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

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

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

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

def readInt():
    return int(input())

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

def solve():
    # Read the number of nodes (n) and the number of rounds (t)
    n, t = readInts()

    # Initialize the tree as an adjacency list
    tree = [[] for _ in range(n+1)]

    # Read the edges and build the tree
    for _ in range(n-1):
        u, v = readInts()
        tree[u].append(v)
        tree[v].append(u)

    # Find the two leaves of the tree
    leaves = [i for i in range(1, n+1) if len(tree[i]) == 1]

    # Start from one leaf and traverse to find the center
    curr = leaves[0]
    par = [-1] * (n+1)

    # BFS to find the path from one leaf to the other
    while tree[curr][0] != leaves[1]:
        for v in tree[curr]:
            if par[v] == -1:
                par[v] = curr
        curr = par[curr]

    # Read the starting node for the stone
    s = readInt()

    # Determine the distance from the starting node to the center
    dist = 0
    while s != curr:
        s = par[s]
        dist += 1

    # Determine the winner based on the parity of the distance
    return "Ron" if dist % 2 == 0 else "Hermione"

print(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 create a challenging test input that exposes the bug in the given code, we need to focus on edge cases or scenarios that might cause the code to fail. Here are the steps to think through this:

1. **Understand the Bug**: The bug in the code is in the BFS part where it tries to find the path from one leaf to the other. The code assumes that the first neighbor of the current node is the next node in the path, which is not always true. This can lead to incorrect paths and distances.

2. **Identify Edge Cases**:
   - **Small Trees**: Test with the smallest possible tree (3 nodes).
   - **Long Paths**: Test with a tree that has a long path between the leaves.
   - **Symmetric Trees**: Test with a symmetric tree where the center is not immediately obvious.

3. **Construct the Test Case**:
   - For a small tree, the input is straightforward and should work correctly.
   - For a long path, we can create a linear tree with many nodes.
   - For a symmetric tree, we can create a tree with multiple levels of nodes.

Let's create a test case for a symmetric tree with 7 nodes:

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

Explanation:
- The tree is symmetric with the center at node 4.
- The leaves are at nodes 1 and 7.
- The starting node is 4, which is the center.

This test case should expose the bug because the code might incorrectly assume the path from one leaf to the other and calculate the wrong distance.

Here is the test input:

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


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

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
import sys
from collections import deque, defaultdict
from itertools import permutations

input = sys.stdin.readline

############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))

############ ---- Output Functions ---- ############
def outlt(a):
    return " ".join(map(str, a))

n, t = inlt()
adj = [-1 for i in range(n + 1)]

leafs = set()
for _ in range(n - 1):
    u, v = inlt()

    leafs.add(str(u))
    if str(v) in leafs:
        leafs.remove(str(v))

    adj[u] = v

curr = int(list(leafs)[0])
imap = [-1 for i in range(n + 1)]

for i in range(n):
    imap[curr] = i
    curr = adj[curr]

i = imap[inp()]

if (n - i) % 2 == 0 and i % 2 == 0:
    print("Hermoine")
else:
    print("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
4 1
1 2
2 3
3 4
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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

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

public class Codechef
{
    static Boolean color[];
    public static void func(List<List<Integer>> tree, int root, int vis[]){
        vis[root] = 1;
        for(int e : tree.get(root)){
            if(vis[e]==-1){
                func(tree, e, vis);
                boolean check = true;
                boolean end = false;
                for(int e1 : tree.get(e)){
                    if(e1!=root){
                        if(color[e1]!=null){
                            check &= color[e1];
                            end = true;
                        }
                    }
                }
                if(!end){
                    color[e] = false;
                }else{
                    color[e] = check;
                }
            }
        }
        // for(int e : tree.get(root)){
        //     if(vis[e]==-1){
        //         func(tree, e, vis);
        //     }
        // }
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
        Scanner scn = new Scanner(System.in);
        int n = scn.nextInt();
        int arr[][] = new int[n][2];
        int edge[] = new int[n+1];
        List<List<Integer>> tree = new ArrayList<>();
        for(int i = 0; i <= n; i++){
            tree.add(new ArrayList<Integer>());
        }
        for(int i = 0; i < n; i++){
            arr[i][0] = scn.nextInt();
            arr[i][1] = scn.nextInt();
            edge[arr[i][0]]++;
            edge[arr[i][1]]++;
            tree.get(arr[i][0]).add(arr[i][1]);
            tree.get(arr[i][1]).add(arr[i][0]);
        }

        int root = -1;
        for(int i = 0; i <= n; i++){
            if(edge[i]==2)root=i;
        }
        int vis[] = new int[n+1];
        Arrays.fill(vis, -1);
        color = new Boolean[n+1];
        Arrays.fill(color, null);
        func(tree, root, vis);

        if(tree.get(root).size()>=2&&!color[tree.get(root).get(0)]&&!color[tree.get(root).get(1)]){
            System.out.println("Ron");
        }else{
            System.out.println("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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
import java.util.*;

public class codeforces {

    static boolean dfs(int v ,int depth, List<List<Integer>> children,  int[] visited ) {
        visited[v] = 1 ;
        if(children.get(v).size() == 1) {
            return depth % 2 != 0;
        }

        boolean pos = false ;
        for(int c : children.get(v)) {
            if(visited[c] == 1) continue;
            pos = pos || dfs(c, depth+1, children, visited )  ;
        }

        return pos ;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int tc = 1 ;
        //        int tc = sc.nextInt();
        while (tc-- > 0) {
            int n = sc.nextInt();
            int rounds = sc.nextInt();

            List<List<Integer>> children = new ArrayList<>(); // List to store children for each vertex
            for (int i = 0; i <= n; i++) {
                children.add(new ArrayList<>());
            }

            int[] visited = new int[n+1] ;
            for (int i = 0; i < n-1; i++) {
                int p = sc.nextInt() , c = sc.nextInt() ;
                children.get(p).add(c) ;
                children.get(c).add(p) ;
            }

            int s = sc.nextInt() ;

            if(children.get(s).size() == 1) {
                if(n%2 == 0) System.out.println("Ron");
                else System.out.println("Hermonie");

                continue;
            }

            if(dfs(s,0,children, visited)) System.out.println("Ron");
            else System.out.println("Hermione");
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the tree structure and the starting node can lead to incorrect results. The bug in the code is that it incorrectly handles the case where the starting node is a leaf node. Specifically, it checks if the number of nodes `n` is even or odd to determine the winner, which is not always correct.

### Step-by-Step Analysis:
1. **Tree with Two Leaves**:
   - The tree is linear and has exactly two leaves.
   - The nodes are connected in a way that forms a path from one leaf to the other.

2. **Starting Node as a Leaf**:
   - If the starting node is a leaf, the first player (Ron) can only move to the next node.
   - The second player (Hermione) then has to move to the next node, and so on.
   - The player who reaches the other leaf node first wins.

3. **Correct Logic**:
   - If the starting node is a leaf and the number of nodes `n` is even, Ron will lose because he will be forced to make the last move.
   - If the starting node is a leaf and the number of nodes `n` is odd, Ron will win because Hermione will be forced to make the last move.

### Test Input:
Let's create a test input where the starting node is a leaf and the number of nodes is odd. This should correctly result in Ron winning, but the buggy code will incorrectly output Hermione.

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

### Explanation:
- **Tree Structure**: The tree has nodes 1, 2, and 3, connected as 1-2-3.
- **Starting Node**: The starting node is 1.
- **Number of Nodes**: There are 3 nodes, which is odd.
- **Expected Output**: Ron should win because he will be the one to make the last move to node 3.

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


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

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b)for(int i=a;i<b;i++)
#define repp(x,arr) for(auto& x:arr)

#define ll              long long
#define ull              unsigned long long
#define int ll
#define db              double
#define pb              push_back
#define all(v)             v.begin(),v.end()
#define F               first
#define S               second
#define mod            1000000007
#define popcount(x)     __builtin_popcountll(x)
#define pll             pair<ll,ll>
#define pii             pair<int,int>
#define nl              "\n"
#define PI              3.1415926536 //acos(-1.0)
#define eps 0.00000001
#define mpp                 make_pair
#define lb lower_bound
#define ub upper_bound
#define PQ priority_queue<pii> //by default provide largest element;
#define RS set <ll, greater <ll> >
#define PQS priority_queue<pll,vector<pll>,greater<pll>>
#define ZERO(a) memset(a,0,sizeof(a))
#define inf 0x3f3f3f3f

using vi = vector<int>;
using vii = vector<pii>;
using vb = vector<bool>;
using vl = vector<ll>;
using vs = vector<string>;
//const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!
//const int dxn[8]={1,0,-1,0,-1,1,-1,1},dyn[8]={0,1,0,-1,-1,1,1,-1};
template <typename A, typename B> A amax (A &a, B b){ if (b > a) a = b ; return a; }
template <typename A, typename B> A amin (A &a, B b){ if (b < a) a = b ; return a; }
template <typename A, typename B> A mod1 (A &a, B b){ a=(a%b+b)%b ; return a; }

int M[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
bool isleap(int by){return (by%4==0 && (by%100!=0 || by%400==0));};
inline int gcd(int x,int y){return (y==0)?x:gcd(y,x%y);}

inline int mul(int a, int b) { return (int) ((long long) (a%mod * b%mod) % mod);}
inline int add(int a,int b){return (a%mod+b%mod)%mod;}
inline int sub(int a,int b){return (a-b+mod)%mod;}
inline int binpow(int a,int b){
	a%=mod;int res=1;
	while(b){if(b&1)res=res*a%mod;a=a*a%mod;b/=2;}
	return res;
}
inline int inv(int a){return binpow(a,mod-2)%mod;}

vector<vector<int>>T;

void solve(){
	int n,t;cin>>n>>t;
	T.resize(n+1);
	rep(i,0,n-1){
		int a,b;cin>>a>>b;
		T[a].pb(b);
		T[b].pb(a);
	}
	int u;cin>>u;
	for(auto x:T[u]){
		if(T[x].size()&1){
			cout<<"Ron\n";return;
		}
	}
	cout<<"Hermione\n";
}

int32_t main()
{
 	#ifndef ONLINE_JUDGE
		freopen("input.txt", "r", stdin);
		freopen("output.txt", "w", stdout);
	#endif

   ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

   int T=1;//cin >> T;
   while(T--)solve();

    return 0;

    }

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider the logic of the game and the specific conditions under which the code might fail. The key points to focus on are:

1. **Tree Structure**: The tree should have exactly two leaves.
2. **Initial Node**: The initial node where the stone is placed.
3. **Node Degrees**: The code checks if the neighbor of the initial node has an odd number of neighbors to determine the winner. This check might be incorrect or incomplete.

### Step-by-Step Analysis

1. **Tree with Two Leaves**:
   - A simple tree with two leaves can be a linear chain. For example, a tree with 5 nodes can be represented as:
     ```
     1 - 2 - 3 - 4 - 5
     ```
   - In this tree, nodes 1 and 5 are the leaves.

2. **Initial Node**:
   - The initial node should be chosen in a way that maximizes the failure probability. For example, starting at node 3 (the middle node) might be interesting because it has two neighbors, both of which have an odd number of neighbors.

3. **Node Degrees**:
   - Node 3 has two neighbors (2 and 4), and both neighbors have an odd number of neighbors (1 and 1, respectively).
   - According to the code, it checks if the neighbor of the initial node has an odd number of neighbors. If so, it prints "Ron". Otherwise, it prints "Hermione".

### Challenging Test Input

Let's create a test input where the initial node is 3, and the tree is a linear chain with 5 nodes:

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

### Explanation

- **Input**:
  - `5 1`: 5 nodes, 1 round.
  - `1 2`: Edge between nodes 1 and 2.
  - `2 3`: Edge between nodes 2 and 3.
  - `3 4`: Edge between nodes 3 and 4.
  - `4 5`: Edge between nodes 4 and 5.
  - `3`: Initial node is 3.

- **Expected Output**:
  - The correct output should be "Hermione" because starting at node 3, both players have an equal number of moves, and Hermione will win if both play optimally.

- **Why This Input is Challenging**:
  - The code incorrectly prints "Ron" because it checks if the neighbor of the initial node has an odd number of neighbors, which is true for both neighbors of node 3. However, this check is not sufficient to determine the winner correctly.

### Final Test Input

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


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

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
// Source: https://usaco.guide/general/io

#include <bits/stdc++.h>
using namespace std;
#include<bits/stdc++.h>
#define ll long long
#define ui unsigned int
#define fai(i,n) for(ll i= 0; i<n; i++)
#define pb push_back
#define mp make_pair
#define bs(v,a) binary_search(v.begin(),v.end(),a)
#define sortt(v) sort(v.begin(),v.end())
#define F first
#define S second
const ll MOD = 1e9+7;
using namespace std;
std::string iTb(ll number) {
    std::bitset<64> bits(number); // Convert integer to 32-bit binary
    return bits.to_string(); // Convert to string
}
bool cC(const std::pair<int, int>& a, const std::pair<int, int>& b) {
    if (a.first == b.first) {
        return a.second > b.second; // Sort by second value in descending order if first values are equal
    }
    return a.first < b.first; // Sort by first value in ascending order
}
void vecin(vector<ll>& v,ll n)
{
	fai(i,n)
	{
		ll temp;
		cin>>temp;
		v.pb(temp);
	}
}
void vecin(vector<pair<ll,ll>>&v, ll n)
{
	fai(i,n)
	{
		ll temp;
		cin>>temp;
		v.pb(mp(temp,i));
	}
}
void preSum(vector<ll>&vs, vector<ll> v)
{
	fai(i,v.size())
	{
		if(i==0)
		{
			vs.pb(v[i]);
		}
		else
		{
			vs.pb(v[i]+vs[i-1]);
		}
	}
}
map<ll, ll> count_elements(vector<ll>& vec) {
    map<ll, ll> element_count;

    for (ll element : vec) {
        // If the element already exists in the map, increment its count
        if (element_count.find(element) != element_count.end()) {
            element_count[element]++;
        } else { // Otherwise, add the element to the map with count 1
            element_count[element] = 1;
        }
    }

    return element_count;
}
struct SvC {
    bool operator()(const std::pair<ll, long long>& a, const std::pair<ll, long long>& b) const {
        return a.second > b.second; // Sort in descending order of the second value
    }
};
ll cPs(ll a, ll b) {

    ll start = ceil(sqrt(a));  // Round up square root of lower bound
    ll end = floor(sqrt(b));   // Round down square root of upper bound
	ll ct=end - start;
	if(floor(sqrt(b))!=sqrt(b))
	ct++;
    return ct ;     // Count of perfect squares
}
bool c2cnds(ll a, ll b) {
    return (a < b) && ((a^b)<4); // Sort in descending order
}
ll gcd(ll a, ll b) {
    while (b != 0) {
        ll temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}
//CHK KARNA KI TEST CASES HAI KI NAHI!!
vector<ll> v;
vector<ll> generateBinaryNumbers(int n) {
    vector<ll> binaryNumbers;
    for (int i = 0; i < (1 << n); ++i) {
        // Convert the number to a binary string with exactly n digits
        if(i!=0&&i!=1)
		{bitset<5> binary(i);
        binaryNumbers.push_back(stoi(binary.to_string()));}
    }
    return binaryNumbers;
}
void solve() {
    ll n,t;
    cin>>n>>t;
    vector<ll> v[n+1],l;
    fai(i,n-1)
    {
        ll u,vs;
        cin>>u>>vs;
        v[u].pb(vs);
        v[vs].pb(u);
    }
    fai(i,n)
    {
        if(v[i].size()==1)
            l.pb(i);
    }
    vector<ll> ar(n);
    ll st=l[0];
    ar[0]=st;
    fai(i,n)
    {
        if(i==0)
        continue;
        ll temp=v[st][0];
        if(temp==st)
        temp=v[st][1];
        st=temp;
        ar[i]=st;
    }
    cin>>st;
    ll temp;
    fai(i,n)
    {
        if(st==ar[i])
        {
            temp=i;
            break;
        }
    }
    //cout<<temp<<endl;
    if(temp%2==1 || (n-1-temp)%2==1)
    cout<<"Ron\n";
    else
    cout<<"Hermione\n";

}
int main()
{
	ios_base::sync_with_stdio(false); cin.tie(NULL);
	int t;
	// cin>>t;
	// //v=generateBinaryNumbers(5);
	// while((t--)>0)
	// {
    //     //cout<<t<<endl;
	// 	solve();
 	// }
	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 consider edge cases and scenarios that can highlight the incorrect logic. Here are the steps to think through this:

1. **Understand the Bug**:
   - The provided code has a logic for determining the winner based on the position of the stone and the length of the path. It checks if the position of the stone is odd or if the remaining path length is odd.
   - The bug might be in how it handles the path and the positions of the nodes.

2. **Edge Cases**:
   - **Small Tree**: A tree with a small number of nodes, such as 3 or 4, can help in quickly validating the logic.
   - **Leaf Nodes**: The tree has exactly two leaves. We should ensure the test case includes a clear path from one leaf to another.
   - **Starting Node**: The starting node should be such that it can test the logic of the path length and the position of the stone.

3. **Construct the Test Case**:
   - **Tree Structure**: A simple path from one leaf to another.
   - **Starting Node**: A node that is not a leaf and is at an odd or even position to test the logic.

Given these considerations, let's construct a test case:

- **Tree with 4 nodes**:
  - Nodes: 1, 2, 3, 4
  - Edges: (1, 2), (2, 3), (3, 4)
  - Starting node: 2 (which is an even position in the path from 1 to 4)

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

This test case should expose the bug if the code incorrectly handles the path length and the position 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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
from math import inf
n, t = map(int, input().split())
adj_list = [[] for _ in range(n)]
for _ in range(n - 1):
    u, v = map(lambda x: x - 1, map(int, input().split()))
    adj_list[u].append(v)
    adj_list[v].append(u)
ends = []
visited = [False] * n
def dfs(src):
    visited[src] = True
    count = 0
    for adj in adj_list[src]:
        if not visited[adj]:
            dfs(adj)
    if count == 0:
        ends.append(src)
dfs(0)
if len(ends) == 1:
    ends.append(0)
def compute_end_dist(end, end_dist):
    for adj in adj_list[end]:
        if end_dist[adj] < inf:
            end_dist[adj] = end_dist[end] + 1
            compute_end_dist(adj, end_dist)
end0_dist = [inf] * n
end0_dist[ends[0]] = 0
compute_end_dist(ends[0], end0_dist)
end1_dist = [inf] * n
end1_dist[ends[1]] = 1
compute_end_dist(ends[1], end1_dist)
for _ in range(t):
    u = int(input())
    u -= 1
    if end0_dist[u] % 2 == 1 or end1_dist[u] % 2 == 1:
        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:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the code and identify potential edge cases or scenarios that could cause it to fail.

### Analysis of the Buggy Code

1. **Tree Construction and DFS**:
   - The code constructs the adjacency list and performs a DFS to find the two leaves of the tree.
   - The DFS function is supposed to count the number of children for each node, but it currently does not update the `count` variable.

2. **Distance Calculation**:
   - The code calculates the distance from each end node to all other nodes.
   - The distance calculation for `end1_dist` is incorrect because it initializes the distance of the second end node to 1 instead of 0.

3. **Winning Condition**:
   - The code determines the winner based on the parity of the distances from the starting node to the two end nodes.

### Edge Cases to Consider

1. **Tree with Only Two Nodes**:
   - This is the smallest valid tree and should be handled correctly.

2. **Tree with a Linear Structure**:
   - A linear tree (a path) can expose issues with the distance calculation.

3. **Tree with a Non-Linear Structure**:
   - A tree with a more complex structure can expose issues in the DFS and distance calculation.

### Challenging Test Input

Let's create a test input that includes a linear tree with an odd number of nodes. This will ensure that the distance from the starting node to one of the end nodes is odd, which should make Ron the winner.

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

### Explanation

- **Tree Structure**:
  - The tree is a linear path: `1 - 2 - 3 - 4 - 5`.
  - The two leaves are nodes 1 and 5.

- **Starting Node**:
  - The starting node is 3.

- **Expected Output**:
  - The distance from node 3 to node 1 is 2 (even).
  - The distance from node 3 to node 5 is 2 (even).
  - Since both distances are even, Hermione should win. However, the buggy code initializes `end1_dist[ends[1]]` to 1, which will make the distance from node 3 to node 5 odd (3), causing the code to incorrectly print "Ron".

This test input should expose the bug in the distance calculation and the incorrect initialization of the second end node's distance.


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

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
def solve():
    n, t = map(int, input().split())
    li = []
    hs = {}
    for i in range(n - 1):
        u, v = map(int, input().split())
        if u not in hs:
            li.append(u)
            hs[u] = 1
        if v not in hs:
            li.append(v)
            hs[v] = 1
    start = int(input())
    startIdx = li.index(start)
    if len(li[:startIdx]) % 2 == 0: return 'Hermione'
    if len(li[startIdx:])-1 % 2 ==0:return 'Hermione'
    return 'Ron'

print(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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class GameOnTree {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

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

        // String[] output = new String[t];

        ArrayList<Integer> tree = new ArrayList<>();

        for (int i = 0; i < n - 1; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();

            if (tree.contains(a)) {
                tree.add(tree.indexOf(a) + 1, b);
            }
            else if (tree.contains(b)) {
                tree.add(tree.indexOf(b) - 1, a);
            }
            else {
                tree.add(a);
                tree.add(b);
            }
        }

        // for (int x = 0; x < t; x++) {

        // }

        int start = sc.nextInt();

            if (tree.indexOf(start) % 2 == 0 && (tree.size() - tree.indexOf(start)) % 2 == 1) {
                System.out.println("Hermione");
            }
            else {
                System.out.println("Ron");
            }

        // for (int i = 0; i < output.length; i++) {
        //     System.out.println(output[i]);
        // }
    }
}

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

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

Think step by step.

Response:
```plaintext
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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.concurrent.atomic.AtomicReference;

public class Scratch {
    public static void main(String[] args) {
        final Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        long t = sc.nextLong();
        final Map<Integer, List<Integer>> tree = new HashMap<>();
        final List<Integer> cases = new ArrayList<>();
        final AtomicReference<Integer> s = new AtomicReference<>(-1);

        for (int j = 1; j < n; j++) {
            Integer u = sc.nextInt();
            Integer v = sc.nextInt();
            if (!tree.containsKey(u)) {
                tree.put(u, new ArrayList<>());
            }
            if (!tree.containsKey(v)) {
                tree.put(v, new ArrayList<>());
            }

            tree.get(u).add(v);
            tree.get(v).add(u);
        }
        for (int i = 0; i < t; i++) {
            cases.add(sc.nextInt());
        }

        if (n % 2 == 0) {
            for (int i = 0; i < t; i++) {
                System.out.println("ron");
            }
            return;
        }

        tree.forEach((k, v) -> {
            if (v.size() == 1) {
                s.set(k);
            }
        });

        final Map<Integer, Integer> indexLocator = new HashMap<>();
        dfs(s.get(), -1, tree, indexLocator, 1);
        cases.forEach(item -> {
            final Integer index = indexLocator.get(item);
            if (index % 2 == 0) {
                System.out.println("Ron");
            } else {
                System.out.println("Hermione");
            }
        });
    }

    private static void dfs(final Integer current,
                            final Integer previous,
                            final Map<Integer, List<Integer>> tree,
                            final Map<Integer, Integer> indexLocator,
                            final Integer index) {
        indexLocator.put(current, index);
        tree.get(current).forEach(v -> {
            if (v == previous) return;

            dfs(v, current, tree, indexLocator, index + 1);
        });
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider the following aspects:

1. **Tree Structure**: The tree should have exactly two leaves, and it should be structured in a way that the buggy logic might fail.
2. **Starting Node**: The starting node should be chosen to maximize the chances of exposing the bug.
3. **Odd and Even Lengths**: The tree should have a length that is either odd or even to test the logic that determines the winner based on the parity of the node indices.

The buggy code has a critical flaw in its logic for determining the winner. Specifically, it assumes that if the number of nodes `n` is even, Ron always wins. This is incorrect because the winner depends on the structure of the tree and the starting node.

Let's create a test case where the tree has an even number of nodes, but the starting node is such that Hermione should win.

### Test Case
- **Tree Structure**: A linear tree with 4 nodes (2, 3, 4, 5).
- **Starting Node**: Node 5.

### Explanation
- The tree is a linear chain: 2 - 3 - 4 - 5.
- The starting node is 5, which is a leaf.
- In a linear tree, if the starting node is a leaf, the first player (Ron) will lose because the second player (Hermione) can always force the first player to move to the only available neighbor, and then the second player will have no valid moves left.

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

This test input should expose the bug in the provided code, as the code incorrectly determines that Ron wins when the number of nodes is even, regardless of the starting node. In this case, Hermione should win.


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

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
import sys
sys.setrecursionlimit(150005)
def root(pos,prev):
    global rg
    for i in g[pos]:
        if(i!=prev):
            rg[pos].append(i)
            root(i,pos)

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

pos = int(input())
rg = [[] for _ in range(n+1)]
root(pos,None)

def check(pos,turn):
    if len(rg[pos])==0:
        return not turn
    elif turn:
        return any([check(i,not turn) for i in rg[pos]])
    else:
        return all([check(i,not turn) for i in rg[pos]])

if(check(pos,True)):
    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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
import java.util.*;

public class Main {

    static class Edge {
        public int u;
        public int v;

        public Edge(int u, int v) {
            this.u = u;
            this.v = v;
        }
    };

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);

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

        List<Edge> edgeList = new ArrayList<>();
        for (int i = 0; i < n - 1; i++) {
            int u = scanner.nextInt();
            int v = scanner.nextInt();
            edgeList.add(new Edge(u,v ));
            edgeList.add(new Edge(v, u ));
        }

        for (int i = 0; i < t; i++) {
            int start = scanner.nextInt();
            Set<Integer> vis = new HashSet<>();
            List<Integer> paths = new ArrayList<>();
            dfs(start, edgeList, paths, 0, vis);
            boolean flag = false;
            for (Integer path : paths) {
                if (path % 2 != 0) {
                    flag = true;
                    break;
                }
            }
            System.out.println(flag  ? "Ron" : "Hermione");
        }
    }

    private static void dfs(int root, List<Edge> edgeList, List<Integer> paths, int len, Set<Integer> vis) {
        if (vis.contains(root)) {
            return;
        }
        for (Integer path : paths) {
            if (path % 2 != 0) {
                return;
            }
        }
        boolean flag = false;
        vis.add(root);
        for (Edge e : edgeList) {
            if (e.u == root && !vis.contains(e.v)) {
                flag = true;
                dfs(e.v, edgeList, paths, len + 1, vis);
            }
        }
        if (!flag) {
            paths.add(len);
        }
    }

}

**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 the following aspects:

1. **Tree Structure**: The tree should have exactly two leaves.
2. **Edge Cases**: Consider the smallest possible tree and trees with specific structures that might cause the DFS to behave unexpectedly.
3. **Starting Node**: Choose a starting node that might lead to incorrect path lengths being recorded.

Let's think through the steps:

1. **Smallest Tree**: The smallest tree with exactly two leaves is a tree with 3 nodes. This tree has the structure:
   ```
   1 - 2 - 3
   ```
   Here, nodes 1 and 3 are leaves.

2. **Starting Node**: The starting node should be the middle node (2) because this is the most straightforward case to verify.

3. **Input Format**: The input should follow the format specified in the problem description.

Given these considerations, the test input should be:

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

This input will test the following:
- The tree has exactly two leaves (1 and 3).
- The starting node is the middle node (2).
- The DFS should correctly determine the path lengths and the winner of the game.

This input is small and valid, making it easy to verify the behavior of the buggy code.


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

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
import java.util.Scanner;
import java.util.Map;
import java.util.Deque;
import java.util.HashMap;
import java.util.Set;
import java.util.ArrayDeque;
import java.util.HashSet;

public class B {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String[] initLine = scanner.nextLine().split(" ");
        int numOfNodes = Integer.valueOf(initLine[0]);
        int numOfRounds = Integer.valueOf(initLine[1]);
        Set<Integer> keyEdges = new HashSet<>();
        Set<Integer> valueEdges = new HashSet<>();
        HashMap<Integer, Integer> adjMap = new HashMap<>();
        int pathStart = -1;
        for (int i = 1; i < numOfNodes; i++) {
            String[] currentEdge = scanner.nextLine().split(" ");
            adjMap.put(Integer.valueOf(currentEdge[0]), Integer.valueOf(currentEdge[1]));
            valueEdges.add(Integer.valueOf(currentEdge[1]));
            keyEdges.add(Integer.valueOf(currentEdge[0]));
        }
        for (Integer key : keyEdges) {
            if (!valueEdges.contains(key)) {
                pathStart = key;
                break;
            }
        }
        Map<Integer, Integer> locationToPathLengthMap = new HashMap<>();
        for (int value : valueEdges) {
            locationToPathLengthMap.put(value, calculatePathLength(value, pathStart, adjMap));
        }
        String stoneLocationsString[] = scanner.nextLine().split(" ");
        for (String stoneLocation : stoneLocationsString) {
            int location = Integer.valueOf(stoneLocation);
            int pathLength = locationToPathLengthMap.containsKey(location) ? locationToPathLengthMap.get(location) : 0;
            if (pathLength % 2 == 1 || ((numOfNodes - pathLength) - 1) % 2 == 1) {
                System.out.println("Ron");
            } else {
                System.out.println("Hermione");
            }
        }
    }

    private static int calculatePathLength(int location, int pathStart, Map<Integer, Integer> adjMap) {
        int pathLength = 0;
        Deque<Integer> deque = new ArrayDeque<>();
        deque.add(pathStart);
        while (deque.size() > 0) {
            int currentNode = deque.removeLast();
            if (currentNode == location) {
                return pathLength;
            } else {
                pathLength++;
                deque.add(adjMap.get(currentNode));
            }
        }
        return pathLength;
    }
}

**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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

std::string game(std::vector<int> E, int root, int n) {
    int index = std::find(E.begin(), E.end(), root) - E.begin();

    if (index % 2 == 0 && (n - 1 - index) % 2 == 0) return "Hermione";
    if (index == n - 1) {
        if (index % 2 == 0) return "Hermione";
    }
    return "Ron";
}

int main() {
    int n, k;
    std::cin >> n >> k;
    std::vector<int> R;
    for (int i = 0; i < n - 1; ++i) {
        int a, b;
        std::cin >> a >> b;
        if (std::find(R.begin(), R.end(), a) == R.end()) R.push_back(a);
        if (std::find(R.begin(), R.end(), b) == R.end()) R.push_back(b);
    }
    int root;
    std::cin >> root;

    std::cout << game(R, root, n) << std::endl;

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Tree Structure**: The tree should have exactly two leaves, and it should be structured in a way that exposes the bug.
2. **Starting Node**: The starting node should be chosen to highlight the incorrect logic in the `game` function.

### Analysis of the Buggy Code
The `game` function has a few issues:
- It uses the index of the root node in the vector `R` to determine the winner, which is incorrect because the vector `R` does not represent the tree structure but rather the order in which nodes are encountered.
- The conditions for determining the winner are not correctly formulated.

### Challenging Test Input
We need a tree with a specific structure and a starting node that will highlight the incorrect logic. Let's consider a linear tree where the nodes are connected in a sequence, and the starting node is in the middle.

#### Example Tree:
- Nodes: 1, 2, 3, 4, 5
- Edges: (1, 2), (2, 3), (3, 4), (4, 5)
- Starting Node: 3

This tree has two leaves (1 and 5), and starting from node 3 should highlight the bug in the `game` function.

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

### Explanation
- **Tree Structure**: The tree is a simple linear chain from 1 to 5.
- **Starting Node**: Node 3 is the starting node.
- **Expected Output**: Hermione should win because the game is symmetric, and the player who starts from the middle node in a linear chain of odd length will lose if both players play optimally.

This input should expose the bug in the `game` function, as it incorrectly uses the index of the node in the vector `R` to 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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;
#define ull unsigned long long
#define ll long long
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define Int int t;cin>>t;while(t--)
#define F1(a,b,c) for(int (a)=(b);(a)<=(c);(a)++)
#define F2(a,b,c) for(int (a)=(b);(a)>=(c);(a)--)
#define ms(a,b) memset((a),(b),sizeof((a)))
#define mc(a,b) memcpy((a),(b),sizeof((a)))
#define l_b(a,b,c) lower_bound((a)+1,(a)+1+(b),(c))-(a)
inline ll Max(ll a, ll b) { return a > b ? a : b; }
inline ll Min(ll a, ll b) { return a > b ? b : a; }
/*-----------------------------------------------------------------*/
const int N = 2e5 + 1;
int nxt[N], head[N], to[N], cnt = 0, d[N], l[N];
inline void add(int u, int v) {
    to[++cnt] = v;
    nxt[cnt] = head[u];
    head[u] = cnt;
}
bool dfs(int u, int fa) {
    d[u] = d[fa] + 1;
    //cout <<u<<":" << d[u] << "\n";
    int ld = 0;
    bool e=1;
    for (int i = head[u]; i != -1; i = nxt[i]) {
        int v = to[i];
        if (v != fa) {
            ld += dfs(v, u);
            e = 0;
        }
    }
    if (e) {
        l[u] = 0;
        //cout << "l["<<u<<"]:" << l[u] << "\n";
        return 1;
    }
    l[u] = ld;
    //cout << "l[" << u << "]:" << l[u] << "\n";
    return l[u]?0:1;
}
bool dc(int u, bool da, bool mde) {
    //cout << u << " " << da << " " << mde << "\n";
    if (l[u] - da)return mde;
    if (u == 1)return mde==0;
    for (int i = head[u]; i != -1; i = nxt[i]) {
        int v = to[i];
        if (d[v] == d[u] - 1)return dc(v, l[u] ? 0 : 1, mde==0);
    }
}
int main() {
    ms(head, -1);
    ms(d, 0);
    IOS;
    int n, t; cin >> n >> t;
    n--;
    while (n--) {
        int u, v; cin >> u >> v;
        add(u, v); add(v, u);
    }
    dfs(1, 0);
    while (t--) {
        int k; cin >> k;
        if (dc(k, 0, 1))cout << "Ron\n";
        else cout << "Hermione\n";
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
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 (Easy)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the easy version of the problem. The difference in this version is that $$$t=1$$$ and we work on an array-like tree.

Ron and Hermione are playing a game on a tree of $$$n$$$ nodes that are initially inactive. This tree is special because it has exactly two leaves. It can thus be seen as an array. 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. It is guaranteed that the tree has exactly two leaves.

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:
3 1
2 3
3 1
3
output:
Ron

input:
5 1
1 2
2 3
3 4
4 5
5
output:
Hermione

Note:

Buggy Code:
def accept_numbers():
    numbers = []

    while True:
        nums = input().split()
        if len(nums) != 2:
            break

        num1 = int(nums[0])
        num2 = int(nums[1])

        numbers.append([num1, num2])
    return numbers

result = accept_numbers()
Ron_co=0
Hib_co=0
for sublist in result:
    num_1 = sublist[0]
    num_2 = sublist[1]
    if num_1 > num_2:
        Ron_co+=1
    elif num_1 == num_2:
        continue
    else:
        Hib_co+=1

if Ron_co>Hib_co:
    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: Bouquet (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference is that in this version, instead of listing the number of petals for each flower, the number of petals and the quantity of flowers in the store is set for all types of flowers.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ different types of flowers in the store, each of which is characterized by the number of petals and the quantity of this type of flower. A flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use to decorate her cake should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of types of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ different integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower type in the store (for different indexes $i \neq j$, it must be $a_i \neq a_j$). The third line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le 10^9$), where $c_i$ is the quantity of the $i$-th flower type in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, print one integer — the maximum possible number of petals in a bouquet that a girl can collect, observing all the conditions listed above.

Examples:
input:
7
3 10
1 2 3
2 2 1
3 1033
206 207 1000
3 4 1
6 20
4 2 7 5 6 1
1 2 1 3 1 7
8 100000
239 30 610 122 24 40 8 2
12 13123 112 1456 124 100 123 10982
6 13
2 4 11 1 3 5
2 2 1 2 2 1
8 10330
206 210 200 201 198 199 222 1000
9 10 11 12 13 14 15 16
2 10000000000
11 12
87312315 753297050
output:
7
1033
19
99990
13
10000
9999999999

Note: In the first test case, some valid bouquets are $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the second test case, you can assemble a valid bouquet with $(206, 206, 207, 207, 207)$ with a sum of $1033$, which is the maximum number of petals the girl can buy. In the third test case, you can assemble a valid bouquet with $(5, 5, 5, 4)$ with a sum of $19$. It can be seen that no valid bouquet can have $20$ petals.

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

using namespace std;

#define int long long

typedef long long ll;
typedef pair<int, int> PII;

#define eb emplace_back
#define ep emplace
#define fir first
#define sec second

#define lowbit(x) ((x) & (-(x)))

inline int read() {
	int x = 0, f = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) f -= (ch == '-') * 2;
	for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return x * f;
}

const int N = 200010;

int n, m, a[N], c[N];
PII tmp[N];

void solve() {
	n = read(), m = read();
	for (int i = 1; i <= n; i ++ ) a[i] = read();
	for (int i = 1; i <= n; i ++ ) tmp[i] = {a[i], read()};
	sort(tmp + 1, tmp + n + 1), sort(a + 1, a + n + 1);
	for (int i = 1; i <= n; i ++ ) c[i] = tmp[i].sec;
	//for (int i = 1; i <= n; i ++ ) printf("%lld %lld\n", a[i], c[i]);
	int ans = 0;
	for (int i = 1; i <= n; i ++ ) {
		int cnt = min(m / a[i], c[i]);
		ans = max(ans, cnt * a[i]);
	}
	for (int i = 2; i <= n; i ++ ) {
		int u = a[i - 1], v = a[i];
		if (v - u > 1) continue;
		//printf("%lld %lld\n", u, v);
		if (u * c[i - 1] <= m) {
			int rem = m - u * c[i - 1];
			int cnt = min(rem / v, c[i]);
			int val = min(rem - cnt * v, c[i] - cnt);
			ans = max(ans, u * c[i - 1] + cnt * v + val);
		}
		else {
			int cntu = min(m / u, c[i - 1]);
			int cntv = min({m % u, c[i], cntu});
			ans = max(ans, u * cntu + cntv);
		}
	}
	printf("%lld\n", ans);
}

signed main() {
	int _ = read();
	while (_ -- ) solve();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input for the given problem, we need to focus on edge cases that can expose the bug in the provided code. Here are the steps to consider:

1. **Edge Case with Limited Coins**: Ensure that the number of coins is just enough to buy a few flowers, but not enough to buy the maximum possible number of petals.
2. **Edge Case with Multiple Flower Types**: Include flower types with a small difference in the number of petals to test the logic for combining flowers.
3. **Edge Case with Large Quantities**: Include large quantities of flowers to test the limits of the code's ability to handle large numbers.
4. **Edge Case with Multiple Test Cases**: Include multiple test cases to ensure the code handles all cases correctly.

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

```plaintext
4
3 10
1 2 3
2 2 1
3 1033
206 207 1000
3 4 1
6 20
4 2 7 5 6 1
1 2 1 3 1 7
8 100000
239 30 610 122 24 40 8 2
12 13123 112 1456 124 100 123 10982
```

This input includes:
- A small number of flower types and coins to test basic functionality.
- A case where the number of coins is just enough to buy a few flowers.
- A case with a large number of coins and a mix of flower types.
- A case with multiple flower types and large quantities to test the handling of large numbers.

This input should help identify any issues in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference is that in this version, instead of listing the number of petals for each flower, the number of petals and the quantity of flowers in the store is set for all types of flowers.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ different types of flowers in the store, each of which is characterized by the number of petals and the quantity of this type of flower. A flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use to decorate her cake should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of types of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ different integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower type in the store (for different indexes $i \neq j$, it must be $a_i \neq a_j$). The third line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le 10^9$), where $c_i$ is the quantity of the $i$-th flower type in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, print one integer — the maximum possible number of petals in a bouquet that a girl can collect, observing all the conditions listed above.

Examples:
input:
7
3 10
1 2 3
2 2 1
3 1033
206 207 1000
3 4 1
6 20
4 2 7 5 6 1
1 2 1 3 1 7
8 100000
239 30 610 122 24 40 8 2
12 13123 112 1456 124 100 123 10982
6 13
2 4 11 1 3 5
2 2 1 2 2 1
8 10330
206 210 200 201 198 199 222 1000
9 10 11 12 13 14 15 16
2 10000000000
11 12
87312315 753297050
output:
7
1033
19
99990
13
10000
9999999999

Note: In the first test case, some valid bouquets are $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the second test case, you can assemble a valid bouquet with $(206, 206, 207, 207, 207)$ with a sum of $1033$, which is the maximum number of petals the girl can buy. In the third test case, you can assemble a valid bouquet with $(5, 5, 5, 4)$ with a sum of $19$. It can be seen that no valid bouquet can have $20$ petals.

Buggy Code:
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define endl '\n'
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define int long long
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N=1e6+10;
const double eps=1e-8;
const int inf=1e18;
const int mod=998244353;
void solve() {
    int n,m;cin>>n>>m;
    vector<int>a(n),b(n);
    unordered_map<int,int>mp;
    int mx=0;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    for(int i=0;i<n;i++){
        cin>>b[i];
        mp[a[i]]=b[i];
    }
    sort(a.begin(),a.end());
    //a.erase(unique(a.begin(),a.end()),a.end());
    for(int i=0;i<n;i++){
        if(i!=n-1){
            if(a[i]<=m){
                int num=min(m/a[i],mp[a[i]]);
                mx=max(mx,num*a[i]);
            }
            if(abs(a[i+1]-a[i])<=1&&a[i+1]<=m){
                int num1=min(m/a[i+1],mp[a[i+1]]);
                int tem=m-a[i+1]*num1;
                int num2=min(tem/a[i],mp[a[i]]);
                mx=max(mx,a[i+1]*num1+a[i]*num2);
                num1=min(m/a[i],mp[a[i]]);
                tem=m-a[i]*num1;
                num2=min(tem/a[i+1],mp[a[i+1]]);
                mx=max(mx,a[i]*num1+a[i+1]*num2);
                int x=m%a[i];
                num1=m/a[i]-x;
                num2=x;
                if(num1>mp[a[i]]&&num2>mp[a[i+1]]){
                    if(mp[a[i]]*a[i]+mp[a[i+1]]*a[i+1]<=m) mx=max(mx,mp[a[i]]*a[i]+mp[a[i+1]]*a[i+1]);
                }
                if(num1<=mp[a[i]]&&num2<=mp[a[i+1]]){
                    if(num1*a[i]+num2*a[i+1]<=m) mx=max(mx,num1*a[i]+num2*a[i+1]);
                }else if(num1>mp[a[i]]){
                    int l=m-mp[a[i]]*a[i];
                    l/=a[i+1];l=min(l,mp[a[i+1]]);
                    if(mp[a[i]]*a[i]+l*a[i+1]<=m) mx=max(mx,mp[a[i]]*a[i]+l*a[i+1]);
                }else{
                    int l=m-mp[a[i+1]]*a[i+1];
                    l/=a[i];l=min(l,mp[a[i]]);
                    if(mp[a[i+1]]*a[i+1]+l*a[i]<=m) mx=max(mx,mp[a[i+1]]*a[i+1]+l*a[i]);
                }
            }
        }else{
            if(a[i]<=m){
                int num=min(m/a[i],mp[a[i]]);
                mx=max(mx,num*a[i]);
            }
        }
    }
    cout<<mx<<endl;
}
signed main(){
    IOS;
    int T=1;
    cin>>T;
    while(T--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference is that in this version, instead of listing the number of petals for each flower, the number of petals and the quantity of flowers in the store is set for all types of flowers.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ different types of flowers in the store, each of which is characterized by the number of petals and the quantity of this type of flower. A flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use to decorate her cake should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of types of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ different integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower type in the store (for different indexes $i \neq j$, it must be $a_i \neq a_j$). The third line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le 10^9$), where $c_i$ is the quantity of the $i$-th flower type in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, print one integer — the maximum possible number of petals in a bouquet that a girl can collect, observing all the conditions listed above.

Examples:
input:
7
3 10
1 2 3
2 2 1
3 1033
206 207 1000
3 4 1
6 20
4 2 7 5 6 1
1 2 1 3 1 7
8 100000
239 30 610 122 24 40 8 2
12 13123 112 1456 124 100 123 10982
6 13
2 4 11 1 3 5
2 2 1 2 2 1
8 10330
206 210 200 201 198 199 222 1000
9 10 11 12 13 14 15 16
2 10000000000
11 12
87312315 753297050
output:
7
1033
19
99990
13
10000
9999999999

Note: In the first test case, some valid bouquets are $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the second test case, you can assemble a valid bouquet with $(206, 206, 207, 207, 207)$ with a sum of $1033$, which is the maximum number of petals the girl can buy. In the third test case, you can assemble a valid bouquet with $(5, 5, 5, 4)$ with a sum of $19$. It can be seen that no valid bouquet can have $20$ petals.

Buggy Code:
#include<bits/stdc++.h>
#include<queue>
#include<map>
typedef long long ll;
const int MAX=1e6;
using namespace std;
priority_queue<int> p;
ll arr[MAX];
struct nod{
	ll flower;
	ll number;
}ji[MAX];
bool cmp(struct nod a,struct nod b){
	return a.flower<b.flower;
}
int main(){
	ll t,b,n,m;
	cin>>t;
	while(t--){
		cin>>n>>m;
		ll maxsum=0;
		for(int i=1;i<=n;i++){
			cin>>ji[i].flower;
		}
		for(int i=1;i<=n;i++){
			cin>>ji[i].number;
			//pp[arr[i]] = b;
		}
		sort(ji+1,ji+n+1,cmp);
		//cout<<ji[2].flower<<" "<<ji[2].number<<"\n";
		for(int i=2;i<=n;i++){
			if(ji[i].flower-ji[i-1].flower==1){
				ll r = min(m/ji[i-1].flower,ji[i-1].number)*ji[i-1].flower,x=0;
				//ll r1 = min(m/ji[i].flower,ji[i].number)*ji[i].flower;
				x=ji[i].number-min((m-r)/ji[i].flower,ji[i].number);
				r+= min((m-r)/ji[i].flower,ji[i].number)*ji[i].flower;

				//if(r>m)r-=ji[i].flower;
				ll min1 =  min(min(m/ji[i-1].flower,ji[i-1].number),x);
				//cout<<min1<<" "<<r<<endl;
				r=min(m,r+min1);

				maxsum =  max(maxsum,r);
				//cout<<r<<"\n";
			}
			else{
				ll r =  ji[i].flower*min(m/ji[i].flower,ji[i].number);
				maxsum = max(maxsum,r);
			}
		}
		cout<<maxsum<<"\n";
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3 10
1 2 3
2 2 1
3 10
1 2 3
1 1 1
6 20
4 2 7 5 6 1
1 2 1 3 1 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Bouquet (Hard Version)
time_limit_ms: 1500
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference is that in this version, instead of listing the number of petals for each flower, the number of petals and the quantity of flowers in the store is set for all types of flowers.

A girl is preparing for her birthday and wants to buy the most beautiful bouquet. There are a total of $n$ different types of flowers in the store, each of which is characterized by the number of petals and the quantity of this type of flower. A flower with $k$ petals costs $k$ coins. The girl has decided that the difference in the number of petals between any two flowers she will use to decorate her cake should not exceed one. At the same time, the girl wants to assemble a bouquet with the maximum possible number of petals. Unfortunately, she only has $m$ coins, and she cannot spend more. What is the maximum total number of petals she can assemble in the bouquet?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. This is followed by descriptions of the test cases.

The first line of each test case contains two integers $n$, $m$ ($1 \le n \le 2 \cdot 10^5, 1 \le m \le 10^{18}$) — the number of types of flowers in the store and the number of coins the girl possesses, respectively. The second line of each test case contains $n$ different integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the number of petals of the $i$-th flower type in the store (for different indexes $i \neq j$, it must be $a_i \neq a_j$). The third line of each test case contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le 10^9$), where $c_i$ is the quantity of the $i$-th flower type in the store.

The sum of $n$ over all test cases does not exceed $2 \cdot {10}^5$.

Output: For each test case, print one integer — the maximum possible number of petals in a bouquet that a girl can collect, observing all the conditions listed above.

Examples:
input:
7
3 10
1 2 3
2 2 1
3 1033
206 207 1000
3 4 1
6 20
4 2 7 5 6 1
1 2 1 3 1 7
8 100000
239 30 610 122 24 40 8 2
12 13123 112 1456 124 100 123 10982
6 13
2 4 11 1 3 5
2 2 1 2 2 1
8 10330
206 210 200 201 198 199 222 1000
9 10 11 12 13 14 15 16
2 10000000000
11 12
87312315 753297050
output:
7
1033
19
99990
13
10000
9999999999

Note: In the first test case, some valid bouquets are $(1, 1, 2, 2), (2, 2, 3), (1, 1), (2, 2)$. The maximum over all valid bouquets not greater than $10$ is $7$ for $(2, 2, 3)$. In the second test case, you can assemble a valid bouquet with $(206, 206, 207, 207, 207)$ with a sum of $1033$, which is the maximum number of petals the girl can buy. In the third test case, you can assemble a valid bouquet with $(5, 5, 5, 4)$ with a sum of $19$. It can be seen that no valid bouquet can have $20$ petals.

Buggy Code:
# Creates a sorted dictionary (sorted by key)
from collections import OrderedDict

def listToString(s):

    # initialize an empty string
    str1 = ""

    # traverse in the string
    for ele in s:
        str1 += str(ele) + ","

    # return string
    return str1

aks = 0
t = int(input())

for i in range(t) :
    n,m = list(map(int,input().split()))
    v =  list(map(int,input().split()))
    c =  list(map(int,input().split()))

    aks += 1
    if aks ==  2876 :
        print(m)

    mp = {}
    for i,x in enumerate(v) :
        mp[x] = c[i]

    mp = dict(OrderedDict(sorted(mp.items())))
    # print(listToString(v))
    # print(mp)

    ans = 0
    l = m
    r = m

    while l <= r :
        mid = (l+r) // 2
        # print("mid ", mid)

        prev_key = -1
        prev_value = -1
        temp = -1
        for key, value in mp.items() :
            # print(key, prev_key)

            if key - prev_key != 1 :
                # fin = key*value
                # print("infin ", fin)

                mul = mid//key
                mul = min(value, mul)
                req = mul*key
                # print("inreq ", req)
                temp = max(temp,req)
                # temp = max(temp,fin)
                # print("intemp ", temp)

                prev_key = key
                prev_value = value
                continue

            have = prev_value + value
            exp = prev_key* have + value

            # print("have exp", have, exp)

            if exp <= mid : temp = max(temp, exp)
            else :
                rem = exp - mid
                m = rem//prev_key
                # print("m ", m)
                prev_value_upd = prev_value - m
                have = prev_value_upd + value
                exp = prev_key* have + value
                # print("have exp", have, exp)
                if exp <= mid : temp = max(temp, exp)
                else :
                    rem = exp - mid
                    if rem <= m : temp = max(temp, exp-rem)
                    else :
                        # print("mar gaya")
                        # rem = exp - mid
                        m += 1
                        # print("m ", m)
                        prev_value_upd = prev_value - m
                        have = prev_value_upd + value
                        exp = prev_key* have + value
                        # print("have exp", have, exp)
                        if exp <= mid : temp = max(temp, exp)
                        else :
                            rem = exp - mid
                            if rem <= m : temp = max(temp, exp-rem)
                            else : print("mar gaya 2")

            prev_key = key
            prev_value = value

            # ct = 0
            # if prev_key : ct = mid//prev_key
            # ct = min(prev_value+value,ct)
            # req = ct*prev_key
            # print("main req ", req)
            # if mid < req :
            #     m = mid//prev_key
            #     m = min(prev_value,m)
            #     t = m*prev_key
            #     remaining = mid - t
            #     c = remaining//key
            #     c = min(key,c)
            #     add = c*key
            #     t += add
            #     print("t ", t)
            #     temp = max(temp, t)
            # else :
            #     rem = mid- req
            #     print("req rem", req, rem)
            #     print("value ", value)
            #     if rem <= value :
            #         print("ch ", ct, prev_value)
            #         extra = max(0,ct - prev_value)
            #         if rem == 0 : rem += extra
            #         req += rem
            #         if req <= m : temp = max(temp,req)
            #     else :
            #         print("ch ", ct, prev_value)
            #         extra = max(0,ct - prev_value)
            #         req += extra
            #         if req <= m : temp = max(temp,req)
            #     print("temp ", temp)
            #     prev_key = key
            #     prev_value = value

        # print("fin temp ", temp)
        if temp >= mid :
            if temp == mid : ans = temp
            l = mid + 1
        else :
            ans = temp
            r = mid - 1

    print(ans)
    print("\n")

    # tot = 87312315 + 753297050
    # print(tot)
    # multi = 11*tot
    # print("mukti", multi)
    # rem = 10000000000 - multi
    # print(rem)

    # print(753297050 - 753296985)

    # print(10000000065 % 11)

    # print(65 % 11)

    # print(56 % 12)

#  1
# 1 1
# 5
# {1: 7, 2: 2, 4: 1, 5: 3, 6: 1, 7: 1}

"""
1
2 10000000000
11 12
87312315 753297050

"""

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

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

Think step by step.

Response:
```plaintext
1
2 10000000000
11 12
87312315 753297050
1 1
```