**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.util.*;

public class G {

    public static int max = 0;

    public static int plushies(int[] g, float[][] a, int v, int idx) {

        if (a[idx][0] == -0.5) {
            return 0;
        }

        if (a[idx][0] < 0) {
            return idx + 1;
        }

        if (a[idx][0] >= v) {
            return (int) -(a[idx][1] + 1);
        }

        if (a[idx][0] > 0 && a[idx][0] < v) {
            // System.out.println(a[idx][1] + " " + a[(int) a[idx][1]][0] + " " +
            // a[idx][0]);
            a[(int) a[idx][1]][0] += v - a[idx][0];
            // System.out.println(a[idx][1] + " " + a[(int) a[idx][1]][0] + " " +
            // a[idx][0]);
            max = (int) Math.max(a[(int) a[idx][1]][0], max);
            return (int) -(a[idx][1] + 1);
        }

        a[idx][0] = -v;
        int temp = plushies(g, a, v + 1, g[idx]);

        if (temp == idx + 1) {
            a[idx][0] = -0.5f;
            return 0;
        } else if (temp == 0) {
            a[idx][0] = v;
            a[idx][1] = idx;
            max = Math.max(max, v);
            return -(idx + 1);
        } else if (temp < 0) {
            a[idx][0] = v;
            a[idx][1] = -temp - 1;
            max = Math.max(max, v);
            return temp;
        } else {
            a[idx][0] = -0.5f;
            return temp;
        }

    }

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

        int tt = sc.nextInt();

        for (int i = 0; i < tt; i++) {
            int n = sc.nextInt();

            int g[] = new int[n];
            float a[][] = new float[n][2];

            for (int j = 0; j < n; j++) {
                int temp = sc.nextInt();
                g[j] = temp - 1;
            }

            for (int j = 0; j < n; j++) {
                if (a[j][0] == 0) {
                    plushies(g, a, 1, j);
                }
            }

            System.out.println(max + 2);
            max = 0;

        }

        sc.close();

    }
}

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

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

Think step by step.

Response:
```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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

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

using namespace std;

#define ll              long long
#define pb              push_back
#define PI              acos(-1.0)
#define MEM(a,x)        memset(a,x,sizeof(a))
#define fast           ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const int mx = 200005;
const int mod = 1e9 + 7;

// ================================== take ip/op like vector,pairs directly!==================================
template<typename typC,typename typD> istream &operator>>(istream &cin,pair<typC,typD> &a) { return cin>>a.first>>a.second; }
template<typename typC> istream &operator>>(istream &cin,vector<typC> &a) { for (auto &x:a) cin>>x; return cin; }
template<typename typC,typename typD> ostream &operator<<(ostream &cout,const pair<typC,typD> &a) { return cout<<a.first<<' '<<a.second; }
template<typename typC,typename typD> ostream &operator<<(ostream &cout,const vector<pair<typC,typD>> &a) { for (auto &x:a) cout<<x<<'\n'; return cout; }
template<typename typC> ostream &operator<<(ostream &cout,const vector<typC> &a) { int n=a.size(); if (!n) return cout; cout<<a[0]; for (int i=1; i<n; i++) cout<<' '<<a[i]; return cout; }
// ===================================END Of the input module ==========================================

// Custom hash function for vector<int>
struct VectorHash {
    size_t operator()(const vector<int>& v) const {
        size_t hash = 0;
        for (int i : v) {
            hash ^= hash * 31 + i;
        }
        return hash;
    }
};

void solve(){
    int n;
    cin >> n;
    vector<int> rec(n);
    for (int i = 0; i < n; ++i) {
        cin >> rec[i];
        rec[i]--;
    }
    vector<int> cur_plu(n, 1);
    unordered_map<vector<int>, int, VectorHash> stt;

    for (int year = 1; ; ++year) {
        if (year > 1 && stt.size() > mx) {
            cout << "-1" << endl;
            return;
        }
        if (year > 1 && stt.count(cur_plu)) {
            cout << year << endl;
            return;
        }
        stt[cur_plu] = year;
        vector<int> nxt_pl(n, 0);
        for (int i = 0; i < n; ++i) {
            if (cur_plu[i] > 0) {
                nxt_pl[rec[i]]++;
            }
        }
        for (int i = 0; i < n; ++i) {
            nxt_pl[i] = min(nxt_pl[i], 1);
        }
        cur_plu = nxt_pl;
    }
}

int main(){
    fast
    int t = 1;
    cin >> t;
    int cse = 0;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that might cause the code to fail. The key aspects to consider are:

1. **Cycles in the Recipient Array**: The recipient array `r` can form cycles, which can lead to complex behavior in the plushie distribution.
2. **Initial Conditions**: The initial state of each spider having one plushie and the constraints on the number of plushies.
3. **Stability Conditions**: The code should correctly identify when the process becomes stable.

Given these considerations, let's create a test case that includes a cycle and a scenario where the process might not become stable quickly.

### Test Case
We will create a test case with a cycle in the recipient array and ensure that the process does not become stable immediately. This will help us check if the code correctly handles the cycles and the stability condition.

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

### Explanation
- **Test Case 1**:
  - **Number of Spiders (n)**: 5
  - **Recipient Array (r)**: [2, 3, 4, 5, 1]
  - **Initial Plushie Distribution**: [1, 1, 1, 1, 1]

### Expected Behavior
- **Year 1**:
  - Plushie distribution after the first year: [1, 1, 1, 1, 1]
- **Year 2**:
  - Plushie distribution after the second year: [1, 1, 1, 1, 1]
- **Year 3**:
  - Plushie distribution after the third year: [1, 1, 1, 1, 1]

Since the distribution remains the same after the first year, the process becomes stable in the second year.

### Output
The expected output for this test case is:
```plaintext
2
```

This test case should expose any issues in the code related to handling cycles and correctly identifying the stability condition.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#include <bits/stdc++.h>
#define int int
#define boost ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;
const int N = 2e5 + 4;
const int inf = 1e9;
const int mod=998244353;
struct edge{
    int x,y,c;
};
vector<int>v;
int dp[N][30];
int dp1[N];
int n;
int dp2[N];
int o=8;
int def(int i,int x){
    int y=i;
    for(int j=o;j>=0;j--){
        if(x>=pow(2,j)){
            x-=pow(2,j);
            y=dp[y][j];
        }
    }
    return y;
}
int ch(int x){
            for(int i=1;i<=n;i++){
                int z=def(i,x);
                int z1=def(i,x-1);
                dp1[z]++;
                dp1[z]=min(dp1[z],1);
                dp2[z1]++;
                dp2[z1]=min(dp2[z1],1);
            }
            int ok=0;
            for(int i=1;i<=n;i++){
                if(dp1[i]!=dp2[i]){
                    ok=1;
                }
                dp1[i]=0;
                dp2[i]=0;
            }
            return ok;
}
signed main(){
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
    boost
    int t;
    cin>>t;
    while(t--){
        cin>>n;
        v.push_back(0);
        for(int i=0;i<n;i++){
            int x;
            cin>>x;
            v.push_back(x);
            dp[i+1][0]=x;
        }
        for(int i=1;i<=o;i++){
            for(int j=1;j<=n;j++){
                dp[j][i]=dp[dp[j][i-1]][i-1];
            }
        }
        int l=1,r=pow(2,o);
        while(l+1<r){
            int mid=(l+r)/2;
            if(ch(mid)==0){
                r=mid;
            }else{
                l=mid;
            }
        }
        if(ch(l)==0){
            r=l;
        }
        cout<<r+1<<"\n";
        for(int i=1;i<=n;i++){
            for(int j=0;j<=o;j++){
                dp[i][j]=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 given code, we need to focus on edge cases that might cause the algorithm to fail. The key points to consider are:

1. **Cycles in the transfer graph**: If the transfer graph forms cycles, the algorithm might not handle it correctly.
2. **Multiple spiders giving to the same spider**: The code should correctly handle scenarios where multiple spiders give plushies to the same spider, but the spider can only keep one.
3. **Initial conditions**: Ensure that the initial conditions are set up correctly, especially for small values of `n`.

Let's create a test input that includes a cycle and multiple spiders giving to the same spider:

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

### Explanation:
1. **Test Case 1**:
   - `n = 3`
   - `r = [2, 3, 1]`
   - This forms a cycle: 1 -> 2 -> 3 -> 1.
   - The process should become stable after 2 years.

2. **Test Case 2**:
   - `n = 4`
   - `r = [2, 3, 4, 1]`
   - This forms a cycle: 1 -> 2 -> 3 -> 4 -> 1.
   - The process should become stable after 2 years.

3. **Test Case 3**:
   - `n = 5`
   - `r = [2, 3, 4, 5, 1]`
   - This forms a cycle: 1 -> 2 -> 3 -> 4 -> 5 -> 1.
   - The process should become stable after 2 years.

These test cases should help expose any issues in the code related to handling cycles and multiple spiders giving to the same spider.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
t = int(input())

for _ in range(t):
    n = int(input())
    r = list(map(int, input().split()))
    prev = [1] * n
    curr = [0] * n

    year = 2
    while True:

        curr = [0] * n

        for i in range(n):
            if prev[i] > 0:
                curr[r[i] - 1] += 1

        for i in range(n):
            if curr[i] > 1:
                curr[i] = 1

        if curr == prev:
            print(year)
            break

        prev = curr[:]
        year += 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
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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
def tarjan(u):

    global idx, scc_count
    index[u] = lowlink[u] = idx
    idx += 1
    st.append(u)
    onStack[u] = 1
    for v in graph[u]:
        if index[v] == -1:
            tarjan(v)
            lowlink[u] = min(lowlink[u], lowlink[v])
        elif onStack[v]:
            lowlink[u] = min(lowlink[u], index[v])
    if lowlink[u] == index[u]:
        scc_nodes = []
        while 1:
            w = st.pop()
            onStack[w] = 0
            scc_id[w] = scc_count
            scc_nodes.append(w)
            if w == u:
                break
        is_cycle = 0
        if len(scc_nodes) > 1:
            is_cycle = 1
        else:
            node = scc_nodes[0]
            if graph[node][0] == node:
                is_cycle = 1
        is_cycle_scc.append(is_cycle)
        scc_count += 1

def dfs(u):
    if depth[u] != -1:
        return depth[u]
    depth[u] = 0
    max_child_depth = 0
    for v in condensed_graph[u]:
        max_child_depth = max(max_child_depth, dfs(v))
    if not is_cycle_scc[u]:
        depth[u] = max_child_depth + 1
    return depth[u]

for t in range(int(input())):
    n = int(input())
    r = list(map(int, input().split()))
    graph = [[] for i in range(n)]
    for i in range(n):
        graph[i].append(r[i]-1)

    index = [-1] * n
    lowlink = [-1] * n
    onStack = [0] * n
    st = []
    idx = 0
    scc_id = [-1] * n
    scc_count = 0
    is_cycle_scc = []

    for i in range(n):
        if index[i] == -1:
            tarjan(i)

    condensed_graph = [[] for i in range(scc_count)]
    for u in range(n):
        for v in graph[u]:
            if scc_id[u] != scc_id[v]:
                condensed_graph[scc_id[u]].append(scc_id[v])

    depth = [-1] * scc_count
    max_depth = 0
    for i in range(scc_count):
        if is_cycle_scc[i]:
            depth[i] = 0
        max_depth = max(max_depth, dfs(i))

    print(max_depth + 2)

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

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

Think step by step.

Response:
```plaintext
1
5
2 3 4 5 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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Solution {

	public static int getCycleLen(int[] arr, int i) {
		int slow = arr[i - 1];
		int fast = arr[arr[i - 1] - 1];

		while (slow != fast) {
			slow = arr[slow - 1];
			fast = arr[arr[fast - 1] - 1];
		}

		int l = 2;
		while (slow != i) {
			slow = arr[slow - 1];
			i = arr[i - 1];
			l++;
		}

		return l;
	}

	public static int solve(int[] arr) {
		int n = arr.length;

		Set<Integer> s = new HashSet<>();
		for (int e : arr)
			s.add(e);

		int maxL = 2;
		for (int i = 1; i <= n; i++) {
			if (!s.contains(i)) {
				int l = getCycleLen(arr, i);
				if (l > maxL) {
					maxL = l;
				}
			}
		}

		return maxL;
	}

	public static void main(String[] args) {
		try (Scanner console = new Scanner(System.in)) {
			int tests = console.nextInt();

			for (int test = 0; test < tests; test++) {
				int n = console.nextInt();

				int[] arr = new int[n];
				for (int i = 0; i < n; i++)
					arr[i] = console.nextInt();

				int r = solve(arr);
				System.out.println(r);
			}
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
3
3
2 3 1
4
2 3 4 1
5
2 3 4 5 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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.util.*;
import java.io.*;
public class Main
{
    // static int M=(int)(1e9+7);
    static int M=998244353;
    static FastReader in  = new FastReader();
    static FastWriter out = new FastWriter();

    static double PI=3.1415926535897932384626433832795;
    @SuppressWarnings("unchecked")

    // हर हर महादेव
    // हरे कृष्णा

    //10^6 space of int is allowed not of long
    // Yesterday was history
    // Tomorrow will be a mystery
    // but today is a gift
    // That is why it is called the present

    /* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
    */

    //Credit Striver
    // Remember you were also a novice when you started,
    // hence never be rude to anyone who wants to learn something
    // Never open a ranklist untill and unless you are done with solving problems, wastes 3/4 minuts
    // Donot treat CP as a placement thing, love it and ejoy it, you will suceed for sure.

    public static void main(String[] args)throws Exception
    {
        // in = new FastReader("input11.txt");
        // out = new FastWriter("mootube.out");
        int testcases=1;

        testcases=in.nextInt();

        // testcases=sc.nextInt();//For interactive problems;
        outer:
        for(int tt=1;tt<=testcases;tt++)
        {
            solve();
            // YES(solve());
            // Yes(solve());
            // out.println("Case #"+tt+": "+solve());
        }
        out.close();

    }

    static ArrayList<ArrayList<Integer>> graph;
    static ArrayList<ArrayList<Integer>> graph1;
    static ArrayList<ArrayList<Edge>> g1;
    static ArrayList<ArrayList<Edge>> g2;

    static boolean visited[];
    static int color[];
    static ArrayList<ArrayList<Pair>> graphh=new ArrayList<>();
    static Scanner sc=new Scanner(System.in);

    @SuppressWarnings("unchecked")
    public static void solve()throws Exception
    {
        int n=in.nextInt();
        int ans=2;
        int ar[]=in.nextIntArray(n);
        for(int i=0;i<n;i++){
            ar[i]--;
        }
        int x[]=new int[n];
        Arrays.fill(x,1);
        int xx[]=new int[n];
        Arrays.fill(xx,1);
        for(int i=0;i<100;i++)
        {
            for(int j=0;j<n;j++)xx[j]=x[j];
            for(int j=0;j<n;j++)
            {
                if(x[j]>=1)
                x[ar[j]]++;
                if(x[j]>=1)
                x[j]--;
            }
            for(int  j=0;j<n;j++)
            {
                if(x[j]>1)x[j]=1;
            }
            boolean ff=true;

            for(int j=0;j<n;j++)
            {
                if(x[j]!=xx[j])
                {
                    ff=false;
                }
            }
            // printArray(x);

            if(ff)
            {
                ans=max(ans,i+2);
                break;
            }
        }
        out.println(ans);
    }

    public static long abs(long a)
    {
        return Math.abs(a);
    }
    public static int abs(int a)
    {
        return Math.abs(a);
    }
    public static double abs(double a)
    {
        return Math.abs(a);
    }
    public static int first(Integer ar[],int k)
    {
        int n=ar.length;
        if(k>ar[n-1])return n;
        int ans=0;
        int min=0;
        int max=n-1;
        while(max>=min)
        {
            int mid=min+(max-min)/2;
            if(ar[mid]>=k)
            {
                ans=mid;
                max=mid-1;
            }
            else min=mid+1;
        }
        return ans;
    }

    public static int last(Integer ar[],int k)
    {
        if(k<ar[0])
        {
            return 0;
        }
        int n=ar.length;
        if(k>=ar[n-1])return n;
        int ans=n-1;
        int min=0;
        int max=n-1;
        while(max>=min)
        {
            int mid=min+(max-min)/2;

            if(ar[mid]>k)
            {
                ans=mid;
                max=mid-1;
            }
            else min=mid+1;
        }

        return ans;
    }

    public static long gcd(long a,long b)//Note a>b
    {
        if(b==0)return a;
        return gcd(b,a%b);
    }

    public static long lcm(long a,long b)
    {
        return a*b/gcd(a,b);
    }

    public static long pow(long a, long n, long M)
    {
        if (n == 0) return 1;
        long ans = 1;
        a %= M;
        while (n != 0) {
            if ((n & 1) == 1) {
                ans = (ans % M * a % M) % M;
            }
            a = (a % M * a % M) % M;
            n /= 2;
        }
        return ans;
    }

    public static int max(int ... a)
    {
        int max = a[0];
        for (int i : a)
        {
            if (i > max)
            {
                max = i;
            }
        }
        return max;
    }

    public static long max(long ... a)
    {
        long max = a[0];
        for (long i : a) {
            if (i > max) {
                max = i;
            }
        }
        return max;
    }

    public static double max(double ... a)
    {
        double max = a[0];
        for (double i : a) {
            if (i > max) {
                max = i;
            }
        }
        return max;
    }

    // Minimum functions
    public static int min(int ... a)
    {
        int min = a[0];
        for (int i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static long min(long ... a)
    {
        long min = a[0];
        for (long i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static double min(double ... a)
    {
        double min = a[0];
        for (double i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static boolean isSorted(long ar[])
    {
        int n=ar.length;
        for(int i=0;i<n-1;i++)
        {
            if(ar[i]>ar[i+1])return false;
        }
        return true;
    }

    public static int maxIntArray(ArrayList<Integer> ar)
    {
        int max=ar.get(0);
        for(int i=0;i<ar.size();i++)max=Math.max(max,ar.get(i));
        return max;
    }

    private static int countDigits(long l)
    {
        if (l >= 1000000000000000000L) return 19;
        if (l >= 100000000000000000L) return 18;
        if (l >= 10000000000000000L) return 17;
        if (l >= 1000000000000000L) return 16;
        if (l >= 100000000000000L) return 15;
        if (l >= 10000000000000L) return 14;
        if (l >= 1000000000000L) return 13;
        if (l >= 100000000000L) return 12;
        if (l >= 10000000000L) return 11;
        if (l >= 1000000000L) return 10;
        if (l >= 100000000L) return 9;
        if (l >= 10000000L) return 8;
        if (l >= 1000000L) return 7;
        if (l >= 100000L) return 6;
        if (l >= 10000L) return 5;
        if (l >= 1000L) return 4;
        if (l >= 100L) return 3;
        if (l >= 10L) return 2;
        return 1;
    }
    public static int minIntArray(ArrayList<Integer> ar)
    {

        int n=ar.size();
        int max=ar.get(0);
        for(int i=0;i<n;i++)max=Math.min(max,ar.get(i));
        return max;
    }
    public static int maxIntArray(int ar[])
    {
        int max=ar[0];
        for(int i=0;i<ar.length;i++)max=Math.max(max,ar[i]);
        return max;
    }

    public static int minIntArray(int ar[])
    {

        int max=ar[0];
        for(int i=0;i<ar.length;i++)max=Math.min(max,ar[i]);
        return max;
    }

    public static void endl()throws Exception
    {
        out.println();
    }
    public static boolean isSorted(int ar[],int n)
    {
        if(n==1)return true;
        int diff=ar[1]-ar[0];
        for(int i=1;i<n;i++)
        {
            diff=Math.min(diff,ar[i]-ar[i-1]);
        }
        if(diff<0)return false;
        return true;
    }
    public static boolean isSorted(Integer ar[],int n)

    {
        if(n==1)return true;
        int diff=ar[1]-ar[0];
        for(int i=1;i<n;i++)
        {
            diff=Math.min(diff,ar[i]-ar[i-1]);
        }
        if(diff<0)return false;
        return true;
    }

    // public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }

    public static <K> void removeMap(Map<K, Integer> map, K element)
    {
        if (!map.containsKey(element)) return;
        int currentCount = map.get(element);
        if (currentCount == 1) {
            map.remove(element);
        } else {
            map.put(element, currentCount - 1);
        }
    }

    public static <K> void addMap(Map<K, Integer> map, K element)
    {
        if (map.containsKey(element)) {
            map.put(element, map.get(element) + 1);
        } else {
            map.put(element, 1);
        }
    }

   // public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }
    public static void removeMap(TreeMap<Integer,Integer> map,int element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(TreeMap<Long,Integer> map,long element)
    {
        if(!map.containsKey(element))return;
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(HashMap<Long,Integer> map,long element)
    {
        if(!map.containsKey(element))return;
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(HashMap<Integer,Integer> map,int element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }

    public static void removeMap(HashMap<String,Integer> map,String element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    // public static void addMap(TreeMap<Object,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }

    public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static void addMap(TreeMap<String,Integer> map,String element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static void addMap(TreeMap<Long,Integer> map,long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Integer,Integer> map,int element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Long,Long> map,long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1L);
    }
    public static void addMap(HashMap<Object,Integer> map,Integer element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Long,Integer> map,Long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<String,Integer> map,String element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static int[] nextIntArray(int n)throws Exception
    {
        int ar[]=new int[n];
        for(int i=0;i<n;i++)ar[i]=in.nextInt();
        return ar;
    }
    public static long[] nextlongArray(int n)throws Exception//For long
    {
        long ar[]=new long[n];
        for(int i=0;i<n;i++)ar[i]=in.nextLong();
        return ar;
    }
    public static Integer[] nextIntegerArray(int n)throws Exception
    {
        Integer ar[]=new Integer[n];
        for(int i=0;i<n;i++)ar[i]=in.nextInt();
        return ar;
    }
    public static Long[] nextLongArray(int n)//For Long array
    {
        Long ar[]=new Long[n];
        for(int i=0;i<n;i++)ar[i]=in.nextLong();
        return ar;
    }
    public static String[] nextStringArray(int n)
    {
        String ar[]=new String[n];
        for(int i=0;i<n;i++)ar[i]=in.next();
        return ar;
    }
    public static void Yes(boolean flag)throws Exception
    {
        if(flag)out.println("Yes");
        else  out.println("No");
    }
    public static void YES(boolean flag)throws Exception
    {
        if(flag)out.println("YES");
        else  out.println("NO");
    }
    public static void YES()throws Exception
    {
        out.println("YES");
    }
    public static void NO()throws Exception
    {
        out.println("NO");
    }
    public static void Yes()throws Exception
    {
        out.println("Yes");
    }
    public static void No()throws Exception
    {
        out.println("No");
    }

    public static void printArrayListPair(ArrayList<Pair> al)throws Exception
    {
        for(int i=0;i<al.size();i++)
        {
            out.println(al.get(i).a+" "+al.get(i).b);
        }
    }

    public static void printArray(boolean ar[])throws Exception{
        for(boolean x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(char ar[])throws Exception{
        for(char x:ar)out.print(x+" ");
        out.println();
    }

    public static void printArray(int ar[])throws Exception
    {
        for(int x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(Object ar[])throws Exception
    {
        for(Object x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(long ar[])throws Exception
    {
        for(long x:ar)out.print(x+" ");
        out.println();
    }
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(String filename) throws FileNotFoundException
        {
            br = new BufferedReader(new FileReader(filename));
        }
        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());
        }
        int[] nextIntArray(int n)
        {
            int ar[]=new int[n];
            for(int i=0;i<n;i++)ar[i]=Integer.parseInt(next());
            return ar;
        }
        long[] nextlongArray(int n)
        {
            long ar[]=new long[n];
            for(int i=0;i<n;i++)ar[i]=Long.parseLong(next());
            return ar;
        }
        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(String filename) throws IOException
        {
            this.bw = new BufferedWriter(new FileWriter(filename));
        }
    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 println() throws IOException
        {
            bw.append("\n");
        }
    public void close() throws IOException
        {
            bw.close();
        }
    }
}

class Edge
{
    int a,b;long weight;
    Edge(int a,int b,long weight)
    {
        this.a=a;
        this.b=b;
        this.weight=weight;
    }
    Edge(int a,long weight)
    {
        this.a=a;
        this.weight=weight;
    }
}

class compareEdge implements Comparator<Edge>
{
    public int compare(Edge p1,Edge p2)
    {
        return Long.compare(p1.weight, p2.weight);
    }
}

class Pair
{
   int a,b,c;
// long a;
// long b;
    int d;
    boolean flag;
    // Pair(int a,int b,boolean flag)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.flag=flag;
    // }
    // Pair(long a,long b)
    // {
    //     this.a=a;
    //     this.b=b;
    // }
    // Pair(long a,long b,long c)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    // }
    // Pair(int a,int b)
    // {
    //     this.a=a;
    //     this.b=b;
    // }
    Pair(int a,int b,int c)
    {
        this.a=a;
        this.b=b;
        this.c=c;
    }
    // Pair (int a,int b,int c)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    // }

    // Pair (int a,int b,int c,int d)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    //     this.d=d;
    // }
    @Override
    public String toString() {
        // return a + " " + b + " " + c;
        return a+" "+b;
    }

    public boolean equals(Object obj) {
        if (this == obj) return true;
        if (obj == null || getClass() != obj.getClass()) return false;
        Pair pair = (Pair) obj;
        // return a == pair.a && b == pair.b && c == pair.c && d == pair.d && flag == pair.flag;

        return a==pair.a && b==pair.b;
    }

    // @Override
    // public int hashCode() {
    //     int result = a;
    //     result = 31 * result + b;
    //     result = 31 * result + c;
    //     result = 31 * result + d;
    //     result = 31 * result + (flag ? 1 : 0);
    //     return result;
    // }
}

class compare1 implements Comparator<Pair>
{
    public int compare(Pair p1,Pair p2)
    {
        // if(p1.a==p2.a && p2.b==p1.b)return 0;
        // return p2.a-p1.a;//decreasing
        // return p1.c-p2.c;
        // if(p1.a!=p2.a)
        // return p1.a-p2.a;//increasing
        // if(p1.a==p2.b)return p1.b-p2.b;

        return p1.a-p2.a;
        // return p1.b-p2.b;
        // return Long.compare(p1.a,p2.a);

        // if(p1.a!=p2.a)
        // {
        //     return p1.a-p2.a;
        // }
        // else
        // return p1.b-p2.b;

        // return Long.compare((p2.a+1)/2*p2.b ,(p1.a+1)/2*p1.b);

        // return p2.a-p1.b;
        // return p1.a-p2.a;
        // return p2.c-p1.c;
        // else
        // return p1.b-p2.b;
        // if(p1.a==p2.a)
        // {
        //     return p2.b-p1.b;
        // }
        // else return +p1.a-p2.a;
        // if(p1.a!=p2.a)
        //     return p1.a-p2.a;
        // else
        //     return p2.b-p1.b;

        // if(p1.a!=p2.a)
        // {
        //     return p2.a-p1.a;
        // }
        // else
        // {
            // return p1.b-p2.b;
        // }
    }

    public boolean equals(Pair p1,Pair p2)
    {
        return p1.a==p2.a && p2.b==p1.b;
    }
}

class MO

{
    int a,b;
    int c;
    static int size;

    MO(int a,int b,int c)
    {
        this.a=a;
        this.b=b;
        this.c=c;
    }
}
class compareMO implements Comparator<MO>
{
    public int compare(MO a,MO b)
    {
        int size=MO.size;
        if(a.a/size!=b.a/size)
        {
            return a.a/size-b.a/size;
        }
        return a.b-b.b;
    }
}

class SegmentTree {
    int[] tree;     // Segment tree array
    int[] arr;      // Original array
    int n;          // Size of the original array

    // Constructor to build the segment tree
    public SegmentTree(int[] input) {
        n = input.length;
        arr = input;
        // Allocate memory for segment tree
        // The maximum size of segment tree array is 2*n - 1 for n leaf nodes
        tree = new int[4 * n];
        build(0, 0, n - 1);
    }

    // Build the segment tree
    private void build(int node, int start, int end) {
        if (start == end) {
            // Leaf node: store the value of the array element
            tree[node] = arr[start];
        } else {
            int mid = (start + end) / 2;
            // Recursively build the left and right subtrees
            build(2 * node + 1, start, mid);
            build(2 * node + 2, mid + 1, end);
            // Internal node: store the minimum of the two children
            tree[node] = Math.min(tree[2 * node + 1], tree[2 * node + 2]);
        }
    }

    // Query for the minimum in range [L, R]
    public int query(int L, int R) {
        return query(0, 0, n - 1, L, R);
    }

    private int query(int node, int start, int end, int L, int R) {
        if (R < start || L > end) {
            // Range [L, R] is completely outside the range [start, end]
            return Integer.MAX_VALUE;
        }
        if (L <= start && end <= R) {
            // Range [L, R] completely covers the range [start, end]
            return tree[node];
        }
        // Partial overlap
        int mid = (start + end) / 2;
        int leftMin = query(2 * node + 1, start, mid, L, R);
        int rightMin = query(2 * node + 2, mid + 1, end, L, R);
        return Math.min(leftMin, rightMin);
    }

    // Update an element in the array
    public void update(int index, int value) {
        update(0, 0, n - 1, index, value);
    }

    private void update(int node, int start, int end, int index, int value) {
        if (start == end) {
            // Leaf node: update the value of the array element
            arr[index] = value;
            tree[node] = value;
        } else {
            int mid = (start + end) / 2;
            if (index <= mid) {
                // Update the left child
                update(2 * node + 1, start, mid, index, value);
            } else {
                // Update the right child
                update(2 * node + 2, mid + 1, end, index, value);
            }
            // Update the current node after updating the children
            tree[node] = Math.min(tree[2 * node + 1], tree[2 * node + 2]);
        }
    }
}

class GCDSegmentTree {
    private int[] tree;
    private int[] arr;

    public GCDSegmentTree(int[] inputArr) {
        int n = inputArr.length;
        arr = inputArr;
        int treeSize = 4 * n;
        tree = new int[treeSize];
        buildTree(0, 0, n - 1);
    }

    // Build segment tree
    private void buildTree(int node, int start, int end) {
        if (start == end) {
            tree[node] = arr[start];  // Leaf node
        } else {
            int mid = (start + end) / 2;
            buildTree(2 * node + 1, start, mid);   // Left child
            buildTree(2 * node + 2, mid + 1, end); // Right child
            tree[node] = gcd(tree[2 * node + 1], tree[2 * node + 2]); // Combine GCD
        }
    }

    // GCD Query
    private int query(int node, int start, int end, int L, int R) {
        if (R < start || end < L) {
            return 0;  // Out of range
        }
        if (L <= start && end <= R) {
            return tree[node];  // Current segment fully in range
        }
        int mid = (start + end) / 2;
        int leftGCD = query(2 * node + 1, start, mid, L, R);
        int rightGCD = query(2 * node + 2, mid + 1, end, L, R);
        return gcd(leftGCD, rightGCD);  // Combine GCD from left and right children
    }

    public int getGCD(int L, int R) {
        return query(0, 0, arr.length - 1, L, R);
    }

    // Utility GCD function
    private int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

}

**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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define fa(i, a, n) for (int i = a; i < n; i++)
#define af(i, a, n) for (int i = n - 1; i >= a; i--)
#define trav(a, x) for (auto &a : x)
#define pb push_back
#define all(x) x.begin(), x.end()
#define el "\n"
#define yes cout << "YES" << el
#define no cout << "NO" << el
const int M = 1e7 + 10;
const int N = 1e5 + 10;
const int MOD = 1e9 + 7;
#define re return;

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

    vector<int> v(n);
    fa(i, 0, n)
    {
        cin >> v[i];
    }

    // vector<int> in(n, 0);
    // for (int i = 0; i < n; i++) {
    //     in[v[i] - 1]++;
    // }

    vector<int> ans(n, 1);
    vector<int> cur(n, 1);

    int year = 1;
    while (true)
    {
        vector<int> next(n, 0);

        fa(i, 0, n)
        {
            if (cur[i])
            {
                next[v[i] - 1]++;
            }
        }

        for (int i = 0; i < n; i++)
        {
            next[i] = min(1ll,next[i]);
        }
        if (next == ans)
        {
            cout << year + 1 << el;
            re;
        }
        cur = next;
        ans = cur;
        year++;
    }
}

signed main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Initial State and Transfers**: Ensure that the initial state and the transfers lead to a scenario where the code's logic fails to correctly identify the stable year.
2. **Edge Cases**: Consider edge cases where the number of plushies might exceed 1 for a spider, and the spider should throw away the extra plushies.
3. **Cycles and Stability**: Create a scenario where the process becomes stable after a few years, but the code might not correctly detect this stability.

Let's construct a test case that fits these criteria:

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

### Explanation:
- **Number of Test Cases**: 1
- **Number of Spiders**: 5
- **Plushie Transfers**:
  - Spider 1 gives to Spider 2
  - Spider 2 gives to Spider 3
  - Spider 3 gives to Spider 4
  - Spider 4 gives to Spider 5
  - Spider 5 gives to Spider 1

### Expected Behavior:
- **Year 1**:
  - Initial state: [1, 1, 1, 1, 1]
  - After transfer: [0, 1, 1, 1, 1]
- **Year 2**:
  - Initial state: [0, 1, 1, 1, 1]
  - After transfer: [1, 0, 1, 1, 1]
- **Year 3**:
  - Initial state: [1, 0, 1, 1, 1]
  - After transfer: [1, 1, 0, 1, 1]
- **Year 4**:
  - Initial state: [1, 1, 0, 1, 1]
  - After transfer: [1, 1, 1, 0, 1]
- **Year 5**:
  - Initial state: [1, 1, 1, 0, 1]
  - After transfer: [1, 1, 1, 1, 0]
- **Year 6**:
  - Initial state: [1, 1, 1, 1, 0]
  - After transfer: [0, 1, 1, 1, 1]
- **Year 7**:
  - Initial state: [0, 1, 1, 1, 1]
  - After transfer: [1, 0, 1, 1, 1]

The process becomes stable at **Year 7** because the state of the spiders' plushies is the same as it was in Year 1.

### Expected Output:
```plaintext
7
```

This test case should expose the bug in the provided code, as the code might not correctly handle the cyclic nature of the plushie transfers and the stability condition.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Medium Demon Problem (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;
typedef long long ll;
typedef __int128_t Big;
typedef vector<ll> vl;
typedef pair<ll,ll> pl;
#define Fcin ios::sync_with_stdio(false); cin.tie(nullptr);cout.tie(nullptr);
const ll oo =0x3f3f3f3f;
#define eb emplace_back
#define pb push_back
#define fs first
#define sc second
const ll MOD=1e9+7;
#define reverse(ds) reverse(ds.begin(),ds.end())
#define Sort(ds) sort(ds.begin(),ds.end())
#define sortc(ds,co) sort(ds.begin(),ds.end(),co)
#define FastFib(n) round((double)(pow((1+sqrt(5)),n)-pow((1-sqrt(5)),n))/(double)((1<<n)*sqrt(5)))
#define show(name) Show(#name,(name))
void Show(char *name,ll value)
{
    cout<<name<<" = "<<value<<endl;
}

struct DSU{
    vector<ll> p,Size;
    ll sz;
    void init(int n)
    {
        sz=n;
        p.resize(n+1);
        Size.resize(n+1);
        for(int i=1;i<=n;i++)p[i]=i,Size[i]=1;
    }
    void clean()
    {
        sz=0;
        p.clear();
        Size.clear();
    }
    int get(int n)
    {
        if(p[n]!=n)return get(p[p[n]]); /** path-compression. **/
        return n;
    }
    bool Union(int a,int b)
    {
        int root_a=get(a),root_b=get(b);
        if(root_a==root_b)return false;
        if(Size[root_a]<Size[root_b])swap(root_a,root_b);
        p[root_b]=root_a;
        Size[root_a]+=Size[root_b];
        return true;
    }
    void add(int n,int v)
    {
        int root=get(n);
    }
    bool IsConnected(int n,int m)
    {
        int root_a=get(n),root_b=get(m);
        return root_a==root_b;
    }
}dsu;

ll get_cycle(vector<ll> &v,vector<bool> &vstd,int node)
{
    if(vstd[node])
    {
        vstd[node]=false;
        return node;
    }
    vstd[node]=true;
    ll ret=get_cycle(v,vstd,v[node]);
    vstd[node]=false;
    return ret;
}

ll bfs(vector<vector<ll>> &tree,vector<bool> &vstd,int node)
{
    queue<pair<ll,ll>> q;
    q.push({node,1});
    pair<ll,ll> p;
    ll res=0;
    while(!q.empty())
    {
        p=q.front();
        q.pop();
        vstd[p.fs]=true;
        res=max(res,p.sc);
        for(auto n:tree[p.fs])
        {
            if(!vstd[n])
            q.push({n,p.sc+1});
        }
    }
    return res;
}

signed main()
{
    Fcin
    ll t,i,j,q,k,x,n,m,y, a, b;
    cin >> t;
    while(t--){
        cin>>n;
        vector<ll> v(n+1);
        vector<bool> vstd(n+1);
        vector<vector<ll>> tree(n+1);
        for(i=1;i<=n;i++)
        {
            cin>>x;
            v[i]=x;
            tree[x].eb(i);
        }
        for(i=1;i<=n;i++)
        {
            ll temp=get_cycle(v,vstd,i);
            while(!vstd[temp])
            {
                vstd[temp]=true;
                temp=v[temp];
            }
        }
        ll ans=2;
        for(i=1;i<=n;i++)
        {
            if(!vstd[i] && vstd[v[i]])
            {
                ans=max(ans,bfs(tree,vstd,i)+2);
            }
        }
        cout<<ans<<endl;
    }
}

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

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

Think step by step.

Response:
```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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    r = list(map(int, input().split()))
    f = [1] * n
    for i in range(1, 100001):
        a = [0] * n
        for j in range(n):
            if f[j] > 0:
                a[r[j] - 1] += 1
        for j in range(n):
            a[j] = min(a[j], 1)
        if a == f:
            print(i + 1)
            break
        f = a
    else:
        print(-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
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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
from sys import stdin,stdout,setrecursionlimit
setrecursionlimit(200002)
from math import *
from collections import *
from pprint import pprint
import random
def read_list_line(elem_func=int):
    return [elem_func(x) for x in stdin.readline().strip().split()]

def read_line(elem_func=str):
    return elem_func(stdin.readline().strip())

def print_list_line(the_list,elem_func=str,sep=' ',end="\n"):
    st=(sep.join([elem_func(x) for x in the_list]))+end
    stdout.write(st)
    return st

def print_line(st):
    stdout.write(str(st)+"\n")
    return st

ntest = read_line(int)

def main():
    n = read_line(int)
    r=[0]+read_list_line(int)
    stp=[r.copy()]
    for i in range(1,int(log2(n*8))):
        stp.append(stp[i-1].copy())
        for j in range(1,n+1):
            stp[i][j]=stp[i-1][stp[i-1][j]]
    stable=[0]*(n+1)
    for i in range(1,1+n):
        stable[stp[-1][i]]=1

    def dfs(u):
        if (stable[u]>=1):
            return
        v = r[u]
        dfs(v)
        stable[u]=stable[v]+1
    for i in range(1,n+1):
        if (stable[i]==0):
            dfs(i)
    print_line(max(stable)+1)
    return 0

for test_case in range(1,1+ntest):
    main()

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

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

Think step by step.

Response:
```plaintext
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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.stream.Collectors;

public class Problem2044G1 {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        var t = Integer.parseInt(br.readLine());
        for (int i = 0; i < t; i++) {
            var n = Integer.parseInt(br.readLine());
            var r = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
            var prevYear = new int[n];
            for (int j = 0; j < n; j++) {
                prevYear[j] = 1;
            }
            var currentYear = Arrays.copyOf(prevYear, n);
            var year = 1;
            do {
                prevYear = Arrays.copyOf(currentYear, n);
                for (int j = 0; j < r.length; j++) {
                    int toIndex = r[j] - 1;
                    if (currentYear[j] != 0) {
                        currentYear[toIndex] += 1;
                        currentYear[j] -= 1;
                    }
                }
                for (int j = 0; j < currentYear.length; j++) {
                    if (currentYear[j] > 1) {
                        currentYear[j] = 1;
                    }
                }
                year++;
            } while (!Arrays.equals(currentYear, prevYear));
            System.out.println(year);
        }
        br.close();
    }

}

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

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

Think step by step.

Response:
```plaintext
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 (easy version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy 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, if any spider has more than $$$1$$$ plushie at any point in time, they will throw all but $$$1$$$ away.

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
4
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: $$$[1, 1, 1, 1, 0]$$$. Then, year $$$2$$$'s exchange happens. Note that even though two spiders gave spider $$$2$$$ plushies, spider $$$2$$$ may only keep one plushie.
- At year $$$3$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 1, 0]$$$. Then, year $$$3$$$'s exchange happens.
- At year $$$4$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Then, year $$$4$$$'s exchange happens.
- At year $$$5$$$, the following array shows the number of plushies each spider has: $$$[1, 1, 0, 0, 0]$$$. Since this array is the same as the previous year, this year is stable.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt(); // Number of test cases

        while (t-- > 0) {
            int n = scanner.nextInt(); // Number of spiders
            int[] r = new int[n]; // Recipients array
            for (int i = 0; i < n; i++) {
                r[i] = scanner.nextInt() - 1; // Convert to 0-based index
            }

            int year = 0;
            int[] plushies = new int[n]; // Initial plushies count
            Arrays.fill(plushies, 1); // Each spider starts with 1 plushie

            while (true) {
                year++;
                int[] newPlushies = new int[n]; // New plushies count after exchange

                // Distribute plushies according to the rules
                for (int i = 0; i < n; i++) {
                    if (plushies[i] > 0) {
                        newPlushies[r[i]]++; // Give plushie to recipient
                    }
                }

                // Each spider can keep at most 1 plushie
                for (int i = 0; i < n; i++) {
                    newPlushies[i] = Math.min(newPlushies[i], 1);
                }

                // Check for stability
                if (Arrays.equals(plushies, newPlushies)) {
                    System.out.println(year+1); // Output the current year
                    break;
                }

                plushies = newPlushies; // Update plushies for the next year
            }
        }

        scanner.close();
    }
}

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

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

Think step by step.

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

1. **Identify the Bug**: The bug in the code is that it incorrectly prints `year + 1` instead of `year` when the process becomes stable. This means the output will always be one year more than it should be.

2. **Edge Cases**:
   - **Single Test Case**: Since the bug affects all test cases, we can focus on a single test case to highlight the issue.
   - **Small Number of Spiders**: To keep the input small and manageable, we can use a small number of spiders, say 5.
   - **Specific Recipient Configuration**: We need a configuration where the process becomes stable in a specific year, so we can easily verify the output.

3. **Example Configuration**:
   - Let's consider a configuration where the process becomes stable in year 2. This will make it clear that the output should be 2, but the buggy code will output 3.

4. **Test Input**:
   - Number of test cases: 1
   - Number of spiders: 5
   - Recipients: 2 3 4 5 1 (This configuration will make the process stable in year 2).

Here is the test input:

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

This input will help expose the bug in the buggy code. The expected output for this input is `2`, but the buggy code will output `3`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Paint a Strip
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have an array of zeros $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform two types of operations on it:

1. Choose an index $$$i$$$ such that $$$1 \le i \le n$$$ and $$$a_i = 0$$$, and assign $$$1$$$ to $$$a_i$$$;
2. Choose a pair of indices $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, $$$a_l = 1$$$, $$$a_r = 1$$$, $$$a_l + \ldots + a_r \ge \lceil\frac{r - l + 1}{2}\rceil$$$, and assign $$$1$$$ to $$$a_i$$$ for all $$$l \le i \le r$$$.

What is the minimum number of operations of the first type needed to make all elements of the array equal to one?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The only line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array.

Note that there is no limit on the sum of $$$n$$$ over all test cases.

Output: For each test case, print one integer — the minimum number of needed operations of first type.

Examples:
input:
4
1
2
4
20
output:
1
2
2
4

Note: In the first test case, you can perform an operation of the $$$1$$$st type with $$$i = 1$$$.

In the second test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 2$$$. After performing this operation, the array will look like this: $$$[1, 1]$$$.

The sequence of operations in the second test case

In the third test case, you can perform the following sequence of operations:

1. Operation of $$$1$$$st type, $$$i = 1$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 0]$$$.
2. Operation of $$$1$$$st type, $$$i = 4$$$. After performing this operation, the array will look like this: $$$[1, 0, 0, 1]$$$.
3. Operation of $$$2$$$nd type, $$$l = 1$$$, $$$r = 4$$$. On this segment, $$$a_l + \ldots + a_r = a_1 + a_2 + a_3 + a_4 = 2$$$, which is not less than $$$\lceil\frac{r - l + 1}{2}\rceil = 2$$$. After performing this operation, the array will look like this: $$$[1, 1, 1, 1]$$$.

The sequence of operations in the third test case

Buggy Code:
#include <iostream>
#include <cmath>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;

        if(n%5==1||n%5==2){
            cout<<n/5+n<<endl;
        }
        else if(n%5==3||n%5==4){
            cout<<n/5+2<<endl;
        }
        else{
            cout<<n/5<<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
4
1
5
6
10
```