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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
t = int(input())
for _ in range(t):
    n,c = [int(i) for i in input().split()]
    lst = [int(i) for i in input().split()]
    graph= [[] for i in range(n)]
    for _ in range(n-1):
        a,b = [int(i) - 1 for i in input().split()]
        graph[a].append(b)
        graph[b].append(a)
    tree = [[] for i in range(n)]
    parent = [-1 for i in range(n)]
    stack = [0]
    visited = set()
    while stack:
        a = stack.pop()
        visited.add(a)
        for b in graph[a]:
            if b not in visited:
                stack.append(b)
                tree[a].append(b)
                parent[b] = a
    dfso = []
    stack = [[0]]
    while stack:
        a = stack.pop()
        dfso.append(a)
        for i in a:
            if tree[i]:
                stack.append(tree[i])
    #print(dfso)
    dp = [-float('inf') for i in range(n)]
    dp[0] = (lst[0],0)
    best = -float('inf')
    for i in range(len(dfso)):
        a = dfso[i]
        if a == [0]:
            continue
        b1 = dp[parent[a[0]]][1] + sum(max(0,lst[a[i]]) for i in range(len(a)))
        b2 = (dp[parent[a[0]]][0] + sum(max(0,lst[a[i]] - 2 * c) for i in range(len(a))))
        for i in range(len(a)):
            if lst[a[i]] < 0:
                dp[a[i]] = (-float('inf'),max(b1,b2))
            elif lst[a[i]] < 2 * c:
                dp[a[i]] = (max(b2 - 2*c + lst[a[i]],b1),max(b2,b1 - lst[a[i]]))
            else:
                dp[a[i]] = (max(b1,b2),max(b1-lst[a[i]],b2 - lst[a[i]] + 2 * c))

        #print(dp)
    print(max(max(i) for i in dp+[[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
-1 0 1
1 2
2 3
3 1
1 1
1
4 2
1 2 3 4
1 2
2 3
3 4
```


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  /*

  5 4
  -1 -4 -5 3 9
  4 3
  5 4
  3 1
  4 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  public K getKey() {
    return key;
  }

  public V getValue() {
    return value;
  }

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

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

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

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

Think step by step.

Response:
To generate a challenging test input for the given buggy code, we need to focus on edge cases that can expose potential issues in the algorithm. Here are some considerations:

1. **Edge Case 1: Single Camp (n = 1)**:
   - This should be straightforward, as there are no neighbors to affect. The result should be the initial gold of the single camp.

2. **Edge Case 2: Linear Structure (n = 3, linear graph)**:
   - This can test the handling of a simple linear structure where each camp has at most two neighbors.

3. **Edge Case 3: Star Graph (n = 5, one central camp with 4 neighbors)**:
   - This can test the handling of a central camp with multiple neighbors and the effect of strengthening the central camp.

4. **Edge Case 4: Negative Initial Gold**:
   - This can test how the algorithm handles negative initial gold values, which can lead to interesting edge cases.

5. **Edge Case 5: High Cost (c) Value**:
   - This can test the algorithm's performance when the cost of strengthening a camp is very high.

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

```plaintext
5
1 1
100
3 1
1 2 3
1 2
2 3
5 1
1 -1 -1 -1 100
1 2
1 3
1 4
1 5
3 1000
1000 2000 3000
1 2
2 3
```

This test input includes:
1. A single camp with initial gold 100.
2. A linear structure of 3 camps with initial golds 1, 2, and 3.
3. A star graph with one central camp and 4 neighbors, where the central camp has initial gold 100 and the neighbors have initial gold -1.
4. A linear structure with high cost (c = 1000) and initial golds 1000, 2000, and 3000.

This should help in exposing any bugs in the algorithm.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

public class Main {
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static Scanner sc = new Scanner(System.in);
    public static PrintWriter out = new PrintWriter(System.out);

    public static void printArr(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            out.print(arr[i] + " ");
        }
        out.println();
    } // change array type as needed

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

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

    public static void readArr(int[] arr) throws IOException {
        int n = arr.length;
        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(st.nextToken());
        }
    }

    public static void readArr(long[] arr) throws IOException {
        int n = arr.length;
        StringTokenizer st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            arr[i] = Long.parseLong(st.nextToken());
        }
    }

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

    public static void sort(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }

    public static void sort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; i++) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }

    public static void solve() throws IOException {
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        long c = Long.parseLong(st.nextToken());
        long[] arr = new long[n];
        readArr(arr);
        ArrayList<HashSet<Integer>> e = new ArrayList<>();
        ArrayList<HashSet<Integer>> removede = new ArrayList<>();
        for(int i = 0; i < n; i++) {
            e.add(new HashSet<>());
            removede.add(new HashSet<>());
        }
        for(int i = 0; i < n-1; i++) {
            st = new StringTokenizer(br.readLine());
            int u = Integer.parseInt(st.nextToken())-1;
            int v = Integer.parseInt(st.nextToken())-1;
            e.get(u).add(v);
            e.get(v).add(u);
            removede.get(u).add(v);
            removede.get(v).add(u);
        }
        Queue<Integer> leaves = new LinkedList<>();
        for(int i = 0; i < n; i++) {
            if(e.get(i).size() == 1) leaves.add(i);
        }
        long[][] dp = new long[n][2];
        long ans = 0;
        while(!leaves.isEmpty()) {
            int curr = leaves.remove();
            int p = -1;
            for(int x : removede.get(curr)) {
                p = x;
            }
            dp[curr][1] = arr[curr];
            for(int x : e.get(curr)) {
                if(x != p) {
                    dp[curr][0] += Math.max(dp[x][0], dp[x][1]);
                    dp[curr][1] += Math.max(dp[x][0], dp[x][1] - 2l*c);
                }
            }
            ans = Math.max(ans, Math.max(dp[curr][0], dp[curr][1]));
            if(p == -1) continue;
            removede.get(p).remove(curr);
            if(removede.get(p).size() == 1) leaves.add(p);
        }
        out.println(ans);
    }

    public static void main(String[] args) throws IOException {
        int t = Integer.parseInt(br.readLine());
        // int t=1;
        while (t-->0) {
            solve();
        }
        out.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
1
4 2
1 2 3 4
1 2
2 3
3 4
```


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll M=1e9+7;
const ll N=1e5;
int n,c;
vector<int>a(N);
vector<vector<int>>adj;
void dfs(int s,vector<int>&vis,vector<vector<ll>>&dp){
    dp[s][0]=0;
    dp[s][1]=a[s];
    for(int ch:adj[s]){
        if(vis[ch]==0){
            vis[ch]=1;
            dfs(ch,vis,dp);
            dp[s][0]=dp[s][0]+max(dp[ch][0],dp[ch][1]);
            dp[s][1]=dp[s][1]+max(dp[ch][0],dp[ch][1]-2*c);
        }
    }
}
void solve(){
    cin>>n>>c;

    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    adj.assign(n,{});
    for(int i=0;i<n-1;i++){
        int u,v;
        cin>>u>>v;
        u--;v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    vector<vector<ll>>dp(n,vector<ll>(2,LLONG_MIN));
    vector<int>vis(n,0);
    vis[0]=1;
    dfs(0,vis,dp);
    cout<<max(dp[0][0],dp[0][1])<<endl;
}
int main() {
    ll t=1;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define LSB(i) ((i) & (-i))
#define ll long long
const int dx[]{-1,1,0,0,-1,-1,1,1};
const int dy[]{0,0,1,-1,-1,1,-1,1};
const int MOD = 1e9+7;
#define int ll

const int N = 1e5+5;

int dp1[N], dp2[N], val[N], n, c;
std::vector<int> G[N];

void dfs(int u, int p){
    dp1[u] = val[u];
    for(int v : G[u]){
        if(v==p) continue;
        dfs(v, u);
        dp1[u] += max(-2 * c + dp1[v], dp2[v]);
        dp2[u] += max(dp1[v], dp2[v]);
    }
}

void solve(int testCase)  {
    cin >> n >> c;
    for(int i = 0; i < n; ++i) dp1[i] = 0, dp2[i] = 0, G[i].clear();
    for(int i = 0; i < n; ++i) {
        cin >> val[i];
    }
    for(int i = 1; i < n; ++i){
        int a, b;
        cin >> a >> b;
        --a, --b;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    dfs(0, 0);
    cout << max(dp1[0], dp2[0]) << '\n';
}

int32_t main(){
    fastio();
    int t = 1;
    cin >> t;
    for(int i = 1; i <= t; ++i){
        solve(i);
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
```


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
from collections import defaultdict

def solve():
  n, d = map(int, input().split())
  gold = list(map(int, input().split()))
  graph = defaultdict(list)
  for _ in range(n - 1):
    a, b = map(int, input().split())
    graph[a - 1].append(b - 1)
    graph[b - 1].append(a - 1)
  f1 = [0] * n
  f2 = [0] * n
  used = [False] * n
  def dfs(node):
    used[node] = True
    children = []
    for v in graph[node]:
      if not used[v]:
        dfs(v)
        children.append(v)
    f2[node] = gold[node]
    for child in children:
      f1[node] += max(f1[child], f2[child])
      f2[node] += max(f1[child], f2[child] - 2 * d)
  dfs(0)
  return max(f1[0], f2[0])

def main():
  t = int(input())
  for _ in range(t):
    print(solve())

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:
To generate a challenging test input that exposes the bug in the given code, let's consider the key aspects of the problem and the potential issues that the code might have:

1. **Edge Cases**: Small graphs, graphs with only one camp, and graphs where the initial gold is zero or negative.
2. **Complex Graphs**: Graphs with multiple paths and varying initial gold values.
3. **Large Deductions**: Large values of `c` that significantly affect the neighboring camps' gold.
4. **Optimal Strengthening**: Cases where the optimal solution involves strengthening multiple camps or none at all.

Given these points, let's create a test input that includes a small, but complex graph with varied initial gold values and a significant deduction value `c`.

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

### Explanation:
- **Number of Test Cases**: 1
- **Number of Camps (n)**: 5
- **Gold Deduction (c)**: 5
- **Initial Gold**: [10, 5, 10, 5, 10]
- ** Trails**:
  - Camp 1 is connected to Camp 2
  - Camp 2 is connected to Camp 3
  - Camp 3 is connected to Camp 4
  - Camp 4 is connected to Camp 5

This test case is designed to:
- Test the algorithm's ability to handle a linear graph structure.
- Ensure that the algorithm correctly calculates the impact of strengthening a camp on its neighbors.
- Verify that the algorithm can handle a deduction value `c` that is significant relative to the initial gold values.

This input should help identify any issues in the faulty solution, particularly in the way it handles the dynamic programming and the recursive function calls.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
import sys; input = lambda: sys.stdin.buffer.readline().decode().strip(); print = lambda *args, **kwargs: sys.stdout.write(" ".join(map(str, args)) + "\n", **kwargs)
def InputList(Dtype=int): return list(map(Dtype, input().split()))
from random import randint; RND = randint(1, 10**9);
from collections import deque, defaultdict
from math import lcm, inf, gcd, ceil
from bisect import bisect_left, bisect_right
from functools import lru_cache, cache
from types import GeneratorType
from heapq import heapify, heappop, heappush

# def bootstrap(f, stack=[]):
# 	def wrappedfunc(*args, **kwargs):
# 		if stack: return f(*args, **kwargs)
# 		else:
# 			to = f(*args, **kwargs)
# 			while True:
# 				if type(to) is GeneratorType:
# 					stack.append(to)
# 					to = next(to)
# 				else:
# 					stack.pop()
# 					if not stack: break
# 					to = stack[-1].send(to)
# 			return to
# 	return wrappedfunc

def solve():
	# print(f"=============================")
	N, K = InputList()
	gold = [0] + InputList()
	edges = [InputList() for _ in range(N-1)]
	root = 1
	adj = [[] for _ in range(N+1)]
	for u, v in edges:
		adj[u].append(v)
		adj[v].append(u)

	@cache
	def dfs(node, parent, parent_taken):
		take = gold[node] + sum(dfs(child, node, True) for child in adj[node] if child != parent)
		if parent_taken: take -= 2*K
		leave = sum(dfs(child, node, False) for child in adj[node] if child != parent)
		return max(take, leave)

	return dfs(root, 0, False)

# print(solve())
for i in range(t:=int(input())):
	ans = solve(); print(ans)

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

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

Think step by step.

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

1. **Small-scale Input**: The test input should be small enough to fit within the constraints and easy to verify manually.
2. **Edge Cases**:
   - **Single Camp**: Test the case where there is only one camp.
   - **Linear Structure**: Test a linear structure where camps are connected in a straight line.
   - **Tree Structure**: Test a tree structure with multiple branches.
   - **High and Low Gold Values**: Test with both high and low initial gold values in camps.
   - **High Cost of Strengthening**: Test with a high cost of strengthening to see if the algorithm handles it correctly.
3. **Optimal Strengthening**: Ensure that the test input has a clear optimal solution that can be easily verified.

Based on these considerations, here is a small-scale test input that should expose potential bugs in the provided code:

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

**Explanation**:
1. **Test Case 1**:
   - **Number of Camps**: 1
   - **Cost of Strengthening**: 1
   - **Initial Gold**: 100
   - **Edges**: None (since there's only one camp)
   - **Expected Output**: 100 (since there's only one camp, it can be strengthened without any cost).

2. **Test Case 2**:
   - **Number of Camps**: 2
   - **Cost of Strengthening**: 1
   - **Initial Gold**: 100, 100
   - **Edges**: 1 2 (a single edge connecting the two camps)
   - **Expected Output**: 198 (strengthen both camps, each taking 1 gold from the other, resulting in 99 + 99).

3. **Test Case 3**:
   - **Number of Camps**: 3
   - **Cost of Strengthening**: 1
   - **Initial Gold**: 1, 2, 3
   - **Edges**: 1 2, 2 3 (a linear structure)
   - **Expected Output**: 6 (strengthen the middle camp, which takes 1 gold from both neighbors, resulting in 0, 2, 2).

This test input covers a range of simple yet challenging scenarios that can help identify issues 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: Sheriff's Defense
time_limit_ms: 2000
memory_limit_mb: 256
Description: "Why, master," quoth Little John, taking the bags and weighing them in his hand, "here is the chink of gold."

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
import java.util.*;

public class F{
    public Pair<Long, Long> func(List<Integer>[] graph, long[] gold, int u, int parent, int c){
        long protectCurr = gold[u], notProtectCurr = 0;

        for(int i=0; i<graph[u].size(); i++){
            int v = graph[u].get(i);
            if(v == parent) continue;

            Pair<Long, Long> child = func(graph, gold, v, u, c);
            long childTaken = child.first();
            long childNotTaken = child.second();

            protectCurr += Math.max(childTaken-(2*c), childNotTaken);
            notProtectCurr += Math.max(childTaken, childNotTaken);
        }

        // System.out.println("u: " + u  +  " gold[u]: " + gold[u] + " protectCurr: " + protectCurr + " notProtCurr: " + notProtectCurr);
        return new Pair<>(protectCurr, notProtectCurr);
    }
    public static void main(String...args){
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while(t > 0){
            t--;

            int n = scanner.nextInt();
            int c = scanner.nextInt();

            List<Integer>[] graph = new ArrayList[n+1];
            long[] gold = new long[n+1];
            for(int i=1; i<=n; i++){
                graph[i] = new ArrayList<>();
            }

            for(int i=1; i<=n; i++){
                gold[i] = scanner.nextInt();
            }

            for(int i=0; i<n-1; i++){
                int u = scanner.nextInt();
                int v = scanner.nextInt();

                graph[u].add(v); graph[v].add(u);
            }
            F obj = new F();
            Pair<Long, Long> resPair = obj.func(graph, gold, 1, 0, c);
            long res = Math.max(resPair.first(), resPair.second());

            System.out.println(res);
        }
    }
    class Pair<T, K>{
        public T first;
        public K second;
        Pair(T t, K k){
            this.first = t;
            this.second = k;
        }
        T first(){
            return this.first;
        }
        K second(){
            return this.second;
        }
    }
}

/*
6
3 1
2 3 1
1 2
2 3
3 1
3 6 3
1 2
2 3
3 1
-2 -3 -1
1 2
2 3
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
4 3
1 5 7 10
4 2
1 2
2 3
*/

/*
10
2 2
-5 -2
2 1
2 5
7 10
2 1
5 4
-1 -4 -5 3 9
4 3
5 4
3 1
4 2
4 3
1 5 7 10
4 2
1 2
2 3
8 1
3 5 2 7 8 5 -3 -4
7 3
1 8
4 3
3 5
7 6
8 7
2 1
2 2
-3 5
2 1
7 2
3 6 7 10 4 1 5
3 1
7 6
4 6
3 7
5 4
2 7
6 1
5 -4 3 6 7 3
4 1
5 1
3 5
3 6
1 2
1 3
-4
8 3
8 4 6 9 9 -5 -1 -2
5 7
4 8
4 3
8 6
3 1
5 2
5 3
5 5
6 1 1 3 5
5 2
4 1
5 4
3 2
8 1
0 5 7 10 0 -5 3 9
2 6
6 5
7 1
8 6
3 4
7 3
6 4
4 5
8 2 6 5
4 3
1 2
4 1
8 1
*/

/*
1
2 5
7 10
2 1
*/

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

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

Think step by step.

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
import java.util.Scanner;

public class Main{
    static void insertEdge(int[][] edges, int u, int v) {
        edges[u][++edges[u][0]] = v;
        edges[v][++edges[v][0]] = u;
    }
    static void dfs(long[][] dp, int[][] edges, long[] vals, int c, int node, int pre) {
        for(int i = 1; i <= edges[node][0]; ++i) {
            int next = edges[node][i];
            if(next == pre) continue;
            dfs(dp, edges, vals, c, next, node);
            dp[0][node] += Math.max(dp[1][next], dp[0][next]);
            dp[1][node] += Math.max(dp[1][next] - 2 * c, dp[0][next]);
        }
        dp[1][node] += vals[node];
    }
    static void solve(Scanner reader) {
        int n, c, u, v;
        n = reader.nextInt();
        c = reader.nextInt();

        long[][] dp = new long[2][n + 1];
        int[][] edges = new int[n + 1][n];
        long[] vals = new long[n + 1];

        for(int i = 1; i <= n; ++i) vals[i] = reader.nextInt();
        for(int i = 1; i < n; ++i) {
            u = reader.nextInt();
            v = reader.nextInt();
            insertEdge(edges, u, v);
        }

        dfs(dp, edges, vals, c, 1, -1);
        System.out.println(Math.max(dp[1][1],dp[0][1]));
    }
    public static void main(String args[]){
        Scanner reader = new Scanner(System.in);
        int t;
        t = reader.nextInt();
        while(t-- > 0) solve(reader);
    }
}

**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
10 10 10 10
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
// Copyright © 2024 Youngmin Park. All rights reserved.
// #pragma GCC optimize("O3")
// #pragma GCC target("avx2")
#include <bits/stdc++.h>
using namespace std;

#pragma region TEMPLATE

using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
using vpi = vector<pii>;
using pll = pair<ll, ll>;
using vl = vector<ll>;
using vpl = vector<pll>;
using ld = long double;
template <typename T, size_t SZ>
using ar = array<T, SZ>;
template <typename T>
using pqg = priority_queue<T, vector<T>, greater<T>>;

#define all(v) (v).begin(), (v).end()
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound

constexpr int INF = 1e9;
constexpr ll LINF = 1e18;
const ld PI = acos((ld)-1.0);
constexpr int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
constexpr bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }
template <typename T>
constexpr bool ckmax(T &a, const T &b) { return b > a ? a = b, 1 : 0; }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }  // divide a by b rounded up
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }  // divide a by b rounded down

#ifdef LOCAL
#include "miscellaneous/debug.h"
#else
#define dbg(...) 42
#endif

#pragma endregion TEMPLATE

void solve() {
    int n;
    cin >> n;
    string c;
    cin >> c;
    {
        int total_red = count(all(c), 'R');
        if (total_red == n || total_red == 0) {
            cout << "YES\n";
            return;
        }
    }
    int st = 0;
    for (int i = 0; i < n; i++) {
        if (c[i] != c[0]) {
            st = i;
            break;
        }
    }
    vi ret;
    for (int i = 0; i < n; i++) {
        int ni = (i + st) % n;
        int cnt = 1;
        while (i + 1 < n && c[(i + 1 + st) % n] == c[ni]) i++, cnt++;
        ret.pb(cnt);
    }
    if (sz(ret) > 4) {
        cout << "NO\n";
        return;
    } else if (sz(ret) == 1) {
        cout << "YES\n";
    } else if (sz(ret) == 2) {
        int v1 = ret[0], v2 = ret[1];
        if (v1 > 1 && v2 > 1) {
            cout << "NO\n";
            return;
        }
        cout << "YES\n";
    } else if (sz(ret) == 4) {
        for (int i = 0; i < 4; i++) {
            int v1 = ret[i], v2 = ret[(i + 1) % 4], v3 = ret[(i + 2) % 4];
            if (v1 > 1 && v2 > 1) {
                cout << "NO\n";
                return;
            }
            if (v1 > 1 && v1 % 2 == v3 % 2) {
                cout << "NO\n";
                return;
            }
        }
        if (*max_element(all(ret)) == 1) {
            cout << "NO\n";
            return;
        }
        cout << "YES\n";
    }
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    int testcase = 1;
    cin >> testcase;
    while (testcase--) {
        solve();
    }
#ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * (double)clock() / CLOCKS_PER_SEC << " s.\n";
#endif
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

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

#define all(x) x.begin(), x.end()
#define int long long

using namespace std;

vector<char> f(vector<char> a, int shift) {
  vector<char> res;
  for (int i = shift + 1; i < a.size(); ++i) {
    res.push_back(a[i]);
  }
  for (int i = 0; i <= shift; ++i) {
    res.push_back(a[i]);
  }
  return res;
}

void sol() {
  int n;
  string s;
  cin >> n >> s;
  vector<char> a;
  int cnt0 = 0;
  int cnt1 = 0;
  for (auto el : s) {
    if (el == 'R') {
      a.push_back(0);
      ++cnt0;
    } else {
      a.push_back(1);
      ++cnt1;
    }
  }
  if (cnt0 <= 1 || cnt1 <= 1) {
    cout << "YES\n";
    return;
  }
  bool flag0 = false;
  bool flag1 = false;
  for (int i = 0; i < n; ++i) {
    if (a[i] == a[(i + 1) % n]) {
      if (a[i] == 0) flag0 = true;
      else flag1 = true;
    }
  }
  if (flag0 && flag1) {
    cout << "NO\n";
    return;
  }
  if (n % 2 == 0) {
    cout << "NO\n";
    return;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] != a[(i + 1) % n]) {
      a = f(a, i);
      break;
    }
  }
  vector<pair<int, int>> aboba{{a[0], 1}};
  for (int i = 1; i < n; ++i) {
    if (a[i] == a[i - 1]) {
      aboba.back().second++;
    } else {
      aboba.push_back({a[i], 1});
    }
  }
  vector<int> kek(2);
  for (int j = 0; j < aboba.size(); ++j) {
    if (aboba[j].second % 2 == 1) {
      kek[j % 2] = 1;
    }
  }
  if (*min_element(all(kek)) > 0) {
    cout << "YES\n";
    return;
  }
  cout << "NO\n";
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) {
    sol();
  }
}

**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
RRR
4
RBRB
5
RBRBR
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    ll T,a,b,c,d,n,m;
    cin>>T;
    for (ll te=0; te<T;te++){
        string str;
        cin>>n>>str;
        bool allEqual=true;
        bool doubleR=false;
        bool doubleE=false;
        for (ll i=0;i<n;i++)
        {
            if (str[i]!=str[0])
            {
                allEqual=false;
            }
            if (str[i]=='B' && str[i]==str[(i+1)%n])
            {
                doubleE=true;
            }
            if (str[i]=='R' && str[i]==str[(i+1)%n])
            {
                doubleR=true;
            }
        }
        bool poss=true;
        if (!allEqual)
        {
            //cout<<doubleE<<" "<<doubleR<<endl;
            if (doubleE && doubleR || !doubleE && !doubleR)
            {
                poss=false;
            }
            else
            {
                ll start=-1;
                for (int i=0;i<n;i++)
                {
                    if (doubleE && str[i]=='R' && str[(i+1)%n]=='B' || doubleR && str[i]=='B' && str[(i+1)%n]=='R')
                    {
                        start=i+1;
                        break;
                    }
                }
                ll last=start;
                ll count=0;
                for (int i=start;i<start+n;i++)
                {
                    if (str[i%n]!=str[(i+1)%n])
                    {
                        ll runLength = i-last+1;
                        if (runLength%2==0)
                            count++;
                        last = i+1;
                    }
                }
                //cout<<count<<endl;
                poss= (count==1);
            }
        }
        if (poss)
        {
            cout<<"YES\n";
        }
        else
        {
            cout<<"NO\n";
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. The key points to consider are:

1. **All Edges of the Same Color**: The code should handle cases where all edges are the same color (all 'R' or all 'B').
2. **Alternating Colors**: The code should handle cases where the colors alternate (e.g., 'RBRBRB').
3. **Single Pair of Consecutive Edges of the Same Color**: The code should handle cases where there is exactly one pair of consecutive edges of the same color.
4. **Multiple Pairs of Consecutive Edges of the Same Color**: The code should handle cases where there are multiple pairs of consecutive edges of the same color.
5. **Mixed Patterns**: The code should handle cases with mixed patterns that are not straightforward.

Given these points, a small-scale test input that includes a mix of these scenarios can be:

```plaintext
4
5
RRRRR
5
RBRBR
6
RBRBRB
6
RRBBRR
```

### Explanation:
1. **Case 1: All Edges of the Same Color (RRRRR)**
   - This should output "YES" because there is a palindrome route between any two vertices.

2. **Case 2: Alternating Colors (RBRBR)**
   - This should output "YES" because there is a palindrome route between any two vertices.

3. **Case 3: Alternating Colors with One Pair of Consecutive Edges of the Same Color (RBRBRB)**
   - This should output "NO" because there is no palindrome route between certain pairs of vertices.

4. **Case 4: Multiple Pairs of Consecutive Edges of the Same Color (RRBBRR)**
   - This should output "NO" because there is no palindrome route between certain pairs of vertices.

This test input is designed to cover a variety of edge cases and should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import sys
import os
from io import BytesIO, IOBase

# region fastio
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

if sys.version_info[0] < 3:
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

file = sys.stdin
if os.environ.get('USER') == "loic":
    file = open("data.in")

line = lambda: file.readline().split()
ui = lambda: int(line()[0])
mi = lambda: map(int,line())
li = lambda: list(mi())
lsi = lambda: list(map(int,line()[0]))
#######################################################################

def solve():

    r = A.count('R')
    b = A.count('B')

    if r <= 1 or b <= 1:
        return "YES"

    if N % 2 == 1:
        if r < b:
            for i in range(N):
                if A[i] == A[i-1] == 'R':
                    return "NO"
            return "YES"

        if b < r:
            for i in range(N):
                if A[i] == A[i-1] == 'B':
                    return "NO"
            return "YES"

    return "NO"

for test in range(1,ui()+1):
    N = ui()
    A = list(line()[0])

    print(solve())

file.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
4
RRRR
4
RRRB
4
RBRB
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import sys
import os
from io import BytesIO, IOBase

# region fastio
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

if sys.version_info[0] < 3:
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

file = sys.stdin
if os.environ.get('USER') == "loic":
    file = open("data.in")

line = lambda: file.readline().split()
ui = lambda: int(line()[0])
mi = lambda: map(int,line())
li = lambda: list(mi())
lsi = lambda: list(map(int,line()[0]))
#######################################################################

def solve():

    r = A.count('R')
    b = A.count('B')

    if r <= 1 or b <= 1:
        return "YES"

    if N % 2 == 1:
        c = 0
        for i in range(N):
            if A[i] == A[i-1]:
                c += 1
        if c == 1:
            return "YES"

        if r == 2 or b == 2:
            val = 'R' if r == 2 else 'B'
            f = A.index(val)
            s = A.index(val, f + 1, N)
            if s - f != 1 and s - f != N - 1:
                return "YES"

    return "NO"

for test in range(1,ui()+1):
    N = ui()
    A = list(line()[0])

    if test == 142:
        print("_".join(v for v in A))

    print(solve())

file.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
1
6
RBRBRB
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import sys

input = lambda: sys.stdin.readline().strip()
printerr = lambda *args, **kwargs: print("\u001B[31m", *args, "\u001B[0m", file=sys.stderr, **kwargs)

def main():
    n = int(input())
    s = input()
    R = s.count('R')
    B = s.count('B')
    r = {i%2 for i in range(n) if s[i]+s[(i+1)%n] == 'RR'}
    r2 = {i%2 for i in range(n) if s[i]+s[(i+1)%n]+s[(i+2)%n] == 'RRB' or s[i]+s[(i+1)%n]+s[(i+2)%n] == 'BRR'}
    b = {i%2 for i in range(n) if s[i]+s[(i+1)%n] == 'BB'}
    b2 = {i%2 for i in range(n) if s[i]+s[(i+1)%n]+s[(i+2)%n] == 'BBR' or s[i]+s[(i+1)%n]+s[(i+2)%n] == 'RBB'}
    print('YES' if (R<=1 or B<=1 or n%2 and (not r-r2 and not b-b2)) and not (r and b) else 'NO')

tcs = int(input())
for tc in range(tcs):
    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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: The code should handle small and large values of \( n \) correctly.
2. **Special Patterns**: The code should handle specific patterns of 'R' and 'B' that might not be correctly evaluated by the current logic.
3. **Balanced and Unbalanced Colors**: The code should handle cases where the number of 'R' and 'B' edges is balanced or unbalanced.

Let's break it down step by step:

1. **Small \( n \)**: Test with small values of \( n \) to ensure the code handles basic cases correctly.
2. **Large \( n \)**: Test with large values of \( n \) to ensure the code does not have performance issues.
3. **Balanced Colors**: Test with equal numbers of 'R' and 'B' edges.
4. **Unbalanced Colors**: Test with a significant imbalance in the number of 'R' and 'B' edges.
5. **Special Patterns**: Test with patterns that might not be correctly evaluated by the current logic, such as alternating 'R' and 'B' or specific sequences like 'RRB' and 'BRR'.

Given these considerations, let's create a test input that includes a variety of cases:

```plaintext
7
3
RRR
3
BBB
4
RBRB
4
BRBR
6
RRBBRR
6
BBRRBB
10
RBRBRBRBRB
```

This test input includes:
- Small values of \( n \) (3 and 4) to test basic cases.
- Large value of \( n \) (10) to test performance.
- Balanced colors (4 and 6).
- Unbalanced colors (3 and 3).
- Special patterns (4 and 6).

This should help expose any bugs in the provided code.

7
3
RRR
3
BBB
4
RBRB
4
BRBR
6
RRBBRR
6
BBRRBB
10
RBRBRBRBRB


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
//make sure to make new file!
import java.io.*;
import java.util.*;

public class F985{

   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);

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

      for(int q = 1; q <= t; q++){

         int n = Integer.parseInt(f.readLine());

         char[] array = f.readLine().toCharArray();
         //check all equal
         boolean allequal = true;
         int rfreq = 0;
         int bfreq = 0;
         if(array[0] == 'R') rfreq++;
         else bfreq++;
         for(int k = 1; k < n; k++){
            if(array[k] != array[k-1]){
               allequal = false;
            }
            if(array[k] == 'R') rfreq++;
            else bfreq++;
         }

         if(allequal || rfreq == 1 || bfreq == 1){
            out.println("Yes");
            continue;
         }

         //only way is if there are equal 4 groupings
         //one color only has groups of 1, other color has 1 even group and 1 odd group
         int[] psize = new int[n];
         psize[0] = 1;
         ArrayList<Integer> alist = new ArrayList<Integer>();
         for(int k = 1; k < n; k++){
            if(array[k] == array[k-1]) psize[k] = psize[k-1] + 1;
            else{
               psize[k] = 1;
               alist.add(psize[k-1]);
            }
         }

         if(array[0] == array[n-1]) alist.set(0,alist.get(0) + psize[n-1]);
         else alist.add(psize[n-1]);

         if(check(alist)){
            out.println("Yes");
         } else {
            out.println("No");
         }

      }

      out.close();
   }

   public static boolean check(ArrayList<Integer> alist){
      if(alist.size() != 4) return false;
      boolean b1 = alist.get(0) == 1 && alist.get(2) == 1 && ((alist.get(1)%2) != (alist.get(3)%2));
      boolean b2 = alist.get(1) == 1 && alist.get(3) == 1 && ((alist.get(0)%2) != (alist.get(2)%2));
      return b1 || b2;
   }

}

**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
6
RBRBRB
```