**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Valuable Cards
time_limit_ms: 4000
memory_limit_mb: 512
Description: In his favorite cafe Kmes once again wanted to try the herring under a fur coat. Previously, it would not have been difficult for him to do this, but the cafe recently introduced a new purchasing policy.

Now, in order to make a purchase, Kmes needs to solve the following problem: $n$ cards with prices for different positions are laid out in front of him, on the $i$-th card there is an integer $a_i$, among these prices there is no whole positive integer $x$.

Kmes is asked to divide these cards into the minimum number of bad segments (so that each card belongs to exactly one segment). A segment is considered bad if it is impossible to select a subset of cards with a product equal to $x$. All segments, in which Kmes will divide the cards, must be bad.

Formally, the segment $(l, r)$ is bad if there are no indices $i_1 < i_2 < \ldots < i_k$ such that $l \le i_1, i_k \le r$, and $a_{i_1} \cdot a_{i_2} \ldots \cdot a_{i_k} = x$.

Help Kmes determine the minimum number of bad segments in order to enjoy his favorite dish.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each set of input data gives you $2$ integers $n$ and $x$ ($1 \le n \le 10^5, 2 \le x \le 10^5$) — the number of cards and the integer, respectively.

The second line of each set of input data contains $n$ integers $a_i$ ($1 \le a_i \le 2 \cdot 10^5, a_i \neq x$) — the prices on the cards.

It is guaranteed that the sum of $n$ over all sets of test data does not exceed $10^5$.

Output: For each set of input data, output the minimum number of bad segments.

Examples:
input:
8
6 4
2 3 6 2 1 2
9 100000
50000 25000 12500 6250 3125 2 4 8 16
5 2
1 1 1 1 1
8 6
4 3 4 3 4 3 4 3
7 12
6 11 1 3 11 10 2
10 5
2 4 4 2 4 4 4 3 1 1
7 8
4 6 5 1 2 4 1
8 27
3 9 17 26 2 20 9 3
output:
3
2
1
1
2
1
3
3

Note:

Buggy Code:
//---------------JAI SRI RAM---------------//
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;

import javax.print.attribute.IntegerSyntax;

public class Main {
    static FastReader sc = new FastReader();
    static int mod=1000000007;
    static String debug="";
    static class pair {
        int a;
        int b;

        pair(int a, int b) {
            this.a = a;
            this.b = b;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o)
                return true;
            if (o == null || getClass() != o.getClass())
                return false;

            pair otherPair = (pair) o;
            return a == otherPair.a && b == otherPair.b;
        }
        @Override
        public int hashCode(){
            return Objects.hash(this.a,this.b);
        }

        public String toString(){
            return "pair(" + a + ", " + b + ")";
        }
    }
    public static void solve(int tt) {
        int n=sc.nextInt();
        int k=sc.nextInt();
        HashSet<Integer> factors=factors(k);
        int arr[]=arrInput(n);
        int count=0;
        HashSet<Integer> cur=new HashSet<>();
        cur.add(1);
        for(int i:arr){
            if(!factors.contains(i))continue;
            if(cur.contains(k/i)){
                cur.clear();
                cur.add(i);
                count++;
                cur.add(1);
                continue;
            }
            HashSet<Integer> nCur=new HashSet<>();
            for(int j:cur){
                nCur.add(j);
                nCur.add(j*i);
            }
            cur=nCur;
        }
        count++;
        System.out.println(count+debug);
    }
    public static boolean isPrime(long num) {
        if (num <= 1) {
            return false;
        }
        for (long i = 2; i * i <= num; ++i) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }
    public static long pow(long a,long b){
        long ans=1;
        while(b>0){
            if(b%2==1){
                ans=(a*ans)%mod;
            }
            b=b>>1;
            a=(a*a)%mod;
            // System.out.println(ans+"=====");
        }
        return ans%mod;
    }
    public static void main(String[] args) {
        if(args.length>0){
            debug = "----";
            System.out.println("offline mode");
        }
        int t =sc.nextInt();
        int i = 1;
        while (t-- > 0) {
            solve(i++);
            if(!debug.isEmpty())
            System.out.println(debug+"tc:"+(i-1));
        }
    }
    static long factorial(long n) {
        long result = 1;
        for (long i = 1; i <= n; i++) {
            result *= i;
            if (result > 998244353) {
                result -= 998244353;
            }
        }
        return result % 998244353;
    }

    static int modFact(int n, int p) {
        if (n >= p)
            return 0;

        int result = 1;
        for (int i = 1; i <= n; i++)
            result = (result * i) % p;

        return result;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

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

    public static int[] arrInput(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = sc.nextInt();
        }
        return arr;
    }

    public static void arrOutput(int[] arr) {
        StringBuilder out = new StringBuilder("");
        for (int i = 0; i < arr.length; i++) {
            out.append(arr[i] + " ");
        }
        System.out.println(out);
    }

    public static HashSet<Integer> factors(int n) {
        HashSet<Integer> h = new HashSet<Integer>();
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                h.add(i);
                h.add(n / i);
            }
        }
        return h;
    }

    public static ArrayList<Integer> primeFactorization(int numbers) {
        int n = numbers;
        ArrayList<Integer> factors = new ArrayList<Integer>();
        for (int i = 2; i <= n / i; i++) {
            while (n % i == 0) {
                factors.add(i);
                n /= i;
            }
        }
        if (n > 1) {
            factors.add(n);
        }
        return factors;
    }

    public static int calculate(int l, int r, int segTree[], int n) {
        l += n - 1;
        r += n - 1;
        int ans = segTree[r];
        while (l <= r) {
            if (r % 2 == 0) {
                ans &= segTree[r];
                r--;
            }
            if (l % 2 == 1) {
                ans &= segTree[l];
                l++;
            }
            r = r >> 1;
            l = l >> 1;
        }
        return ans;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The main issue in the buggy code seems to be the handling of factors and the way it determines if a segment is bad. Specifically, the code might not correctly handle cases where the product of multiple factors equals \( x \).

### Step-by-Step Analysis:
1. **Identify the Bug**: The code uses a `HashSet` to track the current products and checks if any of these products, when multiplied by the current card value, equals \( x \). If it does, it resets the `HashSet` and increments the count. This approach might fail if the factors of \( x \) are not correctly managed.

2. **Edge Cases**:
   - **Single Factor**: If \( x \) is a prime number, the only factor is \( x \) itself. The code should handle this correctly.
   - **Multiple Factors**: If \( x \) has multiple factors, the code might not correctly track the combination of factors.
   - **Repeating Factors**: If the same factor appears multiple times, the code might not handle this correctly.

3. **Challenging Input**:
   - **Input with Multiple Factors**: Choose \( x \) to have multiple factors and ensure the input contains a combination of these factors.
   - **Input with Repeating Factors**: Ensure the input contains the same factor multiple times to test the handling of repeated factors.
   - **Input with Non-Factors**: Include some values that are not factors of \( x \) to ensure the code handles non-factors correctly.

### Challenging Test Input:
```plaintext
1
6 12
2 3 4 6 1 1
```

### Explanation:
- **\( x = 12 \)**: The factors of 12 are 1, 2, 3, 4, 6, and 12.
- **Input Values**:
  - `2`: A factor of 12.
  - `3`: A factor of 12.
  - `4`: A factor of 12.
  - `6`: A factor of 12.
  - `1`: Not a factor of 12.
  - `1`: Not a factor of 12.

This input will test the code's ability to handle multiple factors and the combination of factors that can form 12. The expected output for this input should be 1, as all cards can be combined to form 12 in a single segment. If the code fails to handle this correctly, it will expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
#include <vector>
#include <iostream>
#include <algorithm>
#include <set>
#include <string>
#include <sstream>
#include <map>
#include <iterator>
#include <cmath>
#include <queue>
#include <unordered_map>
#include <algorithm>
#include <cmath>

#define ll long long
#define NE (int)(2e5 + 5)

using namespace std;

void cnt_dfs(vector<vector<int>> &gp, vector<int>& cnt, int u, int p = -1) {
    for (auto v : gp[u]) {
        if (v == p) continue;
        cnt_dfs(gp, cnt, v, u);
        if (cnt[v]) cnt[v]++;
    }
}

void change_root(vector<vector<int>>& gp, vector<int>& cnt, vector<int>& ans, int u, int p = -1) {
    if (cnt[u]) ans[u] = 1;
    for (auto& v : gp[u]) {
        if (v == p) continue;
        if (cnt[v] == 0) cnt[u]--;
        if (cnt[u] == 0) cnt[v]++;
        change_root(gp, cnt, ans, v, u);
        if (cnt[v] == 0) cnt[u]++;
        if (cnt[u] == 0) cnt[v]--;
    }
}

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

    vector<int> cnt(NE);
    vector<int> ans(NE);

    cnt_dfs(gp, cnt, 0);
    change_root(gp, cnt, ans, 0);

    string anss[2] = {"Ron", "Hermione"};

    for (int i = 0; i < t; i++) {
        int v; cin >> v;
        v--;
        cout << anss[ans[v]] << endl;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //int t; 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 test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are some potential edge cases to consider:

1. **Tree with a single branch (linear tree):** This can test the logic for counting and propagating moves.
2. **Tree with a central node and multiple branches:** This can test how the code handles a node with multiple neighbors.
3. **Tree with a cycle (though it's not a valid tree, it can help identify issues in the logic):** This can help identify if the code incorrectly handles cycles.
4. **Tree with a large number of nodes and edges:** This can test the performance and correctness of the algorithm under more complex conditions.

Given the constraints and the need for a small-scale test input, let's focus on a tree with a central node and multiple branches. This will help us test the logic for counting and propagating moves in a more complex structure.

Here's a small-scale test input:

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

**Explanation:**
- The tree has 7 nodes.
- Node 1 is the central node connected to nodes 2, 3, and 4.
- Nodes 2, 3, and 4 each have one child node (5, 6, and 7 respectively).
- The game is played for 3 rounds starting from nodes 1, 2, and 3.

This input will test how the code handles a central node with multiple branches and how it propagates moves correctly. If the code has bugs in handling such structures, this input should expose them.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

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

using namespace std;

#define int long long
#define vi vector<ll>
#define vvi vector<vi>
#define pll pair<ll, ll>
#define vp vector<pll>
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define MOD 1000000007
#define MOD1 998244353
#define INF 1e18
#define endl "\n"
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define PI 3.141592653589793238462
#define set_bits __builtin_popcountll
#define forrr(i,s,e) for(ll i = s; i >= e; i--)
#define forr(i,s,e) for(ll i = s; i <= e; i++)
#define fr(i,s,e) for(ll i = s; i < e; i++)
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
// typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key

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

void _print(ll t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

int random(int s, int e){return s + rand() % (e - s + 1);}
int flip(int n){ forr(i,0,31){ int num = 1<<i; if(num > n) break; n = n xor num;} return n; }
ll power(ll x, ll y){ ll res = 1; while (y > 0){ if (y & 1) res = (ll)(res*x); y = y>>1; x = (ll)(x*x); } return res; }
vp countArr(vi &arr, ll n){ vp v1; forr(i,0,n-1){ ll cnt = 1; while(i < n-1 && arr[i] == arr[i+1]){ cnt++, i++; } v1.pb({arr[i], cnt}); } if((n > 1 && arr[n-1] != arr[n-2]) || n == 1) v1.pb({arr[n-1], 1}); return v1;}
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll lcm(ll a, ll b){return a*b/gcd(a,b);}
vi factors(ll n){ vi fac; for (ll i = 1; i * i <= n; i++){ if (n % i == 0){ fac.pb(i); if (i * i != n)fac.pb(n / i); } } return fac; }
ll maxPow2(ll n){n |= n >> 1; n |= n >> 2; n |= n >> 4; n |= n >> 8; n |= n >> 16; return (n + 1);}
vp primeFactors(ll n){ vp v; for (ll j = 2; j <= sqrtl(n); j++){ ll cnt = 0; while(n%j == 0) n /= j, cnt++; if(cnt)v.pb({j,cnt}); } if(n!=1) v.pb({n,1}); return v;}

const int N=2e5+10;
vi graph[N+1];

void dfs(int i,vi &jeetega,int parent){
    int ans=0;
    for(auto nbr: graph[i]){
        if(nbr == parent) continue;
        if(graph[nbr].size()==1){
            jeetega[nbr]=0;
            ans++;
        }
        else{
            if(jeetega[nbr]==-1)
                dfs(nbr,jeetega,i);
            if(jeetega[nbr]==0) ans++;
        }
    }
    jeetega[i]=ans;
}

void dfs2(int i,int parent,vi &jeetega,vi &visited2,vi &finalAns,int n){
    visited2[i]=1;

    int temp1=jeetega[i],temp2=jeetega[parent];

    if(jeetega[parent]>jeetega[i] && parent != 0) jeetega[i]++;
    jeetega[parent]-=jeetega[i];
    finalAns[i]=jeetega[i];

    for(auto nbr: graph[i]){
        if(visited2[nbr]) continue;
        dfs2(nbr,i,jeetega,visited2,finalAns,n);
    }

    jeetega[i]=temp1;
    jeetega[parent]=temp2;
}

void solve(ll tc){
    ll n,t;cin>>n>>t;

    fr(i,0,n-1){
        int a,b;cin>>a>>b;
        graph[a].pb(b);
        graph[b].pb(a);
    }
    vi jeetega(n+1,-1),ans(n+1,-1),v2(n+1,0);

    dfs(1,jeetega,0);
    dfs2(1,0,jeetega,v2,ans,n);

    debug(ans);
    while(t--){
        int x;cin>>x;
        if(ans[x]) cout<<"Ron"<<endl;
        else cout<<"Hermione"<<endl;
    }
}

signed main() {
    fastio();
    ll t = 1;
    // cin>>t;
    fr(i,0,t) solve(i);
    return 0;
}

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

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

Think step by step.

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

1. **Tree Structure**: The tree should have a structure that can lead to different outcomes depending on the starting node.
2. **Starting Nodes**: The starting nodes should be chosen to test different scenarios, including leaf nodes, nodes with multiple children, and the root node.
3. **Edge Cases**: Consider edge cases such as a tree with a single branch, a tree with a balanced structure, and a tree with a long path.

Given the constraints and the need for a small-scale test input, let's create a test case with a simple but challenging tree structure.

### Test Case Description
- **Tree Structure**: A tree with 6 nodes and a structure that includes a root node with two branches.
- **Starting Nodes**: We will test starting nodes that are the root, a leaf node, and a node with multiple children.

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

### Explanation
- **Node 1**: Root node with two children (2 and 3).
- **Node 2**: Child of 1 with two children (4 and 5).
- **Node 3**: Child of 1 with one child (6).
- **Node 4**: Leaf node.
- **Node 5**: Leaf node.
- **Node 6**: Leaf node.

### Expected Output
- **Starting Node 1**: Ron (since Ron can move to either 2 or 3, and then Hermione will be forced to move to a leaf node).
- **Starting Node 4**: Hermione (since 4 is a leaf node and Ron cannot make a move).
- **Starting Node 6**: Hermione (since 6 is a leaf node and Ron cannot make a move).

This test case should expose the bug in the given code by testing different scenarios and ensuring that the code handles leaf nodes and nodes with multiple children correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
// Online C++ compiler to run C++ program online
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int f(int node,int parent,vector<int> adj[],int turn, map<pair<int,pair<int,int>>,int> &m){
    int ans = true;
    if(turn==0) ans = false;
    for(auto it : adj[node]){
        if(it==parent) continue;
        if(m[{turn,{node,it}}]!=-1){
            if(turn) ans = (ans&m[{turn,{node,it}}]);
            else ans = (ans|m[{turn,{node,it}}]);
            continue;
        }
        int l = f(it,node,adj,abs(turn-1),m);
        m[{turn,{node,it}}] = l;
        if(turn) ans = (ans&m[{turn,{node,it}}]);
        else ans = (ans|m[{turn,{node,it}}]);
    }
    return ans;
}
int main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n,t;
    cin>>n>>t;
    vector<int> adj[n+1];
    map<pair<int,pair<int,int>>,int> mp;
    vector<int> dp2(n+1,-1);
    for(int i = 0; i<n-1; i++){
        int a,b;
        cin>>a>>b;
        adj[a].push_back(b);
        adj[b].push_back(a);
        mp[{0,{a,b}}] = -1;
        mp[{1,{a,b}}] = -1;
        mp[{0,{b,a}}] = -1;
        mp[{1,{b,a}}] = -1;
    }
    while(t--){
        int a;
        scanf("%d", &a);
        int l = 9;
        if(dp2[a]!=-1) l = dp2[a];
        else{
            l = f(a,-1,adj,0,mp);
            dp2[a] = l;
        }
        if(l) 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 3
1 2
2 3
3 4
1 2 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 (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

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

def dfs(node, parent):
    # Initialize a flag to check if the current node is a leaf
    is_leaf = True
    # Traverse all neighbors of the current node
    for nei in adj[node]:
        if nei == parent:
            continue  # Skip the parent node to avoid cycles
        # Recursively check if the neighbor is a leaf
        is_leaf &= dfs(nei, node)
    # If the current node is a leaf, set the result for this node
    if is_leaf:
        ans[node] = True
    return is_leaf

# Read the number of nodes (n) and the number of rounds (t)
n, t = map(int, input().split())
# Initialize the adjacency list for the tree
adj = [[] for _ in range(n + 1)]

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

# Read the starting nodes for each round
start_nodes = list(map(int, input().split()))
# Initialize the answer list for each round
ans = [False] * (n + 1)

# Perform a DFS starting from node 1 (arbitrary starting point)
dfs(1, 0)

# Determine the winner for each round based on the starting node
for start in start_nodes:
    # If the starting node is a leaf, Ron wins
    print("Ron" if ans[start] else "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 3
1 2
2 3
2 4
1 2 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 (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
from collections import defaultdict
n, t = map(int, input().split())
graph = defaultdict(list)
for _ in range(n - 1):
    u, v = map(int, input().split())
    graph[u].append(v)
    graph[v].append(u)
startNode = list(map(int, input().split()))
def iterative_dfs(start, graph, visited):
    stack = [start]
    while stack:
        node = stack.pop()
        if node not in visited:
            visited.add(node)
            stack.extend(neighbor for neighbor in graph[node] if neighbor not in visited)
visited = set()
for node in startNode:
    if node not in visited:
        iterative_dfs(node, graph, visited)
for node in startNode:
    print("Ron" if node in visited else "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 3
1 2
2 3
2 4
1 2 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 (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
from collections import defaultdict
n, t = map(int, input().split())
graph = defaultdict(list)
startNode = list(map(int, input().split()))
for i in range(n - 1):
    u, v = map(int, input().split())
    graph[u].append(v)
    graph[v].append(u)
results = {}
def dfs(node, vis, component):
    vis.add(node)
    if node in startNode:
        results[node] = component
    for v in graph[node]:
        if v not in vis:
            dfs(v, vis, component)
vis = set()
component = 0
for i in range(1, n + 1):
    if i not in vis:
        dfs(i, vis, component)
        component += 1
base_component = results[startNode[0]]
for node in startNode:
    print("Ron" if results[node] == base_component else "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 create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that might cause the code to fail. The buggy code incorrectly assumes that the game outcome depends on the component (connected component) in which the starting node lies, rather than considering the structure of the tree and the optimal play strategy.

### Key Points to Consider:
1. **Tree Structure**: The tree structure can significantly affect the game outcome. For example, a tree with a long path or a tree with multiple branches.
2. **Starting Nodes**: The choice of starting nodes can influence the game outcome. Specifically, starting nodes that are close to each other or far apart can create different scenarios.
3. **Optimal Play**: The code does not account for optimal play strategies, which can lead to incorrect results.

### Challenging Test Input:
Let's create a test input with a tree that has a long path and a starting node that is not in the same component as the first starting node. This will expose the bug in the code.

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

### Explanation:
- **Tree Structure**: The tree is a simple path from node 1 to node 5.
- **Starting Nodes**: The starting nodes are 1, 3, and 5.
- **Expected Output**:
  - For node 1: Ron (since Ron can move to node 2 and force Hermione to move to node 3, and so on).
  - For node 3: Hermione (since both players can only move to nodes 2 or 4, and Hermione can force Ron to lose).
  - For node 5: Ron (since Ron can move to node 4 and force Hermione to move to node 3, and so on).

This test input will expose the bug in the code because the code incorrectly assumes that the game outcome depends on the component, which is not true in this case. The correct output should be:
```
Ron
Hermione
Ron
```

However, the buggy code will output:
```
Ron
Ron
Ron
```

This test input should effectively demonstrate the failure of the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
import java.util.*;

public class gameOnTree {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt();
        int t=sc.nextInt();

        HashMap<Integer , ArrayList<Integer>> map = new HashMap<>();
        for(int i=1;i<=n;i++){
            map.put(i , new ArrayList<>());
        }
        for(int i=0;i<n-1;i++){
            int u=sc.nextInt();
            int v=sc.nextInt();
            map.get(u).add(v);
            map.get(v).add(u);
        }
        int[][] ans = new int[n+1][2];
        int[][] upAns = new int[n+1][2];
        int[][] count = new int[n+1][2];
        dfs1(1,-1,ans,count,map);
        dfs2(1,-1,ans,count,map);

        while(t-->0){
            int node = sc.nextInt();
            if(ans[node][0]==1) System.out.println("Ron");
            else System.out.println("Hermione");
        }
    }

    public static void dfs1(int node, int parent , int[][] ans , int[][] count , HashMap<Integer , ArrayList<Integer>> map){

        int check1=0;
        int check2=0;

        for(int nb : map.get(node)){
            if(nb==parent) continue;
            dfs1(nb , node , ans , count , map);
            if(ans[nb][0]==2) check2++;
            if(ans[nb][1]==1) check1++;
        }

        if(check1 > 0) ans[node][0]=1;
        else ans[node][0]=2;

        if(check2 > 0) ans[node][1]=2;
        else ans[node][1] = 1;

        count[node][0] = check1;
        count[node][1] = check2;
    }

    public static void dfs2(int node ,  int parent , int[][] ans , int[][] count  , HashMap<Integer , ArrayList<Integer>> map){
        if(parent!=-1){

            // if(ans[parent][0]==2) ans[node][1]=2;
            // if(ans[parent][1]==1) ans[node][0]=1;

            int check1 = ans[parent][0];
            int check2 = ans[parent][1];

            if(count[parent][0]==1 && ans[node][1]==1) check1=2;
            if(count[parent][1]==1 && ans[node][0]==2) check2=1;

            if(check1==2) ans[node][1]=2;
            if(check2==1) ans[node][0]=1;

        }

        for(int nb : map.get(node)){
            if(nb==parent) continue;
            dfs2(nb , node , ans  , count, map);

        }
    }
}

**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 3
1 2
2 3
3 4
1 2 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 (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
// package Graph;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class GameOnTree {
    static ArrayList<Integer> tree[];
    static boolean visited[];
    static HashSet<Integer> loseChild[];
    static boolean winPossibleChild[];
    static boolean winPossibleParent[];
    static int parent[];
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder print=new StringBuilder();
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int t=Integer.parseInt(st.nextToken());
        tree=new ArrayList[n+1];
        loseChild=new HashSet[n+1];
        for(int i=1;i<=n;i++){
            tree[i]=new ArrayList<>();
            loseChild[i]=new HashSet<>();
        }
        for(int i=1;i<n;i++){
            st=new StringTokenizer(br.readLine());
            int u=Integer.parseInt(st.nextToken());
            int v=Integer.parseInt(st.nextToken());
            tree[u].add(v);
            tree[v].add(u);
        }
        visited=new boolean[n+1];
        winPossibleChild=new boolean[n+1];
        winPossibleParent=new boolean[n+1];
        parent=new int[n+1];
        checkWinChild(1, 0);
        Queue<Integer> queue=new LinkedList<>();
        queue.add(1);
        visited=new boolean[n+1];
        visited[1]=true;
        while(!queue.isEmpty()){
            int parent=queue.remove();
            for(int child:tree[parent]){
                if(!visited[child]){
                    boolean op1=!winPossibleParent[parent];
                    boolean op2=false;
                    if(loseChild[parent].size()==1&&(loseChild[parent].contains(child))){
                        op2=true;
                    }
                    winPossibleParent[child]=op1||op2;
                    queue.add(child);
                    visited[child]=true;
                }
            }
        }
        st=new StringTokenizer(br.readLine());
        for(int i=1;i<=t;i++){
            int node=Integer.parseInt(st.nextToken());
            if(winPossibleChild[node]||winPossibleParent[node]){
                print.append("Ron\n");
            }
            else{
                print.append("Hermione\n");
            }
        }
        System.out.println(print);
    }
    public static boolean checkWinChild(int v,int p){
        parent[v]=p;
        visited[v]=true;
        boolean hasChild=false;
        boolean isWinnable=false;
        for(int c:tree[v]){
            if(!visited[c]){
                boolean result=checkWinChild(c,v);
                hasChild=true;
                if(!result){
                    isWinnable=true;
                    loseChild[v].add(c);
                }
            }
        }
        winPossibleChild[v]=isWinnable;
        if(!hasChild){
            return false;
        }
        else{
            return isWinnable;
        }
    }

}

**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 might cause the algorithm to fail. Here are the steps to think through the process:

1. **Understanding the Bug**:
   - The code seems to have issues with handling certain tree structures and specific starting nodes.
   - The `checkWinChild` function and the subsequent checks might not correctly propagate the win/lose conditions for all nodes.

2. **Edge Cases**:
   - **Single Child Nodes**: Nodes with only one child might cause issues in the win/lose propagation.
   - **Long Chains**: Long chains of nodes might lead to incorrect win/lose states.
   - **Balanced Trees**: Balanced trees might cause issues with the win/lose conditions.
   - **Multiple Paths to the Same Node**: Trees with multiple paths to the same node might cause issues.

3. **Constructing the Test Case**:
   - We will create a small tree with a specific structure that might expose the bug.
   - For example, a tree with a long chain and a few branches might help.

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

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

### Explanation:
- **Tree Structure**:
  - Node 1 is connected to Node 2.
  - Node 2 is connected to Node 3.
  - Node 3 is connected to Node 4.
  - Node 4 is connected to Node 5.
  - Node 5 is connected to Node 6.
- **Starting Nodes**:
  - The game starts with the stone on Node 1, Node 3, and Node 5.

### Expected Output:
- **Node 1**: Ron
- **Node 3**: Hermione
- **Node 5**: Ron

This test case is designed to check how the code handles a long chain of nodes and specific starting points. It should help expose any issues with the win/lose propagation logic in the 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 (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define in(a, b) for (ll i = (a); i <= (b); i++)                // in using i
#define inj(a, b) for (ll j = (a); j <= (b); j++)               // in using j
#define ink(a, b) for (ll k = (a); k <= (b); k++)               // in using k
#define inl(a, b) for (ll l = (a); l <= (b); l++)               // in using l
#define inr(a, b) for(ll i = (a); i >= (b); i--)                // in reverse
#define inrj(a, b) for(ll j = (a); j >= (b); j--)                // in reverse
#define inrk(a, b) for(ll k = (a); k >= (b); k--)                // in reverse
#define inrl(a, b) for(ll l = (a); l >= (b); l--)                // in reverse
#define tt ll tcs; cin>>tcs; while(tcs--)                       // include test cases
#define ina(arr,n) ll arr[(n+1)]={0}; in(1,n) cin>>arr[i]       // input arr of n elements
#define inv(vec,n) vector<ll> vec(n+1); vec[0]=0; in(1,n) cin>>vec[i]; // input vector of n elements
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define pll pair <ll,ll>
#define vpll vector <pll>
#define sll set <ll>
#define vll vector<ll>
#define mll map <ll,ll>
#define all(x) x.begin(), x.end()
const ll mod=1e9+7;
#define vvll vector<vll>
#define pref(p,a,n) vll p(n+1); in(1,n) p[i]=p[i-1]+a[i];
#define vec2(a,n,m) vvll a(n+1,vll(m+1))
// #define vec2(a,n,m,val) vvll a(n+1,vll(m+1,val))

#define vec3(a,l,m,n) vector<vvll>a(l+1,vvll(m+1,vll(n+1)));
// #define vec3(a,l,m,n,val) vector<vvll>a(l+1,vvll(m+1,vll(n+1,val)));
# define FAST ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

vll par(2e5+10);
vvll adj(2e5+10);
ll root=-1;
struct node{
    // bool allpos=0;
    bool win=0;
};
vll dfsorder;
sll doubtful;;
// T T F

vpll modifyfchild;
vector<node>dp(2e5+10);
void dfs(ll v, ll p){
    par[v]=p;
    dfsorder.push_back(v);
    ll cntchild=0;
    ll cntru=0;
    ll cntfals=0;
    ll falss=-1;
    vll trrru;
    for(auto x:adj[v]){
        if(x!=p){
            cntchild++;
            dfs(x,v);
            if(dp[x].win){
                cntru++;
                trrru.push_back(x);
            }
            else{
                cntfals++;
                falss=x;
            }
        }
    }
    if(cntfals>0){
        dp[v].win=1;
        if(cntru>0){
            for(auto x:trrru){
                doubtful.insert(x);
            }
            doubtful.insert(v);
        }

    }
}

// agar parent ka ek hi false neighbour hua
//har false node ka check if only one false neighbour
int main(){
    FAST
    ll n,t;
    cin>>n>>t;
    in(1,n-1){
        ll u,v;
        cin>>u>>v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    in(1,n){
        if(adj[i].size()>0){
            root=i;
            break;
        }
    }

    dfs(1,0);
    reverse(all(dfsorder));
    for(auto v:dfsorder){
        if(doubtful.count(v)){
            vll fasl;
            vll ttr;
            for(auto x:adj[v]){
                // if(x==par[v]){
                //     continue;
                // }
                if(dp[x].win){
                    ttr.push_back(x);
                }
                else{
                    fasl.push_back(x);
                }
            }
            if(fasl.size()==1&&fasl[0]!=par[v]){
                dp[fasl[0]].win=1;
            }
        }
    }
    while(t--){
        ll x;
        cin>>x;
        if(dp[x].win){
            cout<<"Ron\n";
        }
        else{
            cout<<"Hermione\n";
        }
    }

}

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

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

Think step by step.

Response:
```plaintext
5 3
1 2
1 3
1 4
1 5
2 3 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 (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

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

int const MX = 3e5 + 7;
vector<int> ok(MX), previousPath(MX), vis(MX), dp(MX), g[MX];
vector<vector<int>> dp2(MX, vector<int> (2));

void dfs(int u, int cur, int p){
	vis[u] = 1;
	previousPath[u] = p;
	if (g[u].size() > 2 || (u == 1 && g[u].size() > 1)) p = u;
	if (cur % 2 == 0) ok[u] = 1;
	else dp[u] = 1;
	for (auto v : g[u]){
		if (!vis[v]){
			dfs(v, cur + 1, p);
			dp2[u][dp[v]]++;
			if (ok[u]) dp[u] = max(dp[u], dp[v]);
			else dp[u] = min(dp[u], dp[v]);
		}
	}
}

int main(){
	/*int t;
	cin >> t;
	while (t--){*/
		int n, q, a;
		cin >> n >> q;
		for (int i = 1; i < n; i++){
			int u, v;
			cin >> u >> v;
			g[u].push_back(v);
			g[v].push_back(u);
		}
		dfs(1, 0, 0);
		for (int i = 0; i < q; i++){
			cin >> a;
			if (ok[a]){
				if (dp[a] || (previousPath[a] && ((ok[previousPath[a]] && dp[previousPath[a]]) || (!ok[previousPath[a]] && dp2[previousPath[a]][0] == 1)))) cout << "Ron" << endl;
				else cout << "Hermione" << endl;
			} else {
				if (!dp[a] || !previousPath[a] || (ok[previousPath[a]] && dp2[previousPath[a]][1] == 1) || (!ok[previousPath[a]] && !dp[previousPath[a]])) cout << "Ron" << endl;
				else cout << "Hermione" << endl;
			}
		}
	//}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are some potential edge cases:

1. **Small Trees with Few Edges**: These can help us verify basic logic and edge conditions.
2. **Linear Trees (Chains)**: These can test the alternating moves and the end conditions.
3. **Trees with a Central Node (Star Trees)**: These can test the behavior when the starting node is the central node or one of the leaves.
4. **Trees with Multiple Branches**: These can test more complex scenarios where the game's outcome depends on the structure of the tree.

Given the constraints, a small-scale test input that is complete and valid is:

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

Explanation:
- The tree has 6 nodes and 4 rounds.
- The tree structure is as follows:
  - Node 1 is connected to Node 2.
  - Node 2 is connected to Nodes 3, 4.
  - Node 4 is connected to Nodes 5, 6.
- The starting nodes for each round are 1, 2, 4, and 6.

This test input includes:
- A central node (Node 2) with multiple branches.
- Leaf nodes (Nodes 3, 5, 6).
- A node with a single child (Node 1).

This setup should help expose any issues in the logic for determining the winner based on the tree structure and the starting node.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
n, m = list(map(int, input().split()))
edges=[[] for j in range(n)]
for j in range(n-1):
    a,b=map(int,input().split())
    edges[a-1].append(b-1)
    edges[b-1].append(a-1)

pp=list(map(int,input().split()))

def dfs(nd,par,vis,val,edges,val2,parr):
    vis[nd]=1
    parr[nd]=par
    put=0
    cnt=0
    for ch in edges[nd]:
        if ch==par:
            continue
        if vis[ch]==0:
            dfs(ch,nd,vis,val,edges,val2,parr)
        if val[ch]==0:
            put=1
            cnt+=1
    val[nd]=put
    if cnt>=2:
        val2[nd]=1
    # else:
    #     val2[nd]=1

vis=[0 for i in range(n)]
val=[0 for i in range(n)]
val2=[0 for i in range(n)]
parr=[-1 for i in range(n)]
dfs(0,-1,vis,val,edges,val2,parr)
# print(val)
# print(val2)
for p in pp:
    if p==1:
        if val[p-1]==1:
            print("Ron")
        else:
            print("Hermione")
        continue

    if val[p-1]==1:
        print("Ron")
    else:
        k=parr[p-1]
        cnt=0
        while val2[k]==0:
            k=parr[k]
            cnt+=1
            if k==-1:
                break
        # print(cnt,k)
        milgya=False
        if k!=-1:
            milgya=True
        else:
            if val2[0]==1:
                milgya=True

        if milgya:
            if cnt%2==1:
                print("Ron")
            else:
                print("Hermione")
        else:
            if cnt%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:
```plaintext
4 3
1 2
2 3
3 4
1 2 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 (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
// package Graph;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class GameOnTree {
    static ArrayList<Integer> tree[];
    static boolean visited[];
    static HashSet<Integer> loseChild[];
    static ArrayList<Integer> children[];
    static boolean winPossibleChild[];
    static boolean winPossibleParent[];
    static int parent[];
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder print=new StringBuilder();
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int t=Integer.parseInt(st.nextToken());
        tree=new ArrayList[n+1];
        loseChild=new HashSet[n+1];
        children=new ArrayList[n+1];
        for(int i=1;i<=n;i++){
            tree[i]=new ArrayList<>();
            loseChild[i]=new HashSet<>();
            children[i]=new ArrayList<>();
        }
        for(int i=1;i<n;i++){
            st=new StringTokenizer(br.readLine());
            int u=Integer.parseInt(st.nextToken());
            int v=Integer.parseInt(st.nextToken());
            tree[u].add(v);
            tree[v].add(u);
        }
        visited=new boolean[n+1];
        winPossibleChild=new boolean[n+1];
        winPossibleParent=new boolean[n+1];
        parent=new int[n+1];
        checkWinChild(1, 0);
        Queue<Integer> queue=new LinkedList<>();
        queue.add(1);
        visited=new boolean[n+1];
        visited[1]=true;
        while(!queue.isEmpty()){
            int parent=queue.remove();
            for(int child:tree[parent]){
                if(!visited[child]){
                    boolean op1=winPossibleParent[parent];
                    boolean op2=false;
                    if(children[parent].size()>1){
                        if(loseChild[parent].size()>1){
                            op2=true;
                        }
                    }
                    winPossibleParent[child]=!(op1||op2);
                    queue.add(child);
                    visited[child]=true;
                }
            }
        }
        st=new StringTokenizer(br.readLine());
        for(int i=1;i<=t;i++){
            int node=Integer.parseInt(st.nextToken());
            if(winPossibleChild[node]||winPossibleParent[node]){
                print.append("Ron\n");
            }
            else{
                print.append("Hermione\n");
            }
        }
        System.out.println(print);
    }
    public static boolean checkWinChild(int v,int p){
        parent[v]=p;
        visited[v]=true;
        boolean hasChild=false;
        boolean isWinnable=false;
        for(int c:tree[v]){
            if(!visited[c]){
                children[v].add(c);
                boolean result=checkWinChild(c,v);
                hasChild=true;
                if(!result){
                    isWinnable=true;
                    loseChild[v].add(c);
                }

            }
        }
        winPossibleChild[v]=isWinnable;
        if(!hasChild){
            return false;
        }
        else{
            return isWinnable;
        }
    }

}

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the code seems to be in the logic for determining the win conditions for Ron and Hermione. Specifically, the code might not correctly handle the scenario where a node has multiple children, and the win conditions for the parent and child nodes are not correctly propagated.

2. **Edge Cases**:
   - **Single Path Tree**: A tree where nodes form a single path (e.g., a linear tree).
   - **Star Tree**: A tree where one node is the root and all other nodes are directly connected to the root.
   - **Balanced Tree**: A tree where the structure is balanced, and each node has a similar number of children.
   - **Complex Tree**: A tree with a mix of different structures (e.g., branches and paths).

3. **Challenging Scenario**:
   - A tree where the structure is complex and the win conditions are not straightforward to determine. This can help expose the bug in the win condition logic.

4. **Small-Scale Test Input**:
   - We will use a tree with 7 nodes and 4 rounds to ensure it is small-scale but still complex enough to challenge the code.

Here is the test input:

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

Explanation:
- **Tree Structure**:
  - Node 1 is the root.
  - Node 1 has two children: Node 2 and Node 3.
  - Node 2 has two children: Node 4 and Node 5.
  - Node 3 has two children: Node 6 and Node 7.
- **Rounds**:
  - The stone starts on Node 1, Node 2, Node 3, and Node 4.

This test input should expose any issues in the win condition logic, especially how the code handles multiple children and the propagation of win conditions from parent to child nodes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = false;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static List<Integer>[] G;
    static boolean[] dp;
    static boolean[] res;
    static void solve(PrintWriter o) {
        try {
            int n = nextInt(), t = nextInt();
            G = new ArrayList[n];
            Arrays.setAll(G, key->new ArrayList<>());
            int[] start = new int[t];
            for(int i=0;i<n-1;i++) {
                int u = nextInt(), v = nextInt();
                u--;v--;
                G[u].add(v);
                G[v].add(u);
            }
            dp = new boolean[n];
            res = new boolean[n];
            for(int i=0;i<t;i++) {
                start[i] = nextInt();
                start[i]--;
            }
            dfs(0, 0);
            dfs1(0, 0);
            for(int i=0;i<t;i++) {
                if(res[start[i]]) o.println("Ron");
                else o.println("Hermione");
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    static void dfs(int u, int p) {
        for(int v: G[u]) {
            if(v == p) continue;
            dfs(v, u);
            dp[u] |= !dp[v];
        }
    }
    static void dfs1(int u, int p) {
        int cnt = 0;
        dp[u] = false;
        for(int v: G[u]) if(!dp[v]) {
            dp[u] = true;
            cnt++;
        }
        res[u] = dp[u];
        boolean convert1 = false, convert2 = false;
        for(int v: G[u]) {
            if(v == p) continue;
            if(cnt == 1 && !dp[v]) convert1 = true;
            dp[u] = convert1 ? !dp[u] : dp[u];
            if(!dp[u] && !dp[v]) convert2 = true;
            dp[v] = convert2 ? !dp[v] : dp[v];
            dfs1(v, u);
            dp[v] = convert2 ? !dp[v] : dp[v];
            dp[u] = convert1 ? !dp[u] : dp[u];
        }
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(List<Integer> a, int val){
        int l = 0, r = a.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a.get(mid) < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class DSU {
        int[] parent;
        public DSU(int n) {
            parent = new int[n];
            for(int i=0;i<n;i++) parent[i] = i;
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The provided code seems to handle basic cases correctly, but there are a few potential issues that could be exploited:

1. **Complex Tree Structures**: Trees with specific structures, such as star graphs or linear chains, can sometimes expose bugs in tree traversal algorithms.
2. **Multiple Rounds with the Same Starting Node**: The code should handle multiple rounds where the starting node is the same, as this can affect the state of the tree.
3. **Edge Cases with Small Trees**: Small trees with specific edge configurations can be tricky to handle correctly.

Given these considerations, let's create a test input that includes a small, complex tree structure and multiple rounds with the same starting node.

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

### Explanation
- **Tree Structure**: The tree is structured as follows:
  - Node 1 is connected to nodes 2 and 4.
  - Node 2 is connected to node 3.
  - Node 4 is connected to nodes 5 and 6.
- **Rounds**: The game is played with the starting nodes 1, 1, and 4.
  - The first two rounds start from node 1.
  - The third round starts from node 4.

### Expected Output
- For the first round starting at node 1, Hermione should win because node 1 has multiple children, and Ron can be forced into a losing position.
- For the second round starting at node 1, the result should be the same as the first round.
- For the third round starting at node 4, Ron should win because node 4 has two children, and he can force Hermione into a losing position.

### Why This Input is Challenging
- **Multiple Rounds with the Same Starting Node**: This tests whether the code correctly resets the state of the tree after each round.
- **Complex Tree Structure**: The tree has a non-trivial structure with multiple branching points, which can expose issues in the depth-first search (DFS) and dynamic programming (DP) logic.

This input should help in identifying any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

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

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

q = list(map(int,input().split()))

parent = [0]*(n+1)
state = [0]*(n+1)
ans = [0]*(n+1)

def dfs(i,p=0):

    for j in d[i]:
        if j!=p:
            parent[j] = i
            dfs(j,i)
            state[i] |= (not state[j])
    return

def reroot(i,p=0):

    for j in d[i]:
        if j!=p:
            ans[i] |= (not state[j])

    if not ans[i] and p:
        cur = 0
        c = 0
        for j in d[p]:
            if j!=i and j!=parent[p]:
                cur += state[j]
                c+=1
        if (cur==c and c>0): ans[i] = 1
        if c==0:
            ans[i] =  ans[parent[p]] if parent[p] else 1

    for j in d[i]:
        if j!=p:
            reroot(j,i)

    return

dfs(1)
ans[1] = state[1]
reroot(1)
for x in q:
    print("Ron" if ans[x] else "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
5 3
1 2
1 3
3 4
3 5
1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#pragma GCC targrt("avx,sse,sse2")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define fillWith(a, b) memset(a, b, sizeof(a))
#define test  \
    int t;    \
    cin >> t; \
    while (t--)
#define rep(i, a, b) for (int i = a; i < b; i++)
#define pri(a, b) copy(a, b, ostream_iterator<ll>(cout, " "));
#define ENT cout << "\n"
#define TTOOPP 0x3f3f3f3f3f3f3f3fLL
#define PI 3.14159265370874
#define get_unique(v)                                 \
    {                                                 \
        sort(v.begin(), v.end());                     \
        v.erase(unique(v.begin(), v.end()), v.end()); \
    }
#define mod 1000000007
#define endl '\n'
#define log(x) (63 ^ __builtin_clzll(x)) // stores in cash
//#define int ll

vector<vector<int>> adj;
map<pair<int, pair<int, int>>,int> dp;
bool dfs(int node, int par = -1, int c = 0)
{
    if (adj[node].size() == 1 && par != -1)
    {
        return c % 2;
    }
    if (par != -1 && dp.find({node,{par,c%2}})!=dp.end())
        return dp[{node,{par,c%2}}];
        if (c % 2 == 0)
        {
            bool w = 0;
            for (auto i : adj[node])
            {
                if (i != par)
                {
                    w |= dfs(i, node, c + 1);
                }
            }
            return dp[{node,{par,c%2}}]=w;
        }
        else
        {
            bool w = 1;
            for (auto i : adj[node])
            {
                if (i != par)
                {
                    w &= dfs(i, node, c + 1);
                }
            }
            return dp[{node,{par,c%2}}]=w;
        }
}

void Task()
{
    int n, t;
    cin >> n >> t;
    adj.resize(n + 1);
    rep(i, 0, n - 1)
    {
        int x, y;
        cin >> x >> y;
        adj[x].push_back(y);
        adj[y].push_back(x);
    }
    while(t--){
    int y;
    cin >> y;
    cout << (dfs(y) ? "Ron" : "Hermione")<<endl;
    }
}

signed main()
{

    cin.tie(0)->sync_with_stdio(0);

    // test
    {
        Task();
        cout << endl;
    }
}

**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 aspects:

1. **Edge Cases**: Test cases where the tree structure or the number of rounds can lead to unexpected behavior.
2. **Complex Tree Structures**: Trees with specific properties that can stress the algorithm, such as trees with long paths, trees with many leaves, or trees with a central node connected to many other nodes.
3. **Initial Node Selection**: Choosing initial nodes that can lead to different game outcomes based on the tree structure.

Given the constraints and the nature of the problem, let's consider a tree with a central node connected to many other nodes. This structure can expose issues with the recursive depth and the memoization logic.

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

### Explanation:
- **Tree Structure**: Node 1 is the central node connected to nodes 2, 3, 4, 5, 6, and 7. This structure ensures that the game can be played with multiple paths of the same length.
- **Number of Rounds**: 4 rounds, with initial nodes 1, 2, 3, and 4.
- **Expected Behavior**:
  - **Initial Node 1**: Ron should lose because he cannot make a move (all neighbors are inactive).
  - **Initial Node 2, 3, 4**: Ron should win because he can move to node 1, and Hermione will be forced to move to another leaf node, giving Ron the last move.

This test input should expose any issues with the memoization and recursive logic 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: Game on Tree (Hard)
time_limit_ms: 1000
memory_limit_mb: 256
Description: This is the hard version of the problem. The only difference in this version is the constraint on $$$t$$$.

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

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

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

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

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

Output: The output consists of $$$t$$$ lines, each line being either "Ron" or "Hermione".

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

input:
6 3
1 2
2 3
1 4
4 5
4 6
1 4 6
output:
Hermione
Ron
Hermione

Note:

Buggy Code:
// ffff
// faruk
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define YES cout << "YES" << endl;
#define NO cout << "NO" << endl;
#define SetBit(x, k) (x |= (1ll << k))
#define ClearBit(x, k) (x &= ~(1ll << k))
#define CheckBit(x, k) ((x >> k) & 1)
#define CC(x) cout << "Case " << ++x << ": ";
#define all(x) x.begin(), x.end()
#define pi pair<long long int, long long int>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
// #define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define fastio                  \
  ios_base::sync_with_stdio(0); \
  cin.tie(0);                   \
  cout.tie(0)
const int N = 3e5 + 10;
const int M = 1e9 + 7;
const ll INF = 1e12 + 10;
vector<ll> vdv[N], pdv[N];
// map<ll, ll> vis;
ll i, j, k, m, n, b, c, d, l, q, s;
map<ll, ll> fact;
vector<ll> vk;
map<ll, ll> vis, vis1, vis2;
// ll vis2[N],vis1[N],vis[N];
// ll vis[N],vis1[N];
//  bool vis[N],vis2[N];
//  ll vis1[N];

long long int binE(long long int a, long long int b)
{
  long long res = 1;
  while (b > 0)
  {
    if (b & 1)
    {
      res = (res * a) % M;
    }
    a = (a * a) % M;
    b >>= 1;
  }
  return res;
}

bool cmp(pi p1, pi p2)
{
  if (p1.second == p2.second)
    return p1.first >= p2.second;
  else
    return p1.second < p2.second;
}

vector<ll> isprime(N, 1);
vector<ll> prime;
map<ll,ll>win,fixe,uwin,depth;
void dfs1(ll ver){
  vis[ver]=1;
  for(auto x:vdv[ver]){
    if(vis[x])continue;
    dfs1(x);
    if(!win[x])win[ver]++;
  }
}
void dfs2(ll ver,ll par,ll stat){
   vis[ver]=1;
   if(!win[ver]){
  //  if(ver==4)cout<<par<<" "<<stat<<endl;
    ll diff=abs(depth[par]-depth[ver]);
    if(diff%2){
      if(!stat)uwin[ver]=1;
    }
    else if(stat)uwin[ver]=1;
   }
   for(auto x:vdv[ver]){
    if(vis[x])continue;
    depth[x]=depth[ver]+1;
     if(win[ver]>1||(win[x]&&win[ver]))  dfs2(x,ver,(win[ver]>1||(win[x]&&win[ver])));
     else if(ver==1&&!win[ver])dfs2(x,ver,0);
     else if(ver==1&&win[ver]==1)dfs2(x,ver,0);
     else dfs2(x,par,stat);
   }
}
int main()
{
  fastio;
  // Example: Adding elements to vg

  // for(auto x:vdv[10])cout<<x<<endl;
  ll t;

  // fact[0] = 1;
  // for(ll i=1;i<=N;i++){
  //   fact[i]=(i*fact[i-1])%M;
  // }

  // // ll cft[N]={0};

  // //cout<<t<<endl;
  // ll vs[N][48];
  // for(ll i=2;i<N;i++){
  //  // cout<<i<<" "<<cft[i]<<endl;
  //   for(ll j=1;j<=cft[i];j++){
  //     vs[i][j]=com(cft[i],j);
  //   }
  // }
  // isprime[1] = isprime[0] = 0;
  // for (ll i = 1; i < N; i++)
  // {
  //   if (isprime[i])
  //   {
  //     prime.push_back(i);
  //     for (ll j = 2 * i; j < N; j += i)
  //       isprime[j] = 0;
  //   }
  // }
  // cout<<prime.size()<<endl;
  ll tt = 1;
  const int P = 109, ct = 11630; // 1000000
  ll in = 1014657922693201;
  ll dd = -1, hh = t;
  // ll b;
  t = 1;

  //cin >> t;
  in = t;
  map<pair<int, int>, int> inn;
  tt = 0;
  while (t--)
  {
    ll n,m;
    cin>>n>>m;
    for(ll i=1;i<n;i++){
      ll x,y;
      cin>>x>>y;
      vdv[x].push_back(y);
      vdv[y].push_back(x);
    }
    vis.clear();
    dfs1(1);
    vis.clear();
    dfs2(1,1,win[1]);
   // cout<<uwin[4]<<endl;
    for(ll i=1;i<=m;i++){
      ll l;cin>>l;
      if(win[l]||uwin[l])cout<<"Ron"<<endl;
      else cout<<"Hermione"<<endl;
    }
  }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can highlight potential issues in the algorithm. Here are some key points to consider:

1. **Tree Structure**: The tree should be structured in a way that exposes the bug. For example, a linear chain or a star tree can be good candidates.
2. **Number of Rounds**: The number of rounds should be chosen to ensure that the bug is triggered multiple times if it exists.
3. **Starting Nodes**: The starting nodes should be chosen to test different scenarios, such as starting from the root, leaves, and intermediate nodes.

Given these considerations, let's create a test input:

- **Tree Structure**: A linear chain of 5 nodes (1-2-3-4-5).
- **Number of Rounds**: 5 rounds.
- **Starting Nodes**: 1, 2, 3, 4, 5.

This test input will help us verify the correctness of the algorithm by testing various starting positions on a simple linear chain.

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

This input will:
- Ensure the tree is a linear chain.
- Test the algorithm with all possible starting nodes.
- Check if the algorithm correctly determines the winner for each starting node.