**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yasya and the Mysterious Tree
time_limit_ms: 2500
memory_limit_mb: 512
Description: Yasya was walking in the forest and accidentally found a tree with $n$ vertices. A tree is a connected undirected graph with no cycles.

Next to the tree, the girl found an ancient manuscript with $m$ queries written on it. The queries can be of two types.

The first type of query is described by the integer $y$. The weight of each edge in the tree is replaced by the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of the weight of that edge and the integer $y$.

The second type is described by the vertex $v$ and the integer $x$. Yasya chooses a vertex $u$ ($1 \le u \le n$, $u \neq v$) and mentally draws a bidirectional edge of weight $x$ from $v$ to $u$ in the tree.

Then Yasya finds a simple cycle in the resulting graph and calculates the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of all the edges in it. She wants to choose a vertex $u$ such that the calculated value is maximum. This calculated value will be the answer to the query. It can be shown that such a cycle exists and is unique under the given constraints (independent of the choice of $u$). If an edge between $v$ and $u$ already existed, a simple cycle is the path $v \to u \to v$.

Note that the second type of query is performed mentally, meaning the tree does not change in any way after it.

Help Yasya answer all the queries.

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

The descriptions of the test cases follow.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.

The next $n - 1$ lines of each test case contain three integers $v$, $u$, $w$ ($1 \le v, u \le n$, $1 \le w \le 10^9$) — the ends of some edge in the tree and its weight.

It is guaranteed that the given set of edges forms a tree.

The next $m$ lines of each test case describe the queries:

  * ^ $y$ ($1 \le y \le 10^9$) — parameter of the first type query;    * ? $v$ $x$ ($1 \le v \le n$, $1 \le x \le 10^9$) — parameters of the second type query.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. The same is guaranteed for $m$.

Output: For each test case, output the answers to the queries of the second type.

Examples:
input:
2
3 7
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
^ 6
? 3 1
? 2 9
5 6
1 2 777
3 2 2812
4 1 16
5 3 1000000000
^ 4
? 3 123
? 5 1000000000
^ 1000000000
? 1 908070
? 2 1
output:
13 15 11 10
1000000127 2812 999756331 999999756

input:
3
8 4
8 6 3
6 3 4
2 5 4
7 6 2
7 1 10
4 1 4
5 1 2
^ 4
^ 7
? 7 8
? 4 10
5 6
3 1 4
2 3 9
4 3 6
5 2 10
? 5 7
^ 1
^ 8
? 4 10
? 1 9
? 3 6
4 2
2 1 4
4 3 5
2 3 4
^ 13
? 1 10
output:
14 13
13 8 11 11
10

Note:

Buggy Code:
class TrieNode:
    def __init__(self):
        self.l = None
        self.r = None
        self.cnt = 0
        self.u = -1

def main():

    t = int(input())
    allans = []
    for _ in range(t):
        n, m = readIntArr()
        adj = [[] for __ in range(n + 1)]
        for __ in range(n - 1):
            u, v, w = readIntArr()
            adj[u].append((v, w))
            adj[v].append((u, w))

        root = 1
        xor_to_node = [-1] * (n + 1)
        node_parity = [0] * (n + 1)
        xor_to_node[root] = 0
        st = [root]
        while st:
            u = st.pop()
            for v, w in adj[u]:
                if xor_to_node[v] == -1:
                    xor_to_node[v] = xor_to_node[u] ^ w
                    node_parity[v] = node_parity[u] ^ 1
                    st.append(v)

        even_trie_root = TrieNode()
        odd_trie_root = TrieNode()

        for u in range(1, n + 1):
            xor = xor_to_node[u]
            if node_parity[u] == 0:
                node = even_trie_root
            else:
                node = odd_trie_root
            node.cnt += 1
            node.u = u
            for i in range(29, -1, -1):
                if (xor & (1 << i)) > 0:
                    if node.r is None:
                        node.r = TrieNode()
                    node = node.r
                else:
                    if node.l is None:
                        node.l = TrieNode()
                    node = node.l
                node.cnt += 1
                node.u = u

        def find_max(root, prevalent_xor, v):
            """ return max_xor_value excluding v """
            if root.cnt == 0 or (root.cnt == 1 and root.u == v):  # no available vertex
                return -1
            else:
                node = root
                xor = 0
                for i in range(29, -1, -1):
                    if (prevalent_xor & (1 << i)) > 0:
                        if (node.l is not None) and (node.l.cnt > 1 or (node.l.cnt == 1 and node.l.u != v)):
                            xor ^= (1 << i)
                            node = node.l
                        else:
                            node = node.r
                    else:
                        if (node.r is not None) and (node.r.cnt > 1 or (node.r.cnt == 1 and node.r.u != v)):
                            xor ^= (1 << i)
                            node = node.r
                        else:
                            node = node.l
                return xor

        overall_xor = 0
        ans = []
        for __ in range(m):
            arr = input().split()
            if arr[0] == '^':
                overall_xor ^= int(arr[1])
            else:
                v, x = int(arr[1]), int(arr[2])
                if node_parity[v] == 0:
                    even_root = even_trie_root
                    odd_root = odd_trie_root
                else:
                    even_root = odd_trie_root
                    odd_root = even_trie_root

                # handle even nodes
                prevalent_xor = xor_to_node[v] ^ x
                best_xor1 = find_max(even_root, prevalent_xor, v)

                # handle odd nodes
                prevalent_xor = xor_to_node[v] ^ x ^ overall_xor
                best_xor2 = find_max(odd_root, prevalent_xor, v)

                # print(f'v={v} x={x} overall={overall_xor} xor1={best_xor1} xor2={best_xor2}')

                # assert best_xor1 != best_xor2
                ans.append(max(best_xor1, best_xor2))

        allans.append(ans)
    multiLineArrayOfArraysPrint(allans)

    return

import sys
# input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.

def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))

def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]

def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
    dv=defaultValFactory;da=dimensionArr
    if len(da)==1:return [dv() for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]

def queryInteractive(a, b, c):
    print('? {} {} {}'.format(a, b, c))
    sys.stdout.flush()
    return int(input())

def answerInteractive(x1, x2):
    print('! {} {}'.format(x1, x2))
    sys.stdout.flush()

inf=float('inf')
# MOD=10**9+7
# MOD=998244353

from math import gcd,floor,ceil
import math
# from math import floor,ceil # for Python2

for _abc in range(1):
    main()

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

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

Think step by step.

Response:
```plaintext
1
5 5
1 2 1
2 3 2
3 4 4
4 5 8
^ 3
? 1 10
? 3 5
^ 7
? 2 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yasya and the Mysterious Tree
time_limit_ms: 2500
memory_limit_mb: 512
Description: Yasya was walking in the forest and accidentally found a tree with $n$ vertices. A tree is a connected undirected graph with no cycles.

Next to the tree, the girl found an ancient manuscript with $m$ queries written on it. The queries can be of two types.

The first type of query is described by the integer $y$. The weight of each edge in the tree is replaced by the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of the weight of that edge and the integer $y$.

The second type is described by the vertex $v$ and the integer $x$. Yasya chooses a vertex $u$ ($1 \le u \le n$, $u \neq v$) and mentally draws a bidirectional edge of weight $x$ from $v$ to $u$ in the tree.

Then Yasya finds a simple cycle in the resulting graph and calculates the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of all the edges in it. She wants to choose a vertex $u$ such that the calculated value is maximum. This calculated value will be the answer to the query. It can be shown that such a cycle exists and is unique under the given constraints (independent of the choice of $u$). If an edge between $v$ and $u$ already existed, a simple cycle is the path $v \to u \to v$.

Note that the second type of query is performed mentally, meaning the tree does not change in any way after it.

Help Yasya answer all the queries.

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

The descriptions of the test cases follow.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.

The next $n - 1$ lines of each test case contain three integers $v$, $u$, $w$ ($1 \le v, u \le n$, $1 \le w \le 10^9$) — the ends of some edge in the tree and its weight.

It is guaranteed that the given set of edges forms a tree.

The next $m$ lines of each test case describe the queries:

  * ^ $y$ ($1 \le y \le 10^9$) — parameter of the first type query;    * ? $v$ $x$ ($1 \le v \le n$, $1 \le x \le 10^9$) — parameters of the second type query.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. The same is guaranteed for $m$.

Output: For each test case, output the answers to the queries of the second type.

Examples:
input:
2
3 7
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
^ 6
? 3 1
? 2 9
5 6
1 2 777
3 2 2812
4 1 16
5 3 1000000000
^ 4
? 3 123
? 5 1000000000
^ 1000000000
? 1 908070
? 2 1
output:
13 15 11 10
1000000127 2812 999756331 999999756

input:
3
8 4
8 6 3
6 3 4
2 5 4
7 6 2
7 1 10
4 1 4
5 1 2
^ 4
^ 7
? 7 8
? 4 10
5 6
3 1 4
2 3 9
4 3 6
5 2 10
? 5 7
^ 1
^ 8
? 4 10
? 1 9
? 3 6
4 2
2 1 4
4 3 5
2 3 4
^ 13
? 1 10
output:
14 13
13 8 11 11
10

Note:

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

public class G {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);
	static List<int []> [] graph;
	static int [] depth;
	static int [] xor;

	static Trie oddRoot;
	static Trie evenRoot;

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

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void solve(int t) {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			int n = sc.nextInt();
			int q = sc.nextInt();

			graph = new List[n + 1];

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

			int u, v, w;

			for (int i = 1; i < n; ++i) {
				u = sc.nextInt();
				v = sc.nextInt();
				w = sc.nextInt();

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

			depth = new int [n + 1];
			xor = new int [n + 1];
			dfs(1, 0, 0, 0);

			oddRoot = new Trie();
			evenRoot = new Trie();

			int added = 0;

			char type;

			Trie current;

			for (int i = 1; i <= n; ++i) {
				current = depth[i] % 2 == 0 ? evenRoot : oddRoot;
				addToTrie(xor[i], current, 1);
			}

			int x;

			StringBuilder sb = new StringBuilder();

			Trie same, other;

			for (int i = 0; i < q; ++i) {
				type = sc.next().charAt(0);

				if (type == '^') {
					added ^= sc.nextInt();
					continue;
				}

				v = sc.nextInt();
				x = sc.nextInt();

				current = depth[v] % 2 == 0 ? evenRoot : oddRoot;
				addToTrie(xor[v], current, -1);

				if (depth[v] % 2 == 0) {
					same = evenRoot;
					other = oddRoot;
				} else {
					same = oddRoot;
					other = evenRoot;
				}

				sb.append(Math.max( getMaxXor(x^xor[v], same)  , getMaxXor(x^added^xor[v], other) ));
				sb.append(' ');

				addToTrie(xor[v], current, 1);
			}

			out.println(sb);
		}

		private int getMaxXor(int val, Trie trie) {
			// System.out.println(val);
			Trie current = trie;

			int bit, target;

			int result = 0;

			for (int i = 30; i >= 0; --i) {
				if (current == null)
					return 0;

				bit = ((1 << i) & val) == 0 ? 0 : 1;
				target = 1 ^ bit;

				if (current.children[target] == null
						|| current.children[target].count == 0) {
					current = current.children[bit];
				}else {
					current = current.children[target];
					result ^= 1 << i;
				}
			}

			return result;
		}

		private void dfs(int node, int parent, int d, int x) {
			depth[node] = d;
			xor[node] = x;

			for (int [] child : graph[node]) {
				if (child[0] == parent)
					continue;

				dfs(child[0], node, d + 1, x ^ child[1]);
			}
		}

		private void addToTrie(int val, Trie trie, int add) {
			Trie current = trie;

			int bit;

			for (int i = 30; i >= 0; --i) {
				bit = ((1 << i) & val) == 0 ? 0 : 1;

				if (current.children[bit] == null)
					current.children[bit] = new Trie();

				current = current.children[bit];
				current.count += add;
			}

		}
	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

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

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

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

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

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

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

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

   }

}
class Trie {
	Trie [] children;
	int count;

	Trie() {
		children = new Trie[2];
	}
}

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

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

Think step by step.

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

1. **Small-Scale Test Case**: Since we need a small-scale test input, we can start with a tree of 3 or 4 vertices.
2. **Query Types**: We should include both types of queries: the first type to change the edge weights and the second type to find the maximum XOR value.
3. **Edge Cases**: We should consider edge cases where the tree structure or the queries can lead to unexpected behavior:
   - A tree with a single edge.
   - Queries that involve the root vertex.
   - Queries that involve the leaf vertices.
   - Queries that change the edge weights significantly.

Let's construct a test input that includes these considerations:

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

### Explanation:
- **Tree Structure**: The tree has 3 vertices and 2 edges. The edges are (1, 2, 1) and (2, 3, 2).
- **Queries**:
  - `^ 3`: XOR all edge weights with 3.
  - `? 1 4`: Find the maximum XOR value for a cycle starting from vertex 1 with an edge weight of 4.
  - `? 2 5`: Find the maximum XOR value for a cycle starting from vertex 2 with an edge weight of 5.
  - `? 3 6`: Find the maximum XOR value for a cycle starting from vertex 3 with an edge weight of 6.
  - `^ 4`: XOR all edge weights with 4.
  - `? 1 7`: Find the maximum XOR value for a cycle starting from vertex 1 with an edge weight of 7.

This test input is small and complete, and it includes both types of queries. It also covers edge cases such as queries involving different vertices and changes in edge weights. This should help in exposing any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yasya and the Mysterious Tree
time_limit_ms: 2500
memory_limit_mb: 512
Description: Yasya was walking in the forest and accidentally found a tree with $n$ vertices. A tree is a connected undirected graph with no cycles.

Next to the tree, the girl found an ancient manuscript with $m$ queries written on it. The queries can be of two types.

The first type of query is described by the integer $y$. The weight of each edge in the tree is replaced by the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of the weight of that edge and the integer $y$.

The second type is described by the vertex $v$ and the integer $x$. Yasya chooses a vertex $u$ ($1 \le u \le n$, $u \neq v$) and mentally draws a bidirectional edge of weight $x$ from $v$ to $u$ in the tree.

Then Yasya finds a simple cycle in the resulting graph and calculates the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of all the edges in it. She wants to choose a vertex $u$ such that the calculated value is maximum. This calculated value will be the answer to the query. It can be shown that such a cycle exists and is unique under the given constraints (independent of the choice of $u$). If an edge between $v$ and $u$ already existed, a simple cycle is the path $v \to u \to v$.

Note that the second type of query is performed mentally, meaning the tree does not change in any way after it.

Help Yasya answer all the queries.

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

The descriptions of the test cases follow.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.

The next $n - 1$ lines of each test case contain three integers $v$, $u$, $w$ ($1 \le v, u \le n$, $1 \le w \le 10^9$) — the ends of some edge in the tree and its weight.

It is guaranteed that the given set of edges forms a tree.

The next $m$ lines of each test case describe the queries:

  * ^ $y$ ($1 \le y \le 10^9$) — parameter of the first type query;    * ? $v$ $x$ ($1 \le v \le n$, $1 \le x \le 10^9$) — parameters of the second type query.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. The same is guaranteed for $m$.

Output: For each test case, output the answers to the queries of the second type.

Examples:
input:
2
3 7
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
^ 6
? 3 1
? 2 9
5 6
1 2 777
3 2 2812
4 1 16
5 3 1000000000
^ 4
? 3 123
? 5 1000000000
^ 1000000000
? 1 908070
? 2 1
output:
13 15 11 10
1000000127 2812 999756331 999999756

input:
3
8 4
8 6 3
6 3 4
2 5 4
7 6 2
7 1 10
4 1 4
5 1 2
^ 4
^ 7
? 7 8
? 4 10
5 6
3 1 4
2 3 9
4 3 6
5 2 10
? 5 7
^ 1
^ 8
? 4 10
? 1 9
? 3 6
4 2
2 1 4
4 3 5
2 3 4
^ 13
? 1 10
output:
14 13
13 8 11 11
10

Note:

Buggy Code:
// Generated by Code Flattener.
// https://plugins.jetbrains.com/plugin/9979-idea-code-flattener

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

import static java.lang.Math.max;

public class Main {
    public static void main(String[] args) {
        Solution obj = new Solution();
        obj.invoke();
    }

    private static class Solution {
        private final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        private final static PrintWriter out = new PrintWriter(System.out);
        private StringTokenizer st = new StringTokenizer("");
        private boolean multiCase = true;

        public void invoke() {
            int t = 1;
            if (multiCase) t = nextInt();
            while (t-- > 0) {
                solve();
            }
            out.flush();
        }

        private int N = (int) 1e6;
        private int[][] trie = new int[2][N];
        private int[][] cnt = new int[2][N];
        private int total;

        private int newNode() {
            int x = ++total;
            trie[0][x] = trie[1][x] = 0;
            cnt[0][x] = cnt[1][x] = 0;
            return x;
        }

        private void add(int v, int q, int t) {
            int p = 1;
            cnt[q][p] += t;
            for (int i = 29; i >= 0; i--) {
                int bit = v >> i & 1;
                if (trie[bit][p] == 0) trie[bit][p] = newNode();
                p = trie[bit][p];
                cnt[q][p] += t;
            }
        }

        private int findMax(int v, int q) {
            int ans = 0;
            int p = 1;
            if (cnt[q][p] == 0) return ans;
            for (int i = 29; i >= 0; i--) {
                int bit = v >> i & 1;
                if (cnt[q][trie[bit ^ 1][p]] > 0) {
                    ans |= 1 << i;
                    p = trie[bit ^ 1][p];
                } else {
                    p = trie[bit][p];
                }
            }
            return ans;
        }

        private void solve() {
            int n = nextInt();
            int m = nextInt();
            List<int[]>[] graph = buildGraph(n);
            int[] oddDepth = new int[n];
            int[] xorDp = new int[n];
            dfs(0, 0, oddDepth, xorDp, graph);
            total = 0;
            newNode();
            for (int i = 0; i < n; i++) {
                add(xorDp[i], oddDepth[i], 1);
            }
            int addedXor = 0;
            while (m-- > 0) {
                char ch = nextChar();
                if (ch == '^') {
                    int y = nextInt();
                    addedXor ^= y;
                } else {
                    int v = nextInt() - 1;
                    int x = nextInt();
                    int ans = 0;
                    add(xorDp[v], oddDepth[v], -1);
                    ans = max(ans, max(findMax(xorDp[v] ^ x, oddDepth[v]), findMax(xorDp[v] ^ x ^ addedXor, oddDepth[v] ^ 1)));
                    add(xorDp[v], oddDepth[v], 1);
                    print(ans);
                }
            }
            println();
        }

        private void dfs(int node, int parent, int[] oddDepth, int[] xorDp, List<int[]>[] graph) {
            for (int[] adjNode : graph[node]) {
                int child = adjNode[0], wt = adjNode[1];
                if (child == parent) continue;
                oddDepth[child] = oddDepth[node] ^ 1;
                xorDp[child] = xorDp[node] ^ wt;
                dfs(child, node, oddDepth, xorDp, graph);
            }
        }

        private List<int[]>[] buildGraph(int n) {
            List<int[]>[] graph = new List[n];
            for (int i = 0; i < n; i++) {
                graph[i] = new ArrayList<>();
            }
            for (int i = 1; i < n; i++) {
                int u = nextInt() - 1;
                int v = nextInt() - 1;
                int w = nextInt();
                graph[u].add(new int[]{v, w});
                graph[v].add(new int[]{u, w});
            }
            return graph;
        }

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

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

        public char nextChar() {
            return next().charAt(0);
        }

        public void print(Object... str) {
            for (Object s : str)
                out.print(s + " ");
        }

        public void println(Object... str) {
            for (Object s : str)
                print(s);
            out.println();
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that could maximize the failure probability. Here are some key points to consider:

1. **Tree Structure**: The tree should be structured in a way that maximizes the complexity of the queries. A star-like tree or a linear chain can be particularly challenging.
2. **Query Types**: Mix of both types of queries, especially with the second type of queries that involve finding the maximum XOR value.
3. **Edge Weights and Query Parameters**: Use edge weights and query parameters that can lead to interesting XOR results, especially when combined with the XOR operations in the first type of queries.
4. **Multiple Test Cases**: Include multiple test cases to ensure the code handles different scenarios correctly.

Given these considerations, let's generate a small-scale test input:

```plaintext
3
5 6
1 2 1
2 3 2
3 4 3
4 5 4
^ 1
? 1 5
? 2 6
^ 2
? 3 7
? 4 8
6 8
1 2 1
2 3 2
3 4 3
4 5 4
5 6 5
^ 1
? 1 6
? 2 7
? 3 8
^ 2
? 4 9
? 5 10
4 4
1 2 1
2 3 2
3 4 3
^ 1
? 1 4
? 2 5
^ 2
? 3 6
```

This test input includes:
- Multiple test cases with different tree structures and query sequences.
- A mix of edge weights and query parameters that can lead to interesting XOR results.
- A combination of both types of queries to ensure the code handles the interactions correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    while (t--)
    {
        int n;
        cin>>n;
        int a[n], b[n];
        for (int i = 0; i < n; i++)
        {
            cin>>a[i];
        }
        for (int i = 0; i < n; i++)
        {
            cin>>b[i];
        }
        if(is_sorted(a, a+n) && is_sorted(b,b+n)){
            cout<<"Bob"<<endl;
        }
        else{
            cout<<"Alice"<<endl;
        }

    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define pii pair<ll,ll>
#define uib unordered_map<ll,bool>
#define vi vector<ll>
#define fi first
#define se second
#define endl "\n"
using namespace std;
ll l,r,n,m,i,j,k,x,y,z;
ll mod=1e9+7;
string a="",b="";

void solve(){
    cin>>n;
    a[n];
    b[n];
    for(ll i=0;i<n;i++)cin>>a[i];
    for(ll i=0;i<n;i++)cin>>b[i];
    l=0,r=n-1;
    while(r>=l){
        if(a[l]!=b[r]&&a[l]!=b[l]){
            cout<<"Alice"<<endl;
            return;
        }
        if(a[r]!=b[r]&&a[r]!=b[l]){
            cout<<"Alice"<<endl;
            return;
        }
        l++,r--;
    }
    cout<<"Bob"<<endl;
}

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long int
#define ld long double
#define N 1000006
#define mod 10000007

using namespace std;

void solve()
{
	ll n, i, x, tr=0;
	cin>> n;
	ll a[n+5], b[n+5], t[n+5];
	for(i=1; i<=n; i++)
	{
		cin>> a[i];
		t[a[i]] = min(i, n-i+1);
	}
	for(i=1; i<=n; i++)
	{
		cin>> b[i];
		x = min(i, n-i+1);
		if(t[b[i]]<x)	tr=1;
	}
	if(tr)	cout<< "Alice\n";
	else	cout<< "Bob\n";
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 	ll T = 1, cs = 1, i, k=0;
	cin >> T;
	while(T--)
	{
	    //cout << "Pair #" << cs++ << ": ";
	    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
2
4
1 2 3 4
4 3 2 1
5
1 2 3 4 5
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

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

    index = {}
    for i in range(n):
        index[a[i]] = i + 1

    Alice = False
    for i in range(n):
        if abs(index[b[i]] - (i + 1)) > 1:
            Alice = True
            break

    print("Alice" if Alice else "Bob")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
import sys
import math
import random
from heapq import *
from collections import Counter, defaultdict, deque
from functools import lru_cache
from itertools import permutations, combinations
from decimal import Decimal
from bisect import bisect_right, bisect_left, bisect
import re

input = lambda: sys.stdin.readline().strip()
minput = lambda: map(int, input().split())
linput = lambda: list(minput())
MOD = 1000000007
MOD2 = 998244353
INF = 10 ** 50
primes_N = 0
primes = [1]*(primes_N+1)
primes100 = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97)
primes1000 = (
    2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109,
    113,
    127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241,
    251,
    257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389,
    397,
    401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547,
    557,
    563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691,
    701,
    709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859,
    863,
    877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997)

def factorization_list(n):
    # Разложение числа на простые множители в виде списка чисел
    res = []
    while n % 2 == 0:
        n //= 2
        res.append(2)

    i = 3
    while i**2 <= n:
        while n % i == 0:
            n //= i
            res.append(i)
        i += 2
    if n > 1:
        res.append(n)
    return sorted(res)

def factorization_dict(n):
    # Разложение числа на простые множители в словарь
    res = defaultdict(int)
    while n % 2 == 0:
        res[2] += 1
        n //= 2
    i = 3
    while i**2 <= n:
        while n % i == 0:
            res[i] += 1
            n //= i
        i += 2
    if n > 1:
        res[n] += 1
    return res

def sieve():
    # Решето Эратосфена
    primes[0] = primes[1] = 0
    for j in range(4, primes_N+1, 2):
        primes[j] = 0

    i = 3
    while i**2 <= primes_N:
        if primes[i]:
            for j in range(i**2, primes_N+1, i):
                primes[j] = 0
        i += 2

def solve():
    n = int(input())
    a = linput()
    b = linput()
    if n % 2 == 1:
        if a[n//2+1] == b[n//2+1] and a[n//2-1] == b[n//2-1]:
            print('Bob')
        else:
            print('Alice')
    else:
        if a[n//2] == b[n//2] and a[n//2-1] == b[n//2-1]:
            print('Bob')
        else:
            print('Alice')

if __name__ == "__main__":
    tc = 1
    tc = int(input())
    for _ in range(tc):
        solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
import sys
input = sys.stdin.readline

t = int(input())

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

  isDifferenceFound = 0
  for i in range(0,n):
    if not (a[i] == b[i]):
      isDifferenceFound = 1

  if isDifferenceFound==0:
    print("Bob")
  else:
     for i in range(0,n):
      if not (a[i]==b[n-i-1]):
       isDifferenceFound = 1

     if isDifferenceFound == 0:
       print("Bob")
     else:
       print("Alice")

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

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

Think step by step.

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

1. The code checks if all elements in `a` are the same as the corresponding elements in `b` (i.e., `a[i] == b[i]` for all `i`).
2. If they are all the same, it prints "Bob".
3. Otherwise, it checks if the elements in `a` are the reverse of the elements in `b` (i.e., `a[i] == b[n-i-1]` for all `i`).
4. If they are the reverse, it prints "Bob".
5. If neither condition is met, it prints "Alice".

The bug in the code is that it does not correctly handle the optimal strategy for both players. It only checks two specific conditions (exact match and reverse match) and assumes "Alice" wins if neither condition is met. However, the optimal strategy can be more complex and involve different sequences of removals.

### Challenging Test Input
To expose this bug, we need a test case where the permutations `a` and `b` are such that the optimal strategy leads to a different result than the one determined by the buggy code.

Consider the following test case:

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

### Explanation:
1. **Test Case 1:**
   - `a = [1, 3, 2]`
   - `b = [3, 1, 2]`
   - The elements are not the same and not the reverse of each other. The buggy code will print "Alice".
   - However, if Alice removes `1` first, the arrays become `a = [3, 2]` and `b = [3, 1, 2]`. Bob can then remove `3` from the front, making `b = [1, 2]`. Now, Alice can remove `2`, making `a = [3]` and `b = [1]`. Bob wins because `a[0] != b[0]`.

2. **Test Case 2:**
   - `a = [1, 2, 3, 4]`
   - `b = [4, 3, 2, 1]`
   - The elements are the reverse of each other. The buggy code will print "Bob".
   - However, if Alice and Bob both remove elements optimally, they will end up with `a = [2, 3]` and `b = [3, 2]` after the first two turns. Alice can then remove `2`, making `a = [3]` and `b = [3]`. Bob wins because `a[0] == b[0]`.

This test case should expose the bug in the buggy code by showing that the optimal strategy can lead to different outcomes than the simple checks performed by the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
import java.io.*;
import java.util.*;
public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(System.out);
        int t = Integer.parseInt(reader.readLine());
        while (t-- > 0) {
            int n = Integer.parseInt(reader.readLine());
            String[] aStr = reader.readLine().split(" ");
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = Integer.parseInt(aStr[i]);
            }
            String[] bStr = reader.readLine().split(" ");
            int[] b = new int[n];
            for (int i = 0; i < n; i++) {
                b[i] = Integer.parseInt(bStr[i]);
            }
            writer.println(determineWinner(a, b));
        }
        writer.close();
    }

    private static String determineWinner(int[] a, int[] b) {
        int s1=0,s2=0;
        for (int i = 1; i < a.length - 1; i++) {
            if (a[i] != b[i]) {
                s1=1;
                break;
            }
        }
        for (int i = 1; i < a.length - 1; i++) {
            if (a[i] != b[a.length - 1-i]) {
                s2=1;
                break;
            }
        }
        if(s1==0&&s2==0){
            return "Bob";
        }
        return "Alice";
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on the logic used in the `determineWinner` method. The method checks two conditions:

1. Whether the middle elements of `a` and `b` are the same.
2. Whether the middle elements of `a` and the reverse of `b` are the same.

If both conditions are false, the method returns "Alice". Otherwise, it returns "Bob". This logic is flawed because it does not consider the optimal strategy for both players.

To create a test case that exposes this bug, we need a scenario where the middle elements of `a` and `b` are the same, but the optimal strategy still results in Alice winning. Let's consider a small-scale test case where `n = 4`:

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

### Explanation:
- **Array `a`:** [1, 3, 2, 4]
- **Array `b`:** [1, 4, 2, 3]

Let's analyze the game step by step:

1. **First Turn:**
   - Alice can choose to remove either 1 or 4 from `a`.
   - Bob can choose to remove either 1 or 3 from `b`.

   Suppose Alice removes 1 from `a` and Bob removes 1 from `b`:
   - `a` becomes [3, 2, 4]
   - `b` becomes [4, 2, 3]

2. **Second Turn:**
   - Alice can choose to remove either 3 or 4 from `a`.
   - Bob can choose to remove either 4 or 3 from `b`.

   Suppose Alice removes 4 from `a` and Bob removes 4 from `b`:
   - `a` becomes [3, 2]
   - `b` becomes [2, 3]

3. **Third Turn:**
   - Alice can choose to remove either 3 or 2 from `a`.
   - Bob can choose to remove either 2 or 3 from `b`.

   Suppose Alice removes 3 from `a` and Bob removes 3 from `b`:
   - `a` becomes [2]
   - `b` becomes [2]

4. **Fourth Turn:**
   - Both arrays have only one element left: `a` = [2] and `b` = [2].

Since the remaining elements are the same, Bob wins. However, the optimal strategy for Alice is to avoid this outcome. By removing elements differently, Alice can ensure that the remaining elements are different, leading to her win.

### Optimal Strategy:
- **First Turn:**
  - Alice removes 4 from `a`.
  - Bob removes 1 from `b`.

  - `a` becomes [1, 3, 2]
  - `b` becomes [4, 2, 3]

- **Second Turn:**
  - Alice removes 2 from `a`.
  - Bob removes 3 from `b`.

  - `a` becomes [1, 3]
  - `b` becomes [4, 2]

- **Third Turn:**
  - Alice removes 3 from `a`.
  - Bob removes 4 from `b`.

  - `a` becomes [1]
  - `b` becomes [2]

- **Fourth Turn:**
  - Both arrays have only one element left: `a` = [1] and `b` = [2].

Since the remaining elements are different, Alice wins.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
import java.io.*;
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t=sc.nextInt();
        for(int k=0;k<t;k++){
            int n=sc.nextInt();
            int arr1[]=new int[n];
            for(int i=0;i<n;i++){
                arr1[i]=sc.nextInt();
            }
            int arr2[]=new int[n];
            for(int i=0;i<n;i++){
                arr2[i]=sc.nextInt();
            }
            boolean flag=true;
            for(int i=0;i<n;i++){
                if(arr1[i]!=arr2[i]){
                    flag=false;
                    break;
                }
            }
            if(flag){
                System.out.println("Bob");
            }
            else{
                System.out.println("Alice");
            }
        }

    }
}

**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
1 2 3 4
4 3 2 1
5
1 2 3 4 5
5 4 3 2 1
6
1 3 5 2 4 6
6 4 2 5 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
#define int long long
using namespace std;
constexpr int INF = 0x7fffffff;
void solve() {
    int n; cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; ++ i) cin >> a[i];
    int flag = 0;
    for (int i = 0; i < n; ++ i) cin >> b[i];
    if (a == b) {cout << "Bob\n"; return;}
    reverse(a.begin(), a.end());
    if (a == b) cout << "Bob\b";
    else cout << "Alice\n";
}
signed main() {
    cin.tie(0) -> sync_with_stdio(0);
    int _ = 1; cin >> _;
    while(_ --) solve();
    return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on scenarios where the logic of the code might fail. The buggy code checks if the arrays `a` and `b` are equal or if the reverse of `a` is equal to `b`. However, this logic does not correctly determine the winner in all cases, especially when the permutations are more complex.

### Step-by-Step Reasoning:

1. **Identify the Bug**:
   - The code only checks if `a` is equal to `b` or if the reverse of `a` is equal to `b`.
   - This does not account for the optimal play strategy, where Alice and Bob can make choices that lead to different outcomes.

2. **Edge Cases**:
   - **Small Arrays**: Test with small arrays to ensure the logic is correct.
   - **Symmetric Arrays**: Test with arrays that are symmetric but not exactly equal or their reverse.
   - **Asymmetric Arrays**: Test with arrays that are not symmetric and require more complex reasoning.

3. **Constructing the Test Case**:
   - We need an array where the optimal play strategy does not result in `a` being equal to `b` or its reverse.
   - A good example is an array where the optimal play leads to different elements being left in `a` and `b`.

### Chosen Test Case:
- **Array Size**: 4 (small enough to be manageable but large enough to show the bug).
- **Alice's Array**: [1, 2, 3, 4]
- **Bob's Array**: [1, 3, 2, 4]

### Explanation:
- **Initial Arrays**:
  - Alice: [1, 2, 3, 4]
  - Bob: [1, 3, 2, 4]
- **Optimal Play**:
  - Alice can remove 1 (first element) in the first turn.
  - Bob can remove 1 (first element) in the first turn.
  - Now:
    - Alice: [2, 3, 4]
    - Bob: [3, 2, 4]
  - Alice can remove 4 (last element) in the second turn.
  - Bob can remove 4 (last element) in the second turn.
  - Now:
    - Alice: [2, 3]
    - Bob: [3, 2]
  - Alice can remove 2 (first element) in the third turn.
  - Bob can remove 3 (first element) in the third turn.
  - Now:
    - Alice: [3]
    - Bob: [2]
  - The remaining elements are different, so Alice wins.

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

This test input should expose the bug in the code by showing that the optimal play strategy leads to Alice winning, even though the arrays are not equal or their reverse.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

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

#define ajay_chovatiya                \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define ll long long int
#define vin(v, n) vector<ll> v(n); for (ll i = 0; i < n; i++) cin >> v[i];
#define fir(i, start, end) for (ll i = start; i >= end; i--)
#define print(v, n) for (ll i = 0; i < n; i++) cout << v[i] << " "; cout << endl;

void ajay_chovatiya__()
{
   ll n;
   cin>>n;
   vin(v,n);
   vin(v1,n);
   if(v==v1)
   {
    cout<<"Bob"<<endl;
    return;
   }

   for (int i = 0; i < n/2; i++)
   {
    if(v[i]!=v[n-1-i])
    {
        cout<<"Alice"<<endl;
        return;
    }
   }
   cout<<"Bob"<<endl;

}

int main()
{
    ajay_chovatiya;
    int t=1;
    cin >> t;
    while (t--)
    {
        ajay_chovatiya__();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify the flaw:

1. **Code Flaw Analysis:**
   - The code checks if the two permutations are identical. If they are, it outputs "Bob" (which is correct).
   - Then, it checks if any element in the first half of the array `v` is different from the corresponding element in the second half of the array `v1`. If it finds such a pair, it outputs "Alice".
   - If neither condition is met, it outputs "Bob".

2. **Identifying the Bug:**
   - The code only checks the first half of the array against the second half of the same array, which is incorrect. It should be checking the elements of `v` against the elements of `v1`.

3. **Creating a Test Case:**
   - We need to create a test case where the two permutations are not identical, but the elements in the first half of `v` are the same as the corresponding elements in the second half of `v1`, and vice versa.
   - This will ensure that the code incorrectly outputs "Bob" when it should output "Alice".

4. **Test Case:**
   - Let's create a permutation for Alice (`v`) and a permutation for Bob (`v1`) that meet the above criteria.

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

### Explanation:
- **Input:**
  - `t = 1` (one test case)
  - `n = 4` (length of permutations)
  - `v = [1, 2, 3, 4]` (Alice's permutation)
  - `v1 = [4, 3, 2, 1]` (Bob's permutation)

- **Output:**
  - The correct output should be "Alice" because Alice can always ensure that the remaining elements are different by removing elements optimally.

- **Why this test case works:**
  - The code will check if `v[0] == v[3]` and `v[1] == v[2]`, which is true (1 == 4 and 2 == 3 are false).
  - The code will then incorrectly output "Bob" because it does not correctly compare `v` and `v1`.

This test case should effectively expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
def solve():
    n = int(input())
    a = list(input())
    b = list(input())
    if (a == b or b == reversed(a)):
        print("Bob\n")
    else:
        print("Alice\n")

for _ in range(int(input())):
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
t = int(input())
ans = []
for i in range(t):
    n = int(input())
    li1 = list(map(int,input().split()))
    li2 = list(map(int,input().split()))
    flag = False

    while(len(li1)>=1):
        af = li1[0]
        al = li1[len(li1)-1]
        bf = li2[0]
        bl = li2[len(li2)-1]

        if af != bf or al!=bl:
            flag = True
            break
        else:
            if af == bf:
                li1.pop(0)
                li2.pop(0)
            elif af == bl:
                li1.pop(0)
                li2.pop(len(li2)-1)
            elif bf == al:
                li2.pop(0)
                li1.pop(len(li1)-1)
            else:
                li1.pop(len(li1)-1)
                li2.pop(len(li2)-1)

    if flag ==True:
        ans.append("Alice")
    else:
        ans.append("Bob")

for i in ans:
    print(i)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
import java.util.Scanner;

public class B{
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t=1;
        t=s.nextInt();
        while((t--)>0){
            int n=s.nextInt();
            int[] a = new int[n];
            int[] b  = new int[n];
            for(int i=0;i<n;i++)a[i]=s.nextInt();
            for(int i=0;i<n;i++)b[i]=s.nextInt();
            boolean f=false;
            for(int i=0;i<n;i++){
                if(a[i]==b[i]){
                    f=true;break;
                }
            }
            if(f){
                System.out.println("Bob");
            }else{
                System.out.println("Alice");
            }
        }
        s.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
1
1
1
2
1 2
2 1
4
1 3 2 4
4 2 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

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

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

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

    private static void solve() throws IOException {
        int n = Integer.parseInt(br.readLine());
        int[] nums1 = new int[n], nums2 = new int[n];
        String[] in1 = br.readLine().split(" "), in2 = br.readLine().split(" ");
        for (int i = 0; i < n; i++) {
            nums1[i] = Integer.parseInt(in1[i]);
            nums2[i] = Integer.parseInt(in2[i]);
        }

        if (n <= 2) {
            System.out.println("Bob");
            return;
        }

        boolean found = false;
        for (int i = 0; i < n; i++) {
            if (nums1[i] != nums2[i]) {
                found = true;
                break;
            }
        }

        if (found) {
            System.out.println("Alice");
        } else {
            System.out.println("Bob");
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    a = [int(x) for x in input().split()]
    b = [int(x) for x in input().split()]
    if a[0]==b[-1] or a[0]==b[0]:
        alice_choice = a.pop()
    elif a[-1]==b[-1] or a[-1]==b[0]:
        alice_choice = a.pop(0)
    else:
        alice_choice = a.pop()
    if alice_choice==b[-1]:
        bob_choice = b.pop()
    elif alice_choice==b[0]:
        bob_choice =  b.pop(0)
    else:
        bob_choice = b.pop()

    for i in range(n-2):
        if bob_choice == a[-1]:
            alice_choice = a.pop(0)
        elif bob_choice == a[0]:
            alice_choice = a.pop()
        else:
            alice_choice = a.pop()
        if alice_choice == b[-1]:
            bob_choice = b.pop()
        elif alice_choice == b[0]:
            bob_choice = b.pop(0)
        else:
            bob_choice = b.pop()
    if a[0]==b[0]:
        print("Bob")
    else:
        print("Alice")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
t = int(input())

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

    if a == b or a == reversed(b):
        print("Bob")
    else:
        print("Alice")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define Code ios_base::sync_with_stdio(false);
#define By cin.tie(NULL);
#define Bhavin cout.tie(NULL);
#define big INT_MIN
#define small INT_MAX
#define mod 1000000007
// int gcd(int a, int b)
// {
//     while (b != 0)
//     {
//         int tmp = a % b;
//         a = b;
//         b = tmp;
//     }
//     return a;
// }

// int lcm(int a, int b) {
//     return a * b / gcd(a, b);
// }
void solve()
{
    //Jay Shree Krishna
    int n;
    cin>>n;
    int a[n];
    int b[n];
    for(int i = 0;i<n;i++)
    {
        cin>>a[i];
    }
    for(int i = 0;i<n;i++)
    {
        cin>>b[i];
    }
    int i = 0;
    int j = n-1;
    int k = 0;
    int l = n-1;
    int delt = 0;
    int delt2 = 0;
    while(i < j && k < l)
    {
        // int num1 = a[i];
        // int num2 = a[j];
        // int num3 = b[k];
        // int num4 = b[l];
        //For Alice
        if(delt != a[i] && delt != 0)
        {
            delt2 = a[i];
            i++;
        }
        else if(delt != a[j] && delt != 0)
        {
            delt2 = a[j];
             j--;
        }
        else if(a[i] != b[k] && a[i] != b[l])
        {
            delt2 = a[i];
            i++;
        }
        else if(a[j] != b[k] && a[j] != b[l])
        {
            delt2 = a[j];
            j--;
        }
        else
        {
            delt2 = a[i];
            i++;
        }
        //For Bob
        if(delt2 == b[k])
        {
            delt = b[k];
            k++;
        }
        else if(delt2 == b[l])
        {
            delt = b[l];
            l--;
        }
        // else if(b[l] != a[i] && b[l]  a[j])
        // {
        //     delt = b[l];
        //     l--;
        // }
        // else if(b[k] != a[i] && b[k]  a[j])
        // {
        //     delt = b[k];
        //     k++;
        // }
        else if(b[k] == a[i] || b[k] == a[j])
        {
            delt = b[k];
            k++;
        }
        else
        {
            delt = b[l];
            l--;
        }
    }
    if(a[i] != b[k])
    {
        cout<<"Alice"<<"\n";
    }
    else
    {
        cout<<"Bob"<<"\n";
    }
}
signed main()
{
    Code By Bhavin

    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("abcd.txt", "w", stdout);
    #endif
    int tc;
    cin >> tc;
    while (tc--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The key is to create a situation where the logic in the code might lead to incorrect decisions.

### Step-by-Step Analysis:

1. **Initial Observations:**
   - The code uses a `delt` and `delt2` variable to keep track of the elements removed by Alice and Bob, respectively.
   - The logic for deciding which element to remove is complex and might not handle all edge cases correctly.

2. **Identifying Edge Cases:**
   - **Small Arrays:** Test with the smallest possible arrays to see if the logic breaks.
   - **Symmetric Arrays:** Test with arrays where Alice and Bob have the same permutation.
   - **Asymmetric Arrays:** Test with arrays where Alice and Bob have different permutations.
   - **Edge Elements:** Test with arrays where the first and last elements are crucial for the outcome.

3. **Creating a Test Input:**
   - We will create a test case with a small array to ensure the input is complete and valid.
   - The array should be designed to challenge the logic of the code, especially the part where it decides which element to remove.

### Challenging Test Input:

Let's create a test case where Alice and Bob have different permutations, and the decision logic in the code might lead to an incorrect result.

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

### Explanation:
- **Test Case:**
  - **Number of test cases:** 1
  - **Array size:** 4
  - **Alice's permutation:** [1, 2, 3, 4]
  - **Bob's permutation:** [4, 3, 2, 1]

- **Expected Outcome:**
  - Alice can always win by removing elements in a way that ensures the last remaining elements are different. For example:
    - Alice removes 1 (array becomes [2, 3, 4])
    - Bob removes 4 (array becomes [3, 2, 1])
    - Alice removes 4 (array becomes [2, 3])
    - Bob removes 3 (array becomes [2])
    - Alice removes 2 (array becomes [3])
  - The final elements are 3 (Alice) and 2 (Bob), so Alice wins.

This test case should expose the bug in the provided code by challenging the decision logic used to determine which elements to remove.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Removals Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice got a permutation $a_1, a_2, \ldots, a_n$ of $[1,2,\ldots,n]$, and Bob got another permutation $b_1, b_2, \ldots, b_n$ of $[1,2,\ldots,n]$. They are going to play a game with these arrays.

In each turn, the following events happen in order:

  * Alice chooses either the first or the last element of her array and removes it from the array;    * Bob chooses either the first or the last element of his array and removes it from the array.

The game continues for $n-1$ turns, after which both arrays will have exactly one remaining element: $x$ in the array $a$ and $y$ in the array $b$.

If $x=y$, Bob wins; otherwise, Alice wins. Find which player will win if both players play optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$).

The next line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$, all $a_i$ are distinct) — the permutation of Alice.

The next line contains $n$ integers $b_1,b_2,\ldots,b_n$ ($1\le b_i\le n$, all $b_i$ are distinct) — the permutation of Bob.

It is guaranteed that the sum of all $n$ does not exceed $3\cdot 10^5$.

Output: For each test case, print a single line with the name of the winner, assuming both players play optimally. If Alice wins, print $\texttt{Alice}$; otherwise, print $\texttt{Bob}$.

Examples:
input:
2
2
1 2
1 2
3
1 2 3
2 3 1
output:
Bob
Alice

Note: In the first test case, Bob can win the game by deleting the same element as Alice did.

In the second test case, Alice can delete $3$ in the first turn, and then in the second turn, delete the element that is different from the one Bob deleted in the first turn to win the game.

Buggy Code:
import java.util.*;
public class Main
{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-- >0){
		    int n=sc.nextInt(),i,j;
		    int[] a = new int[n];
		    int[] b = new int[n];
		    for(i=0;i<n;i++) a[i]=sc.nextInt();
		    for(i=0;i<n;i++) b[i]=sc.nextInt();
		    int c=0,c1=0;
		    for(i=0;i<n;i++){
		        if(a[i]!=b[i]) c=1;
		        if(a[i]!=b[n-i-1]) c1=1;
		    }
		    if(c==1&&c1==1) System.out.println("Alice");
		    else System.out.println("Bob");
		}
	}
}

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

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

Think step by step.

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