**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Final Boss
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are facing the final boss in your favorite video game. The boss enemy has $h$ health. Your character has $n$ attacks. The $i$'th attack deals $a_i$ damage to the boss but has a cooldown of $c_i$ turns, meaning the next time you can use this attack is turn $x + c_i$ if your current turn is $x$. Each turn, you can use all attacks that are not currently on cooldown, all at once. If all attacks are on cooldown, you do nothing for the turn and skip to the next turn.

Initially, all attacks are not on cooldown. How many turns will you take to beat the boss? The boss is beaten when its health is $0$ or less.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$)  – the number of test cases.

The first line of each test case contains two integers $h$ and $n$ ($1 \leq h, n \leq 2 \cdot 10^5$) – the health of the boss and the number of attacks you have.

The following line of each test case contains $n$ integers $a_1, a_2, ..., a_n$ ($1 \leq a_i \leq 2 \cdot 10^5$) – the damage of your attacks.

The following line of each test case contains $n$ integers $c_1, c_2, ..., c_n$ ($1 \leq c_i \leq 2 \cdot 10^5$) – the cooldown of your attacks.

It is guaranteed that the sum of $h$ and $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output an integer, the minimum number of turns required to beat the boss.

Examples:
input:
8
3 2
2 1
2 1
5 2
2 1
2 1
50 3
5 6 7
5 6 7
50 3
2 2 2
3 3 3
90000 2
200000 200000
1 1
100000 1
1
200000
6 7
3 2 3 2 3 1 2
6 5 9 5 10 7 7
21 6
1 1 1 1 1 1
5 5 8 10 7 6
output:
1
3
15
25
1
19999800001
1
21

Note: For the first test case, you can use attacks $1$ and $2$ on the first turn, dealing $3$ damage in total, and slaying the boss.

For the second case, you can beat the boss in $3$ turns by using the following attacks:

Turn $1$: Use attacks $1$ and $2$, dealing $3$ damage to the boss. The boss now has $2$ health left.

Turn $2$: Use attack $2$, dealing $1$ damage to the boss. The boss now has $1$ health left.

Turn $3$: Use attack $1$, dealing $2$ damage to the boss. The boss now has $-1$ health left. Since its health is less than or equal to $0$, you beat the boss.

For the sixth test case: remember to use 64-bit integers as the answer can get large.

Buggy Code:
t = int(input())
results = []

for _ in range(t):
    h, n = map(int, input().split())
    a = list(map(int, input().split()))  # Convert input strings to integers
    c = list(map(int, input().split()))

    turn = {0: []}
    pt = 0
    current_health = h

    while current_health > 0:
        if pt == 0:
            current_health -= sum(a)
            for k in range(n):
                new_pt = c[k] + pt
                if new_pt not in turn:
                    turn[new_pt] = [k]
                else:
                    turn[new_pt].append(k)
        else:
            if pt in turn:
                damage = sum(a[klo] for klo in turn[pt])
                current_health -= damage
                for tpt in turn[pt]:
                    new_pt = c[tpt] + pt
                    if new_pt not in turn:
                        turn[new_pt] = [tpt]
                    else:
                        turn[new_pt].append(tpt)

        del turn[pt]

        if current_health > 0:
            pt = min(turn.keys())
        else:
            break

    results.append(pt + 1)

for result in results:
    print(result)

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

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

Think step by step.

Response:
```plaintext
3
1 1
200000
200000
2 2
1 1
2 2
5 3
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: Final Boss
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are facing the final boss in your favorite video game. The boss enemy has $h$ health. Your character has $n$ attacks. The $i$'th attack deals $a_i$ damage to the boss but has a cooldown of $c_i$ turns, meaning the next time you can use this attack is turn $x + c_i$ if your current turn is $x$. Each turn, you can use all attacks that are not currently on cooldown, all at once. If all attacks are on cooldown, you do nothing for the turn and skip to the next turn.

Initially, all attacks are not on cooldown. How many turns will you take to beat the boss? The boss is beaten when its health is $0$ or less.

Input: The first line contains $t$ ($1 \leq t \leq 10^4$)  – the number of test cases.

The first line of each test case contains two integers $h$ and $n$ ($1 \leq h, n \leq 2 \cdot 10^5$) – the health of the boss and the number of attacks you have.

The following line of each test case contains $n$ integers $a_1, a_2, ..., a_n$ ($1 \leq a_i \leq 2 \cdot 10^5$) – the damage of your attacks.

The following line of each test case contains $n$ integers $c_1, c_2, ..., c_n$ ($1 \leq c_i \leq 2 \cdot 10^5$) – the cooldown of your attacks.

It is guaranteed that the sum of $h$ and $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output an integer, the minimum number of turns required to beat the boss.

Examples:
input:
8
3 2
2 1
2 1
5 2
2 1
2 1
50 3
5 6 7
5 6 7
50 3
2 2 2
3 3 3
90000 2
200000 200000
1 1
100000 1
1
200000
6 7
3 2 3 2 3 1 2
6 5 9 5 10 7 7
21 6
1 1 1 1 1 1
5 5 8 10 7 6
output:
1
3
15
25
1
19999800001
1
21

Note: For the first test case, you can use attacks $1$ and $2$ on the first turn, dealing $3$ damage in total, and slaying the boss.

For the second case, you can beat the boss in $3$ turns by using the following attacks:

Turn $1$: Use attacks $1$ and $2$, dealing $3$ damage to the boss. The boss now has $2$ health left.

Turn $2$: Use attack $2$, dealing $1$ damage to the boss. The boss now has $1$ health left.

Turn $3$: Use attack $1$, dealing $2$ damage to the boss. The boss now has $-1$ health left. Since its health is less than or equal to $0$, you beat the boss.

For the sixth test case: remember to use 64-bit integers as the answer can get large.

Buggy Code:
import bisect
t=int(input())
for _ in range(t):
    h,n=(int(x) for x in input().split())
    a=[int(x) for x in input().split()]
    c=[int(x) for x in input().split()]
    d={}
    for i in range(n):
        d[c[i]]=d.get(c[i],0)+a[i]
    n=len(d)
    a=[]
    c=[]
    for k,v in d.items():
        a.append(v)
        c.append(k)
    nums=[]
    for i in range(n):
        nums.append((1,i))
    cnt=0
    while h>0:
        k,i=nums[0]
        cnt=k
        h-=a[i]
        nums.pop(0)
        bisect.insort_left(nums,(k+c[i],i))
    print(cnt)

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

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

Think step by step.

Response:
```plaintext
1
10 3
1 2 3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

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

void solve()
{
    ll n;
    cin >> n;

    if (n > 2)
    {
        cout << "NO" << endl;
    }
    else if ( n == 1)
    {
        cout << "YES" << endl;
    }
    else if ( n == 2)
    {
        ll a, b;
        cin >> a >> b;
        if(abs(a-b) == 1)
        {
            cout << "NO" << endl;
        }
        else
        {
            cout << "YES" << endl;
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
2
1 3
2
4 6
3
1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define MOD 1000000007
#define vi vector<int>
#define vll vector<ll>
#define vii vector<pair<int, int>>
#define vvii vector<vector<pair<int, int>>>
#define vvi vector<vector<int>>
#define pii pair<int, int>
#define pll pair<ll, ll>
#define umap unordered_map
#define uset unordered_set
bool isPrime(int n)
{
    if (n <= 1) return false;
    if (n == 2) return true;
    if (n % 2 == 0) return false;
    for (int i = 3; i * i <= n; i += 2) {
        if (n % i == 0) return false;
    }
    return true;
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        ll n;
        cin>>n;
        vector<ll>v(n);
        for(ll i=0;i<n;i++){
            cin>>v[i];
        }
        sort(v.begin(),v.end());
        ll a=0;
        for(ll i=0;i<n-1;i++){
            if((v[i+1]-v[i])==1){
                a++;
            }
        }
        if(a==0){
            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:
```plaintext
3
2
1 2
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: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define ll long long
#define endl '\n'
#define pb push_back
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define ff(n) for(int i=0; i<n; i++)
#define ff2(n) for(int j=0; j<n; j++)
#define ffr(n) for(int i=n-1; i>=0; i--)
#define y() cout << "YES" << endl
#define n() cout << "NO" << endl
#define gap " "
#define vvi vector<vector<int>>
#define vi vector<int>
#define vc vector<char>
#define vpii vector<pair<int,int>>
#define MOD 1000000007
#define ptr(A) shared_ptr<A>
#define INF64 ((ll) 1 << 60)
#define debug(xlakdffd) cout << #xlakdffd <<"="<< xlakdffd << endl
#define pii pair<int,int>
#define rep(i,o,p) for(int i=o;i<p;i++)
#define F first
#define S second
#define srt(v) sort(all(v))
using namespace std;

int32_t main() {
    int t; cin >> t;
    while(t--){
        int n;
        cin >> n;
        vector<int> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        sort(a.begin(), a.end());
        // Find the minimum difference between consecutive elements
        int min_diff = INT_MAX;
        for (int i = 1; i < n; i++) {
            min_diff = min(min_diff, a[i] - a[i-1]);
        }

        // If the minimum difference is 1, output NO, otherwise YES
        if (min_diff == 1) {
            cout << "NO" << endl;
        } else {
            cout << "YES" << endl;
        }

    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
2
1 2
3
1 2 4
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: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
test_case = int(input())

for _ in range(test_case):
    n = int(input())
    arr = list(map(int, input().split()))

    sm = sum(arr)

    if (sm // n) in arr or (sm + 1) // n in arr:
        print("NO")
    else:
        n1 = sm // n
        n2 = (sm + 1) // n

        for i in range(1, n):
            if arr[i] - arr[i - 1] <= n1 or arr[i] - arr[i - 1] <= n2:
                continue
            else:
                print("NO")
                break
        else:
            print("YES")

**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 the logic of the code and identify scenarios where it might fail. The buggy code checks the average and the average plus one to determine if a new point can be added. However, this approach is flawed because it does not correctly consider the distances between the points.

### Analysis of the Buggy Code:
1. **Sum and Average Check**: The code calculates the sum of the points and checks if the average or the average plus one is in the list. This is not a valid approach to determine if a new point can be the closest to all existing points.
2. **Distance Check**: The code also checks the distances between consecutive points, but it uses the average and average plus one as thresholds, which is incorrect.

### Challenging Test Input:
To expose the bug, we need a test case where the distances between points are such that the average and average plus one logic fails to correctly identify the impossibility of adding a new point.

### Test Input:
Consider a set of points where the distances between consecutive points are large enough to make it impossible to add a new point that is the closest to all existing points.

```plaintext
1
5
1 5 10 15 20
```

### Explanation:
- **Number of Points**: 5
- **Points**: 1, 5, 10, 15, 20

### Why This Test Input is Challenging:
- The distances between consecutive points are 4, 5, 5, and 5, respectively.
- The average of the points is 10, and the average plus one is 11.
- The code will incorrectly check if 10 or 11 is in the list and if the distances are less than or equal to 10 or 11.
- However, no single point can be added that is the closest to all existing points because the distances are too large.

### Expected Output:
- The correct output should be "NO" because it is impossible to add a new point that is the closest to all existing points.
- The buggy code will likely output "YES" due to its flawed logic.

This test input should effectively 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: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
def can_insert_closest_point(test_cases):
    results = []
    for n, points in test_cases:
        points.sort()  # Ensure points are sorted

        min_gap = float('inf')

        # Find the minimum gap between consecutive points
        for i in range(1, n):
            min_gap = min(min_gap, points[i] - points[i-1])

        # If the minimum gap is greater than or equal to 3, it's possible to insert a new point
        if min_gap >= 3:
            results.append("YES")
        else:
            results.append("NO")

    return results

# Input reading
t = int(input())
test_cases = []
for _ in range(t):
    n = int(input())
    points = list(map(int, input().split()))
    test_cases.append((n, points))

# Processing and output
results = can_insert_closest_point(test_cases)
for result in results:
    print(result)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on scenarios where the minimum gap between consecutive points is exactly 2. This is because the buggy code incorrectly assumes that a minimum gap of 3 or more is required to insert a new point, but a gap of 2 can also sometimes allow for a valid insertion.

Let's consider a test case where the points are such that the minimum gap is 2, but it is still possible to insert a new point that becomes the closest to every point in the set.

### Test Case:
- **Number of points**: 4
- **Points**: 1, 3, 5, 7

### Explanation:
- The minimum gap between consecutive points is 2 (e.g., 3-1, 5-3, 7-5).
- However, it is possible to insert the point 4, which will be the closest point to all the existing points:
  - For 1, the closest point is 3 (distance 2), and 4 is also within distance 3.
  - For 3, the closest points are 1 and 5 (distance 2), and 4 is also within distance 1.
  - For 5, the closest points are 3 and 7 (distance 2), and 4 is also within distance 1.
  - For 7, the closest point is 5 (distance 2), and 4 is within distance 3.

This test case should expose the bug in the buggy code, as it will incorrectly output "NO" when the correct output should be "YES".

### Test Input:
```plaintext
1
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: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
t = int(input())  # Number of test cases

for _ in range(t):
    n = int(input())  # Number of points in the set
    points = list(map(int, input().split()))

    # Assume it's possible to add a point
    possible = True

    # Check for the smallest difference between consecutive points
    for i in range(1, n):
        if points[i] - points[i - 1] == 1:
            possible = False
            break

    # Output result for each test case
    if possible:
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

Response:
```plaintext
3
2
1 2
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: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
// import java.util.Arrays;
// import java.util.Collections;
// import java.util.ArrayList;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.jar.Attributes;
import java.util.jar.JarOutputStream;

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

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

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

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

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

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

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

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter(OutputStream out) {
            this.bw = new BufferedWriter(new OutputStreamWriter(out));
        }

        public void print(Object object) throws IOException {
            bw.append(String.valueOf(object));
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    public static boolean isPalindrome(String s) {
        int i = 0, j = s.length() - 1;
        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }

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

    public static int lucky(int n) {
        int luck = 0;
        int a = n;
        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;
        while (a > 0) {
            luck = a % 10;
            a /= 10;
            max = Math.max(max, luck);
            min = Math.min(min, luck);
        }
        return (max - min);
    }

    public static boolean isPrime(int n) {
        int a = (int) Math.sqrt(n);
        if (n == 2) {
            return true;
        }
        for (int i = 2; i <= a; i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static boolean isPrimeDivisor(int n) {
        int cnt = 0;
        if (n == 1) {
            return false;
        }
        for (int i = 2; i <= n; i++) {
            if (n % i == 0 && isPrime(i)) {
                cnt++;
            }
        }
        if (cnt == 2) {
            return true;
        }
        return false;
    }

    public static long ceil(long a, long b) {
        return (a + b - 1) / b;
    }

    public static int[] reverse(int a[]) {
        int start = 0, end = a.length - 1;
        while (start < end) {
            int temp = a[start];
            a[start] = a[end];
            a[end] = temp;
            start++;
            end--;
        }
        return a;
    }

    public static void ConvertToBinary(int n) {
        if (n == 1) {
            System.out.print(1);
            return;
        }
        ConvertToBinary(n / 2);
        System.out.print(n % 2);
    }

    public static int max1(int a[]) {
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < a.length; i++) {
            if (a[i] > max) {
                max = a[i];
            }
        }
        return max;
    }

    public static int min1(int a[]) {
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < a.length; i++) {
            if (a[i] < min) {
                min = a[i];
            }
        }
        return min;
    }

    public static boolean contain(int arr[]) {
        ArrayList<Integer> list = new ArrayList<>();
        list.add(arr[0]);
        for (int i = 1; i < arr.length; i++) {
            if (list.contains(arr[i] - 1)==false || list.contains(arr[i] + 1)==false) {
                return false;
            }
            list.add(arr[i]);
        }
        return true;
    }

    public static void main(String[] args) throws IOException {
        FastReader in = new FastReader(System.in);
        @SuppressWarnings("unused")
        FastWriter out = new FastWriter(System.out);
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int arr[] = new int[n];
            for (int i = 0; i < n; i++){
                arr[i] = in.nextInt();
            }
            boolean Cond=true;
            if(n>2){
                Cond=false;
            }
            if(arr[1]-arr[0]==1) Cond=false;
            else{
                Cond=true;
            }
            if(Cond){
                System.out.println("Yes");
            }else{
                System.out.println("No");
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
import java.util.*;

public class cf_8_15 {

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

        while (t-- > 0) {
            int n = scanner.nextInt();
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = scanner.nextLong();
            }

            long minDifference = Long.MAX_VALUE;
            for (int i = 1; i < n; i++) {
                minDifference = Math.min(minDifference, Math.abs(a[i] - a[i - 1]));
            }

            if (minDifference > 1) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }

        scanner.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases where the code's logic might fail. The code checks if the minimum difference between any two consecutive points is greater than 1. If it is, it prints "YES"; otherwise, it prints "NO".

### Analysis of the Bug:
The bug lies in the fact that the code only checks the minimum difference between consecutive points. However, for a new point to be the closest to every point in the set, it must be within half the minimum distance of each point. The code does not consider this condition.

### Challenging Test Input:
We need to create a test case where the minimum difference between consecutive points is 1, but it is still possible to add a new point that is the closest to every point in the set. This can be achieved by creating a set of points where the minimum difference is 1, but there is a large gap between some points.

### Test Input:
```plaintext
3
2
1 3
3
1 2 4
4
1 2 4 6
```

### Explanation:
1. **Test Case 1:**
   - Points: 1, 3
   - Minimum difference: 2
   - New point 2 is the closest to both 1 and 3.
   - Expected output: "YES"

2. **Test Case 2:**
   - Points: 1, 2, 4
   - Minimum difference: 1
   - No new point can be the closest to all points (2 is not the closest to 4, and 3 is not the closest to 1 or 2).
   - Expected output: "NO"

3. **Test Case 3:**
   - Points: 1, 2, 4, 6
   - Minimum difference: 1
   - No new point can be the closest to all points (3 is not the closest to 1 or 2, and 5 is not the closest to 4 or 6).
   - Expected output: "NO"

This test input will expose the bug in the code, as the code will incorrectly print "NO" for the first test case where it should print "YES".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

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

#define MAX 100

int main(){
    int loop {}, i {};
    int num [MAX];
    cin >> loop;
    while (loop > 0){
        int start {};
        cin >> start;
        while (start > 0){
            cin >> num[i++];
            start--;
        }
        if (i > 2)cout << "NO\n";
        else if (abs(num[0] - num[1]) == 1) cout << "NO\n";
        else cout << "YES\n";
        loop--;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3
1 2 4
4
1 3 5 7
2
10 11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
#include <bits/stdc++.h>
#define endl '\n'
#define int long long
#define gcd __gcd
using namespace std;
typedef pair<int,int> pii;
const int INF = 9e18;
const int inf = 2e9;

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin >> t;
	while( t--)
	{
		int n ;
		cin >> n;
		vector<int> a(n + 1);
		for(int i = 1;i <= n;i ++) cin >> a[i];
		map<int,int> cnt;
		for(int i = 1;i <= n;i ++) cnt[a[i]] = 1;
		bool ok = 0;
		for(int i = -1100;i <= 2100;i ++)
		{
			bool y = 1;
			if(cnt[i]) continue;
			for(int j = 1;j <= n;j ++)
			{
				int len = abs(a[j] - i);
				if(j - 1 >= 1 && a[j] - a[j-1] > len)
				{
					y = 0;
				}
				else if(j + 1 <= n && a[j+1] - a[j] <= len) y = 0;
//				else if(j-1 >= 1 || j + 1 <= n)y = 0;
			}
			if(y)ok = 1;
		}

		if(ok) cout << "YES\n";
		else cout << "NO\n";
	}
}

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

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

Think step by step.

Response:
```plaintext
3
2
1 100
3
1 2 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: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
cases = int(input())
for c in range(cases):
	n = int(input())
	arr = list(map(int, input().split()))

	if n <= 2:
		arr.sort()
		if arr[1] - arr[0] > 1 and (arr[1] - arr[0]) % 2 == 1:
			print("Yes")
		else:
			print("No")
	elif n == 3:
		arr.sort()
		if arr[1] - arr[0] == arr[2] - arr[1]:
			print("Yes")
		else:
			print("No")
	else:
		print("No")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
t= int(input())

for i in range(t):
    n =int(input())
    nums = list(map(int,input().split()))

    sum =0
    for i in range(len(nums)):
        sum += nums[i]

    p = sum // len(nums)
    r= sum % len(nums)

    if r == 0:
        if p in nums : print("NO")
        else : print("YES")

    else :
        if (p + 1 in nums) or (p in nums) : print("NO")
        else: print("YES")

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

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

Think step by step.

Response:
```plaintext
3
2
1 2
4
1 2 3 4
3
1 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
import java.util.*;
public class Main
{
	public static void main(String[] args) {
	    Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		for(int z=0;z<t;z++)
		{
		    int n=sc.nextInt();
		    int[] arr=new int[n];
		    for(int i=0;i<n;i++)
		    {
		        arr[i]=sc.nextInt();
		    }
		    Arrays.sort(arr);
		    int ans=0;
		    for(int i=0;i<n-1;i++)
		    {
		        if(arr[i+1]-arr[i]==1)
		        {
		            ans=1;
		            break;
		        }
		    }
		    if(ans==0)
		    System.out.println("YES");
		    else
		    System.out.println("NO");
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
3
2
1 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: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

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

public class A_Closest_Point {
    public static void main(String[] args) {
        try {
            FastReader in=new FastReader();
            FastWriter out=new FastWriter();
            int testcases=in.nextInt();
            while(testcases-- >0){
                int n=in.nextInt();
                int a[]=new int[n];
                for(int i=0;i<n;i++)a[i]=in.nextInt();
                if(n==2 && (a[1]-a[0])%2!=0 && (a[1]-a[0])!=1){
                    out.println("YES");
                }
                else{
                    out.println("NO");
                }
            }

            out.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    static class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        char nextChar() {
            return next().charAt(0);
        }
        String nextLine(){
            String str="";
            try {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.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 expose the bug in the given buggy code, we need to focus on the conditions under which the code incorrectly determines whether a new point can be added to make it the closest point to every existing point. The code has a specific check for the case when there are exactly two points and the distance between them is odd and not equal to 1. However, this check is insufficient for more complex scenarios.

Let's consider the following edge cases and scenarios:

1. **Two Points with Odd Distance**: The code correctly handles this case.
2. **Two Points with Even Distance**: The code incorrectly always prints "NO" for this case.
3. **More than Two Points**: The code does not handle this case correctly at all.

To create a test input that exposes the bug, we can focus on the scenario where there are more than two points. Specifically, we can use a set of points where the distances between consecutive points are not uniform, and the distances are such that no single point can be added to make it the closest to all points.

Here is a small-scale test input that should expose the bug:

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

Explanation:
1. **Test Case 1**: Two points with an even distance (4 and 6). The correct output should be "YES" because the point 5 can be added to make it the closest to both 4 and 6.
2. **Test Case 2**: Three points (1, 3, 5). The correct output should be "NO" because no single point can be added to make it the closest to all three points.
3. **Test Case 3**: Four points (1, 3, 5, 7). The correct output should be "NO" because no single point can be added to make it the closest to all four points.

This test input should help expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

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

int main() {

int t;
cin>>t;
while(t--) {
vector<int>arr;
int n;
cin>>n;

for (int i = 0; i < n; i++) {
int val;
cin>>val;
arr.push_back(val);
}

sort(arr.begin(), arr.end());
int first = arr[0];
int i=1;

for (; i <n ;i++) {
if (arr[i] == first+1) {
cout<<"NO"<<endl;
break;
}
first = arr[i];
}

if (i == n) {
cout<<"YES"<<endl;
}

}

return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

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

void solve() {
    int n;
    cin>>n;
    if(n>2){
        cout<<"NO"<<endl;
        return;
    }

    int a,b;
    cin>>a>>b;
    if(abs(a-b)==1){
        cout<<"NO"<<endl;
    }else{
        cout<<"YES"<<endl;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;

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

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
2
1 3
2
4 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: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
def closest_point(n,a):
    closest = {}
    a.sort()
    closest[a[0]] = abs(a[1] - a[0])
    for i in range(1, n-1):
        closest[a[i]] = min(abs(a[i+1] - a[i]), abs(a[i-1] - a[i]))
    closest[a[n-1]] = abs(a[n-1] - a[n-2])

    difference = min(closest.values())

    if difference == 1:
        return "NO"
    else:
        return "YES"

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
import random
x = int(input())
for i in range(x):
    y = int(input())
    h = input()
    if y > 2:
        print("NO")
    elif abs(int(h[0])-int(h[2]))>1:
        print("YES")
    else:
        print("NO")

**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 that the code might handle incorrectly. The buggy code has the following issues:

1. It only checks the first two points and does not consider the entire set of points.
2. It incorrectly assumes that if there are more than two points, the answer is always "NO".
3. It does not properly handle the condition where the distance between the first two points is greater than 1.

Given these issues, we can create a test case that:

- Has more than two points.
- Ensures the distance between the first two points is greater than 1.
- Ensures the new point can be the closest to all points, but the code will incorrectly output "NO".

Let's construct such a test case:

1. Number of test cases: 1
2. Number of points: 3
3. Points: 1, 3, 5

For this set of points, the new point 2 can be the closest to all points, but the buggy code will output "NO" because it only checks the first two points and assumes the answer is "NO" for more than two points.

Here is the test input:

```plaintext
1
3
1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
import java.util.*;
public class solution{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            int[] arr = new int[n];
            for(int i=0;i< arr.length;i++){
                arr[i] = sc.nextInt();
            }
            if(n==1){
                System.out.println("YES");
            }
            int ans = 0;
            for(int i=0;i< arr.length;i++){
                for(int j=0;j< arr.length && arr[i]!=arr[j];j++){
                    if(arr[i]-arr[j]<=1){
                        ans++;
                    }
                }
            }
            if(ans>=1){
                System.out.println("NO");
            }else{
                System.out.println("YES");
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Closest Point
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider a set of points on a line. The distance between two points $i$ and $j$ is $|i - j|$.

The point $i$ from the set is the closest to the point $j$ from the set, if there is no other point $k$ in the set such that the distance from $j$ to $k$ is strictly less than the distance from $j$ to $i$. In other words, all other points from the set have distance to $j$ greater or equal to $|i - j|$.

For example, consider a set of points $\\{1, 3, 5, 8\\}$:

  * for the point $1$, the closest point is $3$ (other points have distance greater than $|1-3| = 2$);    * for the point $3$, there are two closest points: $1$ and $5$;    * for the point $5$, the closest point is $3$ (but not $8$, since its distance is greater than $|3-5|$);    * for the point $8$, the closest point is $5$.

You are given a set of points. You have to add an integer point into this set in such a way that it is different from every existing point in the set, and it becomes the closest point to every point in the set. Is it possible?

Input: The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($2 \le n \le 40$) — the number of points in the set;    * the second line contains $n$ integers $x_1, x_2, \dots, x_n$ ($1 \le x_1 < x_2 < \dots < x_n \le 100$) — the points from the set.

Output: For each test case, print YES if it is possible to add a new point according to the conditions from the statement. Otherwise, print NO.

Examples:
input:
3
2
3 8
2
5 6
6
1 2 3 4 5 10
output:
YES
NO
NO

Note: In the first example, the point $7$ will be the closest to both $3$ and $8$.

In the second example, it is impossible to add an integer point so that it becomes the closest to both $5$ and $6$, and is different from both of them.

Buggy Code:
import java.util.Scanner;

public class Problem_A{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            int arr[] = new int[n];
            for(int i=0; i<n; i++){
                arr[i] = sc.nextInt();
            }
            boolean flag = true;
            for(int i=1; i<n; i++){
                int diff = arr[i] - arr[i-1];
                if(diff < 2){
                    flag = false;
                    break;
                }
            }
            if(flag == true){
                System.out.println("YES");
            }
            else{
                System.out.println("NO");
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2
1 2
4
1 3 5 7
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: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
void solve()
{
    int n;cin>>n;
    map<int,int>mp;
    for(int i=1,x;i<=n;i++){
        cin>>x;
        mp[x]++;
    }
    ll a[n+1];
    ll num=0;
    queue<int>q;
    for(auto [x,y]:mp)
    {
        a[++num]=x;
        if(y==1)
        q.push(x);
    }
    ll ans=0;
    bool flag=1;
    ll last=num;
    for(int i=1;i<=num;i++)
    {
        if(flag==1)
        {
            if(mp[a[i]]>0)//下一个还可以继续选
            {
                if(mp[a[i]]==1&&q.size()&&q.front()==a[i])
                q.pop();
                flag=0;
                ans++;
               // cout<<mp[a[i]]<<" ";
              // cout<<a[i]<<" ";
                continue;
            }
        }
        else if(flag==0)
        {
            if(!q.empty())//还有一个个数的可以选
            {
                i--;
               // cout<<q.front()<<" ";
                mp[q.front()]=0;
                q.pop();
                flag=1;
            }
            else
            {
                if(mp[a[last]]>=0)
                {
                    mp[a[last]]--;
                    i--;
                    flag=1;
                }
                else
                last--;
            }
        }
    }
    cout<<ans<<endl;
}
int main(){
    int t;cin>>t;
    while(t--)
    solve();
    return 0;
}

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

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

Think step by step.

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

1. **Repetition of Tastiness Values**: Since the code uses a map to count the occurrences of each tastiness value, we should include cases where there are multiple cakes with the same tastiness value.
2. **Optimal Play**: Both Alice and Bob should play optimally, which means Alice should always choose the cake with the highest possible tastiness value that she hasn't eaten before, and Bob should try to minimize Alice's choices.
3. **Order of Tastiness Values**: The order in which the cakes are presented can affect the game's outcome. We should consider cases where the tastiness values are in increasing, decreasing, or mixed order.

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

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

### Explanation:

1. **Test Case 1: 4 cakes with tastiness values [1, 2, 2, 3]**
   - Alice can eat the cakes with tastiness 1 and 3.
   - Bob will eat the remaining cakes with tastiness 2.
   - Expected output: 2

2. **Test Case 2: 3 cakes with tastiness values [3, 3, 3]**
   - Alice can only eat one cake with tastiness 3.
   - Bob will eat the remaining cakes with tastiness 3.
   - Expected output: 1

3. **Test Case 3: 6 cakes with tastiness values [1, 2, 3, 2, 1, 4]**
   - Alice can eat the cakes with tastiness 1, 2, and 4.
   - Bob will eat the remaining cakes with tastiness 3 and 2.
   - Expected output: 3

4. **Test Case 4: 4 cakes with tastiness values [4, 3, 2, 1]**
   - Alice can eat the cake with tastiness 1.
   - Bob will eat the remaining cakes with tastiness 4, 3, and 2.
   - Expected output: 1

5. **Test Case 5: 5 cakes with tastiness values [1, 1, 2, 2, 3]**
   - Alice can eat the cakes with tastiness 1 and 3.
   - Bob will eat the remaining cakes with tastiness 2.
   - Expected output: 2

This test input covers various scenarios including repeated tastiness values, different orders of tastiness values, and optimal play strategies, which 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: World is Mine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. Initially, there are $n$ cakes, with the $i$-th cake having a tastiness value of $a_i$.

Alice and Bob take turns eating them, with Alice starting first:

  * In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake.    * In his turn, Bob chooses any remaining cake and eats it.

The game ends when the current player can't eat a suitable cake. Let $x$ be the number of cakes that Alice ate. Then, Alice wants to maximize $x$, while Bob wants to minimize $x$.

Find out how many cakes Alice will eat if both players play optimally.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 5000$) — the number of cakes.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le n$) — the tastiness values of the cakes.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5000$.

Output: For each test case, output a single integer — the number of cakes Alice will eat if both players play optimally.

Examples:
input:
9
4
1 4 2 3
3
1 1 1
5
1 4 2 3 4
4
3 4 1 4
1
1
8
4 3 2 5 6 8 3 4
7
6 1 1 3 5 3 1
11
6 11 6 8 7 5 3 11 2 3 5
17
2 6 5 3 9 1 6 2 5 6 3 2 3 9 6 1 6
output:
2
1
3
2
1
3
2
4
4

Note: In the first test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[4, 2, 3]$.    2. Bob eats a cake with a tastiness value of $2$. The remaining cakes are $[4, 3]$.    3. Alice eats a cake with a tastiness of $3$. The remaining cakes are $[4]$.    4. Bob eats a cake with a tastiness value of $4$. The remaining cakes are $[]$.    5. Since there are no more cakes left, the game ends.

In the second test case, one possible sequence of turns is:

  1. Alice eats a cake with a tastiness value of $1$. The remaining cakes are $[1, 1]$.    2. Bob eats a cake with a tastiness value of $1$. The remaining cakes are $[1]$.    3. Since Alice has already eaten a cake with a tastiness value of $1$, she cannot make a turn, so the game ends.

Buggy Code:
#pragma GCC Optimize ("Ofast")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll int
#define N '\n'
#define pb push_back
#define pob pop_back
#define ff first
#define ss second
#define bb begin()
#define ee end()
#define fastIO() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
using namespace std;
using namespace __gnu_pbds;
//order_of_key (k) : Number of items strictly smaller than k .
//find_by_order(k) : K-th element in a set (counting from zero).

vector<ll> v;
ll cnt[5005];
ll m;
ll dp[5005][5005];

int main()
{
    fastIO();
    ll t = 1,it;
    cin >> t;
    for(it = 1;it <= t;it++)
    {
        ll n,q,i,j,k,s = 0,x,y,c = 0,d = 0,f = 0;
        cin >> n;
        ll ara[n];
        memset(cnt,0,sizeof(cnt));
        memset(dp,0,sizeof(dp));
        v.clear();
        ll maxi = 0;
        for(i = 0;i < n;i++)
        {
            cin >> ara[i];
            cnt[ara[i]]++;
            maxi = max(maxi,ara[i]);
        }
        for(i = 1;i <= maxi;i++)
        {
            if(cnt[i] > 0)
            {
                v.pb(i);
            }

        }
        m = v.size();
        ll maxi3 = 0;
        for(i = 1;i < m;i++)
        {
            for(j = 1;j <= i;j++)
            {
                c = cnt[v[i]];
                if(j-c >= 0)
                {
                    dp[i][j] = dp[i-1][j-c-1] + 1;
                    maxi3 = max(maxi3,dp[i][j]);
                }
                else
                {
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        cout << m - maxi3 << N;
    }
    return 0;
}

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

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

Think step by step.

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