**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
t=int(input())
for _ in range(t):
    n,k=map(int,input().split())
    a=list(map(int,input().split()))
    a=sorted(a)
    a=a[::-1]
    alice=0
    for i in range(0,n,2):
        alice+=a[i]
    bob=sum(a)-alice
    bob+=k
    if alice-bob<0:
        print(0)
    else:
        print(alice-bob)

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

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

Think step by step.

Response:
```plaintext
4
2 1000000000
1 1000000000
3 1000000000
1 1000000000 1000000000
4 1000000000
1 1 1000000000 1000000000
2 1000000000
1000000000 1000000000
```

### Explanation:
1. **Case 1:**
   - `n = 2`, `k = 1000000000`
   - `a = [1, 1000000000]`
   - Bob can increase the cost of the first item by up to 1000000000, making the costs `[1000000001, 1000000000]`.
   - Alice will take the highest cost item (1000000001), and Bob will take the second highest (1000000000).
   - The score will be `1000000001 - 1000000000 = 1`.

2. **Case 2:**
   - `n = 3`, `k = 1000000000`
   - `a = [1, 1000000000, 1000000000]`
   - Bob can increase the cost of the first item by up to 1000000000, making the costs `[1000000001, 1000000000, 1000000000]`.
   - Alice will take the highest cost item (1000000001), and Bob will take the second highest (1000000000).
   - The score will be `1000000001 + 1000000000 - 1000000000 = 1000000001`.

3. **Case 3:**
   - `n = 4`, `k = 1000000000`
   - `a = [1, 1, 1000000000, 1000000000]`
   - Bob can increase the cost of the first two items by up to 500000000 each, making the costs `[500000001, 500000001, 1000000000, 1000000000]`.
   - Alice will take the two highest cost items (1000000000, 1000000000), and Bob will take the next two (500000001, 500000001).
   - The score will be `1000000000 + 1000000000 - (500000001 + 500000001) = 999999998`.

4. **Case 4:**
   - `n = 2`, `k = 1000000000`
   - `a = [1000000000, 1000000000]`
   - Bob can increase the cost of the first item by up to 1000000000, making the costs `[2000000000, 1000000000]`.
   - Alice will take the highest cost item (2000000000), and Bob will take the second highest (1000000000).
   - The score will be `2000000000 - 1000000000 = 1000000000`.

These test cases are designed to expose the bug in the buggy code by testing large values of `k` and edge cases where Bob can significantly alter the costs to minimize the score.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

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

        for i in range(n - 1):
            if k == 0:
                break
            if a[i] < a[-1]:
                increase = min(k, a[-1] - a[i])
                a[i] += increase
                k -= increase

        sumOfA = sum(a[i] for i in range(n - 1, -1, -2))
        sumOfB = sum(a[i] for i in range(n - 2, -1, -2))
        print(sumOfA - sumOfB)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

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

public class Main {

    // =================== WELCOME TO THE PLAYGROUND ==================
    public static void solver(int n, int k, int a[]) {

        Arrays.sort(a);
        int i = 0;
        while (i < n && k > 0) {
            int d = a[n - 1] - a[i];
            if (d <= k) {
                a[i] += d;
                k -= d;
            } else {
                a[i] += k;
                k -= d;
            }
            i++;
        }
        // o.println(Arrays.toString(a));
        Arrays.sort(a);
        int alice = 0;
        int bob = 0;
        for (int j = 0; j < n; j++) {
            if (j % 2 == 0)
                alice += a[j];
            else
                bob += a[j];
        }
        int ans = Math.abs(alice - bob);
        o.println(ans);
    }

    // ================================================================

    static BufferedReader r;
    static PrintWriter o;
    static StringTokenizer st;
    static final int MOD = (int) 1e9 + 7;
    static long fact[];
    static long ifact[];

    public static void main(String[] args) throws Exception {
        r = new BufferedReader(new InputStreamReader(System.in));
        o = new PrintWriter(new OutputStreamWriter(System.out));

        int t = readInt();
        while (t-- > 0) {
            // int n = readInt();
            int a[] = readArray();
            int n = a[0];
            int k = a[1];
            a = readArray();
            solver(n, k, a);
        }

        o.close();
        r.close();
    }

    static int[] readArray() throws IOException {
        String s[] = r.readLine().split(" ");
        int n = s.length;
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(s[i]);
        }
        return a;
    }

    static int[] readArray(int n) throws IOException {
        int a[] = new int[n];
        int i = 0;
        while (i < n)
            a[i++] = readInt();
        return a;
    }

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(r.readLine().trim());
        return st.nextToken();
    }

    static long readLong() throws IOException {
        return Long.parseLong(next());
    }

    static int readInt() throws IOException {
        return Integer.parseInt(next());
    }

    static double readDouble() throws IOException {
        return Double.parseDouble(next());
    }

    static char readCharacter() throws IOException {
        return next().charAt(0);
    }

    static String readLine() throws IOException {
        return r.readLine().trim();
    }

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

    static long nck(int n, int k) {
        return n < k ? 0 : ((fact[n] * 1L * ifact[n - k]) % MOD * ifact[k]) % MOD;
    }

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

    static void buildFact(int n) {
        fact = new long[n + 1];
        ifact = new long[n + 1];
        fact[0] = ifact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = (fact[i - 1] * i) % MOD;
            ifact[i] = binexpo(fact[i], MOD - 2);
        }
    }

}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose a bug in the provided buggy code. Here's why:

1. **Initial Array and k Value**:
   - The array is `[1, 2, 3, 4]` and `k` is `5`.
   - The goal is to minimize the score `A - B`.

2. **Optimal Strategy**:
   - Alice and Bob will play optimally. Alice will always take the highest available item, and Bob will try to minimize the score by increasing the costs of items.

3. **Expected Optimal Adjustment**:
   - Bob can increase the cost of the smallest item (1) by 5, making the array `[6, 2, 3, 4]`.
   - After sorting, the array becomes `[2, 3, 4, 6]`.
   - Alice will take the items 6 and 3, and Bob will take the items 4 and 2.
   - The score will be `(6 + 3) - (4 + 2) = 3`.

4. **Buggy Code Analysis**:
   - The buggy code sorts the array and then tries to increase the smallest elements by the available `k`.
   - It then sorts the array again and calculates the score.
   - The issue arises when the code incorrectly handles the distribution of the increased values, leading to an incorrect final score.

5. **Expected Output**:
   - The correct output for this test case should be `3`.

This test case is small and complete, focusing on the edge where the distribution of the increased values can significantly affect the final score. It should help in identifying the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
            for (int i = 0; i < n; i++) {
                pq.add(sc.nextInt());
            }

            long sum = 0, sum1 = 0;
            boolean toggle = true;

            while (!pq.isEmpty()) {
                int x = pq.poll();
                if (toggle) {
                    sum += x;
                } else {
                    sum1 += x;
                }
                toggle = !toggle;
            }
            if (sum1 + k >= sum) {
                System.out.println("0");
            } else {
                System.out.println(sum - (sum1 + k));
            }
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2 0
1 1
4 10
1 2 3 4
2 1
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

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

// Macros for convenience
#define fast_io() ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr)
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int, int>
#define pll pair<long long, long long>
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define endl '\n'

// Constants
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll LINF = 1e18;

// Utility functions
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll mod_pow(ll base, ll exp, ll mod) {
    ll result = 1;
    while (exp > 0) {
        if (exp % 2 == 1) result = (result * base) % mod;
        base = (base * base) % mod;
        exp /= 2;
    }
    return result;
}

// Debugging tools (can be removed in a competitive environment)
template<typename T>
void debug(T x) { cerr << x << " "; }
template<typename T, typename V>
void debug(pair<T, V> p) { cerr << "{" << p.first << "," << p.second << "}"; }
template<typename T>
void debug(vector<T> v) { cerr << "[ "; for (T i : v) { debug(i); } cerr << "]"; }
void debug() { cerr << endl; }

int main() {
    fast_io();

    // Example: Reading input
    int t;
    cin >> t;
    while (t--) {
        int n,k;
        cin>>n>>k;
    vector<int> v(n);
        for(int i=0;i<n;i++){
            cin>>v[i];
        }
        sort(v.begin(),v.end());
        int ans=0;
        for(int i=0;i<n;i++){
            if(i%2==0){
                ans+=v[n-i-1];
            }
            else{
                if(v[n-i]-v[n-i-1]<=k){
                    ans-=v[n-i];
                    continue;
                }
                ans-=(v[n-i-1]+k);
            }
        }
        cout<<ans<<endl;

    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
// #define long long int lli
void comp(){

    long long int n,k;
    cin>>n>>k;

    vector<long long int > a(n);

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

    sort(a.rbegin(),a.rend());

    long long int even=0,odd=0;

    for(int i=0;i<n;i++){
        if(i%2==0)
        even+=a[i];
        else
        odd+=a[i];
    }

    long long int dif=even-odd;

    dif=max((long long int)0,dif-k);
    cout<<dif<<"\n";

}

int main() {

	int t;
	cin>>t;

	while(t--){

	    comp();
	}
	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
2
4 10
1 2 3 4
5 5
1 1 1 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
t = int(input())
while t:
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    a.sort(reverse=True)
    re = 0
    for i in range(1, n, 2):
        if k > 0:
            diff = a[i - 1] - a[i]
            if k > diff:
                a[i] = a[i - 1]
                k -= diff
            elif k < diff:
                a[i] = a[i] + k
                k = 0
        if k == 0:
            re += a[i - 1] - a[i]
    if len(a) % 2 == 1:
        re += a[n - 1]
    print(re)

    t -= 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
1
5 10
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
t = int(input())

for _ in range(t):

    n, k = map(int, input().split())
    v = list(map(int, input().split()))

    v.sort(reverse=True)  # Sort in descending order

    if (t!=4):
        print(v)
    diff = k
    ans = 0

    if n % 2 == 0:
        for i in range(1, n, 2):
            diff = min(k, v[i-1] - v[i])
            v[i] += diff
            ans += v[i-1] - v[i]
        print(ans)

    else:
        for i in range(1, n - 1, 2):
            diff = min(k, v[i-1] - v[i])
            v[i] += diff
            ans += v[i-1] - v[i]
        print(ans + v[n-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
1
4 10
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

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

public class C {

    static int readInt(String s){return Integer.parseInt(s);}
    static long readLong(String s){return Long.parseLong(s);}
    static int[] readIntArr(String s){
        String[] tk = s.split(" ");
        int n = tk.length;
        int[] res = new int[n];
        for(int i =0;i<n;i++)
            res[i] = readInt(tk[i]);
        return res;
    }
    static long[] readLongArr(String s){
        String[] tk = s.split(" ");
        int n = tk.length;
        long[] res = new long[n];
        for(int i =0;i<n;i++){
            res[i] = Long.parseLong(tk[i]);
        }
        return res;
    }

    static <T extends Comparable<T>> T max(T a, T b){
        return a.compareTo(b) > 0 ? a : b;
    }

    static <T extends Comparable<T>> T min(T a, T b){
        return a.compareTo(b) < 0 ? a : b;
    }

    public static void solve() throws IOException{
        PrintWriter pw = new PrintWriter(System.out);
        StringBuilder st = new StringBuilder();
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = readInt(br.readLine());
        while(t-->0){

            String[] tk = br.readLine().split(" ");
            int n = readInt(tk[0]);
            long k = readLong(tk[1]);
            long[] ar = readLongArr(br.readLine());
            long maxE = Long.MIN_VALUE;
            for(long a : ar) maxE = max(a, maxE);
            long diffs = 0;
            for(long a : ar) diffs += maxE-a;
            long ans = 0;
            if(diffs > k){
                //if(k == 0){
                Arrays.sort(ar);
                long A =0, B =0;
                long[] ps = new long[n];
                ps[0]=ar[0];
                int l = n;
                for(int i =1;i<n;i++){
                    ps[i] = ps[i-1] + ar[i];
                }
             for(int i =0;i<n;i++){
                    if(ps[i]>k){
                        l = i;
                        break;
                    }

                }
                for(int i = 0;i< n-l;i++){
                    if(i % 2 == 0)
                        A += ar[n-i-1];
                    else
                        B += ar[n-i-1];
                }
                if(n == 2)
                    ans = ar[1] - ar[0]-k;
                else
                    ans = A-B;

            }
            st.append(ans).append('\n');
        }

        pw.println(st);
        pw.flush();
        pw.close();
    }

    public static void main(String[] args) throws IOException {
        solve();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Splitting Items
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob have $n$ items they'd like to split between them, so they decided to play a game. All items have a cost, and the $i$-th item costs $a_i$. Players move in turns starting from Alice.

In each turn, the player chooses one of the remaining items and takes it. The game goes on until no items are left.

Let's say that $A$ is the total cost of items taken by Alice and $B$ is the total cost of Bob's items. The resulting score of the game then will be equal to $A - B$.

Alice wants to maximize the score, while Bob wants to minimize it. Both Alice and Bob will play optimally.

But the game will take place tomorrow, so today Bob can modify the costs a little. He can increase the costs $a_i$ of several (possibly none or all) items by an integer value (possibly, by the same value or by different values for each item). However, the total increase must be less than or equal to $k$. Otherwise, Alice may suspect something. Note that Bob can't decrease costs, only increase.

What is the minimum possible score Bob can achieve?

Input: The first line contains a single integer $t$ ($1 \le t \le 5000$) — the number of test cases. Then $t$ cases follow.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $0 \le k \le 10^9$) — the number of items and the maximum total increase Bob can make.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the initial costs of the items.

It's guaranteed that the sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print a single integer — the minimum possible score $A - B$ after Bob increases the costs of several (possibly none or all) items.

Examples:
input:
4
2 5
1 10
3 0
10 15 12
4 6
3 1 2 4
2 4
6 9
output:
4
13
0
0

Note: In the first test case, Bob can increase $a_1$ by $5$, making costs equal to $[6, 10]$. Tomorrow, Alice will take $10$ and Bob will take $6$. The total score will be equal to $10 - 6 = 4$, and it's the minimum possible.

In the second test case, Bob can't change costs. So the score will be equal to $(15 + 10) - 12 = 13$, since Alice will take $15$, Bob will take $12$, and Alice — $10$.

In the third test case, Bob, for example, can increase $a_1$ by $1$, $a_2$ by $3$, and $a_3$ by $2$. The total change is equal to $1 + 3 + 2 \le 6$ and costs will be equal to $[4, 4, 4, 4]$. Obviously, the score will be equal to $(4 + 4) - (4 + 4) = 0$.

In the fourth test case, Bob can increase $a_1$ by $3$, making costs equal to $[9, 9]$. The score will be equal to $9 - 9 = 0$.

Buggy Code:
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Collections;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class _169_C {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

         int T=Integer.parseInt(st.nextToken());
         StringBuilder sb=new StringBuilder();
         while(T-->0){
             st = new StringTokenizer(br.readLine());
             int n=Integer.parseInt(st.nextToken());
             int k=Integer.parseInt(st.nextToken());
             int[] arr = readArr(n, br, st);
             PriorityQueue<Integer> pq=new PriorityQueue<>(Collections.reverseOrder());
             for(int i=0;i<n;i++){
                 pq.add(arr[i]);
             }
             long alice=0;
             long bob=0;
             int turn=0;
             while(!pq.isEmpty()){
                 int max=pq.remove();
                 if(turn==0){
                     alice+=max;
                     turn=1;
                 }
                 else{
                     bob+=max;
                     turn=0;
                 }
             }
             bob=bob+k;
             long diff=alice-bob;
              long ans=max(0,diff);
              sb.append(ans+"\n");

         }
          System.out.println(sb);

    }

    public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception {
        int[] arr = new int[N];
        st = new StringTokenizer(infile.readLine());
        for (int i = 0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
        return arr;
    }

    public static long gcd(long a, long b) {
        if (a > b)
            a = (a + b) - (b = a);
        if (a == 0L)
            return b;
        return gcd(b % a, a);
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

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

int ans = 0, mx = 0;
vector<int> stage[200005];
int child[200005];
bool removed[200005];

int GetChild(int cur, int pre = -1)
{
    child[cur] = 1;
    for(int nxt : stage[cur])
    {
        if(nxt == pre || removed[nxt])
            continue;
        child[cur] += GetChild(nxt, cur);
    }

    return child[cur];
}

int GetCent(int cur, int size, int pre = -1)
{
    for(int nxt : stage[cur])
    {
        if(nxt == pre || removed[nxt])
            continue;
        if(child[nxt] > size/2)
            return GetCent(nxt, size, cur);
    }

    return cur;
}

void GetAns(int cur, int val, int cent, bool flag, int pre = -1)
{
    if(!flag)
    {
        ans = max(ans, mx-1 + (val + (int)stage[cur].size()-1));
        ans = max(ans, (int)stage[cent].size()-1 + val + (int)stage[cur].size()-1);
    }
    else
        mx = max(mx, (val + (int)stage[cur].size()-1));

    for(int nxt : stage[cur])
    {
        if(nxt == pre || removed[nxt])
            continue;
        GetAns(nxt, val + (int)stage[cur].size()-2, cent, flag, cur);
    }
}

void BuildCent(int cur)
{
    int cent = GetCent(cur, GetChild(cur));

    removed[cent] = 1;
    ans = max(ans, (int)stage[cent].size());
    for(int nxt : stage[cent])
        if(!removed[nxt])
        {
            GetAns(nxt, 0, cent, 0, cent);
            GetAns(nxt, 0, cent, 1, cent);
        }

    //cout << cent << ' ' << ans << '\n';

    for(int nxt : stage[cent])
        if(!removed[nxt])
            BuildCent(nxt);
}

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

    int t;
    cin >> t;
    while(t--)
    {
        int n;
        cin >> n;

        ans = mx = 0;
        for(int i = 1; i <= n; i++)
        {
            stage[i].clear();
            removed[i] = 0;
        }

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

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

        BuildCent(1);

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

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

// Function to read the tree's edges and calculate the degree of each vertex
void readEdgesAndCalculateDegree(int n, vector<int>& degree, vector<pair<int, int>>& edges) {
    for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        degree[u]++;
        degree[v]++;
        edges.push_back({u, v});
    }
}

// Function to calculate the connected components by removing the path between u and v
int calculateConnectedComponents(int u, int v, const vector<int>& degree) {
    return degree[u] + degree[v] - 2;  // Subtracting 2 because u and v are removed
}

// Function to find the maximum connected components by checking all edges
int findMaxConnectedComponents(const vector<pair<int, int>>& edges, const vector<int>& degree) {
    int max_connected = 1;
    for (const auto& edge : edges) {
        int u = edge.first;
        int v = edge.second;
        int connected_components = calculateConnectedComponents(u, v, degree);
        max_connected = max(max_connected, connected_components);
    }
    return max_connected;
}

// Function to handle each test case
void handleTestCase(int n) {
    vector<int> degree(n + 1, 0);
    vector<pair<int, int>> edges;

    // Read the edges and calculate the degree of each vertex
    readEdgesAndCalculateDegree(n, degree, edges);

    // Find the maximum connected components
    int max_components = findMaxConnectedComponents(edges, degree);

    // Output the result
    cout << max_components << '\n';
}

int main() {
    int t;
    cin >> t;

    // Process each test case
    while (t--) {
        int n;
        cin >> n;
        handleTestCase(n);
    }

    return 0;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `3` vertices with edges `1-2` and `2-3`.
   - **Expected Output:** `1`
   - **Reason:** Removing the path between any two vertices will result in only one connected component remaining.

2. **Test Case 2:**
   - **Input:** `4` vertices with edges `1-2`, `2-3`, and `3-4`.
   - **Expected Output:** `2`
   - **Reason:** Removing the path between `2` and `3` will result in two connected components: `{1}` and `{4}`.

3. **Test Case 3:**
   - **Input:** `6` vertices with edges `1-2`, `2-3`, `3-4`, `4-5`, and `5-6`.
   - **Expected Output:** `3`
   - **Reason:** Removing the path between `3` and `4` will result in three connected components: `{1, 2}`, `{5, 6}`, and the isolated vertex `{3}`.

These test cases are designed to expose the bug in the buggy code by focusing on scenarios where the maximum number of connected components can be achieved by removing a specific path. The buggy code incorrectly calculates the number of connected components by simply adding the degrees of the vertices and subtracting 2, which does not account for the actual structure of the tree.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
/********************
 * what  the  sigma *
 ********************/
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
using namespace std;
#define lgm cin.tie(0)->sync_with_stdio(0);
#define be(x) x.begin(),x.end()
#define ve vector
#define ll long long
#define ld long double
#define db(x) cout << "Debug at " << x << endl;
#define ull unsigned ll
#define f first
#define s second
#define pii pair<int, int>
#define tii tuple<int,int,int>
#define pll pair<ll,ll>
#define sz(x) (int)x.size()
#define pb push_back
const int mod = 1e9+7,maxn=200005;
int deg[maxn];
ve<int> ed[maxn];
ve<int> ans1(maxn),ans2(maxn);
void dfs(int u,int pa) {
    ans1[u]=deg[u];
    ans2[u]=deg[u];
    int mx1=-1,mx2=-1;
    for (auto i:ed[u]) {
        if (i==pa) continue;
        dfs(i,u);
        if (deg[u]+ans1[i]-2>ans1[u]) {
            ans1[u]=deg[u]+ans1[i]-2;
        }
        mx2=max(mx2,ans1[u]);
        if (mx2>mx1) swap(mx1,mx2);
    }
    if (mx2!=-1) {
        ans2[u]=mx1+mx2-deg[u]-4;
    }
}
void test() {
    int n;
    cin >> n;
    for (int i=1;i<=n;i++) {
        deg[i]=0;
        ed[i].clear();
    }
    int x,y;
    for (int i=0;i<n-1;i++) {
        cin >> x >> y;
        ed[x].pb(y);
        ed[y].pb(x);
        deg[x]++;
        deg[y]++;
    }
    dfs(1,1);
    int ans=0;
    for (int i=1;i<=n;i++) {
        ans=max(ans,max(ans1[i],ans2[i]));
    }
    cout << ans << '\n';
}
signed main() {
    lgm;
    int t; cin >> t;
    while (t--) {
        test();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios where the code might fail. The code is intended to find the maximum number of connected components that can be formed after removing a path from a tree. Here are some key points to consider:

1. **Star-shaped Trees**: These trees have one central node connected to all other nodes. Removing the central node should maximize the number of connected components.
2. **Linear Trees**: These trees form a single path. Removing any two nodes on this path should split the tree into the maximum number of components.
3. **Balanced Trees**: These trees have a more uniform distribution of nodes. Removing nodes from the center of such trees can also expose bugs.

Let's focus on a star-shaped tree, as it is a common edge case that can expose issues in the algorithm.

### Test Input
```plaintext
1
6
1 2
1 3
1 4
1 5
1 6
```

### Explanation
- **Test Case 1**:
  - **Tree Structure**: The tree is a star-shaped tree with node 1 as the center and nodes 2, 3, 4, 5, and 6 as the leaves.
  - **Expected Output**: Removing the central node (node 1) should result in 5 connected components, each consisting of a single leaf node.

This test case is designed to check if the algorithm correctly handles the removal of a central node in a star-shaped tree, which should maximize the number of connected components. If the code fails to produce the correct output, it indicates a potential bug in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import sys
import threading

def main():
    import sys

    import sys

    sys.setrecursionlimit(1 << 25)
    t = int(sys.stdin.readline())
    for _ in range(t):
        n = int(sys.stdin.readline())
        degrees = [0] * (n +1)
        edges = []
        for _ in range(n-1):
            u,v = map(int, sys.stdin.readline().split())
            degrees[u] +=1
            degrees[v] +=1
            edges.append( (u,v) )
        max_deg = max(degrees[1:n+1])
        max_edge = 0
        for u,v in edges:
            current = degrees[u] + degrees[v] -2
            if current > max_edge:
                max_edge = current
        answer = max(max_deg, max_edge)
        print(answer)

threading.Thread(target=main).start()

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the code might fail to correctly calculate the maximum number of connected components after removing a path. The code currently calculates the maximum degree and the maximum sum of degrees of an edge minus 2, and then takes the maximum of these two values. However, this approach might not always yield the correct result.

### Step-by-Step Analysis:
1. **Maximum Degree Calculation**:
   - The code correctly calculates the maximum degree of any vertex in the tree.
   - This is a valid approach for trees with a single high-degree vertex, but it might not be optimal for more complex structures.

2. **Maximum Edge Sum Calculation**:
   - The code calculates the sum of degrees of each edge and subtracts 2.
   - This is intended to find the maximum number of components formed by removing the path between two vertices.
   - However, this approach might fail in cases where the path removal does not maximize the number of components.

### Edge Cases to Consider:
- **Star Tree**: A tree where one vertex is connected to all other vertices.
- **Path Tree**: A tree where all vertices form a single path.
- **Balanced Tree**: A tree where the degrees of vertices are more evenly distributed.

### Test Case Generation:
Let's create a test case that exposes the bug. A good candidate is a tree with a central vertex connected to multiple paths. This structure can create more components when the central vertex is removed, but the code might not correctly handle this scenario.

### Test Input:
```plaintext
1
7
1 2
1 3
1 4
2 5
3 6
4 7
```

### Explanation:
- The tree has 7 vertices.
- Vertex 1 is connected to vertices 2, 3, and 4.
- Vertices 2, 3, and 4 are connected to vertices 5, 6, and 7, respectively.
- Removing the vertex 1 (which has the highest degree) should ideally create 4 connected components (vertices 5, 6, 7, and the remaining parts of the tree).

This test case should expose the bug in the code, as the maximum degree approach will not correctly calculate the number of components. The correct answer should be 4, but the code might output a different value.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import sys
import threading

def solve():
    sys.setrecursionlimit(1 << 25)
    input = sys.stdin.read
    data = input().split()

    idx = 0
    t = int(data[idx])
    idx += 1

    results = []
    for _ in range(t):
        n = int(data[idx])
        idx += 1
        degrees = [0] * (n + 1)
        edges = []

        for _ in range(n - 1):
            u, v = int(data[idx]), int(data[idx + 1])
            idx += 2
            degrees[u] += 1
            degrees[v] += 1
            edges.append((u, v))

        max_deg = max(degrees[1:n + 1])
        max_edge = 0

        for u, v in edges:
            current = degrees[u] + degrees[v] - 2
            if current > max_edge:
                max_edge = current

        answer = max(max_deg, max_edge)
        results.append(answer)

    sys.stdout.write('\n'.join(map(str, results)) + '\n')

threading.Thread(target=solve).start()

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic used to determine the maximum number of connected components after removing a path. The code seems to be using the degrees of vertices and the sum of degrees of edges to determine the answer, but this approach might not always be correct.

### Key Points to Consider:
1. **Leaf Nodes and Paths**: The code might not handle paths that involve multiple leaf nodes correctly.
2. **Multiple High-Degree Nodes**: The code might not handle cases where there are multiple high-degree nodes correctly.
3. **Central Node with Many Leaves**: The code might not handle cases where there is a central node connected to many leaves correctly.

### Challenging Test Input:
Let's create a test input that includes a central node with multiple leaf nodes and a path that involves multiple leaf nodes. This should help expose any issues in the code's logic.

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

### Explanation:
1. **First Test Case (4 vertices)**:
   - Central node 1 with three leaf nodes (2, 3, 4).
   - Removing any path involving the central node will result in 3 connected components.

2. **Second Test Case (6 vertices)**:
   - Central node 1 with three leaf nodes (2, 3, 4).
   - Node 4 is connected to two additional leaf nodes (5, 6).
   - Removing the path from 1 to 4 will result in 4 connected components.

3. **Third Test Case (7 vertices)**:
   - Central node 1 with three leaf nodes (2, 3, 4).
   - Node 4 is connected to three additional leaf nodes (5, 6, 7).
   - Removing the path from 1 to 4 will result in 4 connected components.

This test input should help expose any issues in the code's logic, especially in handling paths involving multiple leaf nodes and central nodes with high degrees.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import sys
import threading

def main():
    import sys

    import sys

    sys.setrecursionlimit(1 << 25)
    t = int(sys.stdin.readline())
    for _ in range(t):
        n = int(sys.stdin.readline())
        degrees = [0] * (n +1)
        edges = []
        for _ in range(n-1):
            u,v = map(int, sys.stdin.readline().split())
            degrees[u] +=1
            degrees[v] +=1
            edges.append( (u,v) )
        max_deg = max(degrees[1:n+1])
        max_edge = 0
        for u,v in edges:
            current = degrees[u] + degrees[v] -2
            if current > max_edge:
                max_edge = current
        answer = max(max_deg, max_edge)
        print(answer)

threading.Thread(target=main).start()

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the provided code, let's consider the algorithm's logic and identify where it might fail. The code calculates the maximum number of connected components by considering the maximum degree of any vertex and the sum of degrees of any edge minus 2. However, this approach does not always correctly identify the maximum number of connected components that can be formed by removing a path.

### Key Observations:
1. **Maximum Degree Heuristic**: The code uses the maximum degree of any vertex as one of its criteria. This heuristic might fail if the tree has a structure where removing a path between two low-degree vertices can create more connected components than removing a path involving a high-degree vertex.
2. **Edge Degree Heuristic**: The code also considers the sum of degrees of the vertices connected by an edge minus 2. This might not always be the best strategy, especially in trees with specific structures.

### Test Case Design:
To expose the bug, we need a test case where the maximum number of connected components is not correctly identified by the current logic. Consider a tree where removing a path between two low-degree vertices can create more connected components than removing a path involving a high-degree vertex.

### Example Test Case:
Consider a tree with 6 vertices where the structure is such that removing a path between two low-degree vertices creates more connected components than removing a path involving a high-degree vertex.

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

### Explanation:
- The tree is a simple linear chain: `1 - 2 - 3 - 4 - 5 - 6`.
- The maximum degree of any vertex is 2 (for vertices 2, 3, 4, and 5).
- The maximum edge degree sum minus 2 is also 2 (for any edge in the chain).
- However, removing the path between vertices 1 and 6 will create 4 connected components: {1}, {2}, {3}, {4, 5, 6}.

The correct answer for this test case is 4, but the buggy code will output 2 because it only considers the maximum degree and edge degree heuristics.

### Output:
```plaintext
1
6
1 2
2 3
3 4
4 5
5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

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

public class TreeMaxDegree {
    public static void main(String[] args) {
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

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

            for (int testCase = 0; testCase < t; testCase++) {
                int n = Integer.parseInt(br.readLine());

                // Initialize degrees array
                int[] degrees = new int[n + 1];

                // List to store edges
                List<int[]> edges = new ArrayList<>();

                // Read edges and update degrees
                for (int i = 0; i < n - 1; i++) {
                    String[] edge = br.readLine().split(" ");
                    int u = Integer.parseInt(edge[0]);
                    int v = Integer.parseInt(edge[1]);

                    degrees[u]++;
                    degrees[v]++;

                    edges.add(new int[]{u, v});
                }

                // Find maximum degree
                int maxDeg = 0;
                for (int i = 1; i <= n; i++) {
                    maxDeg = Math.max(maxDeg, degrees[i]);
                }

                // Find max edge combination
                int maxEdge = 0;
                for (int[] edge : edges) {
                    int u = edge[0];
                    int v = edge[1];
                    int current = degrees[u] + degrees[v] - 2;
                    maxEdge = Math.max(maxEdge, current);
                }

                // Calculate final answer
                int answer = Math.max(maxDeg, maxEdge);

                // Output result
                bw.write(answer + "\n");
                bw.flush();
            }

            br.close();
            bw.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on the logic that calculates the maximum number of connected components. The code calculates the maximum degree of any vertex and the maximum value of `degrees[u] + degrees[v] - 2` for any edge `(u, v)`. The final answer is the maximum of these two values.

The bug in the code is that it does not correctly account for the scenario where removing a path between two vertices can split the tree into more connected components than just the maximum degree or the sum of degrees minus 2.

To expose this, we need a test case where the structure of the tree is such that removing a specific path can create more connected components than the maximum degree or the sum of degrees minus 2.

### Test Case Analysis
1. **Star Graph**: In a star graph, the center vertex has the maximum degree, and removing the center vertex will split the graph into the maximum number of components.
2. **Path Graph**: In a path graph, removing any edge will split the graph into two components.
3. **Tree with a Long Path and Branches**: A tree where there is a long path and multiple branches can create a scenario where removing a specific path can create more components than the maximum degree or the sum of degrees minus 2.

### Chosen Test Case
We will use a tree with a long path and multiple branches to create a scenario where the bug is exposed.

```plaintext
1
10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
1 11
2 12
3 13
4 14
5 15
6 16
7 17
8 18
9 19
10 20
```

### Explanation
- The tree has a long path from vertex 1 to vertex 10.
- Each vertex on this path has an additional branch (vertices 11 to 20).
- Removing the path from vertex 1 to vertex 10 will split the tree into 10 connected components (each of the branches).

This test case should expose the bug because the code will calculate the maximum degree as 2 (since each vertex on the path has only 2 edges) and the maximum edge combination as 2 (since each edge connects two vertices with degree 2). However, the correct answer is 10, which is the number of components formed by removing the path from vertex 1 to vertex 10.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
import java.util.*;

public class Main {

    static int dp[];

    static int dfs(int i, int par, ArrayList<ArrayList<Integer>> adj, int deg[]) {

        int max1 = 0;
        int max2 = 0;
        for (int child : adj.get(i)) {
            if (child != par) {
                int x = dfs(child, i, adj, deg);
                if (x > max1) {
                    max2 = max1;
                    max1 = x;
                } else if (x > max2) {
                    max2 = x;
                }
            }
        }
        dp[i] = Math.max(max1 + max2 + deg[i] - 2, max1 + deg[i]);
        return max1 + deg[i] - 2;
    }

    static void solve(Scanner sc) {

        int n = sc.nextInt();
        ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        int deg[] = new int[n];
        for (int i = 0; i < n - 1; i++) {
            int x = sc.nextInt() - 1;
            int y = sc.nextInt() - 1;
            adj.get(x).add(y);
            adj.get(y).add(x);
            deg[x]++;
            deg[y]++;
        }

        dp = new int[n];

        dfs(0, -1, adj, deg);

        // System.out.println(Arrays.toString(dp));

        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans = Math.max(ans, dp[i]);
        }
        System.out.println(ans);

    }

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            solve(sc);
        }

    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that can maximize the failure probability. The bug in the code seems to be related to how it calculates the maximum number of connected components after removing a path. Specifically, the `dfs` function might not correctly handle certain tree structures, especially those with specific connectivity patterns.

Let's consider a test case that can expose this bug:

1. **Star-shaped Tree with a Central Vertex:**
   - A star-shaped tree where one central vertex is connected to all other vertices. Removing the central vertex should create the maximum number of connected components.

2. **Linear Tree:**
   - A linear tree (a path) where removing any vertex (except the endpoints) should split the tree into two components.

3. **Balanced Tree:**
   - A balanced tree where removing the root vertex should split the tree into multiple components.

Given these considerations, let's generate a small-scale test input that includes a star-shaped tree and a linear tree:

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

### Explanation:
1. **First Test Case (Star-shaped Tree):**
   - The tree has 5 vertices: 1, 2, 3, 4, 5.
   - Vertex 1 is the central vertex connected to all other vertices.
   - Removing vertex 1 will create 4 connected components: {2}, {3}, {4}, {5}.
   - Expected output: 4.

2. **Second Test Case (Linear Tree):**
   - The tree has 4 vertices: 1, 2, 3, 4.
   - The vertices are connected in a linear fashion: 1-2-3-4.
   - Removing vertex 2 will split the tree into two components: {1} and {3, 4}.
   - Expected output: 2.

This test input should help expose any issues in the provided code, especially in how it handles the removal of vertices and the calculation of connected components.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
queue<int> q;
int links[200009];
vector<int> linked[200009];
int weight[200009];
int n,u,v;
int now;
int cnt;
int maxi;
int main(){
	int tot;
	cin>>tot;
	for (int t=1;t<=tot;t++){
		cin>>n;
		for (int i=1;i<=n-1;i++){
			cin>>u>>v;
			links[u]++;
			links[v]++;
			linked[u].push_back(v);
			linked[v].push_back(u);
		}
		for (int i=1;i<=n;i++){
			if(links[i]==1){
				q.push(i);
				weight[i]=1;
			}
		}
		while(!q.empty()){
			now=q.front();
			q.pop();
			maxi=0;
			cnt=0;
			for(int nu:linked[now]){
				cnt++;
				links[nu]--;
				if(links[nu]<=1&&weight[nu]>maxi){
					maxi=weight[nu];
				}

				if(links[nu]==1){
					q.push(nu);
				}
			}
			weight[now]=maxi+cnt-1;
			//cout<<" "<< maxi<<" "<<cnt<<endl;
		}
		cout<<weight[now]<<endl;
		for(int i=1;i<=n;i++){
			links[i]=0;
			linked[i].clear();
			weight[i]=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 provided code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the steps to think through this:

1. **Understand the Bug**: The code seems to be using a BFS approach to calculate the weight of each node, which is supposed to represent the maximum number of connected components that can be formed by removing the path from the tree. However, the code might not handle certain edge cases correctly, such as:
   - Trees with multiple leaves.
   - Trees with a long path where removing any part of the path should maximize the number of components.
   - Trees with a central node connected to multiple branches.

2. **Design the Test Case**:
   - **Small Scale**: The test input should be small to ensure it fits within the output limits.
   - **Edge Cases**: Focus on trees that have multiple leaves and a central node with multiple branches.
   - **Complex Structure**: Create a tree that has a mix of long paths and multiple branches to challenge the BFS logic.

3. **Generate the Test Case**:
   - **Tree with Multiple Leaves**: A star-like tree where one central node is connected to multiple leaves.
   - **Tree with Long Path**: A linear tree where removing any part of the path should maximize the number of components.
   - **Tree with Multiple Branches**: A tree with a central node and multiple branches, each with its own leaves.

Here is a test input that combines these elements:

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

**Explanation**:
- **Test Case 1**: A star-like tree with 5 vertices. The central node (1) is connected to 4 leaves (2, 3, 4, 5). The maximum number of components is 4.
- **Test Case 2**: A linear tree with 6 vertices. The path is 1-2-3-4-5-6. Removing any part of this path should maximize the number of components, which is 5.
- **Test Case 3**: A tree with a central node (1) connected to two branches. The first branch (1-2-4-5) and the second branch (1-3-6-7). The maximum number of components is 4.

This test input should expose the bug in the provided code by challenging its ability to handle multiple leaves, long paths, and complex tree structures.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

The first line of each test case contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the size of the tree.

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

Examples:
input:
6
2
1 2
5
1 2
2 3
3 4
3 5
4
1 2
2 3
3 4
5
2 1
3 1
4 1
5 4
6
2 1
3 1
4 1
5 3
6 3
6
2 1
3 2
4 2
5 3
6 4
output:
1
3
2
3
4
3

Note:

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

#define ll long long
const int MAX = 200005;

vector<pair<int, int>> adj;  // Single vector to store all edges (pair of nodes)
int deg[MAX];
bool visited[MAX];

void readInput(int &n) {
    adj.clear();  // Clear the global adj vector
    fill(deg, deg + n + 1, 0);
    fill(visited, visited + n + 1, false);

    for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        adj.emplace_back(u, v);  // Store the edge as a pair (u, v)
        adj.emplace_back(v, u);
         deg[u]++;
            deg[v]++;  // Store the reverse edge (v, u)
    }
}

void dfs(int u) {
    visited[u] = true;
    for (auto &p : adj) {
        int v = p.first;  // Get the neighboring node (first element of the pair)
        if (v == u) continue;  // Skip the reverse direction
        if (!visited[v]) {

            dfs(v);
        }
    }
}

int findMaxDegree(int n) {
    int maxDeg = 0;
    for (int i = 1; i <= n; ++i) {
        maxDeg = max(maxDeg, deg[i]);
    }
    return maxDeg;
}

int computeMaxComponents(int n) {
    int ans = findMaxDegree(n);
    for (int i = 0; i < adj.size(); i++) {
        int u = adj[i].first, v = adj[i].second;
        ans = max(ans, deg[u] + deg[v] - (6 - 4));  // Calculating based on degrees of adjacent nodes
    }
    return ans;
}

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

    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;
        readInput(n);
        dfs(1);

        cout << computeMaxComponents(n) << "\n";
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on the issues that might arise from the incorrect handling of the adjacency list and the degree calculations. The primary issues in the code are:

1. **Incorrect Adjacency List**: The adjacency list is stored in a single vector, which makes it difficult to handle the graph structure correctly.
2. **Degree Calculation**: The degree of each node is not being used correctly to determine the maximum number of connected components.
3. **DFS Usage**: The DFS is not necessary for this problem, as it does not help in determining the maximum number of connected components.

Given these issues, we can create a test case that emphasizes the importance of the adjacency list and degree calculations. A good test case would be a star graph, where one central node is connected to all other nodes. This structure can help expose the incorrect handling of degrees and the adjacency list.

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

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

### Explanation:
- **Test Case**: 1
- **Number of Nodes**: 5
- **Edges**:
  - 1-2
  - 1-3
  - 1-4
  - 1-5

This test case represents a star graph where node 1 is the center and is connected to all other nodes (2, 3, 4, 5). The correct output for this test case should be 4, as removing the path from node 1 to any other node will result in 4 connected components (one for each of the other nodes).

This test case should expose the bug in the buggy code because:
- The incorrect adjacency list handling might lead to incorrect degree calculations.
- The incorrect degree calculations will result in an incorrect number of connected components being computed.

By using this test case, you can verify if the buggy code correctly handles the star graph structure and computes the maximum number of connected components.