**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
n, t = map(int, input().split());

from collections import defaultdict

def compare(a, b):
    return False if a == b and b == True else True;

adj = defaultdict(list);

for i in range (n-1):
    x, y = map(int, input().split());
    adj[x].append(y); adj[y].append(x);

dp = [True for _ in range (n+1)];

vis = [False for _ in range (n+1)];

def dfs(idx):
    if len(adj[idx]) == 0:
        dp[idx] = False;
        return;
    vis[idx] = True;
    for i in adj[idx]:
        if not vis[i]:
            dfs(i);
            dp[idx] = compare(dp[i], dp[idx]);

t = int(input());
dfs(t);
if (dp[t]): print ("Ron");
else: print ("Hermione")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
from collections import defaultdict

def dfs(node, parent, adj_list, visited):
    visited[node] = True
    count = 1  # Count the current node
    for neighbor in adj_list[node]:
        if neighbor != parent and not visited[neighbor]:
            count += dfs(neighbor, node, adj_list, visited)
    return count

def solve(n, t, edges, start_nodes):
    adj_list = defaultdict(list)

    # Build the adjacency list
    for u, v in edges:
        adj_list[u].append(v)
        adj_list[v].append(u)

    for start_node in start_nodes:
        visited = [False] * (n + 1)
        reachable_count = dfs(start_node, -1, adj_list, visited)

        # Determine the winner based on the parity of reachable nodes
        if reachable_count % 2 == 1:
            print("Ron")
        else:
            print("Hermione")

# Input reading
n, t = map(int, input().split())
edges = [tuple(map(int, input().split())) for _ in range(n - 1)]
start_nodes = list(map(int, input().split()))

# Solve the problem
solve(n, t, edges, start_nodes)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider the following aspects:

1. **Tree Structure**: The code should handle various tree structures, including linear, star, and more complex structures.
2. **Starting Node**: The starting node should be chosen to maximize the failure probability. This could be a leaf node, a central node, or a node with a specific degree.
3. **Edge Cases**: The input should include edge cases such as a tree with only two nodes, a tree with a single branch, or a tree with a central node connected to multiple leaf nodes.

Given these considerations, let's create a test input that is small but challenging:

- **Tree Structure**: A simple tree with a central node connected to three leaf nodes.
- **Starting Node**: The central node, which should be a critical point for the game.

Here is the test input:

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

**Explanation**:
- The tree has 4 nodes.
- Node 1 is connected to nodes 2, 3, and 4.
- The starting node is 1.

This input is designed to test the following:
- The central node (1) has three neighbors, which should be reachable.
- The game should end with Ron winning because the number of reachable nodes (3) is odd.

This test input is small and complete, and it should expose any issues in the code's handling of tree structures and the counting of reachable nodes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
import sys
sys.setrecursionlimit(200000)

def dfs(node, parent, depth):
    max_depth = depth
    for neighbor in tree[node]:
        if neighbor != parent:
            max_depth = max(max_depth, dfs(neighbor, node, depth + 1))
    return max_depth

def solve():
    n, t = map(int, input().split())
    global tree
    tree = [[] for _ in range(n + 1)]

    for _ in range(n - 1):
        u, v = map(int, input().split())
        tree[u].append(v)
        tree[v].append(u)

    start_node = int(input())

    # Find the maximum depth from the start node
    max_depth = dfs(start_node, -1, 0)

    # Determine the winner based on the maximum depth
    if max_depth % 2 == 0:
        print("Ron")
    else:
        print("Hermione")

if __name__ == "__main__":
    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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios where the code might fail. The key points to consider are:

1. **Tree Structure**: The code uses a depth-first search (DFS) to determine the maximum depth from the starting node. We need to ensure that the tree structure is such that the DFS might not correctly determine the maximum depth, especially in cases with complex or degenerate structures.
2. **Starting Node**: The starting node should be chosen in such a way that the maximum depth calculation might be incorrect, leading to the wrong determination of the winner.

### Edge Cases to Consider:
1. **Linear Tree**: A tree where all nodes are connected in a straight line. This can cause the DFS to return a depth that might not be optimal.
2. **Star Tree**: A tree where one central node is connected to all other nodes. This can also cause issues with depth calculation.
3. **Balanced Tree**: A tree with a balanced structure. This can be challenging if the starting node is not the root.

### Chosen Test Input:
Let's create a linear tree with 5 nodes, where the nodes are connected in a straight line. This will ensure that the DFS might not correctly determine the maximum depth if the starting node is not the root.

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

### Explanation:
- **Tree Structure**: The tree is a linear chain where each node is connected to the next in a sequence: 1-2-3-4-5.
- **Starting Node**: The starting node is 1, which is one end of the linear chain.
- **Expected Output**: The maximum depth from node 1 is 4 (since the tree is a straight line). Since 4 is even, Ron should win.

This test input will expose the bug if the DFS does not correctly handle the linear structure and returns an incorrect maximum depth. If the code fails to determine the correct maximum depth, it will incorrectly determine the winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

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

public class C_2_Game_on_Tree_Medium {

    int mod = (int) 1e9 + 7;

    public static void main(String[] args) {

            int n = sc.nextInt();
            int tr = sc.nextInt();
            List<Integer>[] tree = new ArrayList[n];
            for(int i=0;i<n;i++) tree[i] = new ArrayList<>();
            for(int i=0;i<n-1;i++){
                int u = sc.nextInt()-1;
                int v = sc.nextInt()-1;
                tree[u].add(v);
                tree[v].add(u);
            }

            int x = dfs(tree,0,-1);
            if(x==1) System.out.println("Ron");
            else System.out.println("Hermione");
    }

    public static int dfs(List<Integer>[] tree, int node, int par){
        for(int x:tree[node]){
            if(x==par) continue;
            int z = dfs(tree,x,node);
            if(z==0) return 1;
        }

        return 0;
    }

    public static void input(Object array) {
        if (array instanceof int[])
            for (int i = 0; i < ((int[]) array).length; i++)
                ((int[]) array)[i] = sc.nextInt();
        else if (array instanceof long[])
            for (int i = 0; i < ((long[]) array).length; i++)
                ((long[]) array)[i] = sc.nextLong();
    }

    public static void printArr(Object array) {
        if (array instanceof int[])
            for (int x : (int[]) array)
                System.out.print(x + " ");
        else if (array instanceof long[])
            for (long x : (long[]) array)
                System.out.print(x + " ");
        System.out.println();
    }

    public static <T extends Number> long gcd(T a, T b) {
        return b.longValue() == 0 ? a.longValue() : gcd(b, a.longValue() % b.longValue());
    }

    static FastReader sc = new FastReader();

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's analyze the problem and the code step by step.

### Problem Analysis:
1. **Tree Structure**: The tree is a collection of nodes connected by edges. Each node can have one or more neighbors.
2. **Game Play**: The game is played on a tree where the players take turns moving a stone from one node to another. The stone can only move to an inactive neighbor.
3. **Winning Condition**: The player who cannot make a move loses.

### Buggy Code Analysis:
1. **DFS Function**: The `dfs` function is supposed to determine if the starting node can lead to a win for Ron or Hermione. However, the current implementation only checks if there is any subtree where Hermione can win, which is not sufficient to determine the overall winner.
2. **Return Value**: The function returns 1 if Hermione can win from any subtree, and 0 otherwise. This logic is flawed because it doesn't consider the entire tree structure and the optimal play strategy.

### Challenging Test Input:
To expose the bug, we need a test case where the current implementation of the `dfs` function will incorrectly determine the winner. A good candidate is a tree where the structure is such that the first move (by Ron) can force the second player (Hermione) into a losing position.

### Example Test Input:
Consider a tree with 4 nodes connected in a linear structure:
- Node 1 is connected to Node 2
- Node 2 is connected to Node 3
- Node 3 is connected to Node 4

If the game starts at Node 1, Ron can force Hermione into a losing position by always moving to the next node. Hermione will be left with no valid moves after Ron's second move.

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

### Explanation:
- **Tree Structure**: The tree is a simple linear structure with nodes 1, 2, 3, and 4 connected in sequence.
- **Starting Node**: The stone starts at Node 1.
- **Expected Output**: "Ron" because Ron can always force Hermione into a losing position by moving the stone to the next node.

This test input should expose the bug in the `dfs` function, as the current implementation will incorrectly determine the winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.File;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.NavigableSet;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

/*
 *   Author : RTE_ELMENATOR
 *
 *   "No one will carry you.
 *    Work harder, and turn your dream into reality."
 *   "
 *      If you're stuck in a problem while solving
 *      Then Think about the Answer (how related to input)
 *   "
 *   " Tree Questions : -> dp on tree
 *    dfs calcualtes the subtree distance
 *    dfs 2 calculates the parent answer and sub tree ans combine we can give
 *     ( Path of max distance can't consider max 2 can consider "
 */

public class CpWala {
	private static void solverA() {
		int[] size = Reading.readIntArray();
		int n = size[ 0 ];
		ArrayList<Integer>[] graph = new ArrayList[ n + 1];
		for(int i = 1; i< n; i++ ) {
			int[] temp = Reading.readIntArray();
			Add( graph , temp [ 0], temp [ 1 ]);
			Add( graph , temp[ 1 ] , temp [ 0 ]);

		}
		int src = Reading.readInt();
		if( dfs( src , graph , -1 , 1  )) {
			out.println("Ron");
		}
		else out.println("Hermione");
	}
	private static boolean dfs( int curNode , ArrayList<Integer>[] graph , int parent  , int depth) {
		if( graph[ curNode ].size() == 1) {

			return ( depth & 1 ) !=0;
		}
		boolean res = false;
		for(int adj : graph [ curNode ]) {
			if( adj == parent ) continue;

		   res = res || dfs( adj , graph , curNode , depth + 1);

		}
		return res;
	}
	private static void Add( ArrayList<Integer>[] graph , int u , int v ) {
		if( graph[ u ] == null ) graph[ u ] = new ArrayList<>();
		graph[ u ].add( v );
	}
  static PrintWriter out = new PrintWriter(System.out);
  // main function
  public static void main(String[] arg) {
    int testNo =  1 ;// Reading.readInt();
    while (testNo-- > 0) {
      solverA();
    }
    out.flush();
    out.close();
  }

}
class Union {
	int[] parent ;
	int[] rank ;
	int n ;
	int[] size ;
	public Union ( int size  ) {
		n  = size + 1;
		this.size = new int[ size + 1 ];
		parent = new int [ size + 1];
	   rank = new int[ size + 1];
	   Arrays.fill( this.size ,1 );
	   Arrays.fill( parent , -1 );
	   Arrays.fill( rank , -1 );
	}
	public void union(int u , int v ) {
		int parent_u = find( u ) ;
		int parent_v = find( v );
		if( parent_u != parent_v ) {
			if( parent_v > parent_u) {
			parent[ parent_u ] = parent_v  ;
			size[ parent_u ] += size[ parent_v ];
			}
			else {
				parent[ parent_v ] = parent_u;
				size[ parent_u ]+=size[ parent_u];
			}
		}
	}
	public int find( int node ) {
		if( parent[ node ] == -1 ) return node ;
		return parent[ node ] = find( parent[ node ]);
	}

}
class TreeNode {
	int  next  ;
	long weight , weight1 ;
	public TreeNode ( int node , int weight , int weight1 ) {
		this.next = node ;
		this.weight = weight;
		this.weight1 = weight1 ;
	}
	public String toString() {
		return "["+next+" , " + weight + " , "+weight1+" ] ";
	}
}
class Reading {
  private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

  public Reading() {
    reader = new BufferedReader(new InputStreamReader(System.in));
  }

  public static int readInt() {
    try {
      return Integer.parseInt(reader.readLine().trim());
    } catch (Exception e) {
      return -1;
    }
  }

  public static int[] readIntArray() {
    try {
      return Arrays.stream(reader.readLine().trim().split("\\s+"))
                   .mapToInt(Integer::parseInt)
                   .toArray();
    } catch (Exception e) {
      return null;
    }
  }

  public static long[] readLongArray() {
    try {
      return Arrays.stream(reader.readLine().trim().split("\\s+"))
                   .mapToLong(Long::parseLong)
                   .toArray();
    } catch (Exception e) {
      return null;
    }
  }

  public static String readString() {
    try {
      return reader.readLine();
    } catch (Exception e) {
      return null;
    }
  }

  public static char[] readToCharArray() {
    try {
      return reader.readLine().replaceAll("\\s+", "").toCharArray();
    } catch (Exception e) {
      return null;
    }
  }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to consider edge cases where the logic of the DFS function might fail. The code checks if a node is a leaf (i.e., has only one neighbor) and returns whether the depth is odd. However, this logic might not be correct in all scenarios, especially in more complex tree structures.

### Step-by-Step Analysis:
1. **Leaf Node Check**: The code assumes that a node with only one neighbor is a leaf. This is generally correct, but we need to ensure that the root node is not incorrectly treated as a leaf.
2. **Depth Calculation**: The code uses the depth to determine the winner. If the depth is odd, Ron wins; if even, Hermione wins. This logic might fail if the tree structure is such that the depth calculation is incorrect.

### Edge Case to Consider:
- **Linear Tree**: A linear tree (a tree where each node has at most one child, forming a chain) can expose issues with the depth calculation.
- **Tree with a Root Node with Only One Child**: This can cause the root node to be incorrectly treated as a leaf.

### Test Input:
Let's create a test input with a linear tree and a root node with only one child to expose the bug.

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

### Explanation:
- **Tree Structure**: The tree is a linear chain: 1 -> 2 -> 3 -> 4.
- **Initial Node**: The game starts with the stone on node 1.
- **Expected Output**: Ron should win because he can always move the stone to an inactive neighbor, and Hermione will eventually be unable to make a move.

This test input should expose the bug in the code if the depth calculation or leaf node check is incorrect.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

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

using namespace std;

#define ll long long

const int MX = 2e5 + 1;
int a[MX];
bool vis[MX];

vector<int> l;

void dfs(int i, int prev, vector<vector<int>>& g) {
	vis[i] = 1;
	if (prev != i) a[i] = a[prev] + 1;
	for (auto& j : g[i])
		if (!vis[j])
			dfs(j, i, g);
}

void F() {
	int n, t;
	cin >> n >> t;

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

	dfs(s, s, g);

	//for (int i = 0; i < n; i++) cout << g[i].size() << ' ';
	//cout << '\n';
	//for (int i = 0; i < n; i++) cout << a[i] << ' ';
	//cout << '\n';

	for (int i = 0; i < n; i++) {
		if (a[i] == 0) continue;
		if (g[i].size() == 1 && a[i] % 2) {
			cout << "Ron\n";
			return;
		}
	}
	cout << "Hermione\n";
}

int main() {
	cin.tie(0)->sync_with_stdio(0);
	int t = 1;
	//cin >> t;
	while (t--) {
		F();
	}
}

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

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

Think step by step.

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

1. **Understanding the Bug:**
   - The code uses a depth-first search (DFS) to calculate the distance from the starting node to all other nodes.
   - It then checks if any leaf node (a node with only one neighbor) has an odd distance from the starting node. If such a node exists, Ron wins; otherwise, Hermione wins.
   - The bug might be in the logic of determining the winner, especially when the tree has a structure that makes the distance check incorrect.

2. **Edge Cases:**
   - A tree with a single path (a linear tree).
   - A tree with a single leaf node.
   - A tree with multiple leaf nodes.
   - A tree where the starting node is a leaf node.
   - A tree where the starting node is a non-leaf node with an odd or even number of children.

3. **Challenging Scenario:**
   - A tree with a structure that makes the distance check fail. For example, a tree with a long path where the starting node is not a leaf but has an odd number of moves to a leaf node.

Let's create a test input that fits the above criteria:

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

### Explanation:
- **Tree Structure:**
  - Node 1 is connected to Node 2.
  - Node 2 is connected to Node 3.
  - Node 3 is connected to Node 4.
  - Node 4 is connected to Node 5.
  - This forms a linear tree where each node has exactly one neighbor except for Node 1 and Node 5, which are leaf nodes.
- **Starting Node:**
  - The starting node is Node 2.
- **Expected Output:**
  - The distance from Node 2 to Node 5 is 3, which is odd. Therefore, Ron should win.

This test input should expose the bug in the code if the distance calculation or the winning condition is not correctly implemented.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define lld long double
#define f(i,a,b) for(int i = a; i < b; i++)
#define fe(i,a,b) for(int i = a; i >= b; i--)
#define vll vector<ll>
#define pb push_back
#define pii pair<int, int>
#define si set<int>
#define sc set<char>
#define mii map<int, int>
#define print(a) for(auto u:a) cout<<u<<" "; cout<<endl;
#define sor(v) sort(v.begin(),v.end())
#define all(v) v.begin(),v.end()
#define ppq priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>>
#define pq priority_queue <int, vector<int>, greater<int>>

#define debugger 1

#ifdef debugger
#define debug(x) cerr << #x<<" "; cerr<<x<<" "; cerr << endl;
#else
#define debug(x);
#endif

int binpow(int a, int b) {
    int res = 1;
    while (b > 0) {
        if(b & 1) res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}

// //unordered_set<pair<int,int>>
// struct pair_hash {
//     template <class T1, class T2>
//     std::size_t operator () (const std::pair<T1, T2>& p) const {
//         auto h1 = std::hash<T1>{}(p.first);
//         auto h2 = std::hash<T2>{}(p.second);
//         // Combine the two hash values using XOR and a bit-shifting operation
//         return h1 ^ (h2 << 1);
//     }
// };

int inv(int a, int m) {
  return a <= 1 ? a : m - (long long)(m/a) * inv(m % a, m) % m;
}

template<typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }
template<typename T> // cin >> vector<T>
istream& operator>>(istream &istream, vector<T> &v) {for (auto &it : v)cin >> it;return istream;}
template<typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second); }
template<typename T> // cout << vector<T>
ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }

bool sorta(const pair<int,int> &a,const pair<int,int> &b) {return (a.second < b.second);}
bool sortd(const pair<int,int> &a,const pair<int,int> &b) {return (a.second > b.second);}

ll gcd(ll a,ll b) { if (b==0) return a; return gcd(b, a%b); }
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

void dfs(unordered_map<int, vll> &adj, vector<bool> &vis, vector<bool> &leaves, int node, bool &check, int cnt) {
    vis[node] = true;
    cnt++;

    for(int i = 0; i < adj[node].size(); i++) {
        if(!vis[adj[node][i]]) {
            dfs(adj,vis,leaves,adj[node][i],check,cnt);
        }
    }

    if(leaves[node] && !(cnt&1)) check = true;
    if(cnt) cnt--;
}

void superSpeed(){
    int n,t; cin >> n >> t;
    unordered_map<int, vll> adj;
    vector<bool> vis(n+1,false),leaves(n+1,true); vis[0] = true;
    f(i, 0, n-1) {
        int u,v; cin >> u >> v;
        adj[u].pb(v);
        adj[v].pb(u);
        leaves[u] = false;
    }
    int initialNode; cin >> initialNode;
    bool check = false; int cnt = 0;
    dfs(adj,vis,leaves,initialNode,check,cnt);
    if(check) cout << "Ron" << endl;
    else cout << "Hermione" << endl;
}

signed main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    // int testCases; cin >> testCases;
    // while(testCases--) superSpeed();
    superSpeed();
}

**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, let's consider the logic of the game and the potential issues that might arise. The key points to consider are:

1. **Tree Structure and Initial Node**: The initial node and the structure of the tree can significantly affect the outcome of the game. We need to create a tree where the bug in the DFS logic can be easily exposed.
2. **DFS Logic**: The DFS function is supposed to determine whether the initial node can lead to a situation where Ron or Hermione wins. The logic in the DFS function might not correctly handle certain cases, especially when the tree has specific structures.

### Analyzing the Buggy Code
The DFS function in the code is designed to traverse the tree and count the number of nodes in a subtree. If the count of nodes in the subtree is even, it sets a flag `check` to `true`. However, this logic might be flawed in certain scenarios, such as when the tree has a specific structure that leads to incorrect counting or flag setting.

### Test Case to Expose the Bug
Let's create a test case where the tree has a structure that can expose the bug. Consider a tree with a linear structure (a path) where the initial node is one of the endpoints. In such a case, the DFS should correctly determine the winner, but if the DFS logic is flawed, it might not.

### Small-Scale Test Input
```plaintext
3 1
1 2
2 3
1
```

### Explanation
- **Tree Structure**: The tree is a simple path with 3 nodes: 1 -> 2 -> 3.
- **Initial Node**: The initial node is 1.
- **Expected Output**: Ron should win because he can move the stone to node 2, and Hermione will be forced to move it to node 3, leaving Ron with no valid move.

This test case should expose the bug if the DFS logic incorrectly counts the nodes or sets the `check` flag. If the code correctly implements the DFS and logic, it should output "Ron". If the code is buggy, it might output "Hermione".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
from collections import defaultdict

x, y = map(int, input().split())
dc = defaultdict(list)
statue = defaultdict(list)
for _ in range(x-1):
    a,b = map(int, input().split())
    dc[a].append(b)
    dc[b].append(a)

z = int(input())
def dfs(dic,item,s,vis = set()):
    if item not in vis:
        vis.add(item)
        s = not s
        for child in dic[item]:
            curr = dfs(dic,child,s,vis)
    statue[item].append(not s)
    return 0

dfs(dc,z,True)
if False in statue[z]:
     print('Ron')
     quit()
print('Hermione')

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases where the code might fail. The main issue in the code is the way the `dfs` function is implemented, particularly how it handles the state and the recursion.

### Steps to Identify a Challenging Test Input:
1. **Tree Structure**: The tree should be structured in a way that the game logic can be tricky to handle. A good candidate is a tree with a long path or a star-like structure.
2. **Initial Node**: The initial node should be chosen such that the game can be influenced by the structure of the tree.

### Challenging Test Input:
Let's consider a tree with a long path and an initial node that is not at the end of the path. This will test the ability of the code to correctly determine the winner based on the structure of the tree.

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

### Explanation:
- **Tree Structure**: The tree is a simple path: `1 - 2 - 3 - 4 - 5`.
- **Initial Node**: The stone is initially placed on node `2`.

### Expected Output:
- **Output**: "Hermione"

### Reasoning:
- If the stone starts on node `2`, Ron can move it to node `1` or node `3`. If Ron moves it to node `1`, Hermione can move it back to node `2` or to node `3`. If Ron moves it to node `3`, Hermione can move it to node `4`. In both cases, Hermione can always force Ron to lose because she can always move the stone to a node with fewer options, eventually leaving Ron with no valid move.

This test input should expose the bug in the provided code, as the code's `dfs` function does not correctly handle the alternating moves and the state transitions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
from __future__ import division, print_function

def main():
    # Template 1.0

    import sys, re, math
    from collections import deque, defaultdict, Counter, OrderedDict
    from bisect import bisect, bisect_left
    from itertools import permutations
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    from fractions import Fraction
    from decimal import Decimal
    def STR():
        return list(input())

    def INT():
        return int(input())

    def MAP():
        return map(int, input().split())

    def LIST():
        return list(map(int, input().split()))

    def list2d(a, b, c):
        return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):
        return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))[::-1]
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7

    def overlap(l1, r1, l2, r2):
        if (l1 < l2):
            return l2 <= r1
        else:
            return l1 <= r2

    def _overlap(as1, ae, bs, be):
        if (bs > ae or as1 > be):
            return None
        else:
            os = max(as1, bs)
            oe = min(ae, be)
            return [os, oe]

    def gcd(a, b):
        if a == 0:
            return b
        return gcd(b % a, a)

    def lcm(a, b):
        return (a * b) // gcd(a, b)

    n, T = MAP()

    graph = defaultdict(list)

    for __ in range(n-1):
        x, y = MAP()
        graph[x].append(y)
        graph[y].append(x)

    u = INT()

    poss = [-1] * (n+1)

    leaves = []
    par = [-1] * (n + 1)

    q = deque([(u, -1)])
    vis = set([u])
    while q:
        node, parent = q.popleft()

        if len(graph[node]) == 1 and node != u:
            leaves.append(node)

        for nei in graph[node]:
            if nei not in vis:
                vis.add(nei)
                par[nei] = node
                q.append((nei, node))

    # print(par, leaves)
    for leaf in leaves:
        curr = 0
        node = leaf
        while node != -1:
            # print(leaf, node, poss)
            if poss[node] == -1:
                poss[node] = curr
            elif poss[node] == 0:
                if curr == 1:
                    poss[node] = 2
            elif poss[node] == 1:
                if curr == 0:
                    poss[node] = 2

            curr ^= 1

            node = par[node]

    # print(poss)
    if poss[u] == 2 or poss[u] == 1:
        print("Ron")
    else:
        print("Hermione")

'''
Observations:
1. Treat u as the root.
2. Game ends on a leaf.
3. Prepare an array that contains 0/1/2 depending on who can win the game

1: whoever comes to this node will win
0: whoever comes to this node will lose
2: anything is possible

'''
######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

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

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# Cout implemented in Python
import sys

class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self

cout = ostream()
endl = '\n'

# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, let's consider the following steps:

1. **Understand the Bug**: The code seems to have a logic issue in how it determines the winner of the game. Specifically, it may not correctly handle the parity of the number of moves or the structure of the tree.

2. **Identify Edge Cases**:
   - **Single Path Tree**: A tree where nodes form a single path. This can test the code's ability to handle linear structures.
   - **Star Tree**: A tree where one node is connected to all other nodes. This can test the code's ability to handle a central hub.
   - **Balanced Tree**: A tree where the depth of all leaves is the same. This can test the code's ability to handle balanced structures.
   - **Unbalanced Tree**: A tree where the depth of leaves varies significantly. This can test the code's ability to handle more complex structures.

3. **Generate a Small-Scale Test Input**:
   - We will create a small tree that is unbalanced and has a specific structure to test the code's handling of the game.

Let's create an unbalanced tree with 5 nodes:

- Node 1 is connected to Node 2.
- Node 1 is connected to Node 3.
- Node 3 is connected to Node 4.
- Node 3 is connected to Node 5.

We will place the stone initially on Node 1.

Here is the test input:

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

This input will help us check if the code correctly handles the unbalanced structure and the initial placement of the stone.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

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

public class C_2_Game_on_Tree_Medium {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static FastReader vt = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        try {
            // int testCases = vt.nextInt();
            int testCases = 1;
            while (testCases-- > 0) {
                // write code here
                solve();
            }
            out.flush();
        } catch (Exception e) {
            return;
        }
    }

    static void solve() {
        int n = vt.nextInt();
        int m = vt.nextInt();
        ArrayList<ArrayList<Integer>> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(new ArrayList<>());
        }
        for (int i = 0; i < n - 1; i++) {
            int u = vt.nextInt();
            int v = vt.nextInt();
            list.get(u - 1).add(v - 1);
            list.get(v - 1).add(u - 1);
        }
        int node = vt.nextInt() - 1;
        boolean ok = solve(node, -1, list, 0);
        out.println(ok ? "Ron" : "Hermione");
    }

    static boolean solve(int node, int parent, ArrayList<ArrayList<Integer>> list, int depth) {
        int count = 0;
        for (int i = 0; i < list.get(node).size(); i++) {
            int child = list.get(node).get(i);
            if (child != parent) {
                count++;
                if (depth % 2 == 0) {
                    if (solve(child, node, list, depth + 1)) {
                        return true;
                    }
                } else {
                    if (!solve(child, node, list, depth + 1)) {
                        return true;
                    }
                }
            }
        }
        if (count == 0 && depth % 2 == 1) {
            return true;
        }
        return false;
    }
}

**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 buggy code, we need to consider edge cases or scenarios that can maximize the failure probability. The code seems to have issues with handling trees where the structure or the starting node can influence the outcome of the game.

### Step-by-Step Analysis:
1. **Tree Structure**: The tree structure can significantly affect the game. For example, a tree with a single leaf node (a star tree) or a balanced tree can lead to different outcomes.
2. **Starting Node**: The starting node is crucial. If the starting node is a leaf node, the first player (Ron) might be at a disadvantage. If it's a central node, the first player might have an advantage.
3. **Depth and Turn Alternation**: The depth of the recursion and the alternation of turns can also influence the result. The code might not correctly handle the depth and turn alternation in certain scenarios.

### Test Input:
Let's consider a simple tree with 4 nodes where the starting node is a leaf node. This should make the game more challenging for Ron.

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

### Explanation:
- **Tree Structure**:
  - Node 1 is connected to Node 2.
  - Node 2 is connected to Node 3 and Node 4.
  - Node 1 is a leaf node.
- **Starting Node**: The game starts with the stone on Node 1.
- **Expected Outcome**: Ron should lose because he starts on a leaf node and cannot make a move, so Hermione wins.

This test input should expose the bug in the code, as the current logic might not correctly handle the scenario where the starting node is a leaf.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

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

public class C_2_Game_on_Tree_Medium {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static FastReader vt = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        try {
            // int testCases = vt.nextInt();
            int testCases = 1;
            while (testCases-- > 0) {
                // write code here
                solve();
            }
            out.flush();
        } catch (Exception e) {
            return;
        }
    }

    static void solve() {
        int n = vt.nextInt();
        int m = vt.nextInt();
        ArrayList<ArrayList<Integer>> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(new ArrayList<>());
        }
        for (int i = 0; i < n - 1; i++) {
            int u = vt.nextInt();
            int v = vt.nextInt();
            list.get(u - 1).add(v - 1);
            list.get(v - 1).add(u - 1);
        }
        int node = vt.nextInt() - 1;
        boolean ok = solve(node, -1, list, 0);
        out.println(ok ? "Ron" : "Hermione");
    }

    static boolean solve(int node, int parent, ArrayList<ArrayList<Integer>> list, int depth) {
        boolean ok = false;
        for (int i = 0; i < list.get(node).size(); i++) {
            int child = list.get(node).get(i);
            if (child != parent) {
                if (depth % 2 == 0) {
                    if (solve(child, node, list, depth + 1)) {
                        ok |= true;
                    }
                } else {
                    if (!solve(child, node, list, depth + 1)) {
                        ok |= true;
                    }
                }
            }
        }
        return ok;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = array<int, 2>;
#define all(x) begin(x), end(x)
#define sz(x) (int) (x).size()

template<class T> bool smax(T &a, T b) {
    return a < b ? a = b, 1 : 0;
}
template<class T> bool smin(T &a, T b) {
    return a > b ? a = b, 1 : 0;
}
int main() {
    cin.tie(0) -> sync_with_stdio(0);
    int n, t;
    cin >> n >> t;
    vector<vector<int>> adj(n);
    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        --u, --v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    int x; cin >> x; --x;
    // 0 = Ron, 1 = Hermione
    vector<array<int, 2>> dp(n, {-1, -1});
    const auto dfs = [&](int u, int p, auto &&self) -> void {
        for (int v : adj[u]) {
            if (v == p) { continue; }
            self(v, u, self);
            if (dp[v][1] == 0) {
                dp[u][0] = 0;
            }
            if (dp[v][0] == 1) {
                dp[u][1] = 1;
            }
        }
        if (adj[u] == vector<int>{p}) {
            dp[u][0] = 1;
            dp[u][1] = 0;
        }
    };
    dfs(x, -1, dfs);
    cout << (!dp[x][0] ? "Ron" : "Hermione") << "\n";
    /*
    for (int i = 0; i < n; i++) {
        cout << "Node " << i + 1 << "\n";
        cout << dp[i][0] << " " << dp[i][1] << "\n";
    }
    */
}

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

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

Think step by step.

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

1. **Identify the Bug**: The buggy code uses a depth-first search (DFS) to determine the winner of the game. The key issue seems to be in the logic used to update the `dp` array, which tracks the possible outcomes for each node.

2. **Edge Cases**:
   - **Leaf Nodes**: If a node is a leaf, the player who starts at that node will lose because there are no valid moves.
   - **Nodes with Multiple Children**: If a node has multiple children, the player who starts at that node can force the opponent into a losing position.
   - **Symmetric Trees**: If the tree is symmetric, the starting node can significantly affect the outcome.

3. **Challenging Scenario**:
   - A tree with a specific structure that forces the player to make suboptimal moves if the starting node is chosen poorly.
   - A tree with a long path and a few branches to create a situation where the first player can force the second player into a losing position.

4. **Small-Scale Test Input**:
   - A small tree with a structure that is easy to visualize but complex enough to challenge the algorithm.
   - For example, a tree with a long path and a single branch.

Let's construct a tree with 7 nodes where node 1 is the root, and it has a long path to node 7, with a single branch at node 4 leading to node 6.

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

This input represents a tree where:
- Node 1 is connected to node 2.
- Node 2 is connected to node 3.
- Node 3 is connected to node 4.
- Node 4 is connected to node 5 and node 6.
- Node 5 is connected to node 7.

The game starts with the stone on node 1. This input should expose the bug in the code by forcing the player to navigate a complex path and a branch, which can lead to incorrect outcomes if the logic is flawed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
// Author: Tejas
// #pragma GCC optimize("Ofast")
// #pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
// find_by_order, order_of_key
/*
pbds A
*A.find_by_order(ind)
A.order_of_key(value) {number of elements less than value
A.erase(val to be removed)
A.insert(val)
*A.upper_bound *A.lower_bound
}

*/
#define ll long long
#define ld long double
#define vvi vector<vector<ll>>
#define lb lower_bound
#define ub upper_bound
#define all(v) v.begin(), v.end()
#define vi vector<long long>
#define fl(i,s,e) for(ll i = s; i < e; i++)
#define ip(v) for(ll i = 0; i < v.size(); i++) cin>>v[i];
#define si set<long long>
#define srt(v) sort(v.begin(), v.end())
#define um map<long long, long long>
#define pb push_back
#define lcm(a,b) (a*b)/__gcd(a,b)
#define vpl vector<pair<ll,ll>>
#define inf 1000000000000000000ll
#define flr(i,e,s) for(ll i = e; i>=s; i--)
#define ops(v) for(ll i= 0; i < v.size(); i++) cout<<v[i]<<' ';
#define F first
#define S second
#define BS binary_search
#define ump gp_hash_table;
#define yeah cout<<"YES"<<endl;
#define nah cout<<"NO"<<endl;

const ll MOD = 1e9 + 7;
using namespace std;
//char b = B|' ' , char B = (b&'_')
// cout<<fixed<<setprecision(6)<<ans;

#define debug(x) cout << #x<<" "; _print(x); cout << endl;
void _print(ll t) {cout << t;}
void _print(int t) {cout << t;}
void _print(string t) {cout << t;}
void _print(char t) {cout << t;}
void _print(ld t) {cout << t;}
void _print(double t) {cout << t;}
void _print(unsigned ll t) {cout << t;}
template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cout << "{"; _print(p.ff); cout << ","; _print(p.ss); cout << "}";}
template <class T> void _print(vector <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T> void _print(set <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T> void _print(multiset <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T, class V> void _print(map <T, V> v) {cout << "[ "; for (auto i : v) {_print(i); cout << " ";} cout << "]";}

constexpr ll bct(ll x) { return __builtin_popcountll(x); } // # of bits set
constexpr ll bsz(ll x) { return x == 0 ? 0 : 31-__builtin_clzll(x); } // floor(log2(x))

template<typename T>T min(const vector<T>&v){return *min_element(v.begin(),v.end());}
template<typename T>T max(const vector<T>&v){return *max_element(v.begin(),v.end());}

const ld PI = acos((ld)-1);

struct chash { const uint64_t C = uint64_t(2e18 * PI) + 71; const uint32_t RANDOM = chrono::steady_clock::now().time_since_epoch().count(); size_t operator()(uint64_t x) const { return __builtin_bswap64((x ^ RANDOM) * C); }};
mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());

void found(ll f){
    if(f) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}

//math

ll binpow(ll a, ll b){
        a %= MOD;   //For larger values of a
        ll res = 1;
        while (b > 0)
        {
            if (b & 1)
                res = res * a % MOD;
            a = a * a % MOD;
            b >>= 1;
        }
        return res;
}

ll extendedgcd(ll a,ll b,ll& x,ll& y){
    // returns gcd and changes x and y
    // such that gcd = ax+by
    x=1,y=0;
    ll x1=0,y1=1,a1=a,b1=b;
    while (b1){
        ll q = a1/b1;
        tie(x,x1) = make_tuple(x1,x-q*x1);
        tie(y,y1) = make_tuple(y1,y-q*y1);
        tie(a1,b1)= make_tuple(b1,a1-q*b1);
    }
    return a1;
}

ll N;
vector<bool> prime(N, true);
vector<ll> hp(N, 1), lp(N, -1);

void Sieve()
{
    //update N
    prime[0] = false;
    prime[1] = false;

    for (int i = 2; i < N; i++)
    {
        if (prime[i])
        {
            hp[i] = i;
            lp[i] = i;
            for (ll j = 2 * i; j < N; j += i)
            {
                prime[j] = false;
                hp[j] = i;
                if (lp[j] == -1) lp[j] = i;
            }
        }
    }
}

ll countDivisors(ll n) {
    ll divisors = 0;
    ll sqrtN = sqrt(n); // Compute square root of n
    for (ll i = 1; i <= sqrtN; i++) {
        if (n % i == 0) {
            // If i divides n, increment divisors count
            divisors++;
            // If i is not the square root of n, increment count for its pair
            if (i*i != n) {
                divisors++;
            }
        }
    }
    return divisors;
}

ll modinv(ll n)
{
        return binpow(n, MOD-2);
}

/*
Points for modulo handling
1) use modinv whenever division is there let it be small number
2) while multiplying take mod of the multiplicands too
*/

ll nCr(int n, int r, int p=MOD)     // faster calculation..
{
    if (n < r)
        return 0;

    if (r == 0)
        return 1;

    vector<int> fac(n+1,0);
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = (fac[i - 1] * i) % p;

    return (fac[n] * modinv(fac[r]) % p * modinv(fac[n - r]) % p) % p;
}

//Range Queries

struct segtree{
vi seg;
vi lazy;
ll id;

ll func(ll a, ll b){
    return a + b;
}

void buildseg(vi& a,ll v,ll tl,ll tr) {
    // First resize seg,lazy to 4*n and assign 0
    // seg[v] gives ans for [tl,tr] 0-indexed
    // Start : v=1,tl=0,tr=n-1
    if (tl == tr) seg[v] = a[tl];
    else {
        ll tm = (tl + tr) / 2;
        buildseg(a, v*2, tl, tm);
        buildseg(a, v*2+1, tm+1, tr);
        seg[v] = func(seg[v*2],seg[v*2+1]); // change function here
    }
}

void push(ll v,ll tl,ll tr){
    // Default : addition operation
    ll tm = (tl+tr)/2;
    seg[2*v]+=(tm-tl+1)*lazy[v];
    lazy[2*v]+=lazy[v];
    seg[2*v+1]+=(tr-tm)*lazy[v];
    lazy[2*v+1]+=lazy[v];
    lazy[v]=id; // change identity here
}

ll queryseg(ll v,ll tl,ll tr,ll l,ll r) {
    // Query for [l,r] 0-indexed
    // seg[v] gives ans for [tl,tr]
    // Start : v=1,tl=0,tr=n-1
    if (l > r) return id; //change identity here
    if (l == tl && r == tr) return seg[v];
    //push(v,tl,tr);
    ll tm = (tl + tr) / 2;
    return func(queryseg(v*2, tl, tm, l, min(r, tm)) ,queryseg(v*2+1, tm+1, tr, max(l, tm+1), r)); //change function here
}

void updateseg(ll v,ll tl,ll tr,ll pos,ll new_val) {
    // Update at index pos (0-indexed)
    // seg[v] gives ans for [tl,tr]
    // Start : v=1,tl=0,tr=n-1
    if (tl == tr) seg[v] = new_val;
    else {
        ll tm = (tl + tr) / 2;
        if (pos <= tm) updateseg(v*2, tl, tm, pos, new_val);
        else updateseg(v*2+1, tm+1, tr, pos, new_val);
        seg[v] = func(seg[v*2] ,seg[v*2+1]); // change function here
    }
}

void upranseg(ll v,ll tl,ll tr,ll l,ll r,ll addend) {
    // Range Update for [l,r] 0-indexed
    // Default: addition update operation, sum query operation
    // seg[v] gives ans for [tl,tr]
    // Start : v=1,tl=0,tr=n-1
    if (l > r) return;
    if (l == tl && tr == r) {
        seg[v] += (tr-tl+1)*addend; // change function here
        lazy[v] += addend; // change function here
    }
    else {
        push(v,tl,tr);
        ll tm = (tl + tr) / 2;
        upranseg(v*2, tl, tm, l, min(r, tm), addend);
        upranseg(v*2+1, tm+1, tr, max(l, tm+1), r, addend);
        seg[v] = func(seg[v*2] ,seg[v*2+1]); // change function here
    }
}
};

//Trees, graphs
// void dfs(ll node, ll par){
//     // take action on the vertex after entering it
//     for(auto it: adj[node]){
//         // take action on child before entering it
//         if(it != par){
//             dfs(it, node);
//         }
//         // take action on child before exiting it
//     }
//     // take action on vertex before exiting vertex
// }

//mlogn
void Dijkstra(vector<vector<pair<ll, ll>>> &a, ll n, ll k, vi &dis, vi &prnt) {
    set<pair<ll, ll>> s;
    s.insert({0, k});
    dis[k] = 0;
    prnt[k] = -1;

    while (!s.empty()) {
        pair<ll, ll> te = *s.begin();
        ll c = te.second;
        ll g = te.first;
        s.erase(s.begin());

        for (auto &edge : a[c]) {
            ll c_v = edge.first;
            ll wt = edge.second;
            if (dis[c] + wt < dis[c_v]) {
                s.erase({dis[c_v], c_v});
                dis[c_v] = dis[c] + wt;
                prnt[c_v] = c;
                s.insert({dis[c_v], c_v});
            }
        }
    }
}

void dfs(ll node, ll par, vvi &adj, vi &nodv, ll nd){
    for(auto it:adj[node]){
        if(it == par) continue;
        dfs(it, node, adj, nodv, nd);
        nodv[node]|=(1-nodv[it]);
    }
    if(adj[node].size() == 1 && node != nd){
        nodv[node] = 0;
    }
}

void solve(){
    ll n, t; cin>>n>>t;
    vvi adj(n+1);
    fl(i,0,n-1){
        ll a, b; cin>>a>>b;
        adj[a].pb(b);
        adj[b].pb(a);
    }
    ll nd; cin>>nd;
    vi nodv(n+1, 0);
    dfs(1, -1, adj, nodv, nd);
    if(nodv[nd] == 1){
        cout<<"Ron"<<endl;
    }
    else{
        cout<<"Hermione"<<endl;
    }

}

/*
1)erase function in set is O(logn)
2)v.size() is unsigned so dont decrement
3)time can be decreased through sets depending on constraints
4)n root n looped can pass 1e5 smtimes
5)n + n/2 + n/3 + ..... -> logn
6)comparator function : a.second < b.second -> a will come before b
7)distance function is linear in non random access like set but O(1) in vectors
8)map -> 1e6 indeed
*/

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll t = 1;
    // cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code seems to have an issue with the DFS logic. Specifically, it might not correctly handle the activation and deactivation of nodes after each round.

2. **Edge Cases**: We need to consider edge cases such as:
   - Trees with a single branch (linear structure).
   - Trees with multiple branches.
   - Trees with a single leaf node.
   - Trees with a balanced structure.

3. **Small-Scale Test Input**: We need to create a small-scale test input that is complete and valid. This will help us focus on the logic and not get overwhelmed by the size of the input.

4. **Example Test Input**:
   - A tree with 5 nodes and a linear structure.
   - A tree with 5 nodes and a balanced structure.
   - A tree with 5 nodes and a single leaf node.

Let's create a test input that is a balanced tree with 5 nodes. This will help us see if the DFS logic correctly handles the activation and deactivation of nodes.

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

This test input represents a tree with the following structure:
- Node 1 is the root.
- Node 1 has two children: Node 2 and Node 3.
- Node 3 has two children: Node 4 and Node 5.

The starting node for the game is Node 1. This structure should be challenging enough to expose any issues in the DFS logic and the activation/deactivation of nodes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
import queue

def dfs(begin_node,fa,lens,edges,win):
    lens+=1
    if len(edges[begin_node])==1:
        return 0
    for i in edges[begin_node]:
        if i!=fa:
            if dfs(i,begin_node,lens,edges,win)==0:
                return 1
    return 0

def solve():
    n=list(map(int,input().split()))[0]
    edges=[[]for i in range(n+1)]
    for i in range(n-1):
        node1,node2=map(int,input().split())
        edges[node1].append(node2)
        edges[node2].append(node1)
    visited = [[0,0] for i in range(n+1)]
    win=[0 for i in range(n+1)]
    begin_node=int(input())
    if dfs(begin_node,0,0,edges,visited)==0:
        print("Hermione")
    else:
        print("Ron")

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

Buggy Code:
n = [int(i) for i in input().split()][0]
adj = [[] for i in range(n+1)]
for i in range(n-1):
    u, v = [int(i) for i in input().split()]
    adj[u].append(v)
    adj[v].append(u)
leaves = set()
for i in range(n+1):
    if len(adj[i])==1:leaves.add(i)
start = int(input())
dfs = [start]
length = 1
vis = [0]*(n+1)
dp = [""]*(n+1)
curr_winner = ""
while dfs:
    if vis[dfs[-1]]==0 or adj[dfs[-1]]:
        vis[dfs[-1]]=1
        if vis[adj[dfs[-1]][-1]]==0:
            turn = "r" if length%2 else "h"
            if turn != dp[dfs[-1]]:
                dp[dfs[-1]] = curr_winner
            dfs.append(adj[dfs[-1]].pop())
            length+=1
        else:
            adj[dfs[-1]].pop()
    else:
        if dfs[-1] in leaves:
            curr_winner = "r" if length%2==0 else "h"
            dp[dfs[-1]]=curr_winner
        else:
            turn = "r" if length%2 else "h"
            if turn != dp[dfs[-1]]:
                dp[dfs[-1]] = curr_winner
            else:
                curr_winner = dp[dfs[-1]]
        dfs.pop()
        length-=1
print("Hermione" if dp[start]=="h" else "Ron")

**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 2
2 3
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the medium version of the problem. The difference in this version is that $$$t=1$$$ and we work on trees.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. The game consists of $$$t$$$ rounds, each of which starts with a stone on exactly one node, which is considered as activated. A move consists of picking an inactive neighbor of the node with a stone on it and moving the stone there (thus activating this neighbor). Ron makes the first move, after which he alternates with Hermione until no valid move is available. The player that cannot make a move loses the round. If both players play optimally, who wins each round of this game?

Note that all the rounds are played with the same tree; only the starting node changes. Moreover, after each round, all active nodes are considered inactive again.

Input: The first line contains integers $$$n$$$ ($$$2 \leq n \leq 2\times 10^5$$$), $$$t$$$ ($$$t=1$$$), the number of nodes in the tree and the number of rounds, respectively.

The next $$$n-1$$$ lines contain two integers $$$1 \leq u, v \leq n$$$ each, corresponding to an edge of the tree.

The next line contains $$$t$$$ integers $$$1 \leq u_1 , \dots, u_t \leq n$$$, corresponding to the node where the stone is initially put.

Output: The output consists of $$$t=1$$$ line which is either "Ron" or "Hermione".

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

Note:

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

public class code {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        PrintWriter pw=new PrintWriter(System.out);
        int t=1;
        while(t-->0){
            HashSet<Integer> h=new HashSet<>();
            long n=sc.nextInt();
            long s=n*(n+1)/2;
            int k=sc.nextInt();
            HashMap<Integer,HashSet<Integer>> x=new HashMap<>();
            for (int i=1;i<=n;i++) {
                x.put(i,new HashSet<>());
            }
            for (int i=0;i<n-1;i++) {
                int a=sc.nextInt();
                int b=sc.nextInt();
                x.get(a).add(b);
                x.get(b).add(a);
            }
            int c=sc.nextInt();

            int f=1;
            boolean bo=Ronwin(f,x,c,0);
            if (bo)
                pw.println("Ron");
            else
                pw.println("Hermione");

        }

        pw.flush();

    }

    public static boolean Ronwin(int f,HashMap<Integer,HashSet<Integer>> x,int p,int pre){
        HashSet<Integer> a=x.get(p);

        if (a.size()==1){
            return f==0;
        }
        boolean ans=false;
        for (int i:a){
            if (i!=pre) {
                ans = ans || Ronwin((f + 1) % 2, x, i, p);
            }
        }
        return ans;
    }

    public int the(ArrayList<Integer> x){
        int min=0;
        int s=0;
        int n=x.size();
        int[] l=new int[n/2+1];
        for (int i=2;i<n;i+=2){
            l[i/2]=l[i/2-1]+(x.get(i-1)-x.get(i-2));}
        int[] r=new int[n/2+1];
        for (int i=n-1;i>0;i-=2)
            r[i/2]=r[i];
        return 0;

    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can highlight the issues in the algorithm. The buggy code has a function `Ronwin` that recursively checks if Ron can win from a given node. The key points to consider are:

1. **Tree Structure**: The tree structure should be such that it can lead to a situation where the recursive calls might not handle all cases correctly.
2. **Initial Node**: The initial node should be chosen to maximize the failure probability of the faulty solution.

A good candidate for such a test input is a tree with a linear structure (a path graph), where the nodes are connected in a straight line. This structure can make it difficult for the recursive function to correctly determine the winner, especially if the starting node is in the middle or near the end of the path.

Let's create a test input with a linear tree of 5 nodes, and the starting node is the middle node (node 3).

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

This input will create a tree where nodes are connected as follows:
1 - 2 - 3 - 4 - 5

Starting from node 3, the game should be symmetric, and the optimal play should result in Hermione winning because she can always mirror Ron's moves. This test input should expose the bug in the recursive logic of the `Ronwin` function.