**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
#include "bits/stdc++.h"
using namespace std;
#include "vector"
#include "map"
#include "set"
#include "queue"
#include "stack"
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define MOD 1000000007
#define MOD1 998244353
#define INF 1e18
#define nline "\n"
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.141592653589793238462
#define set_bits __builtin_popcountll
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
// typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key

#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif

void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

int main() {
fastio();
int tc;cin>>tc;
while(tc--){
 ll n;cin>>n;vector<ll> a(n);
 for(int i=0;i<n;i++)cin>>a[i];

//  sort(a.begin(),a.end());

 ll s=1;ll e=1e18;
 ll fa;
 while(s<=e){
    ll mid=(s+e)/2;
    // ll mid=1;
    int f=1;vector<int> done(n,0);int flag=1;
    for(int i=0;i<n;i++){int fidx=-1;
    if(done[i])continue;
        for(int j=i+1;j<n;j++){
            if(a[j]-a[i]<=mid)fidx=j;
            else break;
        }
        if(fidx==-1){
            if(n%2==0){
                f=0;break;
            }
            if(flag){
                flag=!flag;continue;
            }
            else{
                f=0;break;
            }
        }
        else{
            done[fidx]=1;
        }
    }
    if(f){
        e=mid-1;fa=mid;
    }
    else s=mid+1;

    // break;
 }
 cout<<fa<<endl;
  }
}

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. The code is designed to find the minimum value of \( k \) such that all cells in the list \( a \) can be painted black, with the possibility of painting one additional cell that is not in the list.

### Key Points to Consider:
1. **Single Element List**: If the list contains only one element, the value of \( k \) should be 1, as the only cell can be painted black without needing to paint any other cell.
2. **Consecutive Elements**: If the list contains consecutive elements, the value of \( k \) should be 1.
3. **Large Gaps**: If there are large gaps between elements in the list, the value of \( k \) should be large enough to cover the gaps.
4. **Odd and Even Length Lists**: The code might handle odd and even length lists differently, so we should test both.

### Test Case Generation:
1. **Single Element List**:
   - \( n = 1 \)
   - \( a = [5] \)
   - Expected output: \( 1 \)

2. **Consecutive Elements**:
   - \( n = 2 \)
   - \( a = [1, 2] \)
   - Expected output: \( 1 \)

3. **Large Gaps**:
   - \( n = 3 \)
   - \( a = [1, 1000000000000000000, 1000000000000000001] \)
   - Expected output: \( 999999999999999999 \)

4. **Odd Length List**:
   - \( n = 3 \)
   - \( a = [1, 3, 5] \)
   - Expected output: \( 2 \)

5. **Even Length List**:
   - \( n = 4 \)
   - \( a = [1, 3, 5, 7] \)
   - Expected output: \( 2 \)

### Test Input:
```plaintext
5
1
5
2
1 2
3
1 1000000000000000000 1000000000000000001
3
1 3 5
4
1 3 5 7
```

This test input covers a variety of scenarios that should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
def main():
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))

        if n == 1:
            print(1)
            continue

        if n == 2:
            print(a[-1] - a[0])
            continue

        if n % 2 == 0:
            # For even n, compute differences between consecutive pairs across the list
            d = [a[i + 1] - a[i] for i in range(0, n, 2)]
            print(max(d))
            continue

        # For odd n, two scenarios: paint the last or paint the first element
        d1 = [(a[i + 1] - a[i]) for i in range(0, n - 1, 2)]  # Leaving out the last element
        d2 = [(a[i] - a[i - 1]) for i in range(2, n, 2)]       # Leaving out the first element

        # Take the minimum of the maximum differences in each scenario
        print(min(max(d1), max(d2)))

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
```plaintext
4
1
7
3
2 4 9
5
1 5 8 10 13
6
1 2 3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    l = list(map(int,input().split()))
    d = 0
    if n==1:
        print(1)
    elif n%2==0:
        for i in range(n-1):
            d = max(d,l[i+1]-l[i])
        print(d)
    elif n%2==1:
        c = []
        for i in range(n-1):
            c.append(l[i+1]-l[i])
        c.sort()
        print(c[-2])

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

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

Think step by step.

Response:
```plaintext
5
1
7
3
2 4 9
5
1 5 8 10 13
2
1 3
4
1 3 5 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
for t in range(int(input())):
    n=int(input())
    a=sorted([*map(int,input().split())])
    fs=0
    sc=0
    if n==1:
        print(1)
    elif n==2:
        print(a[1]-a[0])
    else:
        diff=[]
        for i in range(1,n):
            diff.append(a[i]-a[i-1])
        diff.sort()
        print(diff[-2])

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

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

Think step by step.

Response:
```plaintext
4
1
7
3
2 4 9
4
1 3 6 10
5
1 5 8 10 13
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
import java.util.*;

public class B_Black_Cells {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while (T-- > 0) {
            // Your code here
            ////Bunny is topper
            int n=sc.nextInt();
            long arr[]=new long[n];
            for(int i=0;i<n;i++){
                arr[i]=sc.nextLong();
            }
            long min=1;
            long max=(long)Math.pow(10,18);
            // System.out.println(max);
            long ans=0;
            while(min<=max){
                long mid=(min+max)/2;
                if(finder(mid,arr)){
                    ans=mid;
                    max=mid-1;
                }else{
                    min=mid+1;
                }
            }
            System.out.println(ans);
        }
    }
    public static boolean finder(long now,long arr[]){
        int n=arr.length;
        int ans=0;
        for(int i=1;i<n;i++){
            if(arr[i]-arr[i-1]>now){
                 ans++;
            }
            if(ans>=2){
                return false;
            }
        }
        return true;
    }

}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

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

public class CF2026B {
    public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    int T = Integer.parseInt(br.readLine());

    while (T-- > 0) {
        solve(br, pw);
    }

    pw.flush();
    pw.close();
    br.close();
    }
    public static void solve(BufferedReader br, PrintWriter pw) throws IOException {
		int n = Integer.parseInt(br.readLine());
        long[] a = new long[n];
        long k = 1;

        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            a[i] = Long.parseLong(st.nextToken());
        }
        for (int i = 1; i < n; i+=2)  {
            k = Math.max(k, Math.abs(a[i] - a[i - 1]));
        }
        if (n % 2 == 1) {
            long m = 1;
            for (int i = 2; i < n; i+=2) {
                m = Math.max(m, Math.abs(a[i] - a[i - 1]));

            }
            k = Math.min(k, m);
        }
        pw.println(k);
	}
}

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

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

Think step by step.

Response:
```plaintext
1
4
1 3 5 7
```

**Explanation:**
This test case is designed to expose the bug in the provided buggy code. The list of cells to be painted black is `[1, 3, 5, 7]`. The correct minimum value of `k` for this list should be `2` because:

- You can paint cells `(1, 3)` and `(5, 7)` with `k = 2`.

However, the buggy code only considers the differences between cells at odd and even indices separately, and it may not correctly handle the case where the list has an even number of elements. Specifically, the code only updates `k` for the differences between cells at odd indices and then tries to minimize it with the differences between cells at even indices, which is incorrect. This test case will help highlight the issue.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int a[2005];
int main() {
	int T,n;
	cin>>T;
	while(T--) {
		priority_queue<pair<int,int>>p;
		cin>>n;
		for(int i=1; i<=n; i++) {
			cin>>a[i];
		}
		if(n==1) {
			cout<<1<<endl;
			continue;
		}
		for(int i=1; i<=n-1; i++) {
			int v=a[i+1]-a[i];
			p.push({v,i});
		}
		int k=max(a[2]-a[1],a[n]-a[n-1]);
		if(n%2==1&&p.top().second%2==0)p.pop();
		else if(n%2==1&&p.top().first==k)p.pop();
		cout<<p.top().first<<endl;
	}
	return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - Single element list.
   - Two elements with a small difference.
   - Two elements with a large difference.
   - Lists with an odd number of elements.
   - Lists with an even number of elements.

2. **Buggy Code Analysis**:
   - The code uses a priority queue to find the maximum difference between consecutive elements.
   - It handles the case where the list has only one element.
   - It attempts to adjust the result based on the parity of the number of elements and the top element of the priority queue.

3. **Identifying the Bug**:
   - The code incorrectly handles the case where the list has an odd number of elements and the largest difference is between the first or last two elements.
   - The code may also fail to handle the case where the list has an even number of elements and the largest difference is not the top element of the priority queue.

Given these considerations, let's create a small-scale test input that is likely to expose the bug:

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

Explanation:
1. **Test Case 1**: A single element list.
   - Expected output: 1
   - This tests the handling of a single element list.

2. **Test Case 2**: A list with three elements where the largest difference is between the first two elements.
   - Expected output: 1
   - This tests the handling of an odd number of elements and the largest difference being between the first two elements.

3. **Test Case 3**: A list with four elements where the largest difference is between the first and second elements.
   - Expected output: 2
   - This tests the handling of an even number of elements and the largest difference being between the first two elements.

This test input should help identify the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

long long mx = 2147483647;

/*void printNumbers() {
	cout << "i'm heere" << endl;
	for (int i = 0; i < 1000; i++) {
		if (i % 3 == 0 && i % 5 && ((i % 10) + (i / 10) % 10 + (i / 100) < 10)) {
			cout << i << " ";
		}
	}
}
*/
int main() {
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		vector<long long> a(n);
		for (int i = 0; i < n; i++) {
			cin >> a[i];
		}
		if (n == 1) {
			cout << 1 << endl;
			continue;
		}
		if (n == 2) {
			cout << a[1] - a[0] << endl;
			continue;
		}
		if (n == 3) {
			cout << min(a[1] - a[0], a[2] - a[1]) << endl;
			continue;
		}
		long long l = 0, r = 1e18, m, ans;
		while (r - l > 1) {
			m = (r + l) / 2;
			int q = 0;
			for (int i = 1; i < n; i++) {
				if (a[i] - a[i - 1] > m) q++;
			}
			if (q > 2 - ((n + 1) % 2)) {
				l = m;
			}
			else {
				r = m;
			}
		}
		int q = 0;
		m = l;
		for (int i = 1; i < n; i++) {
			if (a[i] - a[i - 1] > m) q++;
		}
		if (q > 1 - (n + 1) % 2) {
			cout << r << endl;
		}
		else {
			cout << l << 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
4
1
1
2
1 1000000000000000000
3
1 2 3
4
1 5 9 13
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    m = list(map(int, input().split()))

    diffs = []
    if n == 1:
        print(1)
    else:
        if n % 2 == 0:
            for i in range(0, n - 1, 2):
                diffs.append(m[i + 1] - m[i])
            print(max(diffs))
            continue
        m1 = m[:]
        m2 = m[:]
        for i in range(n - 1):
            diffs.append(m[i + 1] - m[i])

        p = max(diffs)
        for i in range(n - 1):
            if diffs[i] == p:
                m1.pop(i)
                break

        for i in range(1, n):
            if diffs[i - 1] == p:
                m2.pop(i)
                break
       # print(m1, m2)
        res1 = []
        res2 = []
        for i in range(0,len(m2) - 1, 2):
            res1.append(m1[i + 1] - m1[i])
            res2.append(m2[i + 1] - m2[i])
       # print(res1, res2)
        print(min(max(res1),max(res2)))

**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
1 3 5 7 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    arr = [int(x) for x in input().split()]
    if n < 3:
        print(1)
    else:
        i = 0
        j = n-1

        l_diff = arr[1] - arr[0]
        r_diff = arr[n-1] - arr[n-2]
        if n == 3 or n % 2 == 0:
            print(min(l_diff, r_diff))
        else:
            mid_diff = arr[n-3] - arr[n-4]
            if mid_diff == l_diff:
                print(l_diff)
            else:
                print(r_diff)

**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 buggy code, we need to focus on edge cases and scenarios that challenge the logic of the code. The buggy code has several issues, including:

1. It does not correctly handle the case where the list has more than three elements and the differences between the elements are not consistent.
2. It does not consider the possibility of using an extra black cell to optimize the minimum value of \( k \).

A good test case to expose these issues would be one where the differences between consecutive elements are not consistent, and the list has more than three elements. This will force the code to make incorrect assumptions about the minimum value of \( k \).

Let's construct a test case with \( n = 4 \) and carefully chosen elements:

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

In this test case:
- The differences between consecutive elements are \( 2 \) (i.e., \( 3 - 1 = 2 \), \( 5 - 3 = 2 \), \( 7 - 5 = 2 \)).
- The code should correctly determine that the minimum value of \( k \) is \( 2 \), as it is possible to paint all the given cells black with \( k = 2 \) (e.g., painting cells \( (1, 3) \), \( (5, 7) \)).

However, the buggy code will incorrectly handle this case because it does not consider the possibility of using an extra black cell to optimize \( k \). The code will likely output a larger value for \( k \).

This test case should help expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

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

public class B {

    public void go() throws IOException {
        int zzz = ipar();
        for (int zz = 0; zz < zzz; zz++) {
            int n = ipar();
            long[] arr = lapar(n);
            if (n == 1) {
                out.println(1);
                continue;
            }
            long best = Long.MAX_VALUE;
            for (int i = 0; i <= n; i++) {
                ArrayList<Long> list = new ArrayList<>();
                for (long x : arr) {
                    list.add(x);
                }
                if (i == 0) {
                    list.add(0, arr[i]-1);
                } else if (i == n) {
                    list.add(arr[n-1]+1);
                } else {
                    list.add(i, (arr[i-1]+arr[i]) / 2);
                }
                // out.println(list);
                long max = 0;
                for (int e = 0; e < n+1; e++) {
                    long high = Long.MAX_VALUE;
                    if (e > 0) {
                        high = Math.min(list.get(e) - list.get(e-1), high);
                    }
                    if (e < n) {
                        high = Math.min(list.get(e+1) - list.get(e), high);
                    }
                    max = Math.max(high, max);
                }
                best = Math.min(best, max);
            }
            out.println(best);
        }
    }

    record Triple(long dist, int a, int b) implements Comparable<Triple> {
        public int compareTo(Triple other) {
            if (dist != other.dist) {
                return Long.compare(dist, other.dist);
            }
            return 0;
        }
    }

    class DisjointSet {
        int[] arr;
        int[] size;

        public DisjointSet(int n) {
            arr = new int[n+1];
            size = new int[n+1];
            for (int i = 0; i <= n; i++) {
                arr[i] = i;
                size[i] = 1;
            }
        }

        public int get(int i) {
            return arr[i] = (arr[i] == i ? i : get(arr[i]));
        }

        public boolean merge(int a, int b) {
            a = get(a);
            b = get(b);
            if (a == b) {
                return false;
            }
            if (size[a] > size[b]) {
                arr[b] = a;
                size[a] += size[b];
            } else {
                arr[a] = b;
                size[b] += size[a];
            }
            return true;
        }
    }

    //////////////////////////////////////////////
    //             DEBUG PROPERTIES             //
    //////////////////////////////////////////////

    static final boolean RUN_TIMING = false;
    static final boolean AUTOFLUSH = false;
    static final boolean FILE_INPUT = false;
    static final boolean FILE_OUTPUT = false;

    ///////////////////////////////////////////
    //             TEMPLATE CODE             //
    ///////////////////////////////////////////

    static char[] inputBuffer = new char[1 << 20];
    static PushbackReader in = new PushbackReader(new BufferedReader(new InputStreamReader(System.in)), 1 << 20);
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    public int ipar() throws IOException {
        return Integer.parseInt(spar());
    }

    public int[] iapar(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = ipar();
        }
        return arr;
    }

    public long lpar() throws IOException {
        return Long.parseLong(spar());
    }

    public long[] lapar(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = lpar();
        }
        return arr;
    }

    public double dpar() throws IOException {
        return Double.parseDouble(spar());
    }

    public String spar() throws IOException {
        int len = 0;
        int c;
        do {
            c = in.read();
        } while (Character.isWhitespace(c) && c != -1);
        if (c == -1) {
            throw new NoSuchElementException("Reached EOF");
        }
        do {
            inputBuffer[len] = (char)c;
            len++;
            c = in.read();
        } while (!Character.isWhitespace(c) && c != -1);
        while (c != '\n' && Character.isWhitespace(c) && c != -1) {
            c = in.read();
        }
        if (c != -1 && c != '\n') {
            in.unread(c);
        }
        return new String(inputBuffer, 0, len);
    }

    public String linepar() throws IOException {
        int len = 0;
        int c;
        while ((c = in.read()) != '\n' && c != -1) {
            if (c == '\r') {
                continue;
            }
            inputBuffer[len] = (char)c;
            len++;
        }
        return new String(inputBuffer, 0, len);
    }

    public boolean haspar() throws IOException {
        String line = linepar();
        if (line.isEmpty()) {
            return false;
        }
        in.unread('\n');
        in.unread(line.toCharArray());
        return true;
    }

    public void shuffle(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            int j = (int)(Math.random() * (n-i) + i);
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    public void shuffle(long[] arr) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            int j = (int)(Math.random() * (n-i) + i);
            long temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    public void shuffle(Object[] arr) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            int j = (int)(Math.random() * (n-i) + i);
            Object temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    public static void main(String[] args) throws IOException {
        if (FILE_INPUT) {
            in = new PushbackReader(new BufferedReader(new FileReader(new File("test.txt"))), 1 << 20);
        }
        if (FILE_OUTPUT) {
            out = new PrintWriter(new FileWriter(new File("output.txt")));
        }
        long time = 0;
        time -= System.nanoTime();
        new B().go();
        time += System.nanoTime();
        if (RUN_TIMING) {
            System.out.printf("%.3f ms%n", time / 1000000.0);
        }
        out.flush();
        in.close();
    }
}

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

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

Think step by step.

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

1. **Single Element List**: The code has a special case for `n == 1`, which might be incorrect.
2. **Adjacent Elements**: Test cases where the elements in the list are very close to each other.
3. **Large Gaps**: Test cases where there are large gaps between elements.
4. **Edge Elements**: Test cases where the elements are at the edges of the range.

Let's create a test input that covers these scenarios:

```plaintext
4
1
7
2
1 2
3
2 4 9
5
1 5 8 10 13
```

This test input includes:
1. A single element list.
2. Two adjacent elements.
3. Elements with a small gap.
4. Elements with a larger gap.

This should help in identifying any issues in the code's logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class B {
    public static void main(String[] args) {
        var scan = new Scanner(System.in);
        int t = scan.nextInt();

        while(t-- > 0) {
            int n = scan.nextInt();
            List<BigInteger> sortedNums = new ArrayList<>(n);
            for(int i = 0; i < n; i++) {
                sortedNums.add(scan.nextBigInteger());
            }

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

            BigInteger largestDifference = new BigInteger("-1");
            BigInteger secondLargestDifference = new BigInteger("-1");
            BigInteger last = new BigInteger("-1");

            for(int i = 0; i < n; i++) {
                BigInteger current = sortedNums.get(i);
                if(last.compareTo(BigInteger.ZERO) >= 0) {
                    BigInteger difference = current.subtract(last);
                    if(difference.compareTo(largestDifference) > 0) {
                        secondLargestDifference = largestDifference;
                        largestDifference = difference;
                    } else if (difference.compareTo(secondLargestDifference) > 0) {
                        secondLargestDifference = difference;
                    }
                }
                last = current;
            }

            if(n == 2) {
                System.out.println(largestDifference.toString());
                continue;
            }
            System.out.println(secondLargestDifference.toString());
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic that determines the largest and second-largest differences between consecutive elements in the list. The code incorrectly assumes that the second-largest difference is the correct answer when `n > 2`. This is a flaw because the correct answer should be the smallest value of `k` that allows all cells in the list to be painted black, potentially using an extra cell outside the list.

### Step-by-Step Analysis:

1. **Single Element Case**:
   - The code correctly handles the case where `n = 1` by outputting `1`.

2. **Two Elements Case**:
   - The code correctly handles the case where `n = 2` by outputting the difference between the two elements.

3. **Three or More Elements Case**:
   - The code incorrectly outputs the second-largest difference between consecutive elements.
   - This is a bug because the correct answer should be the smallest `k` that allows all cells to be painted black, which might not necessarily be the second-largest difference.

### Test Case to Expose the Bug:

Consider a test case where the largest difference is much larger than the other differences, and the second-largest difference is not the correct answer. For example:

```plaintext
1
5
1 3 6 8 10
```

### Explanation:

- **Input**:
  - `n = 5`
  - `a = [1, 3, 6, 8, 10]`

- **Expected Output**:
  - The correct answer is `2`, because we can paint the cells as follows:
    - Paint (1, 3) with `k = 2`
    - Paint (6, 8) with `k = 2`
    - Paint (10, 11) with `k = 2` (11 is an extra cell outside the list)

- **Actual Output from Buggy Code**:
  - The largest difference is `4` (between 6 and 10).
  - The second-largest difference is `2` (between 3 and 6, or 8 and 10).
  - The code will output `2`, which is correct in this case, but the logic is flawed.

### Test Case to Further Expose the Bug:

Consider a test case where the second-largest difference is not the correct answer:

```plaintext
1
5
1 3 6 10 12
```

### Explanation:

- **Input**:
  - `n = 5`
  - `a = [1, 3, 6, 10, 12]`

- **Expected Output**:
  - The correct answer is `4`, because we can paint the cells as follows:
    - Paint (1, 3) with `k = 2`
    - Paint (6, 10) with `k = 4`
    - Paint (12, 13) with `k = 4` (13 is an extra cell outside the list)

- **Actual Output from Buggy Code**:
  - The largest difference is `4` (between 6 and 10).
  - The second-largest difference is `2` (between 3 and 6).
  - The code will output `2`, which is incorrect.

### Final Test Input:

```plaintext
1
5
1 3 6 10 12
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

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

int32_t main(){
	int tc;
	cin>>tc;
	while(tc--){
		int n;
		cin>>n;
		vector<int>a(n);
		for(int i=0;i<n;i++) cin>>a[i];

		if(n==1){
			cout<<1<<endl;
			continue;
		}
		else if(n%2==0){
			int val = 0;
			for(int i=0;i<n;i+=2){
				int xx = a[i+1] - a[i];
				val = max(val,xx);
			}
			cout<<val<<endl;
		}
		else{
			vector<int>agye,pich;
			for(int i=0;i<n;i+=2){
				int val = 1;
				if(i!=n-1) val = a[i+1] - a[i];
				agye.push_back(val);
			}

			pich.push_back(1);
			for(int i=n-1;i>=0;i-=2){
				int val = a[i] - a[i-1];
				pich.push_back(val);
			}
			int m = n/2+1;

			int val1 = INT_MIN;
			for(int i=0;i<m;i++) val1 = max(val1,agye[i]);
			int val2 = INT_MIN;
		    for(int i=0;i<m;i++) val2 = max(val2,pich[i]);
		    cout<<min(val1,val2)<<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
4
1
1
2
1 3
3
1 3 5
4
1 3 5 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
#include <cstdio>
#include <algorithm>
using namespace std;
void in(int &n){
	n=0;int f=1;char c=getchar();
	while (c<'0'||'9'<c){if (c=='-')f*=-1;c=getchar();}
	while ('0'<=c&&c<='9'){n=n*10+(c-'0');c=getchar();}
	n*=f;
}
void llin(long long &n){
	n=0;int f=1;char c=getchar();
	while (c<'0'||'9'<c){if (c=='-')f*=-1;c=getchar();}
	while ('0'<=c&&c<='9'){n=n*10+(c-'0');c=getchar();}
	n*=f;
}
const int N=2005;
long long a[N];
int main(){
	/*freopen("CF2026B.in","r",stdin);
	freopen("CF2026B.out","w",stdout);*/
	int t;in(t);
	while (t--){
		int n;in(n);
		for (int i=1;i<=n;i++) llin(a[i]);
		if (n&1){
			if (n==1){puts("1");continue;}
			long long ans=1ll<<60;
			for (int i=1;i<=n;i++){
				long long tmp=0;
				for (int j=2;j<i;j+=2) tmp=max(tmp,a[j]-a[j-1]);
				for (int j=n-1;j>i;j-=2) tmp=max(tmp,a[j+1]-a[j]);
				if (i%2==0) tmp=max(tmp,a[i+1]-a[i-1]);
				long long llim=max(0ll,a[i]-tmp),rlim=a[i]+tmp;
				if (a[i]-llim-i+1>0||rlim-a[i]-n+i>0) {
					ans=min(ans,tmp);//printf ("%d ",tmp);
				}
				else{
					ans=min(ans,min(a[i]-a[1]+1,a[n]-a[i]+1));
					//printf ("%d ",min(a[i]-a[1]+1,a[n]-a[i]+1));
				}

			}
			printf ("%lld\n",ans);
		}
		else{
			long long ans=0;
			for (int i=1;i<=n;i+=2) ans=max(ans,a[i+1]-a[i]);
			printf ("%d\n",ans);
		}
	}
	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 might cause the code to fail. Here are some steps to consider:

1. **Single Cell Case**: Test the case where there is only one cell in the list. The code should return 1.
2. **Two Cells Case**: Test the case where there are two cells. The code should return the difference between the two cells.
3. **Odd Number of Cells**: Test with an odd number of cells where the cells are spread out in a way that the code might fail to find the correct minimum value of \( k \).
4. **Even Number of Cells**: Test with an even number of cells where the cells are spread out in a way that the code might fail to find the correct minimum value of \( k \).
5. **Large Differences**: Test with large differences between consecutive cells to ensure the code handles large numbers correctly.

Given the constraints and the need to create a small-scale test input, here is a test input that covers these cases:

```plaintext
4
1
7
2
1 2
3
2 4 9
5
1 5 8 10 13
```

This input includes:
- A single cell (7) which should return 1.
- Two cells (1, 2) which should return 1.
- Three cells (2, 4, 9) which should return 2.
- Five cells (1, 5, 8, 10, 13) which should return 3.

These test cases should help expose any bugs in the provided code by covering a range of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
def can_paint_all(a, k):
    n = len(a)
    pairs = 0
    i = 0

    while i < n - 1:
        if a[i + 1] - a[i] <= 2 * k:
            pairs += 1
            i += 2  # Skip the next one as it's paired
        else:
            i += 1

    return pairs >= (n - 1) // 2

def solve():
    t = int(input().strip())
    results = []

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

        left, right = 0, 10**18
        answer = right

        while left <= right:
            mid = left + (right - left) // 2

            if can_paint_all(a, mid):
                answer = mid
                right = mid - 1
            else:
                left = mid + 1

        results.append(answer + 1)

    print("\n".join(map(str, results)))

if __name__ == "__main__":
    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
5
1
1
2
1 3
3
1 4 7
4
1 3 5 7
5
1 3 5 7 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    l = list(map(int , input().split()))
    if(n==1):
        print(1)
        continue
    if(n==2):
        print(l[1]-l[0])
        continue
    m1 = max(l[1]-l[0],l[2]-l[1])
    m2 = min(l[1]-l[0],l[2]-l[1])
    for i in range(n-1):
        dif = l[i+1]-l[i]
        if(dif>m1):
            m2 = m1
            m1 = dif
        if(m2<=dif<m1):
            m2 = dif
    if(n%2==0):
        print(m1)
    else:
        print(m2)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code seems to have issues with handling the minimum and maximum differences between consecutive elements in the list. It also has a condition based on whether the number of elements is even or odd, which might not be necessary or correct.

Here are the steps to generate a challenging test input:

1. **Single Element List**: The code incorrectly prints `1` for a single element list, which is not always correct. We need a test case with a single element.
2. **Two Element List**: The code correctly handles two elements by printing the difference between them. This is generally correct, so no special test case is needed here.
3. **Three Element List**: The code calculates the maximum and minimum differences between the first three elements and then iterates through the list. We need a test case where the differences between elements are not straightforward.
4. **Even and Odd Length Lists**: The code has a special condition based on whether the number of elements is even or odd. We need to test both cases to see if this condition is necessary and correct.

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

```plaintext
5
1
7
2
1 2
3
1 3 5
4
1 3 5 7
5
1 3 5 7 9
```

Explanation:
1. **Single Element List**: `1 7` - The code should print `1`, but the correct answer is `1` (since only one cell is given).
2. **Two Element List**: `2 1 2` - The code correctly prints `1`, which is the correct answer.
3. **Three Element List**: `3 1 3 5` - The code should print `2`, which is the correct answer.
4. **Even Length List**: `4 1 3 5 7` - The code should print `2`, which is the correct answer.
5. **Odd Length List**: `5 1 3 5 7 9` - The code should print `2`, which is the correct answer.

This test input will help expose the bug 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: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

Buggy Code:
import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while(t-- > 0){
            int n = in.nextInt();
            int[] a = new int[n];
            for(int i=0; i<n; i++){
                a[i] = in.nextInt();
            }
            int[] f = new int[2];
            for(int i=1; i<n; i++){
                int x = a[i] - a[i-1];
                f[i%2] = Math.max(f[i%2], x);
            }
            f[0] = n < 3 ? Integer.MAX_VALUE : f[0];
            f[1] = n == 1 ? 1 : f[1];
            if(n%2 == 0)
                System.out.println(f[1]);
            else
                System.out.println(Math.min(f[0], f[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, let's consider the following aspects:

1. **Edge Cases:**
   - Single cell in the list.
   - Two cells in the list.
   - Three cells in the list.
   - Four cells in the list.
   - A list with a large gap between cells.

2. **Behavior of the Buggy Code:**
   - The code uses a modulo operation to track the maximum gap between cells for even and odd indices.
   - It has special handling for the case when the number of cells is 1 or 2.
   - It might fail when the number of cells is odd and the gaps between cells are not evenly distributed.

3. **Challenging Scenarios:**
   - A list with a large gap between the first and second cell.
   - A list with a large gap between the second and third cell.
   - A list with a large gap between the third and fourth cell.
   - A list with a small gap between all cells except one large gap.

Based on these considerations, let's create a test input that includes a small number of cells with a large gap between them to challenge the modulo-based logic in the buggy code.

```plaintext
4
1
7
2
1 2
3
1 10 11
4
1 10 11 12
```

Explanation:
- The first test case has a single cell, which should return 1.
- The second test case has two cells with a small gap, which should return 1.
- The third test case has three cells with a large gap between the first and second cells, which should return 9.
- The fourth test case has four cells with a large gap between the first and second cells, which should return 9.

This input should help expose any issues with the buggy code's handling of large gaps and the modulo-based logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Black Cells
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a strip divided into cells, numbered from left to right from $0$ to $10^{18}$. Initially, all cells are white.

You can perform the following operation: choose two white cells $i$ and $j$, such that $i \ne j$ and $|i - j| \le k$, and paint them black.

A list $a$ is given. All cells from this list must be painted black. Additionally, at most one cell that is not in this list can also be painted black. Your task is to determine the minimum value of $k$ for which this is possible.

Input: The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 < a_i < 10^{18}$; $a_i < a_{i + 1}$).

Additional constraint on the input: the sum of $n$ across all test cases does not exceed $2000$.

Output: For each test case, print a single integer — the minimum value of $k$ for which it is possible to paint all the given cells black.

Examples:
input:
4
2
1 2
1
7
3
2 4 9
5
1 5 8 10 13
output:
1
1
2
3

Note: In the first example, with $k=1$, it is possible to paint the cells $(1, 2)$.

In the second example, with $k=1$, it is possible to paint the cells $(7, 8)$.

In the third example, with $k=2$, it is possible to paint the cells $(2, 4)$ and $(8, 9)$.

In the fourth example, with $k=3$, it is possible to paint the cells $(0, 1)$, $(5, 8)$ and $(10, 13)$.

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());
        }

        float nextFloat() {
            return Float.parseFloat(next());
        }

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

        void readArr(int[] ar, int n) {
            for (int i = 0; i < n; i++) {
                ar[i] = nextInt();
            }
        }

        void readArr(long[] ar, int n) {
            for (int i = 0; i < n; i++) {
                ar[i] = nextLong();
            }
        }
    }

    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException {
            din = new DataInputStream(
                    new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException {
            byte[] buf = new byte[BUFFER_SIZE]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    } else {
                        continue;
                    }
                }
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

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

            if (neg)
                return -ret;
            return ret;
        }

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

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

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

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

        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }
    }

    static class p implements Comparable<p> {
        long a;
        long b;
        long c;
        long d;

        p(long a, long b) {
            this.a = a;
            this.b = b;
        }

        p(long a, long b, long c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

        p(long a, long b, long c, long d) {
            this.a = a;
            this.b = b;
            this.c = c;
            this.d = d;
        }

        @Override
        public int compareTo(p other) {
            // if (this.a != other.a) {
            return Long.compare(this.a, other.a);
            // }
            // if (this.b != other.b) {
            // return Long.compare(this.b, other.b);
            // }
            // return Long.compare(this.c, other.c);
        }

        @Override
        public boolean equals(Object o) {
            if (this == o)
                return true;
            if (o == null || getClass() != o.getClass())
                return false;
            p pair = (p) o;
            return a == pair.a && b == pair.b && c == pair.c;
        }

        @Override
        public int hashCode() {
            return Objects.hash(a, b, c);
        }
    }

    public static void solve(Reader fs, FastReader sc, PrintWriter w, StringBuilder sb) throws Exception {
        int n = sc.nextInt();
        ArrayList<Long> al = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            al.add(sc.nextLong());
        }
        if (n == 1) {
            sb.append(1 + "\n");
            return;
        }
        Collections.sort(al);
        ArrayList<Long> diff = new ArrayList<>();
        for (int i = 1; i < al.size(); i += 2) {
            diff.add(al.get(i) - al.get(i - 1));
        }
        Collections.sort(diff);
        long max = n % 2 == 0 ? Long.MIN_VALUE : 1;
        for (int i = 0; i < diff.size() - (n % 2 == 0 ? 1 : 0); i++) {
            max = Math.max(max, diff.get(i));
        }
        if (n % 2 == 0) {
            max = Math.max(max, (long) Math.ceil(diff.get(diff.size() - 1) / 2.0));
        } else {
            long max1 = Long.MIN_VALUE;
            for (int i = 1; i < al.size() - 1; i += 2) {
                max1 = Math.max(max1, al.get(i + 1) - al.get(i));
            }
            max = Math.min(max, max1);
        }
        sb.append(max + "\n");
    }

    public static void main(String[] args) throws Exception {
        Reader sc = new Reader();
        FastReader fs = new FastReader();
        PrintWriter w = new PrintWriter(System.out);
        StringBuilder sb = new StringBuilder();
        long o = fs.nextLong();
        while (o > 0) {
            solve(sc, fs, w, sb);
            o--;
        }
        System.out.print(sb.toString().trim());
        w.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The code has a few potential weaknesses:

1. **Odd and Even Length Lists**: The code handles odd and even lengths differently, so we should test both.
2. **Large Differences**: The code calculates differences between consecutive elements and sorts them, so large differences might cause issues.
3. **Single Element List**: The code has a special case for a single element list, so we should test this.
4. **Consecutive Elements**: The code assumes certain patterns in the differences, so consecutive elements might cause issues.

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

1. A list with a single element.
2. A list with two elements.
3. A list with three elements where the differences are large.
4. A list with four elements where the differences are large and not consecutive.

Here is the test input:

```plaintext
4
1
7
2
1 2
3
2 4 9
4
1 10 100 1000
```

This test input includes:
- A single element list: `7`
- A two-element list: `1 2`
- A three-element list with large differences: `2 4 9`
- A four-element list with large differences and non-consecutive elements: `1 10 100 1000`

This should help expose any bugs in the code's handling of different list lengths and large differences.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Product of Binary Decimals
time_limit_ms: 3000
memory_limit_mb: 256
Description: Let's call a number a binary decimal if it is a positive integer and all digits in its decimal notation are either $$$0$$$ or $$$1$$$. For example, $$$1\,010\,111$$$ is a binary decimal, while $$$10\,201$$$ and $$$787\,788$$$ are not.

Given a number $$$n$$$, you are asked whether or not it is possible to represent $$$n$$$ as a product of some (not necessarily distinct) binary decimals.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$).

Output: For each test case, output "YES" (without quotes) if $$$n$$$ can be represented as a product of binary decimals, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
11
121
1
14641
12221
10110
100000
99
112
2024
12421
1001
output:
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
YES

Note: The first five test cases can be represented as a product of binary decimals as follows:

- $$$121 = 11 \times 11$$$.
- $$$1 = 1$$$ is already a binary decimal.
- $$$14\,641 = 11 \times 11 \times 11 \times 11$$$.
- $$$12\,221 = 11 \times 11 \times 101$$$.
- $$$10\,110 = 10\,110$$$ is already a binary decimal.

Buggy Code:
#include<iostream>
using namespace std;
bool f(int n)
{
	while (n > 0)
	{
		if ((n % 10 == 1) || (n % 10 == 0))
			n /= 10;
		else
			return false;
	}
	return true;
}

bool y(int n)
{
	for (int i = 2; i * i <= n; i++)
	{
		if (n % i == 0)
		{
			if ((f(i) && f(n / i))|| (y(i) && y(n / i)))
				return true;
		}
	}
	return false;
}

int main()
{
	int t; cin >> t;
	int n;
	while (t--)
	{
		cin >> n;
		if (f(n)|| y(n))
			cout << "YES" << endl;
		else
			cout << "NO" << endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios where the code might fail. The primary issue in the code is that it does not correctly handle the decomposition of numbers into binary decimals. Specifically, the function `y` is not correctly checking if a number can be decomposed into binary decimals.

Here are some edge cases to consider:
1. **Small numbers that are binary decimals themselves** (e.g., 1, 11, 101).
2. **Numbers that are products of binary decimals** (e.g., 121, 14641).
3. **Numbers that are not products of binary decimals** (e.g., 99, 112).
4. **Numbers with large factors that are not binary decimals** (e.g., 2024).
5. **Numbers that are binary decimals but have a large number of digits** (e.g., 10110).

Let's create a test input that includes these edge cases:

```plaintext
6
1
121
14641
99
2024
10110
```

This test input includes:
- `1`: A binary decimal itself.
- `121`: A product of binary decimals (11 * 11).
- `14641`: A product of binary decimals (11 * 11 * 11 * 11).
- `99`: Not a product of binary decimals.
- `2024`: Not a product of binary decimals.
- `10110`: A binary decimal itself.

This input should expose the bugs in the given code, as the code might not correctly handle the decomposition of numbers into binary decimals.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Product of Binary Decimals
time_limit_ms: 3000
memory_limit_mb: 256
Description: Let's call a number a binary decimal if it is a positive integer and all digits in its decimal notation are either $$$0$$$ or $$$1$$$. For example, $$$1\,010\,111$$$ is a binary decimal, while $$$10\,201$$$ and $$$787\,788$$$ are not.

Given a number $$$n$$$, you are asked whether or not it is possible to represent $$$n$$$ as a product of some (not necessarily distinct) binary decimals.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$).

Output: For each test case, output "YES" (without quotes) if $$$n$$$ can be represented as a product of binary decimals, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
11
121
1
14641
12221
10110
100000
99
112
2024
12421
1001
output:
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
YES

Note: The first five test cases can be represented as a product of binary decimals as follows:

- $$$121 = 11 \times 11$$$.
- $$$1 = 1$$$ is already a binary decimal.
- $$$14\,641 = 11 \times 11 \times 11 \times 11$$$.
- $$$12\,221 = 11 \times 11 \times 101$$$.
- $$$10\,110 = 10\,110$$$ is already a binary decimal.

Buggy Code:
#include <iostream>
#include<set>
#define int long long
using namespace std ;
int a[1500];
int i;
signed main () {
	int t;
	cin>>t;
	while(t--)
	{
	   for(i=0;i<10;i++)
 	   {
		    a[i]=0;
	   }
		int n;
		cin>>n;
		int m,j;
		m=n%10;
		int f=0,cnt=0;
		j=n;
		while(j!=0)
		{
			a[j%10]++;
			j/=10;
			cnt++;

		}
		for(i=2;i<=9;i++)
		{
			if(a[i]!=0)
			f=1;
		}
		if(f==0)
		{
			cout<<"Yes"<<endl;
			continue;
		}
		else
		{
			if(n%11==0)
		    {
	         if(m==1||m==0)
		     {
			cout<<"Yes"<<endl;
			continue;
		     }
		     else
			    cout<<"No"<<endl;
		    }
	        else
		    cout<<"No"<<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
11
121
1
14641
12221
10110
100000
99
112
2024
12421
1001
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Product of Binary Decimals
time_limit_ms: 3000
memory_limit_mb: 256
Description: Let's call a number a binary decimal if it is a positive integer and all digits in its decimal notation are either $$$0$$$ or $$$1$$$. For example, $$$1\,010\,111$$$ is a binary decimal, while $$$10\,201$$$ and $$$787\,788$$$ are not.

Given a number $$$n$$$, you are asked whether or not it is possible to represent $$$n$$$ as a product of some (not necessarily distinct) binary decimals.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$).

Output: For each test case, output "YES" (without quotes) if $$$n$$$ can be represented as a product of binary decimals, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
11
121
1
14641
12221
10110
100000
99
112
2024
12421
1001
output:
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
YES

Note: The first five test cases can be represented as a product of binary decimals as follows:

- $$$121 = 11 \times 11$$$.
- $$$1 = 1$$$ is already a binary decimal.
- $$$14\,641 = 11 \times 11 \times 11 \times 11$$$.
- $$$12\,221 = 11 \times 11 \times 101$$$.
- $$$10\,110 = 10\,110$$$ is already a binary decimal.

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

bool isMadeOfZerosAndOnes(int num) {
    while (num > 0) {
        int digit = num % 10;
        if (digit != 0 && digit != 1) {
            return false;
        }
        num /= 10;
    }
    return true;
}

vector<int> primeFactors(int n) {
    vector<int> factors;
    for (int i = 2; i * i <= n; ++i) {
        while (n % i == 0) {
            factors.push_back(i);
            n /= i;
        }
    }
    if (n > 1) {
        factors.push_back(n);
    }
    return factors;
}

void solve() {
    int n;
    cin >> n;
    if (n < 1 || n > 100000) {
        cout << "Invalid input" << endl;
        return;
    }
    if (isMadeOfZerosAndOnes(n)) {
        cout << "YES" << endl;
        return;
    }
    vector<int> factors = primeFactors(n);
    bool canBeRepresented = true;
    for (int factor : factors) {
        if (!isMadeOfZerosAndOnes(factor)) {
            canBeRepresented = false;
            break;
        }
    }
    if (canBeRepresented) {
        cout << "YES" << endl;
    } else {
        cout << "NO" << endl;
    }
}

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

    int t;
    cin >> t;
    // if (t < 1 || t > 50000) {
    //     cout << "Invalid input" << endl;
    //     return 0;
    // }

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

    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - Numbers that are already binary decimals (e.g., 1, 10, 11, 101).
   - Numbers that are products of binary decimals (e.g., 121, 14641).
   - Numbers that are not products of binary decimals (e.g., 99, 112).

2. **Special Cases**:
   - Numbers that are prime and not binary decimals (e.g., 97, 101).
   - Numbers that are composite and not products of binary decimals (e.g., 2024, 12421).

3. **Large Numbers**:
   - Numbers close to the upper limit (100000) to test the performance and correctness of the algorithm.

4. **Composite Numbers**:
   - Numbers that have multiple factors, some of which are binary decimals and some are not (e.g., 10010).

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

```plaintext
10
1
11
121
14641
97
101
2024
12421
10010
99999
```

This test input includes:
- Simple binary decimals (1, 11).
- Products of binary decimals (121, 14641).
- Prime numbers that are not binary decimals (97, 101).
- Composite numbers that are not products of binary decimals (2024, 12421).
- A number that is a product of a binary decimal and a non-binary decimal (10010).
- A large number close to the upper limit (99999).

This set of test cases should effectively expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Product of Binary Decimals
time_limit_ms: 3000
memory_limit_mb: 256
Description: Let's call a number a binary decimal if it is a positive integer and all digits in its decimal notation are either $$$0$$$ or $$$1$$$. For example, $$$1\,010\,111$$$ is a binary decimal, while $$$10\,201$$$ and $$$787\,788$$$ are not.

Given a number $$$n$$$, you are asked whether or not it is possible to represent $$$n$$$ as a product of some (not necessarily distinct) binary decimals.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases.

The only line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$).

Output: For each test case, output "YES" (without quotes) if $$$n$$$ can be represented as a product of binary decimals, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
11
121
1
14641
12221
10110
100000
99
112
2024
12421
1001
output:
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
YES

Note: The first five test cases can be represented as a product of binary decimals as follows:

- $$$121 = 11 \times 11$$$.
- $$$1 = 1$$$ is already a binary decimal.
- $$$14\,641 = 11 \times 11 \times 11 \times 11$$$.
- $$$12\,221 = 11 \times 11 \times 101$$$.
- $$$10\,110 = 10\,110$$$ is already a binary decimal.

Buggy Code:
def is_binary(num):
    while num > 0:
        if num % 10 != 0 and num % 10 != 1:
            return False
        num //= 10
    return True

def checkBinary():
    num = int(input())
    orgNum = num
    rem = 0
    result = 0
    while num > 0:
        dig = num % 10
        num //= 10
        rem = rem * 10 + dig
        result += 1
    if is_binary(orgNum):
        print("YES")
    elif rem == orgNum and orgNum % 11 == 0 and result > 2:
        print("YES")
    else:
        print("NO")

testCase = int(input())
for _ in range(testCase):
    checkBinary()

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

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

Think step by step.

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

1. **Binary Decimals**: Numbers that are entirely composed of the digits 0 and 1.
2. **Product of Binary Decimals**: The number can be represented as a product of one or more binary decimals.
3. **Edge Cases**: Numbers that are close to the boundaries of the input range, numbers that are products of binary decimals in non-obvious ways, and numbers that are not binary decimals but might be incorrectly identified as such.

The buggy code has a few issues:
- It checks if the number is a binary decimal.
- It checks if the number is a palindrome and divisible by 11, which is not a general condition for being a product of binary decimals.

To expose these issues, we can use the following test cases:

1. **Binary Decimal**: A number that is already a binary decimal.
2. **Non-Binary Decimal**: A number that is not a binary decimal but should be a product of binary decimals.
3. **Palindrome and Divisible by 11**: A number that is a palindrome and divisible by 11 but not a product of binary decimals.
4. **Large Number**: A large number that is a product of binary decimals.
5. **Small Number**: A small number that is a product of binary decimals.

Here is the test input:

```plaintext
5
10110
121
14641
12221
100011
```

Explanation:
1. **10110**: This is a binary decimal itself.
2. **121**: This is a product of 11 and 11.
3. **14641**: This is a product of 11, 11, 11, and 11.
4. **12221**: This is a product of 11 and 101.
5. **100011**: This is a product of 100011 (itself a binary decimal).

These test cases should help expose the bugs in the given code.