**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

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

void DoIt(){
    ll m,a,b,c;
    cin>>m>>a>>b>>c;
    int row1=m,row2=m,flag=0;
    if(a<=m) flag+=a;
    else flag+=row1;
    if(b<=m) flag+=b;
    else flag+=row2;
    int sit=(2*m)-flag;
    if(c<=sit) flag+=c;
    cout<<flag<<endl;
}
int32_t main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    int t; cin >> t;
    while (t--) {
       DoIt();
    }
    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 create a challenging test input that exposes the bug in the given buggy code, let's analyze the code and identify potential edge cases and scenarios that could cause it to fail.

### Analysis of the Buggy Code
1. **Initialization of `row1` and `row2`**:
   - `row1` and `row2` are both initialized to `m`, which represents the number of seats in each row.
   - The code uses these variables to track the remaining seats in each row, but they are not updated correctly after assigning monkeys to seats.

2. **Assignment of Monkeys**:
   - The code checks if `a` (monkeys that prefer row 1) is less than or equal to `m` and updates `flag` accordingly.
   - Similarly, it checks if `b` (monkeys that prefer row 2) is less than or equal to `m` and updates `flag` accordingly.
   - The remaining seats (`sit`) are calculated as `(2 * m) - flag`, but this calculation is incorrect because it does not account for the actual number of seats left in each row.

3. **Handling Monkeys with No Preference**:
   - The code attempts to seat the remaining monkeys (`c`) in the available seats, but it uses the incorrect `sit` value.

### Edge Cases to Consider
1. **Monkeys that prefer a row exceed the number of seats in that row**:
   - This scenario should result in the maximum number of monkeys being seated, but the code might not handle it correctly.

2. **Monkeys with no preference**:
   - If there are many monkeys with no preference, the code should be able to seat them in the remaining seats, but it might not handle this correctly due to the incorrect calculation of available seats.

### Challenging Test Input
Let's create a test case where the number of monkeys that prefer a row exceeds the number of seats in that row, and there are also many monkeys with no preference.

```plaintext
5
10 15 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
```

### Explanation
1. **Test Case 1**:
   - `m = 10`, `a = 15`, `b = 5`, `c = 10`
   - There are 15 monkeys that prefer row 1, but only 10 seats are available. All 5 monkeys that prefer row 2 can be seated. The remaining 5 seats in row 1 and all 10 seats in row 2 can be used for the 10 monkeys with no preference.
   - Expected output: `20` (10 in row 1, 5 in row 2, and 5 in row 2 from the monkeys with no preference).

2. **Test Case 2**:
   - `m = 3`, `a = 6`, `b = 1`, `c = 1`
   - There are 6 monkeys that prefer row 1, but only 3 seats are available. The 1 monkey that prefers row 2 can be seated. The remaining 1 seat in row 2 can be used for the 1 monkey with no preference.
   - Expected output: `5` (3 in row 1, 1 in row 2, and 1 in row 2 from the monkey with no preference).

3. **Test Case 3**:
   - `m = 15`, `a = 14`, `b = 12`, `c = 4`
   - There are 14 monkeys that prefer row 1 and 12 monkeys that prefer row 2, but only 15 seats are available in each row. The 4 monkeys with no preference can be seated in the remaining seats.
   - Expected output: `30` (14 in row 1, 12 in row 2, and 4 in row 2 from the monkeys with no preference).

4. **Test Case 4**:
   - `m = 1`, `a = 1`, `b = 1`, `c = 1`
   - There is 1 monkey that prefers row 1, 1 monkey that prefers row 2, and 1 monkey with no preference. All monkeys can be seated.
   - Expected output: `2` (1 in row 1, 1 in row 2).

5. **Test Case 5**:
   - `m = 420`, `a = 6`, `b = 9`, `c = 69`
   - There are 6 monkeys that prefer row 1, 9 monkeys that prefer row 2, and 69 monkeys with no preference. All monkeys can be seated.
   - Expected output: `84` (6 in row 1, 9 in row 2, and 69 in row 2 from the monkeys with no preference).

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

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

#define fr(i, n) for (lli i = 0; i < n; i++)
#define pb push_back
#define sz(x) x.size()
#define ff first
#define ss second
#define all(v) v.begin(), v.end()
#define debug(x) cout << '>' << #x << ':' << x << endl;

using lli = long long int;
using ld = long double;
using ii = pair<lli, lli>;
using vii = vector<ii>;
using vi = vector<lli>;
using bl = bool;
using mp = map<lli, lli>;
using ump = unordered_map<lli, lli>;
using str = string;

#pragma GCC optimize("unroll-loops,O3,Ofast")
#pragma GCC target("avx2,avx,fma,bmi,bmi2,lzcnt,popcnt")

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

    lli t;
    cin >> t;

    while (t--)
    {
        int m, a, b, c;
        cin >> m >> a >> b >> c;
        int total = m * 2;
        int cnt = 0;
        if (a > m)
            a -= m;
        else if (b > m)
            b -= m;
        cnt += a + b + c;
        if (cnt > total)
            cnt -= m;
        cout << cnt << endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases where the code might fail. The code has a logic issue in how it handles the preferences of the monkeys and the total number of seats available. Specifically, the code incorrectly subtracts the number of monkeys that can't be seated from the total count.

Let's break down the problem and identify the edge cases:

1. **Case where `a` and `b` are both greater than `m`**:
   - If both `a` and `b` are greater than `m`, the code will incorrectly subtract `m` from `cnt` even if there are enough seats in the other row for the remaining monkeys.

2. **Case where `c` is large and can be used to fill the remaining seats**:
   - If `c` is large, the code might not correctly account for the fact that some of the monkeys with no preference can be used to fill the remaining seats.

3. **Case where `a` and `b` are both less than or equal to `m`**:
   - This is a straightforward case, but we should still test it to ensure the code handles it correctly.

4. **Case where `a` and `b` are both zero**:
   - This tests the scenario where only monkeys with no preference are available.

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

```plaintext
5
10 15 15 10
10 5 5 10
10 10 10 10
10 0 0 20
10 1 1 18
```

Explanation:
1. **10 15 15 10**: Both `a` and `b` are greater than `m`. The code should handle this correctly by seating as many as possible.
2. **10 5 5 10**: A straightforward case where `a` and `b` are both less than or equal to `m`.
3. **10 10 10 10**: Both `a` and `b` are equal to `m`, and there are some monkeys with no preference.
4. **10 0 0 20**: Only monkeys with no preference are available.
5. **10 1 1 18**: A small number of monkeys with specific preferences and a large number of monkeys with no preference.

This test input should help expose any issues in the code's logic for handling the seating preferences and the total number of seats available.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
l=int(input())
for i in range(l):
  f=list(map(int,input().split()))
  j=f[0]
  c=0
  del f[0]
  if f[0]>j:
    c+=f[0]-j
  else:
    c+=f[0]
  if f[1]>j:
    c+=f[1]-j
  else:
    c+=f[1]

  if f[2]>j*2:
    c=j*2
  elif j*2>c:
    c+=f[2]

  print(c)

**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
10 10 10 10
3 1 1 1
15 15 15 15
1 0 0 2
420 420 420 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
def main():
    l = input()
    m, a, b, c = l.split()
    m, a, b, c = int(m), int(a), int(b), int(c)
    a, b = max(a, b), min(a, b)

    if b >= m:
        print(m * 2)

    elif a-b >= c:
        b = b + c
        count = a + b
        if m * 2 - count > 0:
            print(int(count))
        else:
            if b > m:
                count = m
            else:
                count = b

            print(m + count)
    else:
        count = a + b + c
        # a = a + ((c - (a-b)) / 2)
        # count = int(a * 2)
        if m * 2 > count:
            print(count)
        else:
            print(m*2)

t = int(input())
for x in range(t):
    main()

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

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

Think step by step.

Response:
```plaintext
1
10 10 10 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

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

public class Main {
    public static void main(String[] args) throws IOException {
        FastReader fr = new FastReader();
        PrintWriter out = new PrintWriter(System.out);

        int t = fr.nextInt(); // Number of test cases
        while (t-- > 0) {
            // Solve each test case
            int m = fr.nextInt();
            int a = fr.nextInt();
            int b = fr.nextInt();
            int c = fr.nextInt();
            int rem = 2*m;
            int ans =0;
            if(a<=m && b<=m){
                ans = a+b;
                rem = rem-a-b;
                if(rem == 0){
                    ans = a+b;
                }
                else if(rem>=c){
                    ans = a+b+c;
                }else{
                    ans = a+b+c-rem;
                }
            } else if (a>=m && b<=m) {
                ans = a-m+b;
                rem = rem-a+m-b;
                if(rem == 0){
                    ans = a-m+b;
                }
                else if(rem>=c){
                    ans += c ;
                }else{
                    ans = ans+c-rem;
                }
            } else if (a<=m && b>=m) {
                ans = a+b-m;
                rem = rem-a-b+m;
                if(rem == 0){
                    ans = a+b-m;
                }
                else if(rem>=c){
                    ans = ans + c;
                }else{
                    ans = ans + c-rem;
                }
            }else {
                ans = a+b-2*m;
            }
//            int[] arr = new int[n];

            out.println(ans); // Example output
        }
        out.close();
    }

    // Fast input reader
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's consider the following edge cases:

1. **Maximum Preference in One Row**: Test a scenario where the number of monkeys preferring one row is much higher than the available seats in that row.
2. **No Preference Monkeys**: Test a scenario where the number of no preference monkeys is large, and their placement can significantly affect the result.
3. **Balanced Preferences**: Test a scenario where the preferences are balanced, but the total number of monkeys is just more than the available seats.
4. **Minimum Preferences**: Test a scenario with minimal preferences to see if the code handles small values correctly.

Let's construct a test case that combines these edge cases:

```plaintext
4
10 15 5 10
3 6 1 1
15 14 12 4
1 1 1 1
```

Explanation:
1. **10 15 5 10**: Here, 15 monkeys want to sit in row 1, but only 10 seats are available. The code should handle the overflow and use no preference monkeys to maximize the seating.
2. **3 6 1 1**: This is a small-scale test where 6 monkeys want to sit in row 1, but only 3 seats are available. The code should correctly handle the overflow and use the no preference monkeys.
3. **15 14 12 4**: This is a balanced test where the total number of monkeys is just more than the available seats. The code should maximize the seating by using the no preference monkeys.
4. **1 1 1 1**: This is a minimal test case where the preferences are small. The code should handle this correctly and seat all the monkeys.

This test input should expose any issues with the logic in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Hard Problem
time_limit_ms: 1000
memory_limit_mb: 256
Description: Ball is the teacher in Paperfold University. The seats of his classroom are arranged in $$$2$$$ rows with $$$m$$$ seats each.

Ball is teaching $$$a + b + c$$$ monkeys, and he wants to assign as many monkeys to a seat as possible. Ball knows that $$$a$$$ of them only want to sit in row $$$1$$$, $$$b$$$ of them only want to sit in row $$$2$$$, and $$$c$$$ of them have no preference. Only one monkey may sit in each seat, and each monkey's preference must be followed if it is seated.

What is the maximum number of monkeys that Ball can seat?

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

Each test case contains four integers $$$m$$$, $$$a$$$, $$$b$$$, and $$$c$$$ ($$$1 \leq m, a, b, c \leq 10^8$$$).

Output: For each test case, output the maximum number of monkeys you can seat.

Examples:
input:
5
10 5 5 10
3 6 1 1
15 14 12 4
1 1 1 1
420 6 9 69
output:
20
5
30
2
84

Note: In the second test case, $$$6$$$ monkeys want to sit in the front row, but only $$$3$$$ seats are available. The monkeys that have no preference and the monkeys who prefer sitting in the second row can sit in the second row together. Thus, the answer is $$$3+2=5$$$.

Buggy Code:
import java.util.Scanner;

public class C2044 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int m = sc.nextInt();
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();

            int ans = 0;

            if (m > a) {
                ans += a;
                if (c > 0) {

                    if (m - a >= c) {
                        ans += c;
                        c = 0;
                    } else {
                        ans += m - a;
                    }
                }
            } else {
                ans += m;
            }

            if (m > b) {
                ans += b;
                if (c > 0) {
                    if (m - b >= c) {
                        ans += c;
                    } else {
                        ans += m - b;
                    }
                }
            } else {
                ans += m;
            }
            System.out.println(ans);
        }

    }
}

// 4 5 5 5 1 1 2 1
// 4 5 1 2 1 2 2 1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define deb(...)
#define debArr(...)
#endif
#define ll long long
#define ld long double
#define ull unsigned long long
#define all(x) begin(x), end(x)
#define alln(x) begin(x) + 1, begin(x) + 1 + n
#define allr(x) rbegin(x), rend(x)
#define allrn(x) rbegin(x) + sz(x) - n - 1, rbegin(x) + n + sz(x) - n - 1
#define sz(x) (int)(x).size()
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define vll vector<ll>
#define pii pair<int, int>
#define pll pair<ll, ll>
#define FORs(i, a, b) for (int i = a; i <= b; ++i)
#define fi(a, n) for (int i = a; i <= n; ++i)
#define fj(a, n) for (int j = a; j <= n; ++j)
#define fk(a, n) for (int k = a; k <= n; ++k)
#define fl(a, n) for (int l = a; l <= n; ++l)
#define fiz(a, n) for (int i = a; i < n; ++i)
#define trv(x, a) for (auto &x : a)
#define endl '\n'
#define vvi vector<vi>
#define mat(n, m) (n, vector<int>(m, 0))
#define mem(x) memset(x, 0, sizeof x)
#define int ll

bool pair_sort(const pair<ll, ll> &a, const pair<ll, ll> &bb)
{
    if (a.first == bb.first)
    {
        return a.second < bb.second;
    }
    return (a.first > bb.first);
}

int dx[] = {0, 1, -1, 0, 0};
int dy[] = {0, 0, 0, 1, -1};

const int N = 200020;
const int MOD = 1e9 + 7;
const int sz = 1e5 + 5;
const int inf = 1e18;

void solve()
{
    int n;
    cin >> n;
    int c;
    cin >> c;

    vi a(n + 1);
    fi(1, n) cin >> a[i];
    vvi edges(n + 3);
    fi(1, n - 1)
    {
        int u, v;
        cin >> u >> v;
        edges[u].pb(v);
        edges[v].pb(u);
    }

    vector<vector<int>> dp(n + 3, vi(2));
    int ans = 0;

    fi(1, n)
    {
        dp[i][1] = a[i];
    }

    function<void(ll, ll)> dfs = [&](int node, int pr)
    {
        for (auto edge : edges[node])
        {
            if (edge == pr)
                continue;
            dfs(edge, node);

            dp[node][0] += max(dp[edge][1], dp[edge][0]);
            dp[node][1] += max(dp[edge][0],dp[edge][1] - 2 * c);
            ans = max({ans, dp[node][1], dp[node][0]});
        }
    };

    dfs(1, 0);
    cout << ans << endl;
}
bool tests = 1;
int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t = 1;
    if (tests)
        cin >> t;
    fi(1, 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
1
3 1
-1 0 1
1 2
2 3
```

**Explanation:**
- This test case is designed to highlight a potential issue with the algorithm's handling of negative and zero initial gold values.
- The graph consists of 3 nodes with the following initial gold values: -1, 0, and 1.
- The edges connect the nodes in a linear fashion: 1-2 and 2-3.
- The cost to strengthen a camp is 1 gold from each neighboring camp.
- The optimal strategy here is to not strengthen any camp, as strengthening any camp would result in negative gold values, which is not beneficial.
- The expected output is 0, as no camp can be strengthened without causing negative gold in the neighboring camps.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define int ll
typedef unsigned long long ull;
typedef long double lld;
#define all(x) (x).begin(), (x).end()
#define MOD 1000000007
#define endl  '\n'
const double eps = 1e-8, pi = acos(-1);
const int N1 = 1e5, inf = 1e18;
#define ff              first
#define ss              second
#define rep(i,j)        for(int i = 0; i < (j); ++i)

#ifndef ONLINE_JUDGE
#define debug(x) cout << #x <<" "; _print(x); cout << endl;
#define debugarr(x,n) cout << #x <<" "; _print(x,n); cout << endl;
#else
#define debug(x)
#define debugarr(x,n)
#endif
void _print(ll t) {cout << t;}
//void _print(int t) {cout << t;}
void _print(string t) {cout << t;}
void _print(bool t) {cout << (t ? "True" : "False");}
void _print(char t) {cout << t;}
void _print(lld t) {cout << t;}
void _print(double t) {cout << t;}
void _print(ull t) {cout << t;}
template <class T, class V> void _print(pair <T, V> p);
template <class T,class V> void _print(T v[],V n);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T> void _print(multiset <T> v);
template <class T> void _print(unordered_set <T> v);
template <class T> void _print(unordered_multiset <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T, class V> void _print(multimap <T, V> v);
template <class T, class V> void _print(unordered_map <T, V> v);
template <class T, class V> void _print(unordered_multimap <T, V> v);
template <class T, class V> void _print(pair <T, V> p) {cout << "{"; _print(p.first); cout << ","; _print(p.second); cout << "}";}
template <class T,class V> void _print(T v[],V n) {cout << "[ "; for (V i=0;i<n;i++) {_print(v[i]); cout << " ";} cout << "]";}
template <class T> void _print(vector <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T> void _print(set <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T> void _print(multiset <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T> void _print(unordered_set <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T> void _print(unordered_multiset <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T, class V> void _print(map <T, V> v) {cout << "[ "; for (auto i : v) {_print(i); cout << " ";} cout << "]";}
template <class T, class V> void _print(multimap <T, V> v) {cout << "[ "; for (auto i : v) {_print(i); cout << " ";} cout << "]";}
template <class T, class V> void _print(unordered_map <T, V> v) {cout << "[ "; for (auto i : v) {_print(i); cout << " ";} cout << "]";}
template <class T, class V> void _print(unordered_multimap <T, V> v) {cout << "[ "; for (auto i : v) {_print(i); cout << " ";} cout << "]";}

//all global variables
const int N = 100000; // see constraints
// ll factorialNumInverse[N + 1];   //(used in ncr%MOD)
// ll naturalNumInverse[N + 1];
// ll fact[N + 1];

void fast_io()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}

// rec(node,secured) tells the answer for the subtree rooted at node (secured can be 0,1 for root node)
int rec(int node, int secured,int par, vector<int>adj[], vector<int>&points, int &c, vector<vector<int>>&dp)
{
    if((adj[node].size()==1) && (node != 1))
    {
        if(secured==1) {
            return max(0ll,points[node]);
        }
        else
        {
            return 0;
        }
    }

    if(dp[node][secured] != -1) return dp[node][secured];

    int ans = 0;

    if(secured==0)
    {
        for(auto nb : adj[node])
        {
            if(nb==par) continue;

            ans += max(rec(nb,0,node,adj,points,c,dp), rec(nb,1,node,adj,points,c,dp));
        }
    }
    else if(secured==1)
    {
        ans += points[node];
        for(auto nb : adj[node])
        {
            if(nb==par) continue;

            ans += max(rec(nb,0,node,adj,points,c,dp), rec(nb,1,node,adj,points,c,dp)-(2*c));
        }
    }

    return dp[node][secured] = ans;
}

// make set of T.C. with output in notebook
void solve()
{
    int n,c; cin>>n>>c;
    vector<int>points(n+1,0);
    for(int i=0;i<n;i++) {
        cin>>points[i+1];
    }

    vector<int>adj[n+1];
    for(int i=0;i<n-1;i++) {
        int u,v; cin>>u>>v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    // Dp, as no greedy ==> Dp on trees -> Make recursion and give english defination

    if(n==1)
    {
        cout<<points[1]<<endl; return;
    }

    vector<vector<int>>dp(n+1,vector<int>(2,-1));
    int ans = max(rec(1,0,0,adj,points,c,dp), rec(1,1,0,adj,points,c,dp));

    cout<<ans<<endl;

    return;
}

int32_t main()
{
    fast_io();
    ll testcases=1;ll cse = 0;
    cin >> testcases;
    // InverseofNumber(MOD);InverseofFactorial(MOD);factorial(MOD);
    while (testcases--)
    {
        //cse++;
        //cout<<"Case #"<<cse<<": ";
        //cout<<fixed<<setprecision(10)<< <<endl;
        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 create a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that can cause the algorithm to fail. Here are the steps to generate such a test case:

1. **Single Node Case**: This is already handled correctly in the code, so it's not a good candidate for exposing bugs.
2. **Linear Structure**: A linear structure (a chain of nodes) can be challenging because it tests the propagation of the strengthening effect through the chain.
3. **Tree with Multiple Branches**: A tree with multiple branches can test the algorithm's ability to handle complex substructures.
4. **Negative Gold Values**: Negative gold values can cause issues if the algorithm does not handle them correctly.
5. **High Cost of Strengthening**: Setting a high cost for strengthening can make it less beneficial to strengthen some nodes, testing the algorithm's decision-making.

Let's create a test case that combines some of these elements:

- A tree with a linear structure and some branches.
- Some nodes with negative gold values.
- A moderate cost for strengthening to ensure that the algorithm needs to make non-trivial decisions.

Here is the test input:

```plaintext
1
5 2
5 -3 10 1 2
1 2
2 3
3 4
4 5
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Camps and Cost of Strengthening**: 5 camps, cost of strengthening is 2 gold.
- **Initial Gold of Each Camp**: 5, -3, 10, 1, 2
- ** Trails**:
  - Camp 1 is connected to Camp 2
  - Camp 2 is connected to Camp 3
  - Camp 3 is connected to Camp 4
  - Camp 4 is connected to Camp 5

This test case includes:
- A linear structure (chain of nodes).
- A node with negative gold (Camp 2).
- A moderate cost for strengthening (2 gold).

This test case should expose any issues in the algorithm's handling of negative gold values and the cost of strengthening.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

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

using namespace std;
typedef long long ll;

void solve() {
    int n, c;
    cin >> n >> c;

    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }

    vector g(n, vector<int>());
    for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        u--, v--;

        g[u].push_back(v);
        g[v].push_back(u);
    }

    set<pair<int, int>, greater<>> s;
    for (int i = 0; i < n; i++) {
        s.emplace(a[i], i);
    }

    //s存的是[收益, 编号]
    vector<int> val(a);
    vector<int> used(n);
    while (!s.empty()) {
        auto [x, u] = *s.begin();
        s.erase(s.begin());

        if (x < 0) {
            break;
        }

        used[u] = 1;

        for (auto v : g[u]) {
            if (!used[v]) {
                assert(s.count({val[v], v}));
                s.erase({val[v], v});
                val[v] -= 2 * c;
                s.insert({val[v], v});
            }
        }
    }

    ll ans = 0;
    for (int i = 0; i < n; i++) {
        if (used[i]) {
            ans += val[i];
        }
    }

    cout << ans << '\n';
}

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

    int T = 1;
    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
1
4 2
4 2 1 3
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
import functools
import sys
# sys.setrecursionlimit(2000000)
from collections import defaultdict
import collections
#from sortedcontainers import SortedList
import math
import heapq
from collections import deque
import bisect
import sys
input = sys.stdin.readline

def solve(N,C,arr,graph):

    nodes = []
    for i in range(N):
        curr_gold = arr[i]
        neighbor_gold = 0
        for child in graph[i]:
            if arr[child] > 0:
                # if C > arr[child]:
                #     continue
                neighbor_gold += min(arr[child],C)
        nodes.append((curr_gold-neighbor_gold,i))
    nodes.sort(reverse=True)

    ans = 0
    nodes_saved = []
    seen = set()
    for diff,node in nodes:
        curr_gold = arr[node]
        if curr_gold <= 0:
            continue

        neighbor_gold = 0
        for child in graph[i]:
            if arr[child] > 0 or child in seen:
                #neighbor_gold += C
                if child in seen:
                    neighbor_gold += C
                else:
                    neighbor_gold += min(arr[child],C)
        if curr_gold > neighbor_gold:
            #print('node',node,gra)
            for child in graph[node]:
                arr[child] -= C
            nodes_saved.append(node)
            seen.add(node)
    #print(sorted(nodes_saved),arr)
    #print(graph)
    return max(0,sum([arr[node] for node in nodes_saved]))

t = int(input())
for j in range(1 , t +1):
    #S = input()
    #N = int(input())
    #A = input().strip()
    #mat = []
    #A,B = input().split(" ")
    N, C = list(map(int,input().split(" ")))
    arr = list(map(int,input().split(" ")))
    graph = [[] for _ in range(N)]
    mat = []
    for _ in range(N-1):
        u,v = list(map(int,input().split(" ")))
        u -= 1
        v -= 1
        graph[u].append(v)
        graph[v].append(u)

    result = solve(N,C,arr,graph)
    print(result)
    #print("Case #{}: {}".format(j, result))
    ## print('{:.8f}'.format(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
1
5 2
1 2 3 4 5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
def sort1(d,l) :
    if len(l) <= 1:
        return l
    else:

        pivot = l[-1]

        less_than_pivot = [x for x in l[:-1] if d[x]['gold'] <= d[pivot]['gold']]
        greater_than_pivot = [x for x in l[:-1] if d[x]['gold'] > d[pivot]['gold']]

        return sort1(d,greater_than_pivot)  + [pivot] + sort1(d,less_than_pivot)

def distrength(d,i,c):
    for j in d[i]['next'] :
        if d[j]['strengh']==True :
            d[j]['strengh']=False
            d[i]['gold']+=c
            for k in d[j]['next'] :
                d[i]['gold']+=c

out=[]

def rigel(d,i):
    for j in d[i]['next'] :
        d[j]['next'].remove(i)
        d[j]['previous'].append(i)
    for j in d[i]['next'] :
        rigel(d,j)

def ygedrouhou(d,i,c):
    s=0
    for j in d[i]['next']:
        if d[j]['strengh']==True:
            s+=min(max(d[j]['gold']-c,0),c)
    for j in d[i]['previous']:
        s+=min(max(d[j]['gold'],0),c)
    if d[i]['gold']>s:

        return True
    else:
        return False

def strengthen(d,i,c):

    d[i]['strengh']=True
    for j in d[i]['next']:
            d[j]['gold']-=c
    for j in d[i]['previous']:
            d[j]['gold']-=c

def wledou(d,i,c):
    verif=True
    for j in d[i]['next']:
        if d[j]['strengh']==True and d[j]['gold']<2*c:
            verif=False
    return verif

def robinhood(d,i,c):
    if len(d[i]['next'])!=0:

        for j in d[i]['next']:
            robinhood(d,j,c)
        if ygedrouhou(d,i,c) and wledou(d,i,c):

            strengthen(d,i,c)
        elif ygedrouhou(d,i,c) :

            strengthen(d,i,c)
            distrength(d,i,c)
    else:

        a=d[i]['previous'][0]
        if (d[i]['gold']>c or d[a]['gold']<=d[i]['gold']) and d[i]['gold']>0 :

            d[i]['strengh']=True
            d[a]['gold']-=c

t=int(input())
i=0
for _ in range(t):
    i+=1
    n,c=map(int,input().split())
    golds=list(map(int,input().split()))
    L=[]
    d={}
    if i == 43 :
        out.append('20')
    elif n==1 :
        out.append(max(0,golds[0]))
    else :
        for i in range(n):
            d[i+1]={
                'next':[],
                'previous':[],
                'gold':golds[i],

                'strengh':False
                }
        for _ in range(n-1):
            i,j=map(int,input().split())
            d[i]['next'].append(j)
            d[j]['next'].append(i)

        rigel(d,1)
        for i in range(1,n):
            sort1(d,d[i]['next'])

        robinhood(d,1,c)

        s=0
        for k in range(n) :
            if d[k+1]['strengh']==True   :
                s+=d[k+1]['gold']
        out.append(s)
for i in out :
    print(i)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
t = int(input())
for _ in range(t):
    n, c = map(int, input().split())
    gold = list(map(int, input().split()))
    adj = [[] for _ in range(n)]
    for _ in range(n - 1):
        u, v = map(int, input().split())
        adj[u - 1].append(v - 1)
        adj[v - 1].append(u - 1)
    temp = [g for g in gold]
    stren = set()
    for g in range(n):
        if gold[g] > 0:
            stren.add(g)
            for n in adj[g]:
                temp[n] -= c
    final = set()
    for ind in stren:
        if temp[ind] < len(adj[ind]):
            for n in adj[ind]:
                temp[n] += c
        else:
            final.add(ind)
    ans = 0
    for i in final:
        ans += temp[i]
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
import java.util.*;
import java.io.*;
public class cptree {
    static class Data{// implements Comparable<Data>{
        long camped;
        long uncamped;
        public Data(long camped,long uncamped) {
            this.camped = camped;
            this.uncamped = uncamped;
        }
        // @Override
        // public int compareTo(Data p1){
        //     if(this.low-2*this.high-1<=p1.low-2*p1.high-1) return 1;
        //     return -1;
        // }
    }
    static class Edge {
        long val;
        int cr;
        public Edge(long val,int cr){
            this.cr = cr;
            this.val = val;
        }
    }
    public static PP.Reader cs = new PP.Reader(System.in);
    public static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws IOException {
        int tests = cs.nextInt();
        while(tests>0){
            tests--;
            solve(tests);
        }
        writer.flush();
        writer.close();
    }
    public static long mod = 1000000007;
    public static int ptr;
    public static long bad;
    public static void solve(int t) throws IOException{
        //int x = cs.nextInt();
        boolean print = false;
        if(t==9969) print = true;
        // if(print) System.out.println("-----------");
        int n = cs.nextInt();
        //int q = cs.nextInt();
        long c = cs.nextLong();
        long ndvalue[] = new long[n];
        for(int i = 0; i < ndvalue.length; i++) ndvalue[i] = cs.nextLong();
        // boolean[] fix = new boolean[n];
        if(print) System.out.println(n+" "+c);
        ArrayList<ArrayList<Integer>> tree = new ArrayList<>();
        for(int i=0;i<n;i++) tree.add(new ArrayList<Integer>());
        for(int i=0;i<n-1;i++){
            int t1 = cs.nextInt()-1;
            int t2 = cs.nextInt()-1;
            tree.get(t1).add(t2);
            tree.get(t2).add(t1);
            if(print) System.out.println(t1+" "+t2);
        }
        if(print){
            for(int i=0;i<n;i++) System.out.print(ndvalue[i]+" ");
            System.out.println();
        }
        if(print) return;
        if(n==1){
            System.out.println(0);
            return;
        }
        Data fin =  dfstrdep(tree, 0, -1,ndvalue,c);
        System.out.println(Math.max(fin.camped,fin.uncamped));
    }
    //Witte fxn here
    public static Data dfstrdep(ArrayList<ArrayList<Integer>> tree,int curr,int par,long[] ndvalue,long c){
        //System.out.println(curr+" max "+absmax+" min "+absmin);
        if(curr != 0 && tree.get(curr).size() == 1) return new Data(ndvalue[curr],0);
        long camped = ndvalue[curr];
        long uncamped = 0;
        for(Integer nb : tree.get((int)curr)){
            if(nb == par) continue;
            Data child = dfstrdep(tree, nb, curr,ndvalue,c);
            uncamped += Math.max(child.camped,child.uncamped);
            camped += Math.max(child.uncamped,child.camped-2*c);

        }
        return new Data(camped, uncamped);
    }
    public static long dfstrtot(ArrayList<ArrayList<Long>> tree,long curr,long par,long len[]){
        long below = 0;
        for(Long nb : tree.get((int)curr)){
            if(nb == par) continue;
            below += dfstrtot(tree, nb, curr, len);
        }
        len[(int)curr] = below;
        return below+1;
    }

    public static  int binary(int[] arr,int start,int end,int key){
        int mid = start;
        int st = start;
        int en = end;
        while(start<=end){
            mid = (start+end)/2;
            if(arr[mid]==key) return mid;
            else if(arr[mid]>key) end = mid-1;
            else if(arr[mid]<key) start = mid+1;
        }
        //if(end<start){
            if(end<st) end++;
            // if(start>en) start--;
            // if(Math.abs(arr[end]-key)<=Math.abs(arr[start]-key-1)){
                // return start;
            // }
            return end;
        //}
        //return mid+1;
    }

    // public static int invmod(int A, int M)
    // {
//
        // for (int X = 1; X < M; X++)
            // if (((A % M) * (X % M)) % M == 1)
                // return X;
        // return 1;
    // }
    //
    // public static int gcd(int a,int b){
    //     int c = Math.min(a,b);
    //     int d = Math.max(a,b);
    //     if(d%c==0) return c;
    //     return gcd(c,d%c);
    // }
}
class PP {
    static String IN = "%s";
    static String OUT = "%s";

    static class Reader {
        BufferedReader br;
        StringTokenizer st;

        Reader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        String next() {
            try {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(br.readLine());
                return st.nextToken();
            } catch (Exception ignored) {
            }
            return null;
        }

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

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

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
// Problem: F. Sheriff's Defense
// Contest: Codeforces - Codeforces Round 974 (Div. 3)
// URL: https://codeforces.com/contest/2014/problem/F
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Author: Neil Chetty

import java.io.*;
import java.util.*;

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

  private static void solve() {
    // WRITE CODE HERE
    int n = in.nextInt(), c = in.nextInt();
    int[] a = in.readIntArray(n);
    int[] b = new int[n];
    boolean[] visited = new boolean[n];
    List<List<Integer>> store = new ArrayList<>();
    for (int i = 0; i < n; i++) store.add(new ArrayList<>());
    for (int i = 0; i < n - 1; i++) {
      int u = in.nextInt(), v = in.nextInt();
      u--;
      v--;
      store.get(u).add(v);
      store.get(v).add(u);
    }
    int first = -1;
    PriorityQueue<Pair<Integer, Integer>> queue = new PriorityQueue<>(
        (x, y) -> Integer.compare(a[y.getKey()] - y.getValue(), a[x.getKey()] - 2 * x.getValue()));
    for (int i = 0; i < n; i++) queue.offer(new Pair<Integer, Integer>(i, b[i]));
    long result = 0;
    for (int i = 0; i < n; i++) {
      while (!queue.isEmpty()
          && (queue.peek().getValue() != b[queue.peek().getKey()]
              || visited[queue.peek().getKey()]))
        queue.poll();
      if (queue.isEmpty())
        break;
      Pair<Integer, Integer> cur = queue.poll();
      // out.println(cur.getKey() + " " + a[cur.getKey()] + " " + cur.getValue());
      int score = a[cur.getKey()] - 2 * cur.getValue();
      if (score < 0)
        break;
      visited[cur.getKey()] = true;
      result += score;
      for (int edge : store.get(cur.getKey())) b[edge]++;
      for (int edge : store.get(cur.getKey()))
        queue.offer(new Pair<Integer, Integer>(edge, b[edge]));
    }
    // out.println(store);
    out.println(result);
    // out.println();
  }

  // ALL PRE-WRITTEN HELPER FUNCTIONS
  static QuickReader in = new QuickReader();
  static PrintWriter out = new PrintWriter(System.out);
  private final static int MOD = (int) ((1e9) + 7);

  private static boolean[] primeFinder(int n) {
    boolean[] result = new boolean[n + 1];
    Arrays.fill(result, true);
    result[0] = false;
    result[1] = false;
    for (int i = 2; i * i <= n; i++) {
      if (!result[i])
        continue;
      for (int j = i * i; j <= n; j += i) {
        result[j] = false;
      }
    }
    return result;
  }

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

  private static long lcm(long a, long b) {
    return a * b / gcd(a, b);
  }

  private static long power(long a, long b) {
    long result = 1;
    while (b > 0) {
      if ((b & 1) == 1) {
        result = (result * a) % MOD;
      }
      b >>= 1;
      a = (a * a) % MOD;
    }
    return result;
  }

  private static long modInverse(long a) {
    return power(a, MOD - 2);
  }

  private static long nCr(int n, int r) {
    if (n < r)
      return 0;
    if (r == 0)
      return 1;
    long[] dp = factorial(n);
    return ((dp[n] * modInverse(dp[r]) % MOD) * (modInverse(dp[n - r]) % MOD)) % MOD;
  }

  private static long nCr(int n, int r, long[] dp) {
    if (n < r)
      return 0;
    if (r == 0)
      return 1;
    return ((dp[n] * modInverse(dp[r]) % MOD) * (modInverse(dp[n - r]) % MOD)) % MOD;
  }

  private static long nPr(int n, int r) {
    if (n < r)
      return 0;
    if (r == 0)
      return 1;
    long[] dp = factorial(n);
    return (dp[n] * modInverse(dp[n - r])) % MOD;
  }

  private static long nPr(int n, int r, long[] dp) {
    if (n < r)
      return 0;
    if (r == 0)
      return 1;
    return (dp[n] * modInverse(dp[n - r])) % MOD;
  }

  private static long[] factorial(int n) {
    long[] factorial = new long[n + 1];
    factorial[0] = 1;
    for (int i = 1; i <= n; i++) {
      factorial[i] = (factorial[i - 1] * i) % MOD;
    }
    return factorial;
  }

  private static void printArray(int[] a) {
    for (int i : a) out.print(i + " ");
    out.println();
  }

  private static void printlnArray(int[] a) {
    for (int i : a) out.println(i);
  }

  private static void printMatrix(int[][] a) {
    for (int[] i : a) printArray(i);
  }

  private static void sortArray(int[] a, int order) {
    ArrayList<Integer> l = new ArrayList<>();
    for (int i : a) l.add(i);
    if (order == -1)
      Collections.sort(l, Collections.reverseOrder());
    else
      Collections.sort(l);
    for (int i = 0; i < a.length; i++) a[i] = l.get(i);
  }

  private static class QuickReader {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

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

    String next() {
      while (!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());
    }

    int[] readIntArray(int n) {
      int[] a = new int[n];
      for (int i = 0; i < n; i++) a[i] = nextInt();
      return a;
    }

    long[] readLongArray(int n) {
      long[] a = new long[n];
      for (int i = 0; i < n; i++) a[i] = nextLong();
      return a;
    }
  }
}

class Pair<K, V> {
  private K key;
  private V value;

  public Pair(K key, V value) {
    this.key = key;
    this.value = value;
  }

  public K getKey() {
    return key;
  }

  public V getValue() {
    return value;
  }

  public void setKey(K key) {
    this.key = key;
  }

  public void setValue(V value) {
    this.value = value;
  }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
#include "bits/stdc++.h"

using namespace std;

using ll = long long;
using ld = long double;

const ld PI = 3.141592653589793;
const int MOD = 1e9+7;
const int INF = 1e9;
const ll INFLL = 4e18;
const double EPS = 1e-9;
const int MAXN = 1000*1007;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... x) {__print(t); if (sizeof...(x)) cerr << ", "; _print(x...);}

#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#ifdef ONLINE_JUDGE
    #define debug(x...) 42
#endif

void solve(int NT) {
	int n, C; cin >> n >> C;
	vector<ll> a(n + 1);

	ll sum = -2LL * (n - 1) * C;
	for(int i = 1; i <= n; ++i) cin >> a[i], sum += a[i];

	// c[i] - (a[i] - c[i])
	vector<int> c(n + 1);
	vector<vector<int>> g(n + 1);
	for(int i = 1; i < n; ++i) {
		int x, y; cin >> x >> y;
		g[x].emplace_back(y);
		g[y].emplace_back(x);
		c[x]++; c[y]++;
	}

	set<pair<ll, int>> st;

	vector<int> deleted(n + 1);
	auto fix = [&](int i) -> void {
		st.erase({2LL * c[i] * C - a[i], i});
		c[i]--;
		st.insert({2LL * c[i] * C - a[i], i});
	};

	for(int i = 1; i <= n; ++i) {
		st.insert({2LL * c[i] * C - a[i], i});
	}

	while(!st.empty()) {
		debug(st);
		int x = st.rbegin()->second;

		ll val = 2LL * c[x] * C - a[x];
		st.erase({val, x});

		if(val >= 0) {
			sum += val;
			for(int z: g[x]) {
				if(deleted[z] == 0) fix(z);
			}
		}

		deleted[x] = 1;
	}

	cout << sum;
}

#define TESTCASES
int main() {
    cin.tie(0)->sync_with_stdio(0);

    int t = 1;
    #ifdef TESTCASES
        cin >> t;
    #endif

    for(int i = 1; i <= t; ++i){
        solve(i);
        cout << "\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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The key issues in the code include:

1. **Incorrect handling of the sum calculation**: The initial sum calculation might be incorrect.
2. **Inefficient or incorrect use of the set for optimization**: The set might not be used correctly to find the optimal camp to strengthen.
3. **Incorrect handling of deleted camps**: The code might not correctly manage which camps have been deleted.

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

1. **Tests the sum calculation**: Ensure that the sum of gold is correctly calculated.
2. **Tests the set usage**: Ensure that the set is used to find the optimal camp to strengthen.
3. **Tests the handling of deleted camps**: Ensure that the code correctly manages which camps have been deleted.

Here is a small-scale test input that can expose these issues:

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

**Explanation:**

- **Number of test cases**: 1
- **Number of camps (n)**: 4
- **Gold taken from each neighboring camp (c)**: 1
- **Initial gold of each camp**: 5, 2, 3, 4
- **Trails**:
  - 1 is connected to 2
  - 2 is connected to 3
  - 3 is connected to 4

This test case is designed to:

1. **Test the sum calculation**: The initial sum of gold is 5 + 2 + 3 + 4 = 14.
2. **Test the set usage**: The set should correctly identify the camp with the highest value of `2 * c[i] * C - a[i]` to strengthen.
3. **Test the handling of deleted camps**: The code should correctly manage which camps have been deleted and update the set accordingly.

By running this test input, we can observe if the code correctly calculates the maximum gold that can be kept in the surviving camps. If the code fails, it will help identify the specific issue in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
#include<bits/stdc++.h>
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())
#define debug(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "
using namespace std;

typedef unsigned long ull;
typedef long long ll;
const double Pi = acos(- 1.0);
template<typename T>ostream& operator<<(std::ostream& os, const std::vector<T>& vec) {for(T x : vec) cout << x << ' '; cout << endl;return os;}
const int inf = INT_MAX;
const long long INF = LONG_LONG_MAX;
void Yes(bool is = true){ cout << (is ? "Yes" : "No") << endl; }

void solve(){
    int n; cin >> n;
    long long c; cin >> c;
    vector<long long> a(n);
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    vector<int> g[n];
    for(int i=0;i<n-1;i++){
        int u, v;
        cin >> u >> v;
        --u, --v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    function<pair<long long, long long>(int, int)> dfs = [&](int u, int parent){
        long long dp0 = 0;
        long long dp1 = a[u];
        for(auto &v: g[u]){
            if(v == parent) continue;
            auto [vdp0, vdp1] = dfs(v, u);
            dp0 += max(vdp0, vdp1);
            dp1 += max(vdp0, vdp1 - 2 * c);
        }
        return make_pair(dp0, dp1);
    };
    auto [dp0, dp1] = dfs(1, -1);
    cout << max(dp0, dp1) << endl;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
    #endif
    int t = 1;
    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 given code, we need to focus on edge cases and scenarios that might cause the faulty logic to fail. Here are some considerations:

1. **Edge Cases:**
   - Single camp (n = 1).
   - Linear structure (n = 2, 3, etc.).
   - Star structure (one camp connected to all others).
   - Complete binary tree.

2. **Scenarios:**
   - All camps have the same initial gold.
   - Camps have negative initial gold.
   - Camps have very large or very small initial gold.
   - Strengthening a camp results in a large change in neighboring camps' gold.

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

```plaintext
3
1 1
100
3 1
1 2 3
1 2
2 3
4 1
1 2 3 4
1 2
2 3
3 4
```

### Explanation:

1. **Single Camp:**
   - **Input:**
     ```
     1 1
     100
     ```
   - **Expected Output:**
     ```
     100
     ```
   - **Reason:** There is only one camp, so it cannot be strengthened, and the gold remains 100.

2. **Linear Structure:**
   - **Input:**
     ```
     3 1
     1 2 3
     1 2
     2 3
     ```
   - **Expected Output:**
     ```
     5
     ```
   - **Reason:** Strengthening the second camp results in the gold values [0, 2, 2], and the total gold in the surviving camps is 2 + 3 = 5.

3. **Chain Structure:**
   - **Input:**
     ```
     4 1
     1 2 3 4
     1 2
     2 3
     3 4
     ```
   - **Expected Output:**
     ```
     9
     ```
   - **Reason:** Strengthening the third camp results in the gold values [1, 1, 3, 2], and the total gold in the surviving camps is 3 + 4 = 7.

This test input covers various small-scale scenarios and should help expose any bugs 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: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

The folk hero Robin Hood has been troubling Sheriff of Nottingham greatly. Sheriff knows that Robin Hood is about to attack his camps and he wants to be prepared.

Sheriff of Nottingham built the camps with strategy in mind and thus there are exactly $n$ camps numbered from $1$ to $n$ and $n-1$ trails, each connecting two camps. Any camp can be reached from any other camp. Each camp $i$ has initially $a_i$ gold.

As it is now, all camps would be destroyed by Robin. Sheriff can strengthen a camp by subtracting exactly $c$ gold from each of its neighboring camps and use it to build better defenses for that camp. Strengthening a camp doesn't change its gold, only its neighbors' gold. A camp can have negative gold.

After Robin Hood's attack, all camps that have been strengthened survive the attack, all others are destroyed.

What's the maximum gold Sheriff can keep in his surviving camps after Robin Hood's attack if he strengthens his camps optimally?

Camp $a$ is neighboring camp $b$ if and only if there exists a trail connecting $a$ and $b$. Only strengthened camps count towards the answer, as others are destroyed.

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

Each test case begins with two integers $n$, $c$ ($1 \le n \le 2\cdot10^5, 1 \le c \le 10^9$) — the number of camps and the gold taken from each neighboring camp for strengthening.

The second line of each test case contains $n$ integers $a_1,a_2,\dots,a_n$ ($-10^9 \le a_i \le 10^9$) — the initial gold of each camp.

Then follow $n-1$ lines, each with integers $u$, $v$ ($1 \le u, v \le n$, $u \ne v$) — meaning that there is a trail between $u$ and $v$.

The sum of $n$ over all test cases doesn't exceed $2\cdot10^5$.

It is guaranteed that any camp is reachable from any other camp.

Output: Output a single integer, the maximum gold Sheriff of Nottingham can keep in his surviving camps after Robin Hood's attack.

Examples:
input:
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
output:
3
8
0
17
26

Note: In the first test case, it is optimal to strengthen the second base. The final gold at each base is $[1,3,0]$.

In the second test case, it is optimal to strengthen all bases. The final gold at each base is $[2,4,2]$.

In the third test case, it is optimal to not strengthen any base.

Buggy Code:
def sort1(d,l) :
    if len(l) <= 1:
        return l
    else:

        pivot = l[-1]

        less_than_pivot = [x for x in l[:-1] if d[x]['gold'] <= d[pivot]['gold']]
        greater_than_pivot = [x for x in l[:-1] if d[x]['gold'] > d[pivot]['gold']]

        return sort1(d,greater_than_pivot)  + [pivot] + sort1(d,less_than_pivot)

def distrength(d,i,c):
    for j in d[i]['next'] :
        if d[j]['strengh']==True :
            d[j]['strengh']=False
            d[i]['gold']+=c
            for k in d[j]['next'] :
                d[i]['gold']+=c

out=[]

def rigel(d,i):
    for j in d[i]['next'] :
        d[j]['next'].remove(i)
        d[j]['previous'].append(i)
    for j in d[i]['next'] :
        rigel(d,j)

def ygedrouhou(d,i,c):
    s=0
    for j in d[i]['next']:
        if d[j]['strengh']==True:
            s+=min(max(d[j]['gold']-c,0),c)
    for j in d[i]['previous']:
        s+=min(max(d[j]['gold'],0),c)
    if d[i]['gold']>s:

        return True
    else:
        return False

def strengthen(d,i,c):

    d[i]['strengh']=True
    for j in d[i]['next']:
            d[j]['gold']-=c
    for j in d[i]['previous']:
            d[j]['gold']-=c

def wledou(d,i,c):
    verif=True
    for j in d[i]['next']:
        if d[j]['strengh']==True and d[j]['gold']<2*c:
            verif=False
    return verif

def robinhood(d,i,c):
    if len(d[i]['next'])!=0:

        for j in d[i]['next']:
            robinhood(d,j,c)
        if ygedrouhou(d,i,c) and wledou(d,i,c):

            strengthen(d,i,c)
        elif ygedrouhou(d,i,c) :

            strengthen(d,i,c)
            distrength(d,i,c)
    else:

        a=d[i]['previous'][0]
        if (d[i]['gold']>c or d[a]['gold']<=d[i]['gold']) and d[i]['gold']>0 :

            d[i]['strengh']=True
            d[a]['gold']-=c

t=int(input())
i=0
for _ in range(t):
    i+=1
    n,c=map(int,input().split())
    golds=list(map(int,input().split()))
    L=[]
    d={}
    if i == 42 :
        out.append('20')
    elif n==1 :
        out.append(max(0,golds[0]))
    else :
        for i in range(n):
            d[i+1]={
                'next':[],
                'previous':[],
                'gold':golds[i],

                'strengh':False
                }
        for _ in range(n-1):
            i,j=map(int,input().split())
            d[i]['next'].append(j)
            d[j]['next'].append(i)

        rigel(d,1)
        for i in range(1,n):
            sort1(d,d[i]['next'])

        robinhood(d,1,c)

        s=0
        for k in range(n) :
            if d[k+1]['strengh']==True   :
                s+=d[k+1]['gold']
        out.append(s)
for i in out :
    print(i)

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

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

Think step by step.

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