**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.*;
public class Main {
    static FastRead in = new FastRead();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    static Set<Integer>h;
    static List<Integer>[]map;
    static Set<Integer>set=new HashSet<>(),setb=new HashSet<>();
    static int[]s;
    public static void main(String[] args) {
        int t=in.nextInt();
        while (t--!=0){
            int n=in.nextInt();
            int[]arr=new int[n+1];
            int[]r=new int[n+1];
            map=new List[n+1];
            for (int i = 0; i <=n ; i++) {
                map[i]=new ArrayList<>();
            }
            for (int i = 1; i <=n ; i++) {
                arr[i]=in.nextInt();
                map[arr[i]].add(i);
                r[arr[i]]++;
            }
            for (int i = 1; i <=n ; i++) {
                if(r[i]==0){
                    set.add(i);
                }
            }
            long ans=0;
            //统计环
            while (!set.isEmpty()){
                for (Integer i:set){
                    if(--r[arr[i]]==0){
                        setb.add(arr[i]);
                    }
                }
                set=setb;
                setb=new HashSet<>();
            }
            h=new HashSet<>();
            //统计环点
            for (int i = 1; i <=n ; i++) {
                if(r[i]!=0){
                    h.add(i);
                }
            }
            //统计环点权值
//            s=new int[n+1];
//            for (int i = 1; i <=n; i++) {
//                s[arr[i]]++;
//            }
            for (int i = 1; i <=n ; i++) {
                if(r[i]!=0){
                    ans=Math.max(ans,dfs(i)-1);
                }
            }
            out.println(ans+2);
        }
        out.flush();
    }
    public static long dfs(int cu){
        long ans=1;
        for (Integer i:map[cu]){
            if(!h.contains(i)){
                ans+=dfs(i);
//                out.println(ans+" "+i);
            }
        }
        return ans;
    }
}
    class FastRead {
        StringTokenizer st;
        BufferedReader br;
        FastRead() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

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

**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 consider edge cases and scenarios that can cause the algorithm to fail. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code seems to have issues with correctly identifying the cycles and the stability condition. It might not handle cases where the cycles are not straightforward or where the distribution of plushies becomes complex.

2. **Edge Cases**:
   - **Single Cycle**: A scenario where all spiders are in a single cycle.
   - **Multiple Cycles**: A scenario where spiders are in multiple cycles.
   - **Self-loops and Isolated Spiders**: Although not allowed in the problem, ensuring that the code handles isolated spiders correctly.
   - **Complex Distribution**: A scenario where the distribution of plushies becomes complex and the process takes several years to stabilize.

3. **Construct the Test Case**:
   - **Single Cycle**: Create a test case where all spiders are in a single cycle. This should be straightforward and should help verify if the code correctly identifies the cycle.
   - **Multiple Cycles**: Create a test case with multiple cycles to ensure the code can handle more complex scenarios.
   - **Complex Distribution**: Create a test case where the distribution of plushies becomes complex, requiring multiple years to stabilize.

Given the constraints, we will focus on a small-scale test case that is complete and valid.

### Test Case 1: Single Cycle
```plaintext
1
5
2 3 4 5 1
```

### Test Case 2: Multiple Cycles
```plaintext
1
6
2 3 1 5 6 4
```

### Test Case 3: Complex Distribution
```plaintext
1
7
2 3 4 5 6 7 1
```

### Final Test Input
Combining these test cases into a single input:
```plaintext
3
5
2 3 4 5 1
6
2 3 1 5 6 4
7
2 3 4 5 6 7 1
```

This test input includes a single cycle, multiple cycles, and a more complex distribution scenario, which should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.*;
public class Main {
    static FastRead in = new FastRead();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    static Set<Integer>h;
    static List<Integer>[]map;
    static Set<Integer>set=new HashSet<>(),setb=new HashSet<>();
    static int[]s;
    public static void main(String[] args) {
        int t=in.nextInt();
        while (t--!=0){
            int n=in.nextInt();
            int[]arr=new int[n+1];
            int[]r=new int[n+1];
            map=new List[n+1];
            for (int i = 0; i <=n ; i++) {
                map[i]=new ArrayList<>();
            }
            for (int i = 1; i <=n ; i++) {
                arr[i]=in.nextInt();
                map[arr[i]].add(i);
                r[arr[i]]++;
            }
            for (int i = 1; i <=n ; i++) {
                if(r[i]==0){
                    set.add(i);
                }
            }
            long ans=0;
            //统计环
            while (!set.isEmpty()){
                for (Integer i:set){
                    if(--r[arr[i]]==0){
                        setb.add(arr[i]);
                    }
                }
                set=setb;
                setb=new HashSet<>();
            }
            h=new HashSet<>();
            //统计环点
            for (int i = 1; i <=n ; i++) {
                if(r[i]!=0){
                    h.add(i);
                }
            }
            //统计环点权值
//            s=new int[n+1];
//            for (int i = 1; i <=n; i++) {
//                s[arr[i]]++;
//            }
            for (int i = 1; i <=n ; i++) {
                if(r[i]!=0){
                    ans=Math.max(ans,dfs(i));
                }
            }
            out.println(ans+1);
        }
        out.flush();
    }
    public static long dfs(int cu){
        long ans=1;
        for (Integer i:map[cu]){
            if(!h.contains(i)){
                ans+=dfs(i);
            }
        }
        return ans;
    }
}
    class FastRead {
        StringTokenizer st;
        BufferedReader br;
        FastRead() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

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

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose potential issues in the buggy code, particularly in how it handles the distribution and stabilization of plushies among spiders.

1. **Input Setup**:
   - The number of test cases is `1`.
   - The number of spiders is `5`.
   - The recipients of the plushies are `[2, 3, 4, 5, 1]`.

2. **Expected Behavior**:
   - Each spider gives one plushie to the next spider in the sequence.
   - The process should stabilize in year `2` because the distribution of plushies will be the same as the previous year.

3. **Edge Case**:
   - This test case forms a simple cycle where each spider gives a plushie to the next spider in a circular manner.
   - The simplicity of the cycle ensures that any error in the code's handling of cycles or stabilization will be easily exposed.

4. **Stabilization**:
   - After the first year, the distribution of plushies will be `[1, 1, 1, 1, 1]`.
   - After the second year, the distribution will be the same, indicating that the process has stabilized.

This test case is small and valid, making it easy to verify the correctness of the solution while being challenging enough to expose bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define prime_number[21]={11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
///POLICY BASED DATA STRUCTURE..
///order_of_key return number of element which are strictly greater/smaller than x..
///find_by_order return ans iterator corresponding to the xth position of the set..
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
#define ll long long int
#define faster_bro ios_base::sync_with_stdio(false);cin.tie(NULL)
#define esp 0.0000000001
#define ss second
#define ff first
#define nl cout<<endl
//memset(ar,-1,sizeof(ar));
// (1LL<<i) more than efficient pow(2,i);
void fast(){
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
}
const int N = 3e5 + 9;
vector<int> g[N],adj[N];
int col[N], par[N],del[N];
map<int,int>indeg;
vector<int>vc;
bool cycle;
void dfs(int u) {
  col[u] = 1;
  for (auto v: g[u]) {
    if (col[v] == 0) {
      par[v] = u;
      dfs(v);
    }
    else if (col[v] == 1) {
      cycle = true;
      //st.erase(u);
      //vc.push_back(u);
      int p=u;
      del[p]=1;
      //cout<<p<<" ";
    int  k=par[p];

      while(v!=p){
         k=par[p];
         del[k]=1;
         //cout<<k<<" ";
         p=k;

      }
      // you can track the cycle using par array
    }
  }
  col[u] = 2;
}
bool vis[N];
int mx=0;
int cnt=0;
void dfs1(int x){
    vis[x]=true;
    for(auto it:adj[x]){
        if(vis[it]==false && del[it]==0){
            cnt++;

            mx=max(mx,cnt);
            dfs1(it);
        }

    }

    }

int main(){
    fast();
    int tc;
    cin>>tc;
    int h=1;
    while(tc--){
        ll a,b,c,sum=0,sum1=0,sum2=0,mn=LLONG_MAX,mx2=-1e18;
        cin>>a;
        for(int i=1;i<=a;i++){
            vis[i]=false;
            col[i]=0;
            del[i]=0;
            mx=0;
            g[i].clear();
            adj[i].clear();
        }
        int mx1=1;
        for(int i=1;i<=a;i++){
            int x,y;
            cin>>x;
            indeg[x]++;
            g[i].push_back(x);
            adj[x].push_back(i);
           // adj[x].push_back(i);

        }
        for(int i=1;i<=a;i++){
            if(col[i]==0){
                dfs(i);
            }
        }
        vector<int>cy;
         for(int i=1;i<=a;i++){
           if(del[i]==1){
            cy.push_back(i);
           }
        }
          for(auto it:cy){

                cnt=0;
            dfs1(it);
          }
         // cout<<mx<<endl;

        cout<<mx+2<<endl;
//        for(auto it:vc){
//            cout<<it<<" ";
//        }
    }
}

**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
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

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

const int N = 2e5 + 9;
vector<int> g[N], gT[N], G[N];
vector<bool> vis(N, false);
vector<int> order;
int n, roots[N], sz[N], cnt[N];

void dfs(int u) {
  vis[u] = true;
  for (auto v : g[u]) {
    if (!vis[v]) dfs(v);
  }
  order.push_back(u);
}

void dfs2(int u, vector<int> &component) {
  vis[u] = true;
  component.push_back(u);
  for (auto v : gT[u]) {
    if (!vis[v]) dfs2(v, component);
  }
}

void scc() {
  // get order sorted by end time
  order.clear();
  for (int u = 1; u <= n; u++) {
    if (!vis[u]) dfs(u);
  }
  reverse(order.begin(), order.end());
  // transpose the graph
  for (int u = 1; u <= n; u++) {
    for (auto v : g[u]) {
      gT[v].push_back(u);
    }
  }
  // get all components
  for (int i = 1; i <= n; i++) vis[i] = false;
  for (auto u : order) {
    if (!vis[u]) {
      vector<int> component;
      dfs2(u, component);
      sort(component.begin(), component.end());
      for (auto v : component) {
        roots[v] = component.front();
        sz[v] = component.size();
      }
    }
  }
  // add edges to condensation graph
  for (int u = 1; u <= n; u++) {
    for (auto v : g[u]) {
      if (roots[u] != roots[v]) {
        G[roots[v]].push_back(roots[u]);
      }
    }
  }
}

void dfs3(int u) {
  vis[u] = true;
  int cur = 1;
  for (auto v : G[u]) {
    if (!vis[v] and roots[v] == v and sz[v] == 1) {
      dfs3(v);
      cur += cnt[v];
    }
    if (roots[v] == v and sz[v] == 1) {
      cur += cnt[v];
    }
  }
  cnt[u] = cur;
}

void solve() {
  cin >> n;

  for (int i = 0; i <= n; i++) {
    g[i].clear();
    gT[i].clear();
    G[i].clear();
    vis[i] = false;
  }

  for (int i = 1; i <= n; i++) {
    int x; cin >> x;
    g[i].push_back(x);
  }
  scc();

  for (int i = 1; i <= n; i++) {
    vis[i] = false;
  }

  for (auto i : order) {
    if (!vis[i] and roots[i] == i and sz[i] == 1) {
      dfs3(i);
    }
  }

  int mx = 0;
  for (int i = 1; i <= n; i++) {
    // cout << cnt[i] << ' ';
    mx = max(mx, cnt[i]);
  }

  cout << mx + 2 << '\n';
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  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 expose the bug in the provided code, we need to focus on edge cases or specific scenarios that might cause the algorithm to fail. The code involves strongly connected components (SCCs) and a process of counting the number of plushies transferred. The main goal is to find the first year in which the process becomes stable.

### Key Points to Consider:
1. **Cycles in the Graph**: The code should correctly handle cycles, especially when there are multiple cycles or complex cycle structures.
2. **Self-loops and Multiple Edges**: Ensure the code can handle cases where there are self-loops or multiple edges.
3. **Stable States**: The code should correctly identify when the state becomes stable, which means the number of plushies each spider has remains the same as the previous year.

### Test Input:
Let's create a test input that includes a cycle and a more complex structure to see if the code can handle it correctly.

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

### Explanation:
1. **First Test Case (n = 3)**:
   - The cycle is `1 -> 2 -> 3 -> 1`.
   - The number of plushies for each spider will be the same every year, making the process stable from the first year itself. However, since the problem states that year 1 can never be stable, the correct output should be 2.

2. **Second Test Case (n = 4)**:
   - The cycle is `1 -> 2 -> 3 -> 4 -> 1`.
   - Similar to the first test case, the process will be stable from the first year, but the correct output should be 2.

3. **Third Test Case (n = 6)**:
   - The cycle is `1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 1`.
   - Again, the process will be stable from the first year, but the correct output should be 2.

This test input is designed to check if the code can correctly handle cycles and determine the first stable year. If the code fails to handle these cases 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: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
# for I/O for local system
import sys
from os import path
if(path.exists('Input.txt')):
    sys.stdin = open('Input.txt','r')
    sys.stdout = open('Output.txt','w')

# For fast I/O
input = sys.stdin.buffer.readline
# input = sys.stdin.readline
print = sys.stdout.write

# Import libraries here whenever required
from collections import *
from random import randint

# Use this because normal dict can sometimes give TLE
class mydict:
    def __init__(self, func=lambda: 0):
        self.random = randint(0, 1 << 32)
        self.default = func
        self.dict = {}
    def __getitem__(self, key):
        mykey = self.random ^ key
        if mykey not in self.dict:
            self.dict[mykey] = self.default()
        return self.dict[mykey]
    def get(self, key, default):
        mykey = self.random ^ key
        if mykey not in self.dict:
            return default
        return self.dict[mykey]
    def __setitem__(self, key, item):
        mykey = self.random ^ key
        self.dict[mykey] = item
    def getkeys(self):
        return [self.random ^ i for i in self.dict]
    def __str__(self):
        return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'

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

    n = int(input())
    a = [int(x) for x in input().split()]

    ans = 2
    adj = [[] for i in range(n + 1)]
    inn = [0] * (n + 1)
    for i in range(n):
        adj[i + 1].append(a[i])
        inn[a[i]] += 1
    q = deque()
    timee = [1] * (n + 1)
    mx = [0] * (n + 1)
    for i in range(1, n + 1):
        if inn[i] == 0:
            q.append(i)
            timee[i] = 0
    if len(q) == 0:
        print('2\n')
        continue
    v = [1] * (n + 1)
    while q:
        # print(q)
        for i in range(len(q)):
            node = q.popleft()
            v[node] -= mx[node]
            for nbr in adj[node]:
                inn[nbr] -= 1
                v[nbr] += v[node]
                mx[nbr] = max(mx[nbr], v[node])
                timee[nbr] = max(timee[nbr], timee[node] + v[node])
                if inn[nbr] == 0:
                    q.append(nbr)
    ans = max(timee) + 2
    # print(timee)
    print(str(ans) + "\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
1
3
2 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#from math import ceil,log,lcm,gcd,isfinite
from collections import deque
#from bisect import bisect
#from heapq import heappush,heappop
#from fractions import Fraction as fr
#from sortedcontainers import SortedList,SortedSet,SortedDict
#from random import randint

from sys import stdin,stdout
input=lambda:stdin.readline().rstrip()
print=lambda *x,sep=' ',end='\n':stdout.write(sep.join(map(str,x))+end)

def out(l):
    print(' '.join(map(str,l)))

def yes():
    print('Yes')

def no():
    print('No')

def alice():
    print('Alice')

def bob():
    print('Bob')

def topoSort():
    indeg=[0]*(n+1)
    for l in adj:
        for child in l:
            indeg[child]+=1
    #lvl=[0]*(n+1)
    q=deque()
    for i,j in enumerate(indeg):
        if i and not j:
            q.appendleft(i)
            #lvl[i]=1
    #topo=[]
    sub=[0]*(n+1)
    while q:
        parent=q.pop()
        sub[parent]+=1
        #topo.append(parent)
        for child in adj[parent]:
            indeg[child]-=1
            sub[child]+=sub[parent]
            if not indeg[child]:
                q.appendleft(child)
                #lvl[child]=lvl[parent]+1
    return sub

def solve():
    global n,adj
    n=int(input())
    r=list(map(int,input().split()))
    adj=[[] for i in range(n+1)]
    for i,j in enumerate(r):
        adj[i+1].append(j)
    sub=topoSort()
    print(max(sub)+2)

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

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

public class MediumDemonProblemHard {
    static class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str="";
            try {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    public static void main(String[] args) {
        try {
            FastReader in=new FastReader();
            FastWriter out = new FastWriter();
            int t = in.nextInt();
            while (t-- > 0){
                int n = in.nextInt();
                int[] arr = new int[n];
                int[] indegree = new int[n];
                for (int i = 0; i < n; i++) {
                    arr[i] = in.nextInt()-1;
                    indegree[arr[i]]++;
                }

                int[] plushies = new int[n];
                Queue<Integer> que = new LinkedList<>();
                for (int i = 0; i < n; i++){
                    if (indegree[i] == 0) que.add(i);
                }
                Arrays.fill(plushies,1);
                while (!que.isEmpty()){
                    int size = que.size();
                    while (size-- > 0){
                        int curr = que.poll();
                        plushies[arr[curr]]+= plushies[curr];
                        if (--indegree[arr[curr]] == 0) que.add(arr[curr]);
                    }
                }
                int ans = getMax(plushies) + 1;
                out.println(ans);
            }
            out.close();
        } catch (Exception e) {
            return;
        }
    }

    private static int getMax(int[] arr){
        int max = 0;
        for (int val : arr) max = Math.max(max,val);
        return max;
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class main3 {
    static PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
    static FastReader in = new FastReader();
    public static int n;

    public static void main(String[] args) {
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int[] arr = new int[n];
            DisjointUnionSets dsu = new DisjointUnionSets(n);
            for (int i = 0; i < arr.length; i++) {
                arr[i] = in.nextInt() - 1;
                dsu.union(i, arr[i]);
            }
            boolean[] vis2 = new boolean[n];
            int[] score = new int[n];
            Arrays.fill(score, -1);
            boolean[] vis1 = new boolean[n];
            int ans = 0;
            for (int i = 0; i < n; i++) {
                if (!vis2[dsu.find(i)]) {
                    int node = i;
                    vis2[dsu.find(i)] = true;
                    while(!vis1[node]) {
                        vis1[node] = true;
                        node = arr[node];
                    }
                    while(score[node] == -1) {
                        score[node] = 0;
                        node = arr[node];
                    }
                }
            }
            // pw.println(Arrays.toString(score));
            int[] indeg = new int[n];
            for (int i = 0; i < n; i++) {
                int l = arr[i];
                indeg[l]++;
            }
            for (int i = 0; i < vis1.length; i++) {
                if (score[i] == -1) {
                    int steps = 0;
                    int node = i;
                    while(score[node] == -1) {
                        steps += indeg[node];
                        node = arr[node];
                    }
                    steps++;
                    ans = Math.max(ans, steps);
                    node = i;
                    while(score[node] == -1) {
                        score[node] = steps;
                        steps--;
                        node = arr[node];
                    }
                }
            }
            // pw.println(Arrays.toString(score));
            pw.println(ans + 2);
        }
        pw.close();
    }
        static class DisjointUnionSets {
            int[] rank, parent;
            int n;
            int count;

            public DisjointUnionSets(int n) {
                rank = new int[n];
                parent = new int[n];
                count = n;
                this.n = n;
                makeSet();
            }

            void makeSet() {
                for (int i = 0; i < n; i++) {
                    parent[i] = i;
                }
            }

            int find(int x) {
                if (parent[x] != x) {
                    parent[x] = find(parent[x]);
                }

                return parent[x];
            }

            void union(int x, int y) {
                int xRoot = find(x), yRoot = find(y);

                if (xRoot == yRoot)
                    return;

                if (rank[xRoot] < rank[yRoot])
                    parent[xRoot] = yRoot;

                else if (rank[yRoot] < rank[xRoot])
                    parent[yRoot] = xRoot;

                else {
                    parent[yRoot] = xRoot;
                    rank[xRoot] = rank[xRoot] + 1;
                }
                count -= 1;
            }

            int num() {
                return count;
            }
        }
    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 {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define Noushil ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ll long long
#define ld long double
#define pb push_back
#define ppb pop_back
#define mkp make_pair
#define endl "\n"
#define input(n,arr) for(ll i=0; i<n; i++){cin>>arr[i];}
#define fo(i, a, b) for (ll i = a; i < b; i++)
#define print(a)  for(auto x : a) cout << x << " ";
#define printvp(a) for(auto x: a) cout << x.first << " " << x.second << endl
#define printmap(a)  for(auto x : a) cout << x.first << " " << x.second << endl
#define sort(a) sort(a.begin(),a.end())
#define reverse(a) reverse(a.begin(),a.end())
#define revsort(a) sort(a.rbegin(),a.rend())
#define Yes cout<<"YES";
#define No cout<<"NO";

const int N=500000;
const ll int mod1 = 1e9 + 7;
const ll int mod2 = 998244353;

ll int pw(ll int a, ll int b, ll int m=mod1) {
    if(b==0) return 1;

    if(b%2 == 0) {
        ll int t = pw(a, (b/2), m);
        return (1ll*t*t)%m;
    }
    else {
        ll int t = pw(a, (b-1)/2, m);
        t = (1ll*t*t)%m;
        return (1ll*a*t)%m;
    }
}

ll int fact[N], invfact[N];
void init() {
    ll int p = mod1;
    fact[0]=1;
    ll int i=1;
    for(i=1; i<N; i++) {
        fact[i] = (i*fact[i-1])%p;
    }
    i--;
    invfact[i] = pw(fact[i], p-2, p);
    for(i--; i>=0; i--) {
        invfact[i] = (invfact[i+1]*(i+1))%p;
    }
}

ll int ncr(ll int n, ll int r) {
    return (((fact[n]*invfact[r])%mod1)*invfact[n-r])%mod1;
}
ll binarySearch(vector<ll> v, ll To_Find)
{
    int lo = 0, hi = v.size() - 1;
    int mid;
    while (hi - lo > 1) {
        int mid = (hi + lo) / 2;
        if (v[mid] < To_Find) {
            lo = mid + 1;
        }
        else {
            hi = mid;
        }
    }
    if (v[lo] == To_Find) {
        return lo;
    }
    else if (v[hi] == To_Find) {
        return hi;
    }
    else {
        return -1;
    }
}

ll ceilDiv(ll x, ll y){
    if(x%y==0){
        return (x/y);
    }
    else{
        return (x/y)+1;
    }
}

set<ll>pr;
void SieveOfEratosthenes(int n)
{
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));

    for (int p = 2; p * p <= n; p++) {
        if (prime[p] == true) {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
    for (int p = 2; p <= n; p++)
        if (prime[p])
        {
            pr.insert(p);
        }
}

inline void debugMode()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}

int main() {
    Noushil
    debugMode();

    int t;
    cin >> t;

    while (t--) {
        ll n,a,b,c;
        cin>>n;
        vector<ll>v(n);
        input(n,v);
        unordered_map<ll,vector<ll>>adj;
        vector<ll>deg(n);

        for(int i=0; i<n; i++){
            v[i]--;
            adj[i].push_back(v[i]);
            deg[v[i]]++;
        }

        // ll ans = 0;

        queue<ll>q;
        for(int i =0; i<n; i++){
            if(deg[i]==0){
                q.push(i);
            }
        }

        vector<ll>ans(n,1);

        while(!q.empty()){
            ll top = q.front();
            q.pop();
            // cout<<top+1<<" ";
            for(auto it: adj[top]){
                deg[it]--;
                ans[it]+=ans[top];

                if(deg[it]==0){
                    q.push(it);
                }
            }
        }
        sort(ans);
        cout<<(1+ans[n-1])<<endl;

    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
const bool MULTIPLE_TEST_CASES = 1;

#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "/tools-ln/overloads.cc"
#endif
void __SOLVE__();
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);
#ifdef LOCAL
    auto t0 = chrono::high_resolution_clock::now();
#endif
    int CASES = 1;
    if (MULTIPLE_TEST_CASES)
        cin >> CASES;
    while (CASES--)
        __SOLVE__();
#ifdef LOCAL
    auto t = chrono::high_resolution_clock::now();
    auto mus = chrono::duration_cast<std::chrono::microseconds>(t - t0);
    cout << "\n\033[33m[time taken: " << mus.count() << " μs]\033[0m";
#endif
}

const int maxn = 2e5;

void __SOLVE__() {
    int n;
    cin >> n;

    vector<int> r(n+1);
    vector<int> indeg(n+1, 0);
    for (int i = 1; i <= n; i++) {
        int x;
        cin >> x;
        r[i] = x;
        indeg[x]++;
    }

    vector<int> subtreesz(n+1, 1);

    queue<int> q;
    for (int i = 1; i <= n; i++) {
        if (indeg[i] == 0) {
            q.push(i);
        }
    }

    while (!q.empty()) {
        int node = q.front();
        q.pop();

        int succ = r[node];
        subtreesz[succ] += subtreesz[node];

        if (--indeg[succ] == 0) {
            q.push(succ);
        }
    }

    cout << ranges::max(subtreesz) + 1 << '\n';

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

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

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

from collections import deque
def solve():
    n, = map( int , input().strip().split() )
    a = list( map( lambda x:int(x)-1 , input().strip().split() ) )

    inv = [list() for _ in range(n)]
    for i,v in enumerate(a):
        inv[v].append(i)

    done = [False]*n
    out = 0

    def find(s):
        dfs = [s]
        for v in dfs:
            x = a[v]
            if not done[x]:
                dfs.append(x)
                done[x] = True
            else:
                break
        nonlocal out

        for v in dfs:
            for u in inv[v]:
                if not done[u]:
                    dfs.append(u)
        #print(dfs)
        dp = [1]*n
        for v in reversed(dfs):
            if done[v]: break
            done[v] = True
            out = max(out,dp[v])
            p = a[v]
            dp[p] += dp[v]

    def check(s):
        # finds the node in the cycle if any
        slow = a[s]
        fast = a[a[s]]

        while slow != fast:
            slow = a[slow]
            fast = a[a[fast]]

        find(slow)

    for i in range(n):
        if not done[i]:
            check(i)

    print(out+2)

q = 1
q = int(input().strip())
for i in range(q):
    solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
def II() -> int : return int(input())
def MI() -> int : return map(int, input().split())
def TI() -> tuple[int] : return tuple(MI())
def LI() -> list[int] : return list(MI())
#str-input
def SI() -> str : return input()
def MSI() -> str : return input().split()
def SI_L() -> list[str] : return list(SI())
def SI_LI() -> list[int] : return list(map(int, SI()))
#multiple-input
def LLI(n) -> list[list[int]]: return [LI() for _ in range(n)]
def LSI(n) -> list[str]: return [SI() for _ in range(n)]
#1-indexを0-indexでinput
def MI_1() -> int : return map(lambda x:int(x)-1, input().split())
def TI_1() -> tuple[int] : return tuple(MI_1())
def LI_1() -> list[int] : return list(MI_1())

from heapq import heappop,heappush

t = II()
for _ in range(t):
    n = II()
    r = LI_1()

    ecnt = [0]*n
    for i in range(n):
        ecnt[r[i]] += 1

    d = [1]*n
    que = []
    for i in range(n):
        if ecnt[i] == 0:
            que.append((1,i))

    while len(que) != 0:
        cnt,now = heappop(que)
        nxt = r[now]
        d[nxt] += cnt
        ecnt[nxt] -= 1

        if ecnt[nxt] == 0:
            heappush(que,(d[nxt],nxt))

    # print(d)
    print(max(d)+1)

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

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

Think step by step.

Response:
```plaintext
3
2
2 1
3
3 1 2
4
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: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.util.*;
public class G2993 {
    public static void main(String []args){
        try (Scanner sc = new Scanner(System.in)) {
            int t = sc.nextInt();
            while(t-- > 0){
                int n = sc.nextInt();
                int arr[]  = new int[n+1];
                int inDegree[] = new int[n+1];
                for(int i=1;i<=n;i++){
                    arr[i] = sc.nextInt();
                    inDegree[arr[i]]++;
                }

                Queue<Integer> queue = new LinkedList<>();

                for(int i=1;i<=n;i++){
                    if(inDegree[i] == 0){
                        queue.add(i);
                    }
                }
                if(queue.isEmpty()){
                    System.out.println("2");
                }else{
                    int cnt =1;
                    while(!queue.isEmpty()){
                        int len = queue.size();
                        while(len-- > 0){
                            int curr = queue.poll();

                            inDegree[arr[curr]]--;
                            if(inDegree[arr[curr]] == 0){
                                queue.add(arr[curr]);
                            }
                        }
                        if(!queue.isEmpty())
                            cnt++;
                    }
                    int []newDegree = new int[n+1];
                    for(int i=1;i<=n;i++){
                        newDegree[arr[i]]++;
                    }
                    int maxDelay = 0;
                    for(int i=1;i<=n;i++){
                        if(inDegree[i] == 0){
                            maxDelay = Math.max(newDegree[i],maxDelay);
                        }
                    }
                    System.out.println(cnt+maxDelay+1);
                }
            }
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios where the code might fail. The buggy code has a few issues, particularly in how it handles the in-degree and the stability check. Here are some key points to consider:

1. **Cyclic Dependencies**: The code might not handle cyclic dependencies correctly, especially when the spiders form a cycle.
2. **Initial State**: The code might not correctly handle the initial state where all spiders have the same number of plushies.
3. **Stability Check**: The code might not correctly identify when the process becomes stable.

Given these considerations, let's construct a test case that maximizes the probability of exposing these issues.

### Test Case Construction
1. **Cyclic Dependencies**: Create a scenario where spiders form a cycle.
2. **Initial State**: Ensure that the initial state is such that the process does not become stable immediately.
3. **Stability Check**: Create a scenario where the process becomes stable after a few years.

### Example Test Case
```plaintext
1
4
2 3 4 1
```

### Explanation
- **Number of Test Cases**: 1
- **Number of Spiders**: 4
- **Recipient Array**: [2, 3, 4, 1]

### Detailed Steps
1. **Initial State**: All spiders have 1 plushie.
2. **Year 1**:
   - Spider 1 gives to Spider 2.
   - Spider 2 gives to Spider 3.
   - Spider 3 gives to Spider 4.
   - Spider 4 gives to Spider 1.
   - State after Year 1: [0, 1, 1, 1]
3. **Year 2**:
   - Spider 2 gives to Spider 3.
   - Spider 3 gives to Spider 4.
   - Spider 4 gives to Spider 1.
   - State after Year 2: [1, 0, 1, 1]
4. **Year 3**:
   - Spider 3 gives to Spider 4.
   - Spider 4 gives to Spider 1.
   - State after Year 3: [1, 1, 0, 1]
5. **Year 4**:
   - Spider 4 gives to Spider 1.
   - State after Year 4: [1, 1, 1, 0]
6. **Year 5**:
   - State after Year 5: [1, 1, 1, 1]

The process becomes stable in Year 5.

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

This test input should expose the issues in the buggy code by creating a scenario where the process becomes stable after a few years, and the code needs to correctly handle the cyclic dependencies and stability check.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (hard version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The key difference between the two versions is highlighted in bold.

A group of $$$n$$$ spiders has come together to exchange plushies. Initially, each spider has $$$1$$$ plushie. Every year, if spider $$$i$$$ has at least one plushie, he will give exactly one plushie to spider $$$r_i$$$. Otherwise, he will do nothing. Note that all plushie transfers happen at the same time. In this version, each spider is allowed to have more than 1 plushie at any point in time.

The process is stable in the current year if each spider has the same number of plushies (before the current year's exchange) as he did the previous year (before the previous year's exchange). Note that year $$$1$$$ can never be stable.

Find the first year in which the process becomes stable.

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

The first line of each test case contains an integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the number of spiders.

The following line contains $$$n$$$ integers $$$r_1, r_2, \ldots, r_n$$$ ($$$1 \leq r_i \leq n, r_i \neq i$$$) — the recipient of the plushie of each spider.

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

Output: For each test case, output an integer on a new line, the first year in which the process becomes stable.

Examples:
input:
5
2
2 1
5
2 3 4 5 1
5
2 1 4 2 3
5
4 1 1 5 4
10
4 3 9 1 6 7 9 10 10 3
output:
2
2
5
5
5

Note: For the second test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Since this array is the same as the previous year, this year is stable.

For the third test case:

- At year $$$1$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 1, 1, 1]$$$. Then, year $$$1$$$'s exchange happens.
- At year $$$2$$$, the following array shows the number of plushies each spider has: $$$[1, 2, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 3, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 4, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

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

 public class Solution {

   static List<Integer>[] adj; // Adjacency list for the graph
    static boolean[] visited;   // Marks visited nodes during DFS
    static boolean[] inStack;   // Keeps track of nodes currently in the stack
    static Stack<Integer> stack; // Stack for Tarjan's algorithm
    static int[] low;           // Lowest discovery time reachable from a node
    static int[] discovery;     // Discovery time of nodes
    static int time = 0;        // Global timer for discovery times
    static Set<Integer> cycleVertices; // Stores all vertices part of any cycle

    static void solve() {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0) {
            int n = sc.nextInt()+1;

            List<Integer>[] graph = new ArrayList[n];
            for (int i = 0; i < n; i++) {
                graph[i] = new ArrayList<>();
            }

            int sol[] = new int[n];
            for(int i=1;i<n;i++) {
                int x = sc.nextInt();
                graph[i].add(x);
                sol[x]++;
            }

            HashMap<Integer, Integer> map = new HashMap();

            Queue<Integer> queue = new LinkedList<>();
            boolean visit[] = new boolean[n];
            for(int i=1;i<n;i++) {
                if(sol[i] == 0){
                    map.put(i, 1);
                    queue.add(i);
                    visit[i] = true;
                }
            }

            Set<Integer> cVert = findVerticesInCycles(n, graph);

            int max = 0;
            while(!queue.isEmpty()) {
                int tt = queue.poll();
                int val = map.get(tt);
                max = Math.max(val, max);

                for(int nbr: graph[tt]) {
                    if(!cVert.contains(nbr)) {
                            //out.println(nbr+":--");
                        if(!visit[nbr]) {
                            queue.add(nbr);
                            visit[nbr] = true;
                        }
                        map.put(nbr, map.getOrDefault(nbr, 1)+val);
                    }
                }
            }

            out.println(max+2);

        }
    }
    public static Set<Integer> findVerticesInCycles(int n, List<Integer>[] graph) {
        adj = graph;
        visited = new boolean[n];
        inStack = new boolean[n];
        stack = new Stack<>();
        low = new int[n];
        discovery = new int[n];
        cycleVertices = new HashSet<>();

        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                tarjanDFS(i);
            }
        }
        return cycleVertices;
    }

    private static void tarjanDFS(int node) {
        visited[node] = true;
        discovery[node] = low[node] = ++time; // Set discovery and low times
        stack.push(node);
        inStack[node] = true;

        for (int neighbor : adj[node]) {
            if (!visited[neighbor]) {
                // Recur for unvisited neighbors
                tarjanDFS(neighbor);
                low[node] = Math.min(low[node], low[neighbor]);
            } else if (inStack[neighbor]) {
                // Update low[node] if neighbor is in the current stack
                low[node] = Math.min(low[node], discovery[neighbor]);
            }
        }

        // If node is a root of an SCC
        if (low[node] == discovery[node]) {
            List<Integer> scc = new ArrayList<>();
            int w;
            do {
                w = stack.pop();
                inStack[w] = false;
                scc.add(w);
            } while (w != node);

            // If SCC has more than one vertex, all of them belong to a cycle
            if (scc.size() > 1) {
                cycleVertices.addAll(scc);
            } else {
                // Single node forms a cycle only if it has a self-loop
                if (adj[node].contains(node)) {
                    cycleVertices.add(node);
                }
            }
        }
    }

static BufferedReader br;
static PrintWriter out;
static StringTokenizer st;

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

    long c = System.currentTimeMillis();

    solve();

            //System.out.println("time--"+ (System.currentTimeMillis() - c));
    out.close();

}

static String next() throws IOException {
    while (st == null || !st.hasMoreTokens())
        st = new StringTokenizer(br.readLine().trim());
    return st.nextToken();
}

static long readLong() throws IOException {
    return Long.parseLong(next());
}

static int readInt() throws IOException {
    return Integer.parseInt(next());
}

static double readDouble() throws IOException {
    return Double.parseDouble(next());
}

static char readCharacter() throws IOException {
    return next().charAt(0);
}

static String readLine() throws IOException {
    return br.readLine().trim();
}
}

**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 that can cause the code to fail. Here are the steps to think through this:

1. **Identify the Bug**: The buggy code has a few potential issues:
   - The graph is constructed with an extra node (n+1 instead of n), which can cause index out-of-bounds errors.
   - The `sol` array is not properly initialized for all nodes.
   - The cycle detection logic might not handle all edge cases correctly.

2. **Edge Cases to Consider**:
   - **Small Number of Spiders**: Test with the minimum number of spiders (2) to ensure the code handles small inputs correctly.
   - **Cycles in the Graph**: Test with a graph that has cycles to ensure the cycle detection logic works.
   - **Isolated Nodes**: Test with nodes that do not have any incoming or outgoing edges to see how the code handles isolated nodes.
   - **Self-loops**: Test with self-loops to ensure the code handles self-loops correctly.

3. **Construct the Test Input**:
   - We will create a test case with 3 spiders to ensure it is small and manageable.
   - We will create a cycle to test the cycle detection logic.
   - We will include a self-loop to test how the code handles self-loops.

Here is the test input:

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

Explanation:
- **Test Case 1**:
  - **Number of Spiders**: 3
  - **Recipient List**: [2, 3, 2]
    - Spider 1 gives to Spider 2.
    - Spider 2 gives to Spider 3.
    - Spider 3 gives to Spider 2.

This input creates a cycle involving Spiders 2 and 3, and it will test how the code handles cycles and self-loops. The expected output should be 2, as the process becomes stable after the first year.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Genokraken
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is an interactive problem.

Upon clearing the Waterside Area, Gretel has found a monster named Genokraken, and she's keeping it contained for her scientific studies.

The monster's nerve system can be structured as a tree$$$^{\dagger}$$$ of $$$n$$$ nodes (really, everything should stop resembling trees all the time$$$\ldots$$$), numbered from $$$0$$$ to $$$n-1$$$, with node $$$0$$$ as the root.

Gretel's objective is to learn the exact structure of the monster's nerve system — more specifically, she wants to know the values $$$p_1, p_2, \ldots, p_{n-1}$$$ of the tree, where $$$p_i$$$ ($$$0 \le p_i < i$$$) is the direct parent node of node $$$i$$$ ($$$1 \le i \le n - 1$$$).

She doesn't know exactly how the nodes are placed, but she knows a few convenient facts:

- If we remove root node $$$0$$$ and all adjacent edges, this tree will turn into a forest consisting of only paths$$$^{\ddagger}$$$. Each node that was initially adjacent to the node $$$0$$$ will be the end of some path.
- The nodes are indexed in a way that if $$$1 \le x \le y \le n - 1$$$, then $$$p_x \le p_y$$$.
- Node $$$1$$$ has exactly two adjacent nodes (including the node $$$0$$$).

The tree in this picture does not satisfy the condition, because if we remove node $$$0$$$, then node $$$2$$$ (which was initially adjacent to the node $$$0$$$) will not be the end of the path $$$4-2-5$$$.The tree in this picture does not satisfy the condition, because $$$p_3 \le p_4$$$ must hold.The tree in this picture does not satisfy the condition, because node $$$1$$$ has only one adjacent node.

Gretel can make queries to the containment cell:

- "? a b" ($$$1 \le a, b < n$$$, $$$a \ne b$$$) — the cell will check if the simple path between nodes $$$a$$$ and $$$b$$$ contains the node $$$0$$$.

However, to avoid unexpected consequences by overstimulating the creature, Gretel wants to query at most $$$2n - 6$$$ times. Though Gretel is gifted, she can't do everything all at once, so can you give her a helping hand?

$$$^{\dagger}$$$A tree is a connected graph where every pair of distinct nodes has exactly one simple path connecting them.

$$$^{\ddagger}$$$A path is a tree whose vertices can be listed in the order $$$v_1, v_2, \ldots, v_k$$$ such that the edges are $$$(v_i, v_{i+1})$$$ ($$$1 \le i < k$$$).

Input: Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$4 \le n \le 10^4$$$) — the number of nodes in Genokraken's nerve system.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output: None

Examples:
input:
3
4

1

5

1

0

9
output:
? 2 3

! 0 0 1

? 2 3

? 2 4

! 0 0 1 2

! 0 0 0 1 3 5 6 7

Note: In the first test case, Genokraken's nerve system forms the following tree:

- The answer to "? 2 3" is $$$1$$$. This means that the simple path between nodes $$$2$$$ and $$$3$$$ contains node $$$0$$$.

In the second test case, Genokraken's nerve system forms the following tree:

- The answer to "? 2 3" is $$$1$$$. This means that the simple path between nodes $$$2$$$ and $$$3$$$ contains node $$$0$$$.
- The answer to "? 2 4" is $$$0$$$. This means that the simple path between nodes $$$2$$$ and $$$4$$$ doesn't contain node $$$0$$$.

In the third test case, Genokraken's nerve system forms the following tree:

Buggy Code:
#include<iostream>
using namespace std;
const int N=1e4+10;
int p[N],cnt,n;
struct Line
{
	int pre,next;
	int num;
}line[N];
void Init()
{
	if(cnt>1)
	{
		line[cnt]={cnt-1,1,cnt};
		line[1]={cnt,2,1};
	}
	else line[1]={1,1,1};

	for(int i=2;i<cnt;i++)line[i]={i-1,i+1,i};
}
int query(int a,int b)
{
	cout<<"? "<<a<<' '<<b<<endl;
	int x;
	cin>>x;
	if(x==1)return 0;
	else return 1;
}
void solve()
{
	cin>>n;
	if(n==4)
	{
		int a1=query(1,2);
		if(a1)cout<<"! 0 1 2"<<endl;
		else cout<<"! 0 0 1"<<endl;
		return;
	}
    cnt=1;
	p[1]=0;
	for(int i=2;i<n;i++)
	{
		if(!query(1,i))
		{
			cnt++;
			p[i]=0;
		}
		else break;
	}
	Init();
	for(int res=cnt+1,i=1;res<n;res++,i=line[i].next)
	{
		if(query(res,i))
		{
			p[res]=line[i].num;
			line[i].num=res;
		}
		else
		{
			res--;
			line[line[i].pre].next=line[i].next;
			line[line[i].next].pre=line[i].pre;
		}
	}
	cout<<'!';
	for(int i=1;i<n;i++)cout<<' '<<p[i];
	cout<<endl;
	cout.flush();
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	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 challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential edge cases and scenarios where it might fail.

### Analysis of the Buggy Code
1. **Initialization and Query Handling**:
   - The code initializes the tree and handles queries to determine the structure of the tree.
   - It uses a linked list to manage the nodes and their connections.

2. **Handling Small Trees (n=4)**:
   - For a small tree with 4 nodes, it directly queries the path between nodes 1 and 2 and determines the parent based on the result.

3. **General Tree Handling**:
   - It iterates through the nodes to determine which nodes are directly connected to the root node (0).
   - It then uses a linked list to manage the remaining nodes and their connections.

### Potential Bugs
1. **Incorrect Handling of Node 1**:
   - The code assumes that node 1 has exactly two adjacent nodes, including the root node. This is a key condition that must be satisfied.
   - If this condition is not met, the code will fail to correctly determine the structure of the tree.

2. **Incorrect Path Query Handling**:
   - The code uses the `query` function to determine if the path between two nodes contains the root node.
   - If the query results are incorrect or the logic for handling the results is flawed, the tree structure will be incorrect.

### Challenging Test Input
To expose these potential bugs, we can create a test case where:
- Node 1 has exactly two adjacent nodes, including the root node.
- The tree structure is such that the path queries will be challenging to handle correctly.

### Test Input
```plaintext
1
5
```

### Explanation
- **Test Case 1**: A tree with 5 nodes.
- **Tree Structure**:
  - Node 0 (root) is connected to nodes 1 and 2.
  - Node 1 is connected to node 3.
  - Node 2 is connected to node 4.
- **Expected Output**:
  - The path between nodes 3 and 4 should not contain node 0.
  - The path between nodes 1 and 3 should contain node 0.
  - The path between nodes 2 and 4 should contain node 0.

This test input is small-scale and complete, and it should expose the bug in the buggy code by challenging the handling of the path queries and the structure of the tree.