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

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

Think step by step.

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

1. **Test Case with All Elements Equal**:
   - This case should ensure that the minimum range is 0, as no operations are needed.
   - Example: `4 5 5 1 1 1 1`

2. **Test Case with Elements Close to Each Other**:
   - This case should ensure that the minimum range is small, and the operations should be able to equalize the elements.
   - Example: `4 2 3 1 2 3 4`

3. **Test Case with Elements Far Apart**:
   - This case should ensure that the minimum range is large, and the operations should be able to reduce the range.
   - Example: `4 7 7 1 1 2 6`

4. **Test Case with Large Values of `a` and `b`**:
   - This case should ensure that the operations can significantly change the elements.
   - Example: `3 15 9 1 9 5`

5. **Test Case with Different `a` and `b`**:
   - This case should ensure that the operations can be used to minimize the range effectively.
   - Example: `3 18 12 1 4 5`

6. **Test Case with a Single Element**:
   - This case should ensure that the minimum range is 0, as no operations are needed.
   - Example: `1 1 1 1`

7. **Test Case with a Large Number of Elements**:
   - This case should ensure that the operations can be applied efficiently to a large array.
   - Example: `7 27 36 33 13 23 12 35 24 41`

8. **Test Case with a Single Operation**:
   - This case should ensure that a single operation can significantly reduce the range.
   - Example: `2 1 1000000000 1 1000000000`

9. **Test Case with a Large Range**:
   - This case should ensure that the operations can reduce a large range.
   - Example: `6 336718728 709848696 552806726 474775724 15129785 371139304 178408298 13106071`

10. **Test Case with a Small Range**:
    - This case should ensure that the operations can reduce a small range.
    - Example: `6 335734893 671469786 138885253 70095920 456876775 9345665 214704906 375508929`

Combining these test cases, we can create a small-scale test input that is complete and valid:

```plaintext
10
4 5 5 1 1 1 1
4 2 3 1 2 3 4
4 7 7 1 1 2 6
3 15 9 1 9 5
3 18 12 1 4 5
1 1 1 1
7 27 36 33 13 23 12 35 24 41
2 1 1000000000 1 1000000000
6 336718728 709848696 552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786 138885253 70095920 456876775 9345665 214704906 375508929
```

This test input covers a variety of scenarios and should help in exposing any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

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

public final class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		sc.useDelimiter("\\s+");
		int t = sc.nextInt();
		for (int tt = 0; tt < t; tt++) {
		    int n = sc.nextInt();
		    int a = sc.nextInt();
		    int b = sc.nextInt();
		    int temp;
		    int max = Integer.MIN_VALUE;
		    if (b < a) {
		        a = a + b;
		        b = a - b;
		        a = a - b;
		    }
		    List<Integer> array = new ArrayList<>();
		    for (int i = 0; i < n; i++) {
		        temp = sc.nextInt();
		        array.add(temp);
		        max = max > temp ? max : temp;
		    }
		    int ans = Collections.max(array) - Collections.min(array);
		    while (b % a != 0) {
		        temp = b % a;
		        b = a;
		        a = temp;
		    }
		    for (int i = 0; i < n; i++) {
		        temp = (((max - array.get(i))/a) * a) + array.get(i);
		        array.set(i, temp);
		    }
		    ans = Math.min(max - Collections.min(array), ans);
		    for (int i = 0; i < n; i++) {
		        if (array.get(i) + a - max < max - array.get(i)) {
		            array.set(i, array.get(i) + a);
		        }
		    }
		    ans = Math.min(ans, Collections.max(array) - Collections.min(array));
		  //  array.forEach(value -> {
		  //      System.out.print(value);
		  //      System.out.print(" ");
		  //  });
		    for (int i = 0; i < n; i++) {
		        if (array.get(i) < max) {
		            array.set(i, array.get(i) + a);
		        }
		    }
		    ans = Math.min(ans, Collections.max(array) - max);
		    System.out.println(ans);
		}

	}
}

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

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

Think step by step.

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

**Explanation:**
- This test case is designed to challenge the buggy code's handling of the greatest common divisor (GCD) and the way it modifies the array elements.
- The input values are small and simple, making it easy to verify the expected output.
- The array `[1, 2, 3, 4]` with `a = 2` and `b = 3` should ideally be transformed to minimize the range. The correct minimum range is `1`, which can be achieved by incrementing the smallest element (1) by `a` (2) to make it `3`, resulting in the array `[3, 2, 3, 4]` with a range of `2 - 1 = 1`.
- The buggy code might fail to correctly handle the GCD and the adjustments to the array elements, leading to an incorrect range.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

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

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str="";
            try {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static void sort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    public static void sort(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }
    static class DisjointSet {
        List<Integer> rank = new ArrayList<>();
        List<Integer> parent = new ArrayList<>();
        List<Integer> size = new ArrayList<>();
        public DisjointSet(int n) {
            for (int i = 0; i <= n; i++) {
                rank.add(0);
                parent.add(i);
                size.add(1);
            }
        }

        public int findUPar(int node) {
            if (node == parent.get(node)) {
                return node;
            }
            int ulp = findUPar(parent.get(node));
            parent.set(node, ulp);
            return parent.get(node);
        }

        public void unionBySize(int u, int v) {
            int ulp_u = findUPar(u);
            int ulp_v = findUPar(v);
            if (ulp_u == ulp_v) return;
            if (size.get(ulp_u) < size.get(ulp_v)) {
                parent.set(ulp_u, ulp_v);
                size.set(ulp_v, size.get(ulp_v) + size.get(ulp_u));
            } else {
                parent.set(ulp_v, ulp_u);
                size.set(ulp_u, size.get(ulp_u) + size.get(ulp_v));
            }
        }
    }
    static long power(long a, long b, long m) {
        a %= m;
        long res = 1;
        while (b > 0) {
            if ((b & 1) == 1) res = (res * a) % m;
            a = (a * a) % m;
            b >>= 1;
        }
        return res;
    }
    static int gcd(int a, int b)  {
        return b == 0? a: gcd(b, a % b);
    }
    public static int lcm_of_array(int a[]) {
        int lcm = a[0];
        for (int i = 1; i < a.length; i++) {
            int num1 = lcm;
            int num2 = a[i];
            int gcd_val = gcd(num1, num2);
            lcm = (lcm * a[i]) / gcd_val;
        }
        return lcm;
    }
    static class pair {
        int height;
        int diameter;
        pair(int x, int y) {
            this.height = x;
            this.diameter = y;
        }
    }
    public static void main(String args[]) throws IOException {
        FastReader sc = new FastReader();
        int T = sc.nextInt();
        while (T-- > 0) {
            int n = sc.nextInt();
            int a = sc.nextInt();
            int b = sc.nextInt();

            long arr[] = new long[n];
            for(int i=0; i<n; i++) {
                arr[i] = sc.nextLong();
            }

            int g = gcd(a, b);

            sort(arr);
            long min = arr[n-1];
            long max = arr[n-1];
            for(int i=n-2; i>=0; i--) {
                long l = (min-arr[i]+g-1)/g;
                long r = (max-arr[i])/g;

                if(l <= r) continue;
                long l1 = (min-arr[i])/g;
                long r1 = (max-arr[i]+g-1)/g;
                long n1 = arr[i] + Math.max(l1*g, r*g);
                long n2 = arr[i] + Math.min(r1*g, l*g);

                if(Math.abs(n1-max) < Math.abs(min-n2)) {
                    min = n1;
                } else {
                    max = n2;
                }
            }

            System.out.println((max-min));
        }

    }
}

**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
5 2 3
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

Buggy Code:
import heapq
import math
from collections import defaultdict, deque
'''
БИНАРНЫЙ ПОИСК
for i in range(strlen):
        left = 0
        right = alphlen
        while (left < right):
            mid = (left + right) // 2
            if alphabet[mid] < b[i]:
                left = mid + 1
            else:
                right = mid
'''

for _ in range(int(input())):
    n, a, b = tuple(map(int, input().split()))
    arr = list(map(int, input().split()))
    gcdab = math.gcd(a, b)
    arr = list(map(lambda x: x % gcdab, arr))
    #print(arr)
    diff1 = max(arr) - min(arr)
    arr1 = list(map(lambda x: x if x > gcdab // 2 else x + gcdab, arr))
    arr = list(map(lambda x: x if x >= gcdab // 2 else x + gcdab, arr))
    #print(arr)
    diff2 = max(arr) - min(arr)
    diff3 = max(arr1) - min(arr1)
    print(min(diff1, diff2, diff3))

**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
5 2 3
1 4 7 10 13
```

**Explanation:**
- **Test Case 1:**
  - **n = 5, a = 2, b = 3**
  - **Initial Array: [1, 4, 7, 10, 13]**

This test case is designed to expose the bug in the buggy code by creating a situation where the array elements are not uniformly distributed. The goal is to check how the code handles the minimum and maximum values after applying the operations.

- **Expected Output:**
  - The minimum possible range after any number of operations is **1**.

The input is small enough to be manageable and clearly demonstrates the issue with the buggy code. The elements of the array are chosen such that they are spaced out by multiples of the GCD of `a` and `b` (which is 1 in this case), making it a good test case to verify the correctness of the solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

Buggy Code:
import math as m
from collections import defaultdict as dd

def egcd(a,b):
    if(a > b):
        return egcd(b,a)
    else:
        if(a == 0):
            return b,0,1
        else:
            g,x,y = egcd(b%a,a)
            return g,y-(b//a)*x,x

# print(egcd(336718728,709848696))

def solve():
    n,a,b = tuple(map(int,input().split()))
    C = list(set(map(int,input().split())))
    C.sort()

    if(len(C)==1):
        print(0)
        return

    g = m.gcd(a,b)
    gaps = [C[i+1]-C[i] for i in range(len(C)-1)]
    l = []
    t = 0
    for i in range(len(C)-1):
        t = (t+(gaps[i]))%g
        l.append(t)

    l.sort()

    cl = [0]
    for i in range(len(l)):
        cl.append(cl[-1]+l[i])

    v = g-l[0]

    for i in range(1,len(l)):
        cv = g-l[i] + cl[i]
        if(cv > g):
            print(min(v,l[-1]))
            return
        else:
            v = min(cv,v)
    else:
        print(min(v,l[-1]))
        return

if(__name__ == "__main__"):
    for i in range(int(input())):
        solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

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

public class Main {
    static int broken = -1;

    static void solution() {
        int n = in.nextInt();
        int a = in.nextInt();
        int b = in.nextInt();
        int g = gcd(a, b);
        Integer[] c = new Integer[n];
        for (int i = 0; i < n; i++) {
            c[i] = in.nextInt();
        }
        Arrays.sort(c);
        for (int i = 0; i < n; i++) {
            c[i] %= g;
        }
        Arrays.sort(c);
        int ans = c[n - 1] - c[0];
        for (int i = 0; i < n; i++) {
            if (Math.abs(c[i] - g) < c[i]) {
                c[i] -= g;
            }
        }
        Arrays.sort(c);
        ans = Math.min(c[n - 1] - c[0], ans);
        for (int i = 0; i < n; i++) {
            if (c[i] > 0) {
                c[i] -= g;
            }
        }
        Arrays.sort(c);
        ans = Math.min(c[n - 1] - c[0], ans);
        out.println(ans);
    }

    static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    public static void main(String[] args) {
        in = new Scuffer();
        int n = in.nextInt();
        while (n-- > 0) {
            solution();
        }
        out.close();
    }

    public static PrintWriter out;
    public static Scuffer in;

    public static class Scuffer {
        BufferedReader br;
        StringTokenizer st;

        public Scuffer() {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new BufferedOutputStream(System.out));
        }

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

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

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

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

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

    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

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

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, a, b;
        cin >> n >> a >> b;
        int c = __gcd(a, b);

        int ans[10000];

        for (int i = 1; i <= n; i++) {
            cin >> ans[i];
            ans[i] = ans[i]%c;
        }

        sort(ans + 1, ans + n + 1);
        int result = ans[n] - ans[1];

        for (int i = 1; i < n; i++) {
            result = min(result, ans[i] + c - ans[i + 1]);
        }

        cout << result << 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
5 2 3
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

Buggy Code:
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast,no-stack-protector")
#include <bits/stdc++.h>
#define int long long
#define all(x) x.begin(),x.end()

using namespace std ;

const int inf = 0x3f3f3f3f ;
const int LLinf = 0x3f3f3f3f3f3f3f3f ;

/*
y1 = x * a + z * b + C1 ;
y2 = x * a + z * b + C2 ;
...
yn = x * a + z * b + Cn ;
( x , z ∈ N )
若 x == y :
y1 = x * a + C1
...
yn = x * a + Cn
相当于一堆斜率相同截距不同的直线各选出一个点后构成的序列极差最小。
*/

void solve()
{
	int n , a , b ;
	cin >> n >> a >> b ;
	a = std::__gcd(a , b) ;
	std::vector<int> c(n);
	for(auto & p : c) {
		cin >> p ;
		p %= a ;
	}
	int ans = *max_element(all(c)) - *min_element(all(c)) ;
	for(int i = a ; i <= a + 3000 ; ++ i) {
		int mx = 0 , mn = LLinf ;
		for(auto p : c)
		{
			int t = (i - p) / a ;
			if ( t <= 0 ) {
				mx = max(mx , p) ;
				mn = min(mn , p) ;
			} else {
				mx = max(mx , ((i - t * a - p) < ((t + 1) * a + p - i) ? t * a + p : (t + 1) * a + p) ) ;
				mn = min(mn , ((i - t * a - p) < ((t + 1) * a + p - i) ? t * a + p : (t + 1) * a + p) ) ;
			}
		}
		ans = min(ans , mx - mn) ;
	}
	std::cout << ans << '\n' ;
}

signed main()
{
	cin.tie(nullptr)->ios::sync_with_stdio(false) ;

	int t = 1 ;
	cin >> t ;

	for(int i = 1 ; i <= t ; i++)
	{
		// cout << "case #" << i << " : " << '\n' ;
		solve() ;
	}
	return 0 ;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases where the code's assumptions or logic might break. Here are the key points to consider:

1. **Small-scale test case**: The input should be small but effective in exposing the bug.
2. **Edge cases**: Focus on scenarios where `a` and `b` are small, or where the array `c` has elements that are close to each other but not exactly the same.
3. **Special values**: Use values for `a` and `b` that are not coprime, as the code uses the greatest common divisor (GCD) in its logic.

Let's construct a test case step by step:

1. **Number of test cases**: Set `t = 1` for simplicity.
2. **Array length**: Choose a small `n` (e.g., `n = 4`).
3. **Constants `a` and `b`**: Use values that are not coprime, such as `a = 2` and `b = 4`.
4. **Array elements**: Choose elements that are close to each other but not exactly the same, such as `c = [1, 3, 5, 7]`.

Here is the test input:

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

This test case is designed to check if the code correctly handles the situation where the elements of the array are not exactly the same but can be made closer by using the operations with `a` and `b`. The expected output for this test case is `2`, as the minimum range can be achieved by making all elements equal to `3` or `5`.

Let's verify the expected output:

- Initial array: `[1, 3, 5, 7]`
- We can increase `1` by `2` to get `[3, 3, 5, 7]`.
- We can increase `5` by `2` to get `[3, 3, 7, 7]`.
- The range is `7 - 3 = 4`.

However, we can also:
- Increase `1` by `4` to get `[5, 3, 5, 7]`.
- The range is `7 - 3 = 4`.

To achieve the minimum range of `2`:
- Increase `1` by `2` to get `[3, 3, 5, 7]`.
- Increase `5` by `2` to get `[3, 3, 7, 7]`.
- The range is `7 - 5 = 2`.

This test case should expose the bug if the code does not handle the operations correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dora and C++
time_limit_ms: 2000
memory_limit_mb: 256
Description: Dora has just learned the programming language C++!

However, she has completely misunderstood the meaning of C++. She considers it as two kinds of adding operations on the array $c$ with $n$ elements. Dora has two integers $a$ and $b$. In one operation, she can choose one of the following things to do.

  * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $a$.    * Choose an integer $i$ such that $1 \leq i \leq n$, and increase $c_i$ by $b$.

Note that $a$ and $b$ are constants, and they can be the same.

Let's define a range of array $d$ as $\max(d_i) - \min(d_i)$. For instance, the range of the array $[1, 2, 3, 4]$ is $4 - 1 = 3$, the range of the array $[5, 2, 8, 2, 2, 1]$ is $8 - 1 = 7$, and the range of the array $[3, 3, 3]$ is $3 - 3 = 0$.

After any number of operations (possibly, $0$), Dora calculates the range of the new array. You need to help Dora minimize this value, but since Dora loves exploring all by herself, you only need to tell her the minimized value.

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

The first line of each test case contains three integers $n$, $a$, and $b$ ($1 \leq n \leq 10^5$, $1 \leq a, b \leq 10^9$) — the length of the array $c$ and the constant values, respectively.

The second line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 10^9$) — the initial elements of the array $c$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the minimum possible range of the array after any number of operations.

Examples:
input:
10
4 5 5
1 3 4 4
4 2 3
1 3 4 6
4 7 7
1 1 2 6
3 15 9
1 9 5
3 18 12
1 4 5
7 27 36
33 13 23 12 35 24 41
10 6 9
15 5 6 9 8 2 12 15 3 8
2 1 1000000000
1 1000000000
6 336718728 709848696
552806726 474775724 15129785 371139304 178408298 13106071
6 335734893 671469786
138885253 70095920 456876775 9345665 214704906 375508929
output:
3
0
3
2
3
5
1
0
17
205359241

Note: In the first test case, we can increase $c_1 = 1$ by $a = 5$. The array $c$ will become $[6, 3, 4, 4]$, and the range is $3$. Note that there is more than one way to reach the answer.

In the second test case, we can increase $c_1 = 1$ by $a = 2$ and then increase $c_1 = 3$ by $b = 3$. Also, we can increase $c_2 = 3$ by $b = 3$ and increase $c_3 = 4$ by $a = 2$. The array $c$ will become $[6, 6, 6, 6]$, and the range is $0$.

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

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    public static void main(String[] args) {
        FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(System.out);

        int tt = sc.nextInt();
        while (tt-- > 0) {
            long n = sc.nextLong();
            long a = sc.nextLong();
            long b = sc.nextLong();
            long k = gcd(a, b);

            long[] v = new long[(int) n];
            for (int i = 0; i < n; i++) {
                v[i] = sc.nextLong() % k;
            }
            Arrays.sort(v);

            long ans = Long.MAX_VALUE;
            for (int i = 0; i < n; i++) {
                ans = Math.min(ans, v[v.length - 1] - v[i]);
                v = Arrays.copyOf(v, v.length + 1);
                v[v.length - 1] = v[i] + k;
            }
            out.println(ans);
        }
        out.flush();
        out.close();
    }

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

**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
3 2 3
1 4 7
```

**Explanation:**

1. **Input Structure**: The input follows the structure specified in the problem description, with a single test case.
2. **Test Case Design**:
   - **Array Size (n)**: Chosen to be 3, which is a small size to ensure the test input is manageable.
   - **Constants (a, b)**: Chosen to be 2 and 3, respectively. This combination is chosen because the greatest common divisor (GCD) of 2 and 3 is 1, which means any integer can be represented as a linear combination of 2 and 3.
   - **Initial Array (c)**: Chosen to be [1, 4, 7]. This array has a range of 6 (7 - 1 = 6), which is relatively large and provides a good opportunity to test the algorithm's ability to minimize the range.

3. **Expected Output**:
   - The minimum possible range for the array [1, 4, 7] using operations with a = 2 and b = 3 should be 1. This is because we can adjust the array to [3, 3, 3] by performing the following operations:
     - Increase c[0] by 2: [1 + 2, 4, 7] = [3, 4, 7]
     - Increase c[1] by 3: [3, 4 + 3, 7] = [3, 7, 7]
     - Decrease c[2] by 2: [3, 7, 7 - 2] = [3, 7, 5]
     - Decrease c[1] by 2: [3, 7 - 2, 5] = [3, 5, 5]
     - Decrease c[1] by 2: [3, 5 - 2, 5] = [3, 3, 5]
     - Decrease c[2] by 2: [3, 3, 5 - 2] = [3, 3, 3]

This test case is designed to expose any issues in the algorithm's handling of operations and the calculation of the minimum range.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Increase/Decrease/Copy
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two integer arrays: array $a$ of length $n$ and array $b$ of length $n+1$.

You can perform the following operations any number of times in any order:

  * choose any element of the array $a$ and increase it by $1$;    * choose any element of the array $a$ and decrease it by $1$;    * choose any element of the array $a$, copy it and append the copy to the end of the array $a$.

Your task is to calculate the minimum number of aforementioned operations (possibly zero) required to transform the array $a$ into the array $b$. It can be shown that under the constraints of the problem, it is always possible.

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

Each test case consists of three lines:

  * the first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$);    * the third line contains $n + 1$ integers $b_1, b_2, \dots, b_{n + 1}$ ($1 \le b_i \le 10^9$).

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum number of operations (possibly zero) required to transform the array $a$ into the array $b$.

Examples:
input:
3
1
2
1 3
2
3 3
3 3 3
4
4 2 1 2
2 1 5 2 3
output:
3
1
8

Note: In the first example, you can transform $a$ into $b$ as follows: $[2] \rightarrow [2, 2] \rightarrow [1, 2] \rightarrow [1, 3]$.

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

#define sz(x) (int)(x).size()
#define all(x) (x).begin(),(x).end()
#define rep(i, a, b) for (int i = a; i < (b); i++)
#define rev(i, a, b) for (int i = a; i > (b); i--)
#define fr first
#define sc second
#define pb push_back
#define endl "\n"

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

void solve_task() {
    int n;cin >> n;
    vector<ll>a(n),b(n + 1);

    for(auto &i : a) cin >> i;
    for(auto &i : b) cin >> i;

    ll ans = 0;
    bool ok = 0;
    for(int i = 0;i < n;i++) {
        if(a[i] >= b[n] && ok == 0) {
            ans++;
            ok = 1;
        }
        ans += abs(a[i] - b[i]);
    }
    if(ok) {
        cout << ans << endl;
        return;
    }
    ll aux = 1e9 + 10;
    for(int i = 0;i < n;i++) {
        aux = min(aux,b[n] - a[i] + 1);
    }
    cout << ans + aux << endl;
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int tc = 1;cin >> tc;
    while(tc--) solve_task();

    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
3
1 2 3
4 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Increase/Decrease/Copy
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two integer arrays: array $a$ of length $n$ and array $b$ of length $n+1$.

You can perform the following operations any number of times in any order:

  * choose any element of the array $a$ and increase it by $1$;    * choose any element of the array $a$ and decrease it by $1$;    * choose any element of the array $a$, copy it and append the copy to the end of the array $a$.

Your task is to calculate the minimum number of aforementioned operations (possibly zero) required to transform the array $a$ into the array $b$. It can be shown that under the constraints of the problem, it is always possible.

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

Each test case consists of three lines:

  * the first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$);    * the third line contains $n + 1$ integers $b_1, b_2, \dots, b_{n + 1}$ ($1 \le b_i \le 10^9$).

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum number of operations (possibly zero) required to transform the array $a$ into the array $b$.

Examples:
input:
3
1
2
1 3
2
3 3
3 3 3
4
4 2 1 2
2 1 5 2 3
output:
3
1
8

Note: In the first example, you can transform $a$ into $b$ as follows: $[2] \rightarrow [2, 2] \rightarrow [1, 2] \rightarrow [1, 3]$.

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

#define mod 1000000007

#define ll long long
#define pii pair<int, int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vb vector<bool>
#define vvi vector<vector<int>>
#define vvb vector<vector<bool>>
#define vvc vector<vector<char>>
#define vvll vector<vector<long long int>>
#define vc vector<char>
#define vll vector<long long>
#define mii map<int, int>
#define umii unordered_map<int, int>
#define si set<int>
#define rep(i, s, e) for(ll i = s; i <= e; i++)
#define ren(i, s, e) for(ll i = s; i >= e; i--)

void solve(){
    ll n;
    cin>>n;
    vector<ll> a(n),b(n+1);
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    rep(i, 0, n){
        cin>>b[i];
    }
    ll last = b[n];
    ll res = 1;
    ll least = INT_MAX;
    bool can = true;
    rep(i, 0, n-1){
        if(can && ((a[i]<= last && b[i]>=last) || (a[i]>=last && b[i]<=last))){
            can = false;
        }
        res += abs(a[i]-b[i]);
    }
    if(can){
        rep(i, 0, n-1){
            least = min(abs(a[i]-last),abs(b[i]-last));
        }
        res += least;
    }
    cout<<res<<endl;
    return;
}

int main(){
   ios_base::sync_with_stdio(false);
   cin.tie(NULL);
   cout.tie(NULL);
   int t;
   cin >> t;
   while(t--){
      solve();
   }
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose the bug in the given code. Here's how it works:

1. **Initial Arrays:**
   - `a = [1, 2, 3]`
   - `b = [4, 2, 1, 3]`

2. **Expected Transformations:**
   - To transform `a` into `b`, we need to:
     - Increase `a[0]` from 1 to 4 (3 operations).
     - Keep `a[1]` as 2 (0 operations).
     - Decrease `a[2]` from 3 to 1 (2 operations).
     - Append `3` to the end of `a` (1 operation).

3. **Total Operations:**
   - The total number of operations required is `3 (increase 1 to 4) + 0 (keep 2 as 2) + 2 (decrease 3 to 1) + 1 (append 3) = 6`.

4. **Buggy Code Analysis:**
   - The buggy code calculates the operations by iterating through the elements of `a` and `b` and summing the absolute differences.
   - It also includes a special case where it checks if any element in `a` can be transformed to the last element of `b` (`b[n]`).
   - However, the logic in the code is flawed because it does not correctly handle the appending operation, which is crucial for transforming `a` into `b`.

5. **Expected Output:**
   - The correct output for this test case should be `6`.

This test case should reveal the bug in the code, as the incorrect handling of the appending operation and the special case logic will likely lead to an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Increase/Decrease/Copy
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two integer arrays: array $a$ of length $n$ and array $b$ of length $n+1$.

You can perform the following operations any number of times in any order:

  * choose any element of the array $a$ and increase it by $1$;    * choose any element of the array $a$ and decrease it by $1$;    * choose any element of the array $a$, copy it and append the copy to the end of the array $a$.

Your task is to calculate the minimum number of aforementioned operations (possibly zero) required to transform the array $a$ into the array $b$. It can be shown that under the constraints of the problem, it is always possible.

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

Each test case consists of three lines:

  * the first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$);    * the third line contains $n + 1$ integers $b_1, b_2, \dots, b_{n + 1}$ ($1 \le b_i \le 10^9$).

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum number of operations (possibly zero) required to transform the array $a$ into the array $b$.

Examples:
input:
3
1
2
1 3
2
3 3
3 3 3
4
4 2 1 2
2 1 5 2 3
output:
3
1
8

Note: In the first example, you can transform $a$ into $b$ as follows: $[2] \rightarrow [2, 2] \rightarrow [1, 2] \rightarrow [1, 3]$.

Buggy Code:
// ██   ██  █████  ██     ██  █████  ███    ███ ██████  ███████ ██    ██  █████      ███    ██  ██████      ██
// ██   ██ ██   ██ ██     ██ ██   ██ ████  ████ ██   ██ ██       ██  ██  ██   ██     ████   ██ ██    ██   ████
// ███████ ███████ ██  █  ██ ███████ ██ ████ ██ ██   ██ █████     ████   ███████     ██ ██  ██ ██    ██     ██
// ██   ██ ██   ██ ██ ███ ██ ██   ██ ██  ██  ██ ██   ██ ██         ██    ██   ██     ██  ██ ██ ██    ██     ██
// ██   ██ ██   ██  ███ ███  ██   ██ ██      ██ ██████  ███████    ██    ██   ██     ██   ████  ██████  ██  ██
#include "bits/stdc++.h"
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define ll long long
#define rv return void
#define brk(x) x break;

ll LCM(ll a, ll b);
ll GCD(ll a, ll b);
bool isPrime(ll x);

void gg()
{
    int n;
    cin >> n;

    int a[n], b[n + 1];
    for (auto &it : a)
        cin >> it;
    for (auto &it : b)
        cin >> it;

    int c = 0;
    int ans = 0, flag = 1;
    for (int i = 0; i < n; i++)
    {
        ans += abs(a[i] - b[i]);
        c = min(abs(b[n] - a[i]), abs(b[n] - b[i]));
        if (b[n] <= max(a[i], b[i]) && b[n] >= min(a[i], b[i]))
            flag = 0;
    }
    if (flag)
        ans += c;

    cout << ++ans << '\n';

    return;
}

/*
    3
    1 1 3
    1 1 3 2

 */

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int t = 1;
    cin >> t;

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

    return 0;
}

ll LCM(ll a, ll b) { return (a / GCD(a, b)) * b; }

ll GCD(ll a, ll b) { return (!b ? a : GCD(b, a % b)); }

bool isPrime(ll x)
{
    for (ll i = 2; i * i <= x; i++)
        if (x % i == 0)
            return 0;
    return (x == 1 ? 0 : 1);
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to craft a test input that highlights the issue with the logic in the `gg` function. The code seems to be incorrectly handling the transformation of array `a` into array `b`, particularly when dealing with the last element of `b`.

### Analysis of the Buggy Code:
1. **Incorrect Calculation of `c` and `flag`**:
   - The code calculates `c` as the minimum absolute difference between the last element of `b` and the current elements of `a` and `b`.
   - The `flag` is set to 0 if the last element of `b` is within the range of the current elements of `a` and `b`.
   - The final answer is incremented by `c` if `flag` is still 1.

2. **Incorrect Final Answer**:
   - The final answer is incremented by 1, which might not be necessary in all cases.

### Edge Cases to Consider:
1. **Array `a` and `b` with identical elements**:
   - This should require 0 operations, but the code might incorrectly add 1 to the final answer.

2. **Array `a` and `b` with a single element difference**:
   - This should require a minimal number of operations, but the code might incorrectly handle the last element of `b`.

3. **Array `a` and `b` where the last element of `b` is not within the range of `a` and `b`**:
   - This should require a specific number of operations, but the code might incorrectly add 1 to the final answer.

### Test Input to Expose the Bug:
Let's create a test input where the last element of `b` is not within the range of the current elements of `a` and `b`, and the transformation requires a specific number of operations.

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

### Explanation:
- **Test Case 1**:
  - `n = 3`
  - `a = [1, 2, 3]`
  - `b = [1, 2, 3, 4]`

The correct transformation steps are:
1. Copy the element 3 from `a` to the end of `a` to get `[1, 2, 3, 3]`.
2. Increase the last element of `a` by 1 to get `[1, 2, 3, 4]`.

The minimum number of operations required is 2.

However, the provided code will incorrectly calculate the answer as 3 because it increments the final answer by 1 unnecessarily.

This test input should expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Increase/Decrease/Copy
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two integer arrays: array $a$ of length $n$ and array $b$ of length $n+1$.

You can perform the following operations any number of times in any order:

  * choose any element of the array $a$ and increase it by $1$;    * choose any element of the array $a$ and decrease it by $1$;    * choose any element of the array $a$, copy it and append the copy to the end of the array $a$.

Your task is to calculate the minimum number of aforementioned operations (possibly zero) required to transform the array $a$ into the array $b$. It can be shown that under the constraints of the problem, it is always possible.

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

Each test case consists of three lines:

  * the first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$);    * the second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$);    * the third line contains $n + 1$ integers $b_1, b_2, \dots, b_{n + 1}$ ($1 \le b_i \le 10^9$).

Additional constraint on the input: the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum number of operations (possibly zero) required to transform the array $a$ into the array $b$.

Examples:
input:
3
1
2
1 3
2
3 3
3 3 3
4
4 2 1 2
2 1 5 2 3
output:
3
1
8

Note: In the first example, you can transform $a$ into $b$ as follows: $[2] \rightarrow [2, 2] \rightarrow [1, 2] \rightarrow [1, 3]$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    a = [int(x) for x in input().split()]
    b = [int(x) for x in input().split()]
    #a = sorted(a)
    #b = sorted(b)
    ans = 0
    flag = False
    near = 999999999999
    for r in range(n):
        ans += abs(a[r] - b[r])
        near = min(near , abs(a[r] - b[-1]))
        if (a[r] <= b[-1] and b[r] >= b[-1]) or (a[r] >= b[-1] and b[r] <= b[-1]):
            flag = True

    if not flag:
        ans += near + 1
    else:
        ans += 1
    print(ans)

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

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

Think step by step.

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