**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

Buggy Code:
def solve():
    n = int(input())
    a = list(map(int, input().split()))
    c = 0

    for i in range(n):
        if a[i]-1!=i and a[a[i]-1]-1 != i:
            c += 1
    return (c // 4) + (c % 2)

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on the logic of the code and identify scenarios where it might fail. The buggy code counts the number of elements that do not satisfy the conditions of being simple and then calculates the number of operations based on this count. The key issue is that it does not correctly handle the transformation of the permutation into a simple permutation.

### Analysis of the Buggy Code:
1. **Counting Non-Simple Elements**: The code iterates through the permutation and counts elements that do not satisfy either of the conditions `p_i = i` or `p_{p_i} = i`.
2. **Calculation of Operations**: The code then calculates the number of operations as `(c // 4) + (c % 2)`, which is incorrect. The correct approach should consider the cycles in the permutation and the minimum number of swaps needed to break these cycles into simple structures.

### Challenging Test Input:
To expose the bug, we need a permutation that has a structure where the simple conditions are not easily met, and the number of operations required to make it simple is not accurately calculated by the buggy code.

### Example Test Input:
Consider a permutation with a cycle that requires multiple swaps to break into a simple permutation. For example, a permutation with a 3-cycle that needs to be broken down into 1-cycles or 2-cycles.

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

### Explanation:
- **Permutation**: `[2, 3, 4, 5, 6, 1]`
- **Cycles**: This permutation has a single 6-cycle: `1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 1`.
- **Simple Condition**: None of the elements satisfy `p_i = i` or `p_{p_i} = i`.
- **Correct Solution**: To make this permutation simple, we need to break the 6-cycle into 1-cycles or 2-cycles. The minimum number of swaps required is 3 (one for each pair of elements to break the cycle).

### Expected Output:
The correct output for this test case should be `3`. However, the buggy code will incorrectly calculate the number of operations, likely returning a different value.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

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

public class Main {
    private static final FastIO sc = new FastIO();

    public static void main(String[] args) {
        int t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    static class Pair {
        int start, end;

        Pair(int start, int end) {
            this.start = start;
            this.end = end;
        }
    }

    static void solve() {
        int n = sc.nextInt();
        int [] a=new int [n+1];
        ArrayList<Integer> arr=new ArrayList<>();
        boolean [] vi=new boolean [n+1];
        int i=0;
        long ans=0L;
        for(i=1;i<=n;i++)
        a[i]=sc.nextInt();
        for(i=1;i<=n;i++)
        {
            if(a[i]==i || vi[i] || vi[a[i]])
            {
                vi[i]=true;
                continue;
            }
            if(a[a[i]]==i)
            {

                vi[a[i]]=true;
                continue;
            }
            arr.add(a[i]);
            dfs(a,a[i],arr,vi);
            ans+=arr.size()/2;
          /*   for(int j=0;j<arr.size();j++)
            System.out.print(arr.get(j)+" ");
            System.out.println();*/
            arr.clear();
        }
        System.out.println(ans);
    }
    static void dfs(int a[],int i,ArrayList<Integer> arr,boolean vi[])
    {
        if(a[i]==i || vi[i] || vi[a[i]])
            {
                vi[i]=true;
              return;
            }
            if(a[a[i]]==i)
            {

                vi[a[i]]=true;
               return;
            }
            arr.add(a[i]);
            vi[a[i]]=true;
            dfs(a,a[i],arr,vi);

    }

}
class FastIO extends PrintWriter {
    private BufferedReader br;
    private StringTokenizer st;

    public FastIO() {
        this(System.in, System.out);
    }

    public FastIO(InputStream in, OutputStream out) {
        super(out);
        br = new BufferedReader(new InputStreamReader(in));
    }

    public FastIO(String input, String output) throws FileNotFoundException {
        super(output);
        br = new BufferedReader(new FileReader(input));
    }

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

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

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

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

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

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

Think step by step.

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

1. **Permutations that are already simple**: The code should handle these correctly and output 0.
2. **Permutations that require the minimum number of swaps**: The code should correctly calculate the minimum number of swaps.
3. **Permutations with complex cycles**: The code should handle cycles that require multiple swaps to simplify the permutation.

Let's create a test input that includes these scenarios:

1. A permutation that is already simple.
2. A permutation that requires exactly one swap to become simple.
3. A permutation with a complex cycle that requires multiple swaps to become simple.

Here is the test input:

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

Explanation:
1. The first test case is a simple permutation `[1, 2, 3, 4, 5]`, which is already simple, so the output should be `0`.
2. The second test case is a permutation `[2, 3, 4, 5, 1]`, which can be made simple with one swap (swapping `p_1` and `p_5`), so the output should be `1`.
3. The third test case is a permutation `[2, 3, 1, 5, 6, 7, 4]`, which has a complex cycle and requires two swaps to become simple (swapping `p_1` and `p_3`, and then `p_4` and `p_7`), so the output should be `2`.

This test input should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int n = sc.nextInt();
            int[] p = new int[n + 1];

            for (int i = 1; i <= n; i++) {
                p[i] = sc.nextInt();
            }

            int operations = 0;
            boolean[] fixed = new boolean[n + 1];

            for (int i = 1; i <= n; i++) {
                if (p[i] == i) {
                    fixed[i] = true;
                }
            }

            for (int i = 1; i <= n; i++) {
                if (fixed[i]) continue;

                if (p[p[i]] != i) {
                    boolean found = false;
                    for (int j = i + 1; j <= n; j++) {
                        if (fixed[j]) continue;

                        int pi = p[i], pj = p[j];

                        if ((pi == j && pj == i) ||
                            (p[pi] == j && p[pj] == i) ||
                            (pi == j && p[pj] == i) ||
                            (pj == i && p[pi] == j)) {

                            int temp = p[i];
                            p[i] = p[j];
                            p[j] = temp;
                            operations++;
                            found = true;
                            break;
                        }
                    }

                    if (!found) {
                        if (i < n) {
                            int temp = p[i];
                            p[i] = p[i + 1];
                            p[i + 1] = temp;
                            operations++;
                        }
                    }
                }
            }

            System.out.println(operations);
        }
        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
6
3 1 2 6 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

Buggy Code:
#include<iostream>
#include<map>
#include<vector>
#include<set>
#include<cmath>
#include<cassert>
#include<algorithm>
#include <iterator>
#include <random>
using namespace std;

typedef long long ll;

void debug(vector<int> & v) {
    cout << "debug: \n";
    for(int i = 1; i < v.size(); ++i) {
        cout << v[i] << ' ';
    }
    cout << "\n\n";
}

int process(vector<int> & v, vector<int> & t, vector<int> & re, int cur) {
    if(v[cur] == cur || v[v[cur]] == cur) return 0;
    vector<int> c(2), g(2);
    c[0] = re[cur], c[1] = re[re[cur]], g[0] = v[cur], g[1] = v[v[cur]];
    for(int i = 0; i < 2; ++i) {
        for(int j = 0; j < 2; ++j) {
            if(c[i] != g[j]) continue;
            // cout << "swap: " << cur << ' ' << c[i] << "\n";
            swap(v[cur], v[c[i]]);
            t[cur] = 1;
            t[v[i]] = 1;
            return 1;
        }
    }
    return 0;
}

int main()
{
    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        vector<int> v(n + 1), t(n + 1, 0), l, re(n + 1, 0);
        for(int i = 1; i <= n; ++i) {
            cin >> v[i];
            re[v[i]] = i;
            if(v[i] == i || v[v[i]] == i) {
                assert(v[v[v[i]]] == v[i]);
                t[i] = 1;
            }
            else l.push_back(i);
        }
        int cnt = 0;
        for(int i = 1; i <= n; ++i) {
            cnt += process(v, t, re, i);
        }
        // debug(v);
        while(l.size() > 0) {
            int cur = l.back();
            l.pop_back();
            if(v[cur] == cur || v[v[cur]] == cur) continue;
            // cout << "cur: " << cur << "\n";
            // cout << "swap: " << cur << ' ' << v[v[cur]] << '\n';
            if (t[cur] == 1) continue;
            swap(v[v[v[cur]]], v[cur]);
            assert(t[cur] == 0);
            // assert(t[v[v[cur]]] == 0);
            t[v[v[cur]]] == 1;
            t[cur] == 1;
            cnt += 1;
            // debug(v);
            cnt += process(v, t, re, cur);
            // debug(v);
            // cout << "cnt: " << cnt << "\n\n\n";
        }
        cout << cnt << "\n";
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long

using namespace std;

ll mod = (ll)(1e9 + 7LL);

ll gcd(ll a, ll b)
{
    if (b == 0LL)
        return a;
    return gcd(b, a % b);
}

// Function to return LCM of two numbers
long long lcm(int a, int b)
{
    return (a / gcd(a, b)) * b;
}

ll ncr(ll n, ll r)
{

    double sum = 1;
    for (ll i = 1; i <= r; i++)
    {
        sum = sum * (n - r + i) / i;
    }
    return (ll)sum;
}

ll power(ll a, ll b)
{
    ll ans = 1LL;

    while (b > 0LL)
    {
        if (b & 1LL)
            (ans *= a) %= mod;
        b >>= 1LL;

        (a *= a) %= mod;
    }

    return ans % mod;
}

ll sumTill(ll n)
{
    return (n * (n + 1LL)) / 2LL;
}

class UnionSet {
    private:
        vector<ll> parent;
        vector<ll> rank;

    public:
        UnionSet(ll n) {
            parent.assign(n+1LL, 0LL);
            rank.assign(n+1LL, 0LL);
            for (ll i=0LL; i<=n; i++) {
                parent[i] = i;
            }
        }

        void print() {
            for (ll i=0LL; i<parent.size(); i++) {
                cout << parent[i] << " ";
            }
            cout << endl;
            for (ll i=0LL; i<rank.size(); i++) {
                cout << rank[i] << " ";
            }
            cout << endl;
        }

        ll findParent(ll n) {
            return (parent[n] == n) ? n : parent[n] = findParent(parent[n]);
        }

        void unionSet(ll a, ll b) {
            ll parentA = findParent(a);
            ll parentB = findParent(b);

            if (parentA == parentB) return;

            if (rank[parentA] == rank[parentB]) {
                parent[parentA] = parentB;
                rank[parentB]++;
            }
            else if (rank[parentA] > rank[parentB]) {
                parent[parentB] = parentA;
                // rank[parentA]++;
            }
            else {
                parent[parentA] = parentB;
                // rank[parentB]++;
            }
        }
};

class SegmentTree {
    private:
    ll n;
    vector<ll> st, lazy;

    public:
    SegmentTree(ll n) {
        this->n = n;
        st.assign(4 * n, LLONG_MAX);
        lazy.assign(4 * n, LLONG_MAX);
    }

    void push(ll left, ll right, ll ind) {
        if (lazy[ind] != LLONG_MAX) {
            st[ind] = min(st[ind], lazy[ind]);

            if (left != right) {
                lazy[2 * ind + 1] = min(lazy[2 * ind + 1], lazy[ind]);
                lazy[2 * ind + 2] = min(lazy[2 * ind + 2], lazy[ind]);
            }
            lazy[ind] = LLONG_MAX;
        }
    }

    ll query(ll start, ll end, ll left, ll right, ll ind) {
        push(start, end, ind);

        if (start > right || end < left) return LLONG_MAX;

        if (left <= start && end <= right) return st[ind];

        ll mid = (start + end) / 2;
        return min(query(start, mid, left, right, 2 * ind + 1), query(mid + 1, end, left, right, 2 * ind + 2));
    }

    void update(ll start, ll end, ll left, ll right, ll val, ll ind) {
        push(start, end, ind);

        if (start > right || end < left) return;

        if (left <= start && end <= right) {
            lazy[ind] = min(lazy[ind], val);
            push(start, end, ind);
            return;
        }

        ll mid = (start + end) / 2;
        update(start, mid, left, right, val, 2 * ind + 1);
        update(mid + 1, end, left, right, val, 2 * ind + 2);

        st[ind] = min(st[2 * ind + 1], st[2 * ind + 2]);

        return;
    }

    void print() {
        for (ll i = 0; i < st.size(); i++) {
            cout << st[i] << " ";
        }
        cout << endl;
        for (ll i = 0; i < lazy.size(); i++) {
            cout << lazy[i] << " ";
        }
        cout << endl;
    }
};

// ceilVal = (a + b - 1) / b

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

    vector<int> arr(n+1);
    for (int i=1; i<=n; i++) {
        cin >> arr[i];
    }

    int ind = 1, cnt = 0, a, b, c;

    while (ind <= n) {
        if (arr[ind] == ind || arr[arr[ind]] == ind) ind++;
        else {
            // cout << ind << " " << arr[ind] << " " << arr[arr[ind]] << endl;
            a = arr[ind];
            b = arr[arr[ind]];
            c = arr[arr[arr[ind]]];
            arr[ind] = c;
            arr[b] = a;
            ind++;
            // arr[ind] = arr[arr[arr[ind]]];
            // arr[arr[arr[ind]]] = temp;
            // swap(arr[ind], arr[arr[ind]]);
            // for (int i=1; i<=n; i++) {
            //     cout << arr[i] << " ";
            // }
            // cout << endl;
            cnt++;
        }
    }
    cout << cnt << "\n";
    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    // vector<ll> fact(200010, 1);
    // vector<ll> invfact(200010, 1);
    // for (ll i=1LL; i<=200009LL; i++) {
    //     fact[i] = (fact[i-1LL] * i)%mod;
    //     invfact[i] = power(fact[i], mod - 2LL)%mod;
    // }

    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to consider the logic of the code and identify where it might fail. The code aims to make a permutation simple by performing swaps. A permutation is simple if for every index \(i\), one of the following conditions holds:
- \(p_i = i\)
- \(p_{p_i} = i\)

The code uses a while loop to iterate through the permutation and performs swaps when neither of these conditions is met. However, the logic for swapping elements and incrementing the index might not always correctly handle all cases, especially in complex cycles or when the permutation has multiple cycles of length greater than 2.

### Step-by-Step Analysis:
1. **Identify the Bug**:
   - The code might fail when it encounters a cycle of length greater than 2, and it doesn't handle the swaps correctly to break down these cycles into cycles of length 2 or 1.

2. **Construct a Test Case**:
   - We need a permutation with a cycle of length greater than 2, such as a 3-cycle or a 4-cycle, to see if the code correctly breaks it down.

3. **Small-Scale Test Input**:
   - A permutation of length 4 with a 3-cycle and one fixed point might be a good candidate.

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

### Explanation:
- **Permutation**: [2, 3, 4, 1]
- **Cycle Analysis**:
  - The cycle is (1, 2, 3, 4) with a length of 4.
  - The code should ideally break this cycle into cycles of length 2 or 1.
- **Expected Output**:
  - The minimum number of operations needed to make this permutation simple is 1. One possible swap is to swap \(p_2\) and \(p_4\), resulting in the permutation [2, 1, 4, 3], which is simple.

This test input should help expose any issues in the code's handling of cycles of length greater than 2.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

Buggy Code:
#To configure ctrl+shift+p, user snippets-> python
'''
1. Math: sieve, gcdlcm
2. Binary search: sip
3. Segment Trees: segmenttreesum, segmenttree
4. Sorted List: sorted_list
5. Disjoint Set Union: dsurank
6. Strings: strings
7. Testcase generator: testcase
'''
import sys
def get(): return map(int,sys.stdin.readline().strip().split())
def getarr(): return list(map(int,sys.stdin.readline().strip().split()))
mod=int(1e9)+7
def gets(): return map(str,sys.stdin.readline().strip().split())
def pow(x,y,p=int(1e9)+7):
    res=1
    while(y>0):
        y-=1
        res=res*x%p
    return res % p
def fact(n,mod):
    ans=1
    for i in range(1,n+1):
        ans=ans*i%mod
    return ans
sys.setrecursionlimit(10**5)
def solution():
    n,=get()
    arr=getarr()
    for i in range(n):
        arr[i]-=1
    g=[[] for i in range(n)]
    for i in arr:
        if(arr[i]==i):
            pass
        else:
            g[arr[i]].append(i)
    c=[0]
    vis=[0]*n
    def dfs(i):
        for x in g[i]:
            if(not vis[x]):
                vis[x]=1
                c[0]+=1
                dfs(x)
    tot=0
    for i in range(n):
        if(not vis[i]):
            c[0]=1
            vis[i]=1
            dfs(i)
            if(c[0]>2):
                tot+=((c[0]-1)//2)
            c[0]=0
    print(tot)

    pass

t=int(input())
for _ in range(t):
    solution()

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `5 1 2 3 4 5`
   - **Output:** `0`
   - **Reason:** The permutation is already simple.

2. **Test Case 2:**
   - **Input:** `5 5 4 3 2 1`
   - **Output:** `0`
   - **Reason:** The permutation is already simple.

3. **Test Case 3:**
   - **Input:** `5 2 3 4 5 1`
   - **Output:** `2`
   - **Reason:** This permutation is not simple. We need to perform two swaps to make it simple. For example, swap `p_1` and `p_5`, then swap `p_2` and `p_3`.

4. **Test Case 4:**
   - **Input:** `4 2 3 4 1`
   - **Output:** `1`
   - **Reason:** This permutation is not simple. We need to perform one swap to make it simple. For example, swap `p_1` and `p_4`.

5. **Test Case 5:**
   - **Input:** `3 1 3 2`
   - **Output:** `0`
   - **Reason:** The permutation is already simple.

6. **Test Case 6:**
   - **Input:** `7 2 3 1 5 6 7 4`
   - **Output:** `2`
   - **Reason:** This permutation is not simple. We need to perform two swaps to make it simple. For example, swap `p_1` and `p_3`, then swap `p_4` and `p_7`.

### Why These Test Cases?
- **Simple Permutations:** Test cases 1, 2, and 5 are already simple to ensure that the algorithm correctly identifies and outputs `0` operations.
- **Non-Simple Permutations:** Test cases 3, 4, and 6 are not simple and require a specific number of operations to become simple. These test cases are designed to expose any issues in the algorithm's logic for counting the minimum number of operations.
- **Edge Cases:** The test cases cover a range of permutation lengths and structures to ensure the algorithm handles different scenarios correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

Buggy Code:
import sys
sys.setrecursionlimit(10**9)

tn = int(input())
for ti in range(tn):
    n = int(input())
    a = tuple(map(int, input().split()))

    """
    Case1: p_i = i means permutation is sorted (in right position)
    Case2: p_{p_i} = i means... or only two things point to each other

    so... if something is Case1 or Case2, then leave it alone
    so you're left with a bunch of unpaired / wrong position elements

    you can always do Case1 (prog=1)
    optimal Case1 does not exist (prog=2) because that is Case2
    can you do Case2? yes! you know can match some arbitrary thing (prog=2)
    can you do better? apparently so

    2 3 1 5 6 7 4
    $   $          (prog=3) -- Case1 + Case2
    1 3.2 5 6 7 4
            $   $  (prog=4) -- Case2
    1 3.2 5.4 7.6

    prog=4 is optimal, but how do you recognize it?
    p_4 = 5 -> p_5 = 6
          6 -> p_6 = 7
    p_7 = 4
    a cycle of 4!! -- you can flip the cycle and make 4 happy

    prog=3 is also optimal, but how? i think its a cycle of 3 such that one of the element
    {p_i} goes over i

    oh actually... cycle1 and cycle2 is good
    you can turn      cycle3 -> c1 + c2 with one swap
    you can also turn cycle4 -> c2 + c2 with one swap
    cycle 5 ... uhh, c2 + c3 ? yes should be possible
    more generally, if X+Y=Z, you can split cZ into cX and cY

    3 cost 1
    4 cost 1
    5 cost 2 (3+2)
    6 cost 2 (4+2) -> [4]1+1 cost
    7 cost 3 (5+2) -> [5]2+1 cost
    8 cost 3 (6+2) -> [6]2+1 cost
             (5+3) -> [5]2+[3]1+1 = 4 cost
             (4+4) -> [4]1+[4]1+1
    cost = max(0, cyclesize - 1) // 2

    also since p is a permutation, then it MUST be a cycle somewhere
    """
    a = [x-1 for x in a]

    cycles = []

    dvis = [-1 for x in a] # dfs vis
    cvis = [False for x in a] # cycle vis
    def cdfs(u, depth=0):
        if cvis[u]:
            cycles.append(depth)
            return depth
        cvis[u] = True
        return cdfs(a[u], depth+1)
    def dfs(u, st):
        if dvis[u] == st: # cycle?
            cdfs(u)
        if dvis[u] == -1: # not yet seen
            dvis[u] = st
            dfs(a[u], st)

    for i in range(n): dfs(i, i)
    # print(cycles)
    ans = 0
    for x in cycles: ans += max(0, x-1) // 2
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

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

public class SakuroKosukeAndPermutation {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t= sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n =sc.nextInt();
            int a[] = new int[n];
            HashMap<Integer,Integer> map = new HashMap<>();
            for (int j = 0; j < n; j++) {
                a[j] = sc.nextInt()-1;
                map.put(a[j],j);
            }
            int operations = 0;
            for (int j = 0; j < n; j++) {
                if(a[j] != j && a[a[j]]!=j){
                    int x = a[j];
                    int y = map.get(j);
                    int temp = a[x];
                    a[x] = a[y];
                    a[y] = temp;
                    map.put(a[x],x);
                    map.put(a[y],y);
                    operations++;
                }
            }
            System.out.println(operations);
        }
    }
}

**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
6
2 3 4 5 6 1
4
2 4 1 3
5
3 1 2 5 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Collection;
import java.util.StringTokenizer;

public class E_Sakurako_Kosuke_and_the_Permutation {

    public static void main(String args[]) throws IOException {
        int cases = StdIO.nextInt();
        Solution solution = new Solution();
        for (int i = 1; i <= cases; i++) {
            try {
                solution.solve();
            } catch (Throwable e) {
                System.err.println(e.toString() + " for testcase: " + i);
                e.printStackTrace();
            }
        }
        StdIO.out.close();
    }

}

class Solution extends StandardFunctions {

    int globalCount = Integer.MAX_VALUE;

    public void solve() {
        int n = nextInt();

        int[] arr = new int[n + 1];

        for (int i = 1; i <= n; i++) {
            arr[i] = nextInt();
        }

        globalCount = Integer.MAX_VALUE;

        dfs(1, arr, 0);

        out.println(globalCount);
    }

    private void dfs(int node, int[] arr, int count) {

        if (count > globalCount)
            return;
        if (node == arr.length) {
            globalCount = Math.min(count, globalCount);
            return;
        }

        if (arr[node] == node || arr[arr[node]] == node) {
            dfs(node + 1, arr, count);
            return;
        }

        int index1 = find(node, arr);

        int index2 = arr[node];
        if (arr[index2] != index2 && arr[arr[index2]] != index2) {
            swap(index2, index1, arr);

            dfs(node + 1, arr, count + 1);

        } else {
            swap(index1, node, arr);

            dfs(node + 1, arr, count + 1);
        }
    }

    private static void swap(int i, int j, int[] arr) {
        int temp = arr[i];

        arr[i] = arr[j];
        arr[j] = temp;
    }

    private static int find(int val, int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            if (val == arr[i])
                return i;
        }
        return -1;
    }

}

class StandardFunctions extends StdIO {
    public static int upperBound(int[] arr, int s, int e, int k) {
        while (s <= e) {
            int m = (s + e) / 2;

            if (arr[m] > k) {
                e = m - 1;
            } else
                s = m + 1;
        }

        return s;
    }

    public static int lowerBound(int[] arr, int s, int e, int k) {
        while (s <= e) {
            int m = (s + e) / 2;

            if (arr[m] >= k) {
                e = m - 1;
            } else
                s = m + 1;
        }

        return s;
    }
}

class StdIO {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st = new StringTokenizer("");
    static PrintWriter out = new PrintWriter(System.out);

    static String nextString() {
        while (!st.hasMoreTokens())
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
            }
        return st.nextToken();
    }

    static int nextInt() {
        return Integer.parseInt(nextString());
    }

    static long nextLong() {
        return Long.parseLong(nextString());
    }

    static double nextDouble() {
        return Double.parseDouble(nextString());
    }

    static float nextFloat() {
        return Float.parseFloat(nextString());
    }

    static boolean nextBoolean() {
        return Boolean.parseBoolean(nextString());
    }

    static char nextChar() throws IOException {
        return (char) br.read();
    }

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

    static int[] nextIntArray(int length) {
        int[] output = new int[length];
        for (int i = 0; i < length; i++)
            output[i] = nextInt();
        return output;
    }

    static long[] nextLongArray(int length) {
        long[] output = new long[length];
        for (int i = 0; i < length; i++)
            output[i] = nextLong();
        return output;
    }

    static double[] nextDoubleArray(int length) {
        double[] output = new double[length];
        for (int i = 0; i < length; i++)
            output[i] = nextDouble();
        return output;
    }

    static float[] nextFloatArray(int length) {
        float[] output = new float[length];
        for (int i = 0; i < length; i++)
            output[i] = nextFloat();
        return output;
    }

    static boolean[] nextBooleanArray(int length) {
        boolean[] output = new boolean[length];
        for (int i = 0; i < length; i++)
            output[i] = nextBoolean();
        return output;
    }

    static char[] nextCharArray(int length) throws IOException {
        char[] output = new char[length];
        for (int i = 0; i < length; i++)
            output[i] = nextChar();
        return output;
    }

    static String[] nextStringArray(int length) throws IOException {
        String[] output = new String[length];
        for (int i = 0; i < length; i++)
            output[i] = nextLine();
        return output;
    }
}

class Stack<T> extends ArrayDeque<T> {
    Stack(Collection<T> collection) {
        super(collection);
    }

    public T pop() {
        return pollLast();
    }

    public void push(T val) {
        offerLast(val);
    }
}

class SegmentTreeForSlots {
    int size;
    int[] tree;

    SegmentTreeForSlots(int size) {
        this.size = size;
        this.tree = new int[4 * size];
        buildTree(0, size - 1, 0);
    }

    private void buildTree(int start, int end, int index) {
        if (start == end) {
            tree[index] = 1;
            return;
        }

        int left = 2 * index + 1;
        int right = 2 * index + 2;
        int mid = (end - start) / 2 + start;

        buildTree(start, mid, left);
        buildTree(mid + 1, end, right);
        tree[index] = tree[left] + tree[right];
    }

    public int getPositionAndUpdate(int k) {
        return getPositionAndUpdate(0, size - 1, 0, k);
    }

    private int getPositionAndUpdate(int start, int end, int index, int k) {
        if (start == end) {
            tree[index] = 0;
            return start;
        }
        int left = 2 * index + 1;
        int right = 2 * index + 2;
        int mid = (end - start) / 2 + start;
        int returnVal = -1;
        if (tree[left] >= k) {
            returnVal = getPositionAndUpdate(start, mid, left, k);
        } else {
            returnVal = getPositionAndUpdate(mid + 1, end, right, k - tree[left]);
        }
        tree[index] = tree[left] + tree[right];
        return returnVal;
    }
}

class DisjointSet {
    int[] parent;
    int[] size;

    DisjointSet(int n) {
        this.parent = new int[n];
        this.size = new int[n];

        for (int i = 0; i < n; i++)
            parent[i] = i;
        Arrays.fill(size, 1);
    }

    public void union(int u, int v) {
        int parU = getParent(u);

        int parV = getParent(v);

        if (parU == parV)
            return;

        if (size[parU] > size[parV]) {
            parent[parV] = parU;
            size[parU] = size[parU] + size[parV];
        } else {
            parent[parU] = parV;
            size[parV] = size[parU] + size[parV];
        }
    }

    public int getParent(int u) {
        if (parent[u] == u)
            return u;
        parent[u] = getParent(parent[u]);
        return parent[u];
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

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

using i64 = long long;
const int N = 2e5 + 7;
int n;
struct DSU {
    int fa[N];
    inline void init() {
        for (int i = 1;i <= n;i++) fa[i] = i;
    }
    inline int find(int x) {
        return x == fa[x] ? x : fa[x] = find(fa[x]);
    }
    inline void uni(int x, int y) {
        int fx = find(x), fy = find(y);
        if (fx != fy) fa[fx] = fy;
    }
    inline bool same(int x, int y) {
        return find(x) == find(y);
    }
} d;
void solve() {
	cin >> n;

	vector<int> a(n + 1);
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}

	d.init();
	for (int i = 1; i <= n; i++) {
		d.uni(a[i], i);
	}

	int c = 0;
	vector<int> sum(n + 1);
	for (int i = 1; i <= n; i++) {
		sum[d.find(i)]++;
	}

	int ans = 0;
	for (int i = 1; i <= n; i++) {
		if (sum[i]) {
			ans += max(0, (sum[i] - 1) / 2);
		}
	}
	cout << ans << '\n';
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int t;
	cin >> t;

	while(t--) {
		solve();
	}

	return 0;
}

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

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

Think step by step.

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

1. **Cycles and Self-loops**: The code uses a Disjoint Set Union (DSU) to group elements that are in the same cycle. We need to ensure that the code correctly handles cycles of different lengths, including self-loops (where \( p_i = i \)) and 2-cycles (where \( p_{p_i} = i \)).

2. **Complex Cycles**: The code should correctly count the number of operations needed to break down larger cycles into smaller ones that satisfy the simple permutation condition.

3. **Edge Cases**: Small permutations with specific patterns that might not be handled correctly by the code.

Given these points, let's construct a test input that includes a permutation with a mix of self-loops, 2-cycles, and a larger cycle that needs to be broken down.

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

### Explanation
1. **First Test Case (1 2 3 4 5)**:
   - This is a simple permutation where \( p_i = i \) for all \( i \).
   - Expected output: 0 (no operations needed).

2. **Second Test Case (5 4 3 2 1)**:
   - This is a simple permutation where \( p_{p_i} = i \) for all \( i \).
   - Expected output: 0 (no operations needed).

3. **Third Test Case (2 3 1 5 6 7 4)**:
   - This permutation has a 3-cycle (1, 2, 3) and a 4-cycle (4, 5, 6, 7).
   - To make it simple, we need to break the 3-cycle and 4-cycle into 2-cycles and self-loops.
   - Expected output: 2 (one operation to break the 3-cycle and one to break the 4-cycle).

4. **Fourth Test Case (3 4 5 1 6 2)**:
   - This permutation has a 5-cycle (1, 3, 5, 6, 2) and a 1-cycle (4).
   - To make it simple, we need to break the 5-cycle into 2-cycles and self-loops.
   - Expected output: 2 (one operation to break the 5-cycle into two 2-cycles).

This test input should help expose any issues in the buggy code, particularly in how it handles cycles and the calculation of the minimum number of operations needed to make the permutation simple.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako, Kosuke, and the Permutation
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako's exams are over, and she did excellently. As a reward, she received a permutation $p$. Kosuke was not entirely satisfied because he failed one exam and did not receive a gift. He decided to sneak into her room (thanks to the code for her lock) and spoil the permutation so that it becomes simple.

A permutation $p$ is considered simple if for every $i$ $(1\le i \le n)$ one of the following conditions holds:

  * $p_i=i$    * $p_{p_i}=i$

For example, the permutations $[1, 2, 3, 4]$, $[5, 2, 4, 3, 1]$, and $[2, 1]$ are simple, while $[2, 3, 1]$ and $[5, 2, 1, 4, 3]$ are not.

In one operation, Kosuke can choose indices $i,j$ $(1\le i,j\le n)$ and swap the elements $p_i$ and $p_j$.

Sakurako is about to return home. Your task is to calculate the minimum number of operations that Kosuke needs to perform to make the permutation simple.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($1\le n \le 10^6$) — the length of the permutation $p$.    * The second line contains $n$ integers $p_i$ ($1\le p_i\le n$) — the elements of the permutation $p$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^6$.

It is guaranteed that $p$ is a permutation.

Output: For each test case, output the minimum number of operations that Kosuke needs to perform to make the permutation simple.

Examples:
input:
6
5
1 2 3 4 5
5
5 4 3 2 1
5
2 3 4 5 1
4
2 3 4 1
3
1 3 2
7
2 3 1 5 6 7 4
output:
0
0
2
1
0
2

Note: In the first and second examples, the permutations are already simple.

In the fourth example, it is sufficient to swap $p_2$ and $p_4$. Thus, the permutation will become $[2, 1, 4, 3]$ in $1$ operation.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef vector<int> vi;
typedef vector<long long int> vlli;
typedef pair<int , int> pii;
typedef pair<long , long> pll;
typedef vector<pair<int ,int>> vpii;
typedef vector<pair<long, long>> vpll;
typedef vector<long double> vld;

int main()
{
    ios_base::sync_with_stdio(false);
        cin.tie(NULL);

    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        vi p(n+1);
        map<int,int> ind; //{element, index}
        for(int i=1;i<=n;i++)
        {
            cin>>p[i];
            ind[p[i]]=i;
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
           if(p[i]!=i && p[p[i]]!=i)
           {
            //swap element at ith index with the value
            int id=ind[i];
            int t1=p[i];
            int ti=ind[id];
            swap(p[i],p[ind[id]]);
            ind[t1]=ti;
            ind[p[i]]=i;
            ans++;
           }
        }
        cout<<ans<<"\n";
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Job Interview
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp is opening his own IT company. He wants to hire $n$ programmers and $m$ testers.

There are $n+m+1$ candidates, numbered from $1$ to $n+m+1$ in chronological order of their arriving time. The $i$-th candidate has programming skill $a_i$ and testing skill $b_i$ (a person's programming skill is different from their testing skill). The skill of the team is the sum of the programming skills of all candidates hired as programmers, and the sum of the testing skills of all candidates hired as testers.

When a candidate arrives to interview, Monocarp tries to assign them to the most suitable position for them (if their programming skill is higher, then he hires them as a programmer, otherwise as a tester). If all slots for that position are filled, Monocarp assigns them to the other position.

Your task is, for each candidate, calculate the skill of the team if everyone except them comes to interview. Note that it means that exactly $n+m$ candidates will arrive, so all $n+m$ positions in the company will be filled.

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

Each test case consists of three lines:

  * the first line contains two integers $n$ and $m$ ($0 \le n, m \le 2 \cdot 10^5$; $2 \le n + m + 1 \le 2 \cdot 10^5$) — the number of programmers and the number of testers Monocarp wants to hire, respectively;    * the second line contains $n + m + 1$ integers $a_1, a_2, \dots, a_{n+m+1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the programming skill of the $i$-th candidate;    * the third line contains $n + m + 1$ integers $b_1, b_2, \dots, b_{n+m+1}$ ($1 \le b_i \le 10^9$; $b_i \ne a_i$), where $b_i$ is the testing skill of the $i$-th candidate.

Additional constraint on the input: the sum of $(n + m + 1)$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print $n + m + 1$ integers, where the $i$-th integer should be equal to the skill of the team if everyone except the $i$-th candidate comes to interview.

Examples:
input:
4
1 0
2 1
1 2
0 2
4 5 5
5 4 1
1 2
2 1 5 4
5 2 3 1
3 1
4 3 3 4 1
5 5 4 5 2
output:
1 2
5 6 9
8 11 11 12
13 13 13 12 15

Note: Let's consider the third test case of the example:

  * if the $1$-st candidate does not arrive, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $2 + 5 + 1 = 8$;    * if the $2$-nd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $5 + 5 + 1 = 11$;    * if the $3$-rd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $4$-th candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 4 = 11$;    * if the $4$-th candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 5 = 12$.

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

ll bs(ll l, ll r, vector<ll>arr, ll f){

    ll ans = l;
    //cout<<"&"<<" "<<l<<" "<<r<<endl;
    while(l<=r){
         ll mid = (l+(r-l)/2);
        if(arr[mid]>f){
            r = mid-1;
        }
        else if(arr[mid]<f){
            l = mid+1;
        }
        else{
            ans = mid;
            r=mid-1;
        }
    }
    return ans;
}

void solve(){
    ll n,m; cin>>n>>m;
    ll k = n+m+1;
    vector<ll>a(k), b(k);
    for(ll i=0;i<k;i++) cin>>a[i];
    for(ll i=0;i<k;i++) cin>>b[i];
    vector<ll>sump(k), sumt(k), vp(k), vt(k), np(k), nt(k);
    sump[0] = a[0], sumt[0] = b[0], np[0] = (a[0]>b[0]), nt[0] = !np[0], vp[0] = np[0]==1?a[0]:0, vt[0]= nt[0]==1?b[0]:0;
    for(ll i=1;i<k;i++){
        sump[i] = sump[i-1]+a[i];
        sumt[i] = sumt[i-1]+b[i];
        if(a[i]>b[i]){
            vp[i] = vp[i-1]+a[i];
            vt[i] = vt[i-1];
            np[i] = 1+np[i-1];
            nt[i] = nt[i-1];
        }
        else {
            vt[i] = vt[i-1]+b[i];
            vp[i] = vp[i-1];
            nt[i] = 1+nt[i-1];
            np[i] = np[i-1];
        }
    }

    for(ll i=0;i<k;i++){
        ll ans = 0;
        if(i>=1 && np[i-1]>=n){
            ll ind = bs(0,i-1,np,n);
            ans+=vp[ind]+vt[ind];
            ans+=sumt[i-1]-sumt[ind]+sumt[k-1]-sumt[i];
            cout<<ans<<" ";
            continue;
        }
        else if(i>=1 && nt[i-1]>=m){
            ll ind = bs(0,i-1,nt,m);
            ans+=vt[ind]+vp[ind];
            ans+=sump[i-1]-sump[ind]+sump[k-1]-sump[i];
            cout<<ans<<" ";
            continue;
        }
        ll n1 = n, m1 = m;
        if(i>=1){
            ans+=(vp[i-1]+vt[i-1]);
            n1 = n-np[i-1], m1 = m-nt[i-1];
        }
        if((np[k-1]-np[i])>=n1){
            ll ind = 0;
            if(n1!=0) ind = bs(i+1,k-1,np,n1+np[i]);
            ans+=vp[ind]-vp[i];
            ans+=vt[ind]-vt[i];
            ans+=sumt[k-1]-sumt[ind];
            cout<<ans<<" ";
            continue;
        }
        else if(nt[k-1]-nt[i]>=m1){
            ll ind = 0;
            if(m1!=0) ind= bs(i+1,k-1,np,m1+nt[i]);
            ans+=vt[ind]-vt[i];
            ans+=vp[ind]-vp[i];
            ans+=sump[k-1]-sump[ind];
            cout<<ans<<" ";
            continue;
        }
    }
    return;
}

int main(){
    ll t; cin>>t;
    while(t--){
        solve();
        cout<<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
1
2 2
3 1 5 2 4
2 4 1 3 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Job Interview
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp is opening his own IT company. He wants to hire $n$ programmers and $m$ testers.

There are $n+m+1$ candidates, numbered from $1$ to $n+m+1$ in chronological order of their arriving time. The $i$-th candidate has programming skill $a_i$ and testing skill $b_i$ (a person's programming skill is different from their testing skill). The skill of the team is the sum of the programming skills of all candidates hired as programmers, and the sum of the testing skills of all candidates hired as testers.

When a candidate arrives to interview, Monocarp tries to assign them to the most suitable position for them (if their programming skill is higher, then he hires them as a programmer, otherwise as a tester). If all slots for that position are filled, Monocarp assigns them to the other position.

Your task is, for each candidate, calculate the skill of the team if everyone except them comes to interview. Note that it means that exactly $n+m$ candidates will arrive, so all $n+m$ positions in the company will be filled.

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

Each test case consists of three lines:

  * the first line contains two integers $n$ and $m$ ($0 \le n, m \le 2 \cdot 10^5$; $2 \le n + m + 1 \le 2 \cdot 10^5$) — the number of programmers and the number of testers Monocarp wants to hire, respectively;    * the second line contains $n + m + 1$ integers $a_1, a_2, \dots, a_{n+m+1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the programming skill of the $i$-th candidate;    * the third line contains $n + m + 1$ integers $b_1, b_2, \dots, b_{n+m+1}$ ($1 \le b_i \le 10^9$; $b_i \ne a_i$), where $b_i$ is the testing skill of the $i$-th candidate.

Additional constraint on the input: the sum of $(n + m + 1)$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print $n + m + 1$ integers, where the $i$-th integer should be equal to the skill of the team if everyone except the $i$-th candidate comes to interview.

Examples:
input:
4
1 0
2 1
1 2
0 2
4 5 5
5 4 1
1 2
2 1 5 4
5 2 3 1
3 1
4 3 3 4 1
5 5 4 5 2
output:
1 2
5 6 9
8 11 11 12
13 13 13 12 15

Note: Let's consider the third test case of the example:

  * if the $1$-st candidate does not arrive, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $2 + 5 + 1 = 8$;    * if the $2$-nd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $5 + 5 + 1 = 11$;    * if the $3$-rd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $4$-th candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 4 = 11$;    * if the $4$-th candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 5 = 12$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int __int128
#define i64 long long
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define V vector
#define A array
#define F function
#define sz(x) (int)x.size()
const int N=2e5+5,inf=1e18;

struct BIT{
    int n;V<int> a;
    BIT(int n):n(n),a(n+1,0){};
#define lb(x) x&(-x)
    void upd(int x,int k){
        for(int i=x+1;i<=n;i+=lb(i)){
            a[i]+=k;
        }
    }
    int qry(int x){
        int res=0;
        for(int i=x+1;i;i-=lb(i)){
            res+=a[i];
        }
        return res;
    }
};

void sol(){
    i64 n,ll,rr;
    cin>>ll>>rr;
    n=ll+rr+1;
    V<i64> a(n),b(n);
    int sma=0,smb=0;
    rep(i,0,n-1) cin>>a[i],sma+=a[i];
    rep(i,0,n-1) cin>>b[i],smb+=b[i];
    BIT t1(n+5),t2(n+5),v1(n+5),v2(n+5);
    rep(i,0,n-1){
        if(a[i]>b[i]){
            t1.upd(i,1);
            v1.upd(i,a[i]-b[i]);
        }
        else{
            t2.upd(i,1);
            v2.upd(i,b[i]-a[i]);
        }
    }
    auto get1=[&](int pos){
        int l=0,r=n-1,res=n;
        if(t1.qry(n-1)<pos) return res;
        while(l<=r){
            int mid=(l+r)/2;
            if(t1.qry(mid)>=pos) res=mid,r=mid-1;
            else l=mid+1;
        }
        return res;
    };
    auto get2=[&](int pos){
        int l=0,r=n-1,res=n;
        if(t2.qry(n-1)<pos) return res;
        while(l<=r){
            int mid=(l+r)/2;
            if(t2.qry(mid)>=pos) res=mid,r=mid-1;
            else l=mid+1;
        }
        return res;
    };
    rep(i,0,n-1){
        if(a[i]>b[i]){
            sma-=a[i];
            smb-=b[i];
            t1.upd(i,-1);
            v1.upd(i,-a[i]+b[i]);
            int pos1=get1(ll),pos2=get2(rr);
            if(pos1<pos2){
                cout<<(i64)(v1.qry(pos1)+smb)<<' ';
            }
            else{
                cout<<(i64)(v2.qry(pos2)+sma)<<' ';
            }
            v1.upd(i,a[i]-b[i]);
            t1.upd(i,1);
            sma+=a[i];
            smb+=b[i];
        }
        else{
            sma-=a[i];
            smb-=b[i];
            t2.upd(i,-1);
            v2.upd(i,a[i]-b[i]);
            int pos1=get1(ll),pos2=get2(rr);
            if(pos1<pos2){
                cout<<(i64)(v1.qry(pos1)+smb)<<' ';
            }
            else{
                cout<<(i64)(v2.qry(pos2)+sma)<<' ';
            }
            v2.upd(i,-a[i]+b[i]);
            t2.upd(i,1);
            sma+=a[i];
            smb+=b[i];
        }
    }
    cout<<'\n';
}

int32_t main() {
    std::ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    i64 tt=1;
    cin>>tt;
    while(tt--) sol();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Job Interview
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp is opening his own IT company. He wants to hire $n$ programmers and $m$ testers.

There are $n+m+1$ candidates, numbered from $1$ to $n+m+1$ in chronological order of their arriving time. The $i$-th candidate has programming skill $a_i$ and testing skill $b_i$ (a person's programming skill is different from their testing skill). The skill of the team is the sum of the programming skills of all candidates hired as programmers, and the sum of the testing skills of all candidates hired as testers.

When a candidate arrives to interview, Monocarp tries to assign them to the most suitable position for them (if their programming skill is higher, then he hires them as a programmer, otherwise as a tester). If all slots for that position are filled, Monocarp assigns them to the other position.

Your task is, for each candidate, calculate the skill of the team if everyone except them comes to interview. Note that it means that exactly $n+m$ candidates will arrive, so all $n+m$ positions in the company will be filled.

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

Each test case consists of three lines:

  * the first line contains two integers $n$ and $m$ ($0 \le n, m \le 2 \cdot 10^5$; $2 \le n + m + 1 \le 2 \cdot 10^5$) — the number of programmers and the number of testers Monocarp wants to hire, respectively;    * the second line contains $n + m + 1$ integers $a_1, a_2, \dots, a_{n+m+1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the programming skill of the $i$-th candidate;    * the third line contains $n + m + 1$ integers $b_1, b_2, \dots, b_{n+m+1}$ ($1 \le b_i \le 10^9$; $b_i \ne a_i$), where $b_i$ is the testing skill of the $i$-th candidate.

Additional constraint on the input: the sum of $(n + m + 1)$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print $n + m + 1$ integers, where the $i$-th integer should be equal to the skill of the team if everyone except the $i$-th candidate comes to interview.

Examples:
input:
4
1 0
2 1
1 2
0 2
4 5 5
5 4 1
1 2
2 1 5 4
5 2 3 1
3 1
4 3 3 4 1
5 5 4 5 2
output:
1 2
5 6 9
8 11 11 12
13 13 13 12 15

Note: Let's consider the third test case of the example:

  * if the $1$-st candidate does not arrive, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $2 + 5 + 1 = 8$;    * if the $2$-nd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $5 + 5 + 1 = 11$;    * if the $3$-rd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $4$-th candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 4 = 11$;    * if the $4$-th candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 5 = 12$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int T,n,m,now,nowo,out,fh,cnt[2],outs[2],p[200010],b[200010];
long long sum,a[200010][2],f[200010];
multiset<int>s[2];
char cc;
int read()
{
	out=0,fh=1,cc=getchar();
	while(cc<48||cc>57){if(cc==45){fh=-1;}cc=getchar();}
	while(cc>=48&&cc<=57){out=(out<<3)+(out<<1)+(cc&15),cc=getchar();}
	return out*fh;
}
void Clear()
{
	sum=cnt[0]=cnt[1]=0;
	for(int i=1;i<=n+m+1;++i){p[i]=b[i]=f[i]=0;}
	s[0].clear();s[1].clear();
}
void Solve()
{
	n=read(),m=read();
	for(int i=1;i<=n+m+1;++i){a[i][0]=read();}
	for(int i=1;i<=n+m+1;++i){a[i][1]=read();}
	for(int i=1;i<=n+m;++i)
	{
		if(cnt[0]==n)
		{
			++cnt[1],p[i]=1,sum+=a[i][1],b[i]=1;
			s[1].insert(i);
			continue;
		}
		if(cnt[1]==m)
		{
			++cnt[0],p[i]=0,sum+=a[i][0],b[i]=1;
			s[0].insert(i);
			continue;
		}
		if(a[i][0]>a[i][1])
		{
			++cnt[0],p[i]=0,sum+=a[i][0];
			s[0].insert(i);
		}
		else
		{
			++cnt[1],p[i]=1,sum+=a[i][1];
			s[1].insert(i);
		}
	}
	f[n+m+1]=sum;
	if(n==0||m==0)
	{
		for(int i=1;i<=n+m;++i)
		{
			if(!n){f[i]=sum+a[m+1][1]-a[i][1];}
			else{f[i]=sum+a[n+1][0]-a[i][0];}
		}
		for(int i=1;i<=n+m+1;++i){printf("%lld ",f[i]);}
		puts("");
		Clear();
		return;
	}
	for(int i=n+m;i;--i)
	{
		if(b[i])
		{
			sum+=a[i+1][p[i]]-a[i][p[i]],
			f[i]=sum;
			s[p[i]].erase(s[p[i]].find(i));
			s[p[i]].insert(i+1);
			continue;
		}
		now=a[i+1][0]<a[i+1][1];
		if(now==p[i])
		{
			sum+=a[i+1][p[i]]-a[i][p[i]],
			f[i]=sum;
			s[p[i]].erase(s[p[i]].find(i));
			s[p[i]].insert(i+1);
		}
		else
		{
			outs[now]=*s[now].rbegin();
			outs[p[i]]=*s[p[i]].rbegin();
			s[p[i]].erase(s[p[i]].find(i));
			s[now].insert(i+1);
			s[now].erase(s[now].find(outs[now]));
			s[p[i]].insert(outs[now]);
			sum+=a[i+1][now]-a[i][p[i]]+a[outs[now]][p[i]]-a[outs[now]][now];
			if(outs[now]<outs[p[i]])
			{
				nowo=a[outs[now]][0]<a[outs[now]][1];
				if(nowo==p[i])
				{
					s[p[i]].erase(s[p[i]].find(outs[p[i]]));
					s[now].insert(outs[p[i]]);
					s[now].erase(s[now].find(outs[now]));
					s[p[i]].insert(outs[now]);
					sum+=a[outs[now]][p[i]]-a[outs[now]][now]+a[outs[p[i]]][now]-a[outs[p[i]]][p[i]];
				}
			}
			f[i]=sum;
		}
	}
	for(int i=1;i<=n+m+1;++i){printf("%lld ",f[i]);}
	puts("");
	Clear();
}
int main()
{
	T=read();
	while(T--){Solve();}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
2 2
3 5 7 9 11
2 4 6 8 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Job Interview
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp is opening his own IT company. He wants to hire $n$ programmers and $m$ testers.

There are $n+m+1$ candidates, numbered from $1$ to $n+m+1$ in chronological order of their arriving time. The $i$-th candidate has programming skill $a_i$ and testing skill $b_i$ (a person's programming skill is different from their testing skill). The skill of the team is the sum of the programming skills of all candidates hired as programmers, and the sum of the testing skills of all candidates hired as testers.

When a candidate arrives to interview, Monocarp tries to assign them to the most suitable position for them (if their programming skill is higher, then he hires them as a programmer, otherwise as a tester). If all slots for that position are filled, Monocarp assigns them to the other position.

Your task is, for each candidate, calculate the skill of the team if everyone except them comes to interview. Note that it means that exactly $n+m$ candidates will arrive, so all $n+m$ positions in the company will be filled.

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

Each test case consists of three lines:

  * the first line contains two integers $n$ and $m$ ($0 \le n, m \le 2 \cdot 10^5$; $2 \le n + m + 1 \le 2 \cdot 10^5$) — the number of programmers and the number of testers Monocarp wants to hire, respectively;    * the second line contains $n + m + 1$ integers $a_1, a_2, \dots, a_{n+m+1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the programming skill of the $i$-th candidate;    * the third line contains $n + m + 1$ integers $b_1, b_2, \dots, b_{n+m+1}$ ($1 \le b_i \le 10^9$; $b_i \ne a_i$), where $b_i$ is the testing skill of the $i$-th candidate.

Additional constraint on the input: the sum of $(n + m + 1)$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print $n + m + 1$ integers, where the $i$-th integer should be equal to the skill of the team if everyone except the $i$-th candidate comes to interview.

Examples:
input:
4
1 0
2 1
1 2
0 2
4 5 5
5 4 1
1 2
2 1 5 4
5 2 3 1
3 1
4 3 3 4 1
5 5 4 5 2
output:
1 2
5 6 9
8 11 11 12
13 13 13 12 15

Note: Let's consider the third test case of the example:

  * if the $1$-st candidate does not arrive, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $2 + 5 + 1 = 8$;    * if the $2$-nd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $5 + 5 + 1 = 11$;    * if the $3$-rd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $4$-th candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 4 = 11$;    * if the $4$-th candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 5 = 12$.

Buggy Code:
# https://codeforces.com/contest/1976

import sys
from collections import deque

input = lambda: sys.stdin.readline().rstrip()  # faster!

def solve_case():
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    ans = [0] * (n + m + 1)

    a_a, b_b, a_b, b_a = deque(), deque(), deque(), deque()
    cnt_a, cnt_b = 0, 0
    for i in range(1, n + m + 1):
        if a[i] > b[i]:
            if cnt_a < n:
                cnt_a += 1
                ans[0] += a[i]
                a_a.append(i)
            else:
                cnt_b += 1
                ans[0] += b[i]
                a_b.append(i)
        else:
            if cnt_b < m:
                cnt_b += 1
                ans[0] += b[i]
                b_b.append(i)
            else:
                cnt_a += 1
                ans[0] += a[i]
                b_a.append(i)

    cnt_a, cnt_b = 0, 0
    for i in range(1, n + m + 1):
        ans[i] = ans[i - 1]

        if a[i - 1] > b[i - 1]:
            if cnt_a < n:
                cnt_a += 1
                ans[i] += a[i - 1]
            else:
                cnt_b += 1
                ans[i] += b[i - 1]
        else:
            if cnt_b < m:
                cnt_b += 1
                ans[i] += b[i - 1]
            else:
                cnt_a += 1
                ans[i] += a[i - 1]

        if a_a and a_a[0] == i:
            ans[i] -= a[i]
            a_a.popleft()
        elif a_b and a_b[0] == i:
            ans[i] -= b[i]
            a_b.popleft()
        elif b_b and b_b[0] == i:
            ans[i] -= b[i]
            b_b.popleft()
        elif b_a and b_a[0] == i:
            ans[i] -= a[i]
            b_a.popleft()

        if cnt_a + len(a_a) + len(b_a) == n:
            pass
        elif cnt_a + len(a_a) + len(b_a) < n:
            if len(a_b) > 0:
                j = a_b.popleft()
                ans[i] -= b[j]
                a_a.append(j)
                ans[i] += a[j]
            elif len(b_b) > 0:
                j = b_b.pop()
                ans[i] -= b[j]
                b_a.append(j)
                ans[i] += a[j]
            else:
                assert False

        elif cnt_b + len(b_b) + len(a_b) < m:
            if len(b_a) > 0:
                j = b_a.popleft()
                ans[i] -= a[j]
                b_b.append(j)
                ans[i] += b[j]
            elif len(a_a) > 0:
                j = a_a.pop()
                ans[i] -= a[j]
                a_b.append(j)
                ans[i] += b[j]
            else:
                assert False

        else:
            assert False

    return " ".join(map(str, ans))

ans = []
for _ in range(int(input())):
    ans += [str(solve_case())]
print("\n".join(ans))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Job Interview
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp is opening his own IT company. He wants to hire $n$ programmers and $m$ testers.

There are $n+m+1$ candidates, numbered from $1$ to $n+m+1$ in chronological order of their arriving time. The $i$-th candidate has programming skill $a_i$ and testing skill $b_i$ (a person's programming skill is different from their testing skill). The skill of the team is the sum of the programming skills of all candidates hired as programmers, and the sum of the testing skills of all candidates hired as testers.

When a candidate arrives to interview, Monocarp tries to assign them to the most suitable position for them (if their programming skill is higher, then he hires them as a programmer, otherwise as a tester). If all slots for that position are filled, Monocarp assigns them to the other position.

Your task is, for each candidate, calculate the skill of the team if everyone except them comes to interview. Note that it means that exactly $n+m$ candidates will arrive, so all $n+m$ positions in the company will be filled.

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

Each test case consists of three lines:

  * the first line contains two integers $n$ and $m$ ($0 \le n, m \le 2 \cdot 10^5$; $2 \le n + m + 1 \le 2 \cdot 10^5$) — the number of programmers and the number of testers Monocarp wants to hire, respectively;    * the second line contains $n + m + 1$ integers $a_1, a_2, \dots, a_{n+m+1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the programming skill of the $i$-th candidate;    * the third line contains $n + m + 1$ integers $b_1, b_2, \dots, b_{n+m+1}$ ($1 \le b_i \le 10^9$; $b_i \ne a_i$), where $b_i$ is the testing skill of the $i$-th candidate.

Additional constraint on the input: the sum of $(n + m + 1)$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print $n + m + 1$ integers, where the $i$-th integer should be equal to the skill of the team if everyone except the $i$-th candidate comes to interview.

Examples:
input:
4
1 0
2 1
1 2
0 2
4 5 5
5 4 1
1 2
2 1 5 4
5 2 3 1
3 1
4 3 3 4 1
5 5 4 5 2
output:
1 2
5 6 9
8 11 11 12
13 13 13 12 15

Note: Let's consider the third test case of the example:

  * if the $1$-st candidate does not arrive, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $2 + 5 + 1 = 8$;    * if the $2$-nd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $5 + 5 + 1 = 11$;    * if the $3$-rd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $4$-th candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 4 = 11$;    * if the $4$-th candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 5 = 12$.

Buggy Code:
class SegmentTree():
    def __init__(self,tab):
        self.N = 1
        n = len(tab)
        while self.N < n:
            self.N *= 2
        self.tab = [0] * (2 * self.N)
        for i in range(n):
            self.tab[self.N + i] = tab[i]
        for i in range(self.N-1,0,-1):
            self.tab[i] = self.tab[2*i] + self.tab[2*i + 1]

    def sum_(self,a,b,l,r,i):
        if a <= l <= r <= b:
            return self.tab[i]
        if r >= a and l <= b:
            m = (l+r)//2
            return self.sum_(a,b,l,m,2*i) + self.sum_(a,b,m+1,r,2*i+1)
        return 0

    def sum(self,a,b):
        return self.sum_(a,b,0,self.N-1,1)

    def prefix_sum(self, k, interdit):
        s = self.sum_(0,k,0,self.N-1,1)
        if interdit <= k:
            s -= self.tab[interdit+self.N]
        return s

    def first_depass(self, n, interdit):
        i = 0
        j = self.N-1
        while j-i > 0:
            m = (i+j)//2
            s = self.prefix_sum(m,interdit)
            if s < n:
                i = m+1
            else:
                j = m
        return i

t = int(input())
p = []

for _ in range(t):
    n,m = map(int,input().split())
    a = list(map(int,input().split()))
    b = list(map(int,input().split()))
    a_pris = []
    b_pris = []
    acc_score = 0
    best_acc_score = []
    for i in range(0,n+m+1):
        if a[i] > b[i]:
            a_pris.append(1)
            b_pris.append(0)
            acc_score += a[i]
        else:
            a_pris.append(0)
            b_pris.append(1)
            acc_score += b[i]
        best_acc_score.append(acc_score)

    ta = SegmentTree(a_pris)
    tb = SegmentTree(b_pris)

    suffix_b = [0] * (n+m+2)
    suffix_a = [0] * (n+m+2)
    for i in range(n+m,-1,-1):
        suffix_a[i] = suffix_a[i+1] + a[i]
        suffix_b[i] = suffix_b[i+1] + b[i]

    sol = []
    for interdit in range(n+m+1):
        fa = ta.first_depass(n,interdit)
        fb = tb.first_depass(m,interdit)
        if fa <= fb:
            stop = fa
            score = best_acc_score[stop] + suffix_b[stop+1]
            if interdit <= stop:
                score -= (max(a[interdit],b[interdit]))
            else:
                score -= b[interdit]
            sol.append(score)
        else:
            stop = fb
            score = best_acc_score[stop] + suffix_a[stop+1]
            if interdit <= stop:
                score -= (max(a[interdit],b[interdit]))
            else:
                score -= a[interdit]
            sol.append(score)
    p.append(sol)

for x in p:
    print(*x)

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

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

Think step by step.

Response:
```plaintext
1
2 1
10 20 30
30 10 20
```