**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define pb push_back
#define lp(i, a, b) for (int i = a; i < b; i++)
#define rlp(i, a, b) for (int i = a; i >= b; i--)
#define yes cout << "Yes" << "\n"
#define no cout << "No" << "\n"
#define F first
#define S second
#define Fast ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
#define nl "\n"
#define all(v) v.begin(), v.end()
int INF = 1e9 + 7;

void solve()
{
    int n,m;
    cin>>n>>m;
    vector<int> a(n),b(m);
    lp(i,0,n) cin>>a[i];
    lp(i,0,m) cin>>b[i];
    sort(all(b));
    int cur=a[0];
    sort(all(a));
    vector<int> val;
    lp(i,0,m)
    {
        int pos=(lower_bound(a.begin(),a.end(),b[i])-a.begin());
        if(pos==n)
        {
            val.pb(1);
        }
        else
        {
            if(a[pos]<=cur)
            {
                val.pb(1);
            }
            else
            {
                val.pb(n-pos+1);
            }
        }

    }

    sort(all(val));

lp(j,1,m+1)
{
    int cur=0;
    for(int l=j-1;l<m;l+=j)
    {
       cur+=val[l];
    }
    cout<<cur<<" ";
}
cout<<nl;
}

int main()
{
    std::cout << std::setprecision(10) << std::fixed;
    Fast
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 6
10 20 30
1 2 3 4 5 6
```

**Explanation:**

- **Number of Test Cases (t):** 1
- **Number of Participants (n):** 3
- **Number of Problems (m):** 6
- **Participant Ratings (a):** [10, 20, 30]
- **Problem Difficulties (b):** [1, 2, 3, 4, 5, 6]

This test case is designed to expose potential issues in the buggy code:

1. **Sorted Participant Ratings and Problem Difficulties:**
   - The ratings and difficulties are sorted in increasing order, which can help in identifying issues with the logic that relies on the order of elements.

2. **Kevin's Rating (a[0]):**
   - Kevin's rating is 10, which is the lowest among the participants. This means that Kevin will only be able to solve problems with difficulties less than or equal to 10.

3. **Problem Difficulties:**
   - The problem difficulties are [1, 2, 3, 4, 5, 6], which means that Kevin can solve the first 3 problems (1, 2, 3) and will struggle with the rest.

4. **Contest Distribution:**
   - For \( k = 1 \), each contest has only one problem, and Kevin will have a rank of 1 for the first 3 problems and a higher rank for the rest.
   - For \( k = 2 \), each contest has 2 problems. Kevin can solve both problems in the first contest (1, 2) but will have a higher rank in subsequent contests.
   - For \( k = 3 \), each contest has 3 problems. Kevin will solve all problems in the first contest (1, 2, 3) and have a rank of 1.
   - For \( k = 4 \), each contest has 4 problems. Kevin will solve only the first 3 problems and have a higher rank.
   - For \( k = 5 \), each contest has 5 problems. Kevin will solve the first 3 problems and have a higher rank.
   - For \( k = 6 \), there is only one contest with all 6 problems. Kevin will solve the first 3 problems and have a higher rank.

This test case should help in identifying any issues with the logic used to calculate the minimum sum of Kevin's ranks for different values of \( k \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

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

#define watch(x) cerr << "\n" << (#x) << " is " << (x) << "\n";
// #define int             long long
#define ffor(i, a, b)   for(int i=a;i<b;i++)
#define rfor(i, a, b)   for(int i=a-1;i>=b;i--)
#define all(v)          v.begin(), v.end()
#define rall(v)         v.rbegin(), v.rend()
#define pb              push_back
#define fi              first
#define se              second
#define pii             pair<int, int>
#define vi              vector<int>
#define vvi             vector<vi>
#define show(a)         for(auto e : a) {cout<<e<<" ";} cout<<"\n";
#define pr(a, b, c)     ffor(i, b, c) {cout<<a[i]<<" ";} cout<<"\n";
// #define cerr            if(false)cerr

const int N = 2e5 + 10;
int Mod = 1e9 + 7, BASE = 7, INF = 1e9 + 11, Block = 555;

void solve() {

    int n, m; cin>>n>>m;

    int f; cin>>f;
    vi r;
    ffor(i, 1, n) {
        int x; cin>>x;
        if(x>f) r.pb(x);
    }

    vi d;
    ffor(i, 0, m) {
        int x; cin>>x;
        if(x <= f) x = INF;
        d.pb(x);
    }

    sort(all(d)); sort(all(r));

    int c[m];
    int rr = (int)r.size() - 1;

    rfor(i, d.size(), 0) {
        while(rr >= 0 && r[rr] >= d[i]) {
            rr--;
        }
        c[i] = (int)r.size() - 1 - rr;
    }

    reverse(c, c + m);

    int ans[m + 1] = {0};
    ffor(k, 1, m + 1) {
        int idx = k - 1;
        while(idx < m) {
            ans[k] += c[idx] + 1; idx += k;
        }
    }

    pr(ans, 1, m + 1);
}

signed main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t = 1; cin>>t;

    while(t--) solve();

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

Buggy Code:
import math
t = int(input())
for  _ in range(t):
    n, m =  map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    rank_per_question = []
    for i in b:
        if i <= a[0]:
            rank_per_question.append(1)
        elif i > a[0]:
            count = 0
            for num in a[:]:
                if num < a[0]:
                    a.remove(num)
                if num >= i:
                    count += 1
            rank_per_question.append(count + 1)

    rank_list = []
    rank_per_question.sort(reverse=False)
    for i in range(1, m+1):
        number_of_contest = math.floor(m / i)
        index = 0
        rank = 0
        while number_of_contest > 0:
            number_of_contest -= 1
            rank += rank_per_question[index+i-1]
            index += i
        rank_list.append(rank)

    result = " ".join(map(str, rank_list))
    print(result)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

Buggy Code:
#let's optimize this pretty slow code!
from functools import lru_cache

@lru_cache(maxsize=None)
def magic_func(testcase, i):
    return len([p for p in a_prime if p>=hard_ps[i]])

t = int(input())

for testcase in range(t):
    n, m = [int(d) for d in input().split()]
    a = [int(d) for d in input().split()]
    b = [int(d) for d in input().split()]
    kevin = a[0]
    a_prime = [p for p in a if p>kevin]
    hard_ps = [p for p in b if p>kevin]
    hard_ps.sort(reverse=True)
    n_easy_ps = m - len(hard_ps)
    # so easy ones are no problems
    # but the one kevin doesn't get - no one else should get them
    for k in range(1, m+1):
        print(m//k + sum([magic_func(testcase, i) for i in range(k-(n_easy_ps%k)-1, len(hard_ps), k)]), end=' ')
    print('')

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

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

Think step by step.

Response:
```plaintext
1
3 6
10 8 12
1 2 3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

Buggy Code:
// Author: Swarnadip Kar
// IIT Bhilai

import java.io.*; // imports everything
import java.lang.*;
import java.util.*;

public class D_Kevin_and_Competition_Memories {

	static FastScanner sc = new FastScanner();
	static long MOD=1000000007;

	public static void main(String[] args) throws IOException {
		int caseNo, i, j;
		int N = sc.nextInt();
		for (caseNo = 0; caseNo < N; caseNo++) {
			int n = sc.nextInt();
			int m = sc.nextInt();
			long[] a = new long[n];
			long[] b = new long[m];

			// Read the array elements
			for (i = 0; i < n; i++) {
				a[i] = sc.nextLong();
			}
			long kevin=a[0];
			for (i = 0; i < m; i++) {
				b[i] = sc.nextLong();
			}
			Arrays.sort(a);
			Arrays.sort(b);

			long[] p = new long[m+1];
			j=1;
			for (i = 0; i < m; i++) {
				if(b[i] <= kevin)
				{
					p[j++]=b[i];
				}
				else
				{
					break;
				}
			}

			int high=(j-1);//highest problem solvable by kevin
			for(int idx=m-1;idx>=i;idx--)
			{
				p[j++]=b[idx];
			}

			//debug("a",a);

			//debug("p",p);
			for(int k=1;k<=m;k++)
			{
				int sum=0;
				int con=m/k;
				int cn=high/k;
				sum+=Math.min(con,cn); //rank 1;
				cn++;

				//debug("sum",sum);

				while(cn<=con)
				{
					long pp=p[cn*k];

					//debug("k",k);

					//debug("cn",cn);

					int z=binarySearch(0, n-1, a, pp);//how many people solve more than the problem rated p
					//debug("z",z);
					if(z==n)
					{
						z=0;
					}
					else
					{
						z=n-z;
					}

					sum+=z+1;
					cn++;
				}
				System.out.print(sum+" ");
			}
			System.out.println();

		}
	}

	//Binary Search Template for Finding Minimum Value in Array Satisfying a Condition
	static int binarySearch(int lo, int hi, long p[],long key) { //lowest indx whose value is greater than or equl tp key
		int ans = hi+1; // Initialize the answer
		while (lo <= hi) {
			int mid = lo + (hi - lo) / 2; // Avoid overflow
			if (p[mid]>=key) { //update the arguments here{
				ans = mid;
				hi = mid - 1;
				 // Update answer to current mid
				 // Move left to minimize the value
			} else {
				lo = mid + 1; // Move right if condition is false
			}
		}
		return ans; // Returns -1 if no valid value exists
	}
	// FastScanner Class
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

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

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

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

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

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

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

// Debugging Funtions
	// use debug("a",a);

	// Toggle this flag for debug mode (set to false for online judge)
	private static final boolean DEBUG_MODE = !isOnlineJudge();
	// private static final boolean isOnlineJudge =

	private static boolean isOnlineJudge() {
		// System.out.println("Working Directory: " + System.getProperty("user.dir"));
		return !new File("FileForDebugging.txt").exists();
	}

	// Debug method to log variable name and value
	public static void debug(String varName, Object value) {
		if (DEBUG_MODE) {
			System.err.println("DEBUG: " + varName + " = " + value);
		}
	}

	// Overloaded debug method for collections and maps
	public static void debug(String varName, Collection<?> value) {
		if (DEBUG_MODE) {
			System.err.print("DEBUG: " + varName + " = [ ");
			for (Object item : value) {
				System.err.print(item + " ");
			}
			System.err.println("]");
		}
	}

	public static void debug(String varName, Map<?, ?> value) {
		if (DEBUG_MODE) {
			System.err.print("DEBUG: " + varName + " = { ");
			for (Map.Entry<?, ?> entry : value.entrySet()) {
				System.err.print(entry.getKey() + ": " + entry.getValue() + ", ");
			}
			System.err.println("}");
		}
	}

	public static void debug(String varName, Object[] value) {
		if (DEBUG_MODE) {
			System.err.print("DEBUG: " + varName + " = [ ");
			for (Object item : value) {
				System.err.print(item + " ");
			}
			System.err.println("]");
		}
	}

	public static void debug(String varName, int[] value) {
		if (DEBUG_MODE) {
			System.err.print("DEBUG: " + varName + " = [ ");
			for (int item : value) {
				System.err.print(item + " ");
			}
			System.err.println("]");
		}
	}

	public static void debug(String varName, long[] value) {
		if (DEBUG_MODE) {
			System.err.print("DEBUG: " + varName + " = [ ");
			for (long item : value) {
				System.err.print(item + " ");
			}
			System.err.println("]");
		}
	}

	public static void debug(String varName, float[] value) {
		if (DEBUG_MODE) {
			System.err.print("DEBUG: " + varName + " = [ ");
			for (float item : value) {
				System.err.print(item + " ");
			}
			System.err.println("]");
		}
	}

	public static void debug(String varName, String[] value) {
		if (DEBUG_MODE) {
			System.err.print("DEBUG: " + varName + " = [ ");
			for (String item : value) {
				System.err.print(item + " ");
			}
			System.err.println("]");
		}
	}

	// //Example
	// public static void main(String[] args) {
	// // Example variables
	// int number = 42;
	// String text = "Hello, Debug!";
	// List<Integer> list = Arrays.asList(1, 2, 3, 4);
	// Map<String, Integer> map = new HashMap<>();
	// map.put("One", 1);
	// map.put("Two", 2);

	// // Debugging the variables
	// debug("number", number); // DEBUG: number = 42
	// debug("text", text); // DEBUG: text = Hello, Debug!
	// debug("list", list); // DEBUG: list = [ 1 2 3 4 ]
	// debug("map", map); // DEBUG: map = { One: 1, Two: 2, }
	// }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to create such a test input:

1. **Small-scale Test Case**: Ensure the test case is small enough to be manageable and to fit within the constraints.
2. **Edge Cases**:
   - **Single Participant**: Test cases with only one participant.
   - **Single Problem**: Test cases with only one problem.
   - **Equal Ratings and Difficulties**: Test cases where all participants have the same rating or all problems have the same difficulty.
   - **Kevin's Rating Lower than All Problems**: Test cases where Kevin's rating is lower than the difficulty of all problems.
   - **Kevin's Rating Higher than All Problems**: Test cases where Kevin's rating is higher than the difficulty of all problems.
3. **Complex Distribution**:
   - Test cases where the distribution of problems and participants' ratings are such that the ranking calculation can be tricky.

Given these considerations, here is a small-scale test input that covers various edge cases:

```plaintext
5
1 1
5
10
2 2
5 10
1 2
3 3
5 5 10
1 5 10
4 4
1 2 3 4
1 2 3 4
5 5
0 1 2 3 4
0 1 2 3 4
```

### Explanation of Test Cases:
1. **Test Case 1**:
   - **Participants**: 1
   - **Problems**: 1
   - **Ratings**: [5]
   - **Difficulties**: [10]
   - **Expected Output**: 1
   - **Reason**: Kevin cannot solve any problems, so his rank is 1.

2. **Test Case 2**:
   - **Participants**: 2
   - **Problems**: 2
   - **Ratings**: [5, 10]
   - **Difficulties**: [1, 2]
   - **Expected Output**: 2 2
   - **Reason**: Kevin can solve both problems, but there is another participant who can also solve both, so his rank in both contests is 1.

3. **Test Case 3**:
   - **Participants**: 3
   - **Problems**: 3
   - **Ratings**: [5, 5, 10]
   - **Difficulties**: [1, 5, 10]
   - **Expected Output**: 3 2 1
   - **Reason**: Kevin can solve problems with difficulty 1 and 5, and there is another participant who can solve all problems, so his rank varies based on the number of problems per contest.

4. **Test Case 4**:
   - **Participants**: 4
   - **Problems**: 4
   - **Ratings**: [1, 2, 3, 4]
   - **Difficulties**: [1, 2, 3, 4]
   - **Expected Output**: 4 3 2 1
   - **Reason**: Each participant can solve problems up to their rating, so Kevin's rank varies based on the number of problems per contest.

5. **Test Case 5**:
   - **Participants**: 5
   - **Problems**: 5
   - **Ratings**: [0, 1, 2, 3, 4]
   - **Difficulties**: [0, 1, 2, 3, 4]
   - **Expected Output**: 5 4 3 2 1
   - **Reason**: Each participant can solve problems up to their rating, so Kevin's rank varies based on the number of problems per contest.

This test input should help expose any bugs in the provided code by covering a variety of edge cases and complex distributions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kevin and Competition Memories
time_limit_ms: 2000
memory_limit_mb: 256
Description: Kevin used to get into Rio's Memories, and in Rio's Memories, a series of contests was once held. Kevin remembers all the participants and all the contest problems from that time, but he has forgotten the specific rounds, the distribution of problems, and the exact rankings.

There are $$$ m $$$ problems in total, with the $$$ i $$$-th problem having a difficulty of $$$ b_i $$$. Let each contest consist of $$$ k $$$ problems, resulting in a total of $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. This means that you select exactly $$$ \lfloor \frac{m}{k} \rfloor \cdot k $$$ problems for the contests in any combination you want, with each problem being selected at most once, and the remaining $$$m\bmod k$$$ problems are left unused. For example, if $$$m = 17$$$ and $$$k = 3$$$, you should create exactly $$$5$$$ contests consisting of $$$3$$$ problems each, and exactly $$$2$$$ problems will be left unused.

There are $$$ n $$$ participants in the contests, with Kevin being the $$$1$$$-st participant. The $$$ i $$$-th participant has a rating of $$$ a_i $$$. During the contests, each participant solves all problems with a difficulty not exceeding their rating, meaning the $$$ i $$$-th participant solves the $$$ j $$$-th problem if and only if $$$ a_i \geq b_j $$$. In each contest, Kevin's rank is one plus the number of participants who solve more problems than he does.

For each $$$ k = 1, 2, \ldots, m $$$, Kevin wants to know the minimum sum of his ranks across all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. In other words, for some value of $$$k$$$, after selecting the problems for each contest, you calculate the rank of Kevin in each contest and sum up these ranks over all $$$ \lfloor \frac{m}{k} \rfloor $$$ contests. Your goal is to minimize this value.

Note that contests for different values of $$$k$$$ are independent. It means that for different values of $$$k$$$, you can select the distribution of problems into the contests independently.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$ t $$$ ($$$ 1 \le t \le 5\cdot 10^4 $$$).

The first line of each test case contains two integers $$$ n $$$ and $$$ m $$$ ($$$ 1 \le n, m \leq 3\cdot 10^5 $$$) — the number of participants and the number of problems.

The second line of each test case contains $$$ n $$$ integers $$$ a_1, a_2, \ldots, a_n $$$ ($$$ 0 \le a_i \le 10^9 $$$) — the rating of each participant.

The third line of each test case contains $$$ m $$$ integers $$$ b_1, b_2, \ldots, b_m $$$ ($$$ 0 \le b_i \le 10^9 $$$) — the difficulty of each problem.

It is guaranteed that both the sum of $$$ n $$$ and the sum of $$$ m $$$ over all test cases do not exceed $$$ 3 \cdot 10^5 $$$.

Output: For each test case, output $$$m$$$ integers — the minimum sum of Kevin's ranks for each $$$ k = 1, 2, \ldots, m$$$.

Examples:
input:
4
4 4
4 3 7 5
2 5 4 6
5 5
5 0 4 8 6
1 3 9 2 7
6 7
1 1 4 5 1 4
1 9 1 9 8 1 0
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
output:
7 4 2 3
6 2 1 1 2
7 3 2 1 1 1 1
15 9 5 4 4 4

Note: For the first test case:

When $$$k=1$$$, since each contest only contains one problem, the distribution is in fact unique. For example, in the contest which only includes the third problem (which has a difficulty of $$$4$$$), all participants except the $$$2$$$-nd can solve it. Since no one solves strictly more problems than Kevin, his ranking in this contest is $$$1$$$. Similarly, in all $$$4$$$ contests, Kevin's rankings are $$$1,3,1,2$$$, and the sum is $$$7$$$.

When $$$k=2$$$, one optimal way is to choose the $$$1$$$-st and the $$$3$$$-rd problem to form a contest, while the $$$2$$$-nd and $$$4$$$-th for another. In the former contest, $$$4$$$ participants respectively solve $$$2,1,2,2$$$ problems, so Kevin's ranking is $$$1$$$; in the latter one, they respectively solve $$$0,0,2,1$$$, since there are $$$2$$$ participants ($$$3$$$-rd and $$$4$$$-th) solve more problems than Kevin, his ranking is $$$1+2=3$$$. Thus the answer is $$$1+3=4$$$. It can be proven that there's no way to achieve a lower sum.

When $$$k=3$$$, we can simply choose the $$$1$$$-st, the $$$3$$$-rd, and the $$$4$$$-th problem to make a contest, and Kevin has a ranking of $$$2$$$, which is optimal.

When $$$k=4$$$, since there's only one contest, the distribution is also unique, and Kevin's ranking is $$$3$$$.

Buggy Code:
import java.util.*;

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

        while (t-- > 0) {
            int N = sc.nextInt();
            int N2 = sc.nextInt();

            ArrayList<Integer> l = new ArrayList<>();
            for (int i = 0; i < N; i++) {
                int inputValue = sc.nextInt();
                if (l.isEmpty() || inputValue > l.get(0)) {
                    l.add(inputValue);
                }
            }

            int[] arr = new int[N2];
            for (int i = 0; i < N2; i++) {
                arr[i] = sc.nextInt();
            }

            Collections.sort(l);
            Arrays.sort(arr);

            List<Integer> dff = new ArrayList<>();
            int ptr = 0;

            for (int i = 0; i < N2; i++) {
                while (ptr < l.size() && l.get(ptr) < arr[i]) {
                    ptr++;
                }
                if (ptr > 0 && ptr < l.size()) {
                    dff.add(l.size() - ptr);
                }
            }

            dff.sort(Collections.reverseOrder());
            int len = dff.size();
            for (int j = 1; j <= N2; j++  ) {
                long sum = N2 / j;
                int rem = N2 % j;
                for (int k = rem; k < len; k += j)  sum += dff.get(k);
                System.out.print(sum + " ");
            }
            System.out.println();
        }
    }
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to challenge the buggy code by focusing on the edge case where the distribution of problems and the participants' ratings are such that the ranking of Kevin (the first participant) can vary significantly based on the value of \( k \).

- **Participants and Ratings**: There are 4 participants with ratings 4, 3, 7, and 5.
- **Problems and Difficulties**: There are 4 problems with difficulties 2, 5, 4, and 6.

### Analysis:
1. **When \( k = 1 \)**:
   - Each contest has 1 problem.
   - Kevin's rankings in each contest are 1, 3, 1, 2.
   - Sum of ranks: 1 + 3 + 1 + 2 = 7.

2. **When \( k = 2 \)**:
   - Each contest has 2 problems.
   - One optimal distribution is:
     - Contest 1: Problems 1 and 3 (difficulties 2 and 4)
     - Contest 2: Problems 2 and 4 (difficulties 5 and 6)
   - In Contest 1: Kevin solves both problems (rank 1).
   - In Contest 2: Kevin solves 1 problem (rank 3).
   - Sum of ranks: 1 + 3 = 4.

3. **When \( k = 3 \)**:
   - Each contest has 3 problems.
   - One optimal distribution is:
     - Contest 1: Problems 1, 3, and 4 (difficulties 2, 4, 6)
   - Kevin solves 2 problems (rank 2).
   - Sum of ranks: 2.

4. **When \( k = 4 \)**:
   - Each contest has 4 problems.
   - Only one contest.
   - Kevin solves 2 problems (rank 3).
   - Sum of ranks: 3.

This test case is small and valid, and it should expose the bug in the buggy code, which incorrectly handles the distribution of problems and the calculation of Kevin's ranks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Hard Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is the hard version of the problem. The only difference is that in this version $n \leq 5 \cdot 10^5$ and the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, in which each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 5 \cdot 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
// LUOGU_RID: 164426733
// Author: XZC(L_Wave)
// Language: Cpp/G++14
// Problem: Permutation Problem (Hard Version)
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF1986G2
// Memory Limit: 125 MB
// Time Limit: 3000 ms
// Create Time: not 2024-07-07 15:31:39, but 1926-08-17 11:45:14
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = (a), i##ABRACADABRA = (b); i <= i##ABRACADABRA; i++)
#define drep(i, a, b) for (int i = (a), i##ABRACADABRA = (b); i >= i##ABRACADABRA; i--)
using namespace std;
using ll = long long;
constexpr int N=5e5+10;

vector<int>fac[N],P[N],Q[N];
int n,p[N],q[N],cnt[N];

void solve(){
  ll res=0;
  scanf("%d",&n);
  rep(i,1,n){
    scanf("%d",&p[i]);
    q[i]=i;
    res-=p[i]%q[i]==0;
    int d=__gcd(p[i],q[i]);
    p[i]/=d,q[i]/=d;
  }
  // p[i] % q[j] == 0
  rep(i,1,n)P[i].clear(),Q[i].clear();
  rep(i,1,n)P[p[i]].push_back(i);
  rep(i,1,n)Q[q[i]].push_back(i);
  rep(i,1,n)if (Q[i].size()){
    for (auto x:Q[i])for (auto t:fac[p[x]])++cnt[t];
    for (int j=1;1ll*i*j<=N-5;j++) // p[j]
      for (auto x:P[i*j])res+=cnt[q[x]];
    for (auto x:Q[i])for (auto t:fac[p[x]])--cnt[t];
  }
  printf("%lld\n",res>>1);
}

int main() {
  rep(i,1,N-5)for (int j=1;1ll*i*j<=N-5;j++)
    fac[i*j].push_back(i);
  int tt;
  scanf("%d",&tt);
  while (tt--)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 create a challenging test input that exposes potential bugs in the provided code, we need to consider the following aspects:

1. **Edge Cases:**
   - Single element permutation.
   - Permutations where all elements are in increasing or decreasing order.
   - Permutations where elements are in a pattern that maximizes the number of valid pairs.

2. **Large Input:**
   - Test with the maximum allowed input size to check for performance issues.

3. **Specific Patterns:**
   - Test with permutations that have specific patterns to ensure the code handles all cases correctly.

Given the constraints and the need for a small-scale test input, let's focus on creating a test case that includes a permutation with a pattern that maximizes the number of valid pairs. This will help in identifying any issues in the code's logic.

### Chosen Test Input:
- **Permutation:** `[1, 2, 3, 4, 5]`
- **Reasoning:** This permutation is in increasing order and has a simple structure. It should generate a few valid pairs, making it a good candidate for testing the code's logic.

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

### Explanation:
- **t = 1:** One set of input data.
- **n = 5:** The length of the permutation.
- **p = [1, 2, 3, 4, 5]:** The permutation itself.

### Expected Output:
- The expected output for this input is `3`, as there are three valid pairs: `(1, 2)`, `(1, 4)`, and `(2, 4)`.

This test input is small enough to be manageable and complete, while still being challenging enough to expose potential issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Hard Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is the hard version of the problem. The only difference is that in this version $n \leq 5 \cdot 10^5$ and the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, in which each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 5 \cdot 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
import math
from sys import stdin

class FastIO:
    def __init__(self):
        self.random_seed = 0
        self.flush = False
        self.inf = 1 << 32
        return

    @staticmethod
    def read_int():
        return int(stdin.readline().rstrip())

    @staticmethod
    def read_float():
        return float(stdin.readline().rstrip())

    @staticmethod
    def read_list_ints():
        return list(map(int, stdin.readline().rstrip().split()))

    @staticmethod
    def read_list_ints_minus_one():
        return list(map(lambda x: int(x) - 1, stdin.readline().rstrip().split()))

    @staticmethod
    def read_str():
        return stdin.readline().rstrip()

    @staticmethod
    def read_list_strs():
        return stdin.readline().rstrip().split()

    def get_random_seed(self):
        import random
        self.random_seed = random.randint(0, 10 ** 9 + 7)
        return

    def st(self, x):
        return print(x, flush=self.flush)

    def yes(self, s=None):
        self.st("Yes" if not s else s)
        return

    def no(self, s=None):
        self.st("No" if not s else s)
        return

    def lst(self, x):
        return print(*x, flush=self.flush)

    def flatten(self, lst):
        self.st("\n".join(str(x) for x in lst))
        return

    @staticmethod
    def max(a, b):
        return a if a > b else b

    @staticmethod
    def min(a, b):
        return a if a < b else b

    @staticmethod
    def ceil(a, b):
        return a // b + int(a % b != 0)

    @staticmethod
    def accumulate(nums):
        n = len(nums)
        pre = [0] * (n + 1)
        for i in range(n):
            pre[i + 1] = pre[i] + nums[i]
        return pre

class NumFactor:
    def __init__(self):
        return

    @staticmethod
    def get_all_factor(num):  # faster when 1 <= num <= 10**6!
        """Obtain all factors of an integer, including 1 and itself"""
        assert num >= 1
        factor = set()
        for i in range(1, int(math.sqrt(num)) + 1):
            if num % i == 0:
                factor.add(i)
                factor.add(num // i)
        return factor

class Solution:
    def __init__(self):
        return

    @staticmethod
    def main(ac=FastIO()):
        """
        url: url of the problem
        tag: algorithm tag
        """
        res = []
        for _ in range(ac.read_int()):
            ac.read_int()
            nums = ac.read_list_ints()
            need = dict()

            ans = 0
            for i, num in enumerate(nums):
                i += 1
                g = math.gcd(num, i)
                xx = num // g
                cur = set()
                for w in range(1, int(math.sqrt(xx)) + 1):
                    if num % w == 0:
                        cur.add(w)
                        cur.add(num // w)
                for w in cur:
                    if w in need:
                        for k in need[w]:
                            if k % (i // g) == 0 and k in need[w]:
                                ans += need[w][k]
                if i // g not in need:
                    need[i // g] = dict()
                need[i // g][num // g] = need[i // g].get(num // g, 0) + 1
            res.append(ans)

        for a in res:
            ac.st(a)
        return

Solution().main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that can maximize the failure probability. The key aspects to focus on are:

1. **Edge Cases**:
   - Single-element permutations.
   - Permutations of small size (e.g., 2, 3).
   - Permutations with specific patterns (e.g., all elements in ascending or descending order).

2. **Complex Cases**:
   - Permutations with large values of \( n \) (up to \( 5 \cdot 10^5 \)).
   - Permutations with many pairs that satisfy the condition \( p_i \cdot p_j \) being divisible by \( i \cdot j \).

3. **Special Cases**:
   - Permutations where the values of \( p_i \) and \( i \) have specific relationships (e.g., \( p_i \) is a multiple of \( i \)).

Let's construct a test input that includes a mix of these cases to ensure the code is thoroughly tested.

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

### Explanation:
1. **Single-element permutation**:
   - Test case 1: `1 1` (No pairs, expected output: 0)

2. **Small permutations**:
   - Test case 2: `2 1 2` (One valid pair (1, 2), expected output: 1)
   - Test case 3: `3 2 3 1` (One valid pair (1, 2), expected output: 1)

3. **Larger permutation**:
   - Test case 4: `5 2 4 1 3 5` (Three valid pairs (1, 2), (1, 5), (2, 5), expected output: 3)

4. **Descending order permutation**:
   - Test case 5: `10 10 9 8 7 6 5 4 3 2 1` (Many valid pairs, expected output: 9)

This test input covers a range of scenarios, including edge cases and more complex permutations, to help identify any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Hard Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is the hard version of the problem. The only difference is that in this version $n \leq 5 \cdot 10^5$ and the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, in which each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 5 \cdot 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
import math
from sys import stdin

class FastIO:
    def __init__(self):
        self.random_seed = 0
        self.flush = False
        self.inf = 1 << 32
        return

    @staticmethod
    def read_int():
        return int(stdin.readline().rstrip())

    @staticmethod
    def read_float():
        return float(stdin.readline().rstrip())

    @staticmethod
    def read_list_ints():
        return list(map(int, stdin.readline().rstrip().split()))

    @staticmethod
    def read_list_ints_minus_one():
        return list(map(lambda x: int(x) - 1, stdin.readline().rstrip().split()))

    @staticmethod
    def read_str():
        return stdin.readline().rstrip()

    @staticmethod
    def read_list_strs():
        return stdin.readline().rstrip().split()

    def get_random_seed(self):
        import random
        self.random_seed = random.randint(0, 10 ** 9 + 7)
        return

    def st(self, x):
        return print(x, flush=self.flush)

    def yes(self, s=None):
        self.st("Yes" if not s else s)
        return

    def no(self, s=None):
        self.st("No" if not s else s)
        return

    def lst(self, x):
        return print(*x, flush=self.flush)

    def flatten(self, lst):
        self.st("\n".join(str(x) for x in lst))
        return

    @staticmethod
    def max(a, b):
        return a if a > b else b

    @staticmethod
    def min(a, b):
        return a if a < b else b

    @staticmethod
    def ceil(a, b):
        return a // b + int(a % b != 0)

    @staticmethod
    def accumulate(nums):
        n = len(nums)
        pre = [0] * (n + 1)
        for i in range(n):
            pre[i + 1] = pre[i] + nums[i]
        return pre

class NumFactor:
    def __init__(self):
        return

    @staticmethod
    def get_all_factor(num):  # faster when 1 <= num <= 10**6!
        """Obtain all factors of an integer, including 1 and itself"""
        assert num >= 1
        factor = set()
        for i in range(1, int(math.sqrt(num)) + 1):
            if num % i == 0:
                factor.add(i)
                factor.add(num // i)
        return factor

class Solution:
    def __init__(self):
        return

    @staticmethod
    def main(ac=FastIO()):
        """
        url: url of the problem
        tag: algorithm tag
        """
        n = 5 * 10 ** 5
        # calculate the minimum prime factor for all numbers from 1 to n
        # all_factor = [[], [1]] + [[1, i] for i in range(2, n + 1)]
        # # complexity is O(nlogn)
        # for i in range(2, n + 1):
        #     for j in range(i * i, n + 1, i):
        #         all_factor[j].append(i)
        #         if j > i * i:
        #             all_factor[j].append(j // i)

        # nums = [random.randint(1, n) for _ in range(n)]
        res = []
        for _ in range(ac.read_int()):
            ac.read_int()
            nums = ac.read_list_ints()
            need = dict()

            ans = 0
            for i, num in enumerate(nums):
                i += 1
                g = math.gcd(num, i)
                xx = num // g
                for w in range(1, int(math.sqrt(xx)) + 1):
                    if num % w == 0:
                        if w in need:
                            for k in need[w]:
                                if k % (i // g) == 0 and k in need[w]:
                                    ans += need[w][k]
                        if num // w != w:
                            w = num // w
                            if w in need:
                                for k in need[w]:
                                    if k % (i // g) == 0 and k in need[w]:
                                        ans += need[w][k]
                if i // g not in need:
                    need[i // g] = dict()
                need[i // g][num // g] = need[i // g].get(num // g, 0) + 1
            res.append(ans)

        for a in res:
            ac.st(a)
        return

Solution().main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that maximize the failure probability in faulty solutions. Here are the steps to think through this:

1. **Small Scale but Challenging**: Since the output length is limited, we need to create a small-scale test case that still challenges the algorithm. A permutation of length around 10-20 can be effective.

2. **Edge Cases**:
   - **Single Element Permutation**: This is a trivial case but important to ensure the code handles it correctly.
   - **Two Element Permutation**: This is another simple case but can help verify the basic logic.
   - **Permutation with Many Valid Pairs**: This can help test the algorithm's ability to handle multiple valid pairs.
   - **Permutation with No Valid Pairs**: This can help test the algorithm's ability to handle cases where no pairs are valid.

3. **Complex Permutations**:
   - **Permutations with Specific Patterns**: For example, permutations where the product of elements is divisible by the product of indices in multiple ways.

4. **Memory and Time Efficiency**:
   - **Permutations that Test Memory Usage**: While the test case is small, we can still try to make it challenging by ensuring it requires careful handling of data structures.

Based on these considerations, here is a small-scale test input that includes various edge cases:

```plaintext
5
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
10
8 9 7 12 1 10 6 3 2 4
```

This test input includes:
- A single element permutation.
- A two-element permutation.
- A three-element permutation with one valid pair.
- A five-element permutation with three valid pairs.
- A ten-element permutation with multiple valid pairs.

This should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Hard Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is the hard version of the problem. The only difference is that in this version $n \leq 5 \cdot 10^5$ and the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, in which each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 5 \cdot 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
// package com.company;

import java.util.*;
import java.lang.*;
import java.io.*;

public class Rough_Work {
    public static void main(String[] args) throws IOException {
        new Rough_Work().TestCases();
    }
    void TestCases() throws IOException{
        precompute_factors((int) 5e5);
        int t = 1;
        t = nextInt();
        for (int tc = 1; tc <= t; tc++) {
//            out.print("#"+tc+" ");
            solve();
        }
        out.flush();
    } // Put tc = 1, for NO testcases // out.flush() is written in this method

    // Use ArrayList instead of array..
    // Array give TLE in worst case for JAVA 14 and prior but now worst case Time for sorting is NlogN for JAVA 17.
    // Especially in codeforces.
    void solve () throws IOException {
        int n = nextInt();
        int[] arr = nextIntArray(n);

        int[] a = new int[n], b = new int[n];
        for (int i = 0; i < n; i++) {
            int e = arr[i];

            a[i] = (int) (e / gcd(e,i+1));
            b[i] = (int) ((i+1) / gcd(e,i+1));
        }

//        printIntArray(a);
//        printIntArray(b);

        long ans = 0;
        HashMap<Integer,HashMap<Integer,Integer>> x = new HashMap<>();
        for (int i = 0; i < n; i++) {
            for (int z: factors[a[i]]) {
                if (x.containsKey(z)) {
                    ans += x.get(z).getOrDefault(b[i],0);
                }
            }

            for (int z: factors[a[i]]) {
                x.putIfAbsent(b[i],new HashMap<>());
                x.get(b[i]).put(z,x.get(b[i]).getOrDefault(z,0)+1);
            }
        }

        out.println(ans);
    }

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);
    StringTokenizer st;

    void printIntArray (int[] arr) {
        for (int z : arr) out.print(z+" ");
        out.println();
    }
    void printIntList (ArrayList<Integer> arr) {
        for (int z : arr) out.print(z+" ");
        out.println();
    }
    void printLongArray (long[] arr) {
        for (long z : arr) out.print(z+" ");
        out.println();
    }
    void printLongList (ArrayList<Long> arr) {
        for (long z : arr) out.print(z+" ");
        out.println();
    }
    String next () throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    int nextInt () throws IOException {
        return Integer.parseInt(next());
    }
    int[] nextIntArray (int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) arr[i] = nextInt();
        return arr;
    }
    long nextLong () throws IOException {
        return Long.parseLong(next());
    }
    long[] nextLongArray (int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) arr[i] = nextLong();
        return arr;
    }
    ArrayList<Integer> nextIntList (int n) throws IOException {
        ArrayList<Integer> lis = new ArrayList<>(n);
        for (int i = 0; i < n; i++) lis.add(nextInt());
        return lis;
    }
    ArrayList<Long> nextLongList (int n) throws IOException {
        ArrayList<Long> lis = new ArrayList<>(n);
        for (int i = 0; i < n; i++) lis.add(nextLong());
        return lis;
    }
    double nextDouble () throws IOException {
        return Double.parseDouble(next());
    }
    char nextChar () throws IOException {
        return next().charAt(0);
    }
    String nextLine () throws IOException {
        return br.readLine().trim();
    }

    boolean is_Sorted_int (int[] arr, int l, int r) {
        for (int i = l; i < r-1; i++) if (arr[i] > arr[i+1]) return false;
        return true;
    }
    boolean is_rev_Sorted_int (int[] arr, int l, int r) {
        for (int i = l; i < r-1; i++) if (arr[i] < arr[i+1]) return false;
        return true;
    }
    boolean is_Sorted_long (long[] arr) {
        for (int i = 0; i < arr.length-1; i++) if (arr[i] > arr[i+1]) return false;
        return true;
    }

    long gcd(long a, long b) { return (b==0?a:gcd(b,a%b)); }
    long lcm(long a, long b) { return (a / gcd(a, b)) * b; }

    ArrayList<Integer>[] factors;
    void precompute_factors (int n) {
        factors = new ArrayList[n+1];
        for (int i = 0; i <= n; i++) factors[i] = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            for (int j = i; j <= n; j+=i) {
                factors[j].add(i);
            }
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or specific scenarios that can highlight potential issues. Here are the steps to think through this:

1. **Understand the Bug**: The provided code seems to have issues with how it handles the factors and the gcd calculations. Specifically, the way it updates the `HashMap` and checks for valid pairs might not be optimal.

2. **Edge Cases**:
   - **Single Element Permutations**: These should always result in 0 valid pairs.
   - **Small Permutations**: Test with small permutations to ensure the logic is correct.
   - **Large Permutations**: Test with large permutations to ensure the code handles large inputs efficiently.
   - **Specific Patterns**: Test with permutations that have specific patterns, such as sorted, reverse sorted, or permutations with specific gcd properties.

3. **Challenging Input**:
   - **Permutations with Specific GCD Properties**: Create a permutation where the gcd of `p_i` and `i` and `p_j` and `j` have specific properties to test the gcd calculations.
   - **Permutations with Repeated Patterns**: Create a permutation with repeated patterns to test the handling of repeated factors.

Given these considerations, a challenging small-scale test input could be:

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

This input includes:
- A permutation of length 1, which should result in 0 valid pairs.
- A permutation of length 2, which should result in 1 valid pair.
- A permutation of length 4, which has a specific pattern to test the gcd and factor handling.

This test input is small enough to be manageable but complex enough to potentially expose 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: Permutation Problem (Hard Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is the hard version of the problem. The only difference is that in this version $n \leq 5 \cdot 10^5$ and the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, in which each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 5 \cdot 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

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

#define int long long

using namespace std;

const int N = 500010;

int T;
int n;
int a[N];
vector<int> s1[N], s2[N];
map<pair<int, int>, int> p;

int gcd(int a, int b)
{
	return b? gcd(b, a % b) : a;
}

signed main()
{
	for (int i = 1; i <= 500000; i++)
		for (int j = i; j <= 500000; j += i)
			s1[j].push_back(i);

	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> T;
	while (T--)
	{
		p.clear();

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

		int res = 0;
		for (int i = 1; i <= n; i++)
		{
			int g = gcd(a[i], i);
			int x = a[i] / g, y = i / g;
			for (int j = 0; j < s1[x].size(); j++)
			{
				if (s1[x][j] < 10) res += p[{s1[x][j], y}];
				else
					for (int k = 0; k < s2[s1[x][j]].size(); k++)
						res += s2[s1[x][j]][k] == y;
			}
			for (int j = 0; j < s1[x].size(); j++)
			{
				if (s1[x][j] < 10) p[{y, s1[x][j]}]++;
				else s2[y].push_back(s1[x][j]);
			}
		}

		cout << res << endl;

		for (int i = 1; i <= n; i++) s2[i].clear();
	}

	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 and scenarios that are likely to cause issues. The code involves calculating the greatest common divisor (GCD) and using a map to count valid pairs. Here are some considerations:

1. **Permutations with specific patterns**: Certain patterns in the permutation might expose issues in the GCD calculations or the map usage.
2. **Large values of \( n \)**: Testing with the maximum allowed \( n \) can help identify performance issues or logical errors.
3. **Edge cases**: Permutations of very small length, permutations with specific values, and permutations with a large number of valid pairs.

Given these considerations, let's construct a small-scale test input that is complete and valid:

1. **Small permutation with a specific pattern**: A permutation where the values are in a specific order to maximize the number of valid pairs.
2. **Permutation with large values**: A permutation where the values are large but still within the allowed range.

Here is a small-scale test input that should challenge the buggy code:

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

Explanation:
- The first set of input data is a simple permutation of length 5. The valid pairs are (1, 2), (1, 3), (1, 4), (1, 5), (2, 4), and (3, 4).
- The second set of input data is a permutation of length 6. The valid pairs include (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6), (3, 4), and (3, 6).

This test input should expose any issues in the GCD calculations and the map usage 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: Permutation Problem (Hard Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is the hard version of the problem. The only difference is that in this version $n \leq 5 \cdot 10^5$ and the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, in which each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 5 \cdot 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
import io
import sys
from collections import defaultdict, deque, Counter
from itertools import permutations, combinations, accumulate
from heapq import heappush, heappop
from bisect import bisect_right, bisect_left
from math import gcd
import math

_INPUT = """\
6
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
"""

def input():
  return sys.stdin.readline()[:-1]

divisor=[[] for _ in range(5*10**5)]
for i in range(5*10**5):
  m=i+1
  for j in range((5*10**5)//m):
    divisor[m*(j+1)-1].append(m)
# print(divisor[:20])

def solve(test):
  t=int(input())
  for _ in range(t):
    n=int(input())
    p=list(map(int, input().split()))
    a=[p[i]//gcd(i+1,p[i]) for i in range(n)]
    b=[(i+1)//gcd(i+1,p[i]) for i in range(n)]
    d1=[[] for _ in range(n)]
    d2=[[] for _ in range(n)]
    ans=0
    for i in range(n):
      d1[a[i]-1].append(b[i])
      d2[b[i]-1].append(a[i])
    for i in range(n):
      d3=defaultdict(int)
      for j in range(n//(i+1)):
        for k in d1[(i+1)*(j+1)-1]:
          d3[k]+=1
      # print(d3)
      for j in d2[i]:
        # if j==a[i]: continue
        for div in divisor[j-1]:
          ans+=d3[div]
    for i in range(n):
      if p[i]%(i+1)==0: ans-=1
    print(ans//2)

def random_input():
  from random import randint,shuffle
  N=randint(1,10)
  M=randint(1,N)
  A=list(range(1,M+1))+[randint(1,M) for _ in range(N-M)]
  shuffle(A)
  return (" ".join(map(str, [N,M]))+"\n"+" ".join(map(str, A))+"\n")*3

def simple_solve():
  return []

def main(test):
  if test==0:
    solve(0)
  elif test==1:
    sys.stdin = io.StringIO(_INPUT)
    case_no=int(input())
    for _ in range(case_no):
      solve(0)
  else:
    for i in range(1000):
      sys.stdin = io.StringIO(random_input())
      x=solve(1)
      y=simple_solve()
      if x!=y:
        print(i,x,y)
        print(*[line for line in sys.stdin],sep='')
        break

#0:提出用、1:与えられたテスト用、2:ストレステスト用
main(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 buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are some steps to consider:

1. **Small Permutations with Specific Patterns**: Small permutations can help us test the correctness of the algorithm without overwhelming the input size.
2. **Permutations with Repeated GCD Patterns**: Permutations where the GCD of \(p_i\) and \(i\) has specific patterns can expose issues in the GCD calculations.
3. **Permutations with Large GCD Values**: Permutations where \(p_i\) and \(i\) have large GCD values can test the limits of the algorithm.
4. **Permutations with No Valid Pairs**: Permutations where no valid pairs exist can test the correctness of the base case handling.

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

1. **Test Case 1**: A permutation of length 1, which should have 0 valid pairs.
2. **Test Case 2**: A permutation of length 2, which should have 1 valid pair.
3. **Test Case 3**: A permutation of length 3, which should have 1 valid pair.
4. **Test Case 4**: A permutation of length 5, which should have 3 valid pairs.
5. **Test Case 5**: A permutation of length 10, which should have a specific number of valid pairs.

Here is the test input:

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

This test input includes a variety of small permutations to ensure that the algorithm handles different patterns and edge cases correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Hard Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is the hard version of the problem. The only difference is that in this version $n \leq 5 \cdot 10^5$ and the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, in which each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 5 \cdot 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
import java.util.Scanner;

public class Main {
    static long count=0;
    static void f(int a[],int i,int j){
        if(i>=a.length){
            System.out.println(count);
            return;
        }
        if(j<a.length){if ((a[i]*a[j])%((i+1)*(j+1))==0)
            count+=1;
            f(a,i,j+1);
        }
        if(j>=a.length){
            f(a,i+1,i+2);
        }
        return;

    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t=sc.nextInt();
        for(int i=0;i<t;i++){
            int s=sc.nextInt();
            int a[]=new int[s];
            for(int j=0;j<s;j++){
                a[j]=sc.nextInt();
            }
            count=0;
            f(a,0,1);
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios where the code might fail. The code iterates through the array and checks pairs of indices to see if the product of the elements at those indices is divisible by the product of the indices themselves. The main issue in the code is that it uses a global `count` variable, which is not reset for each test case, leading to incorrect results for subsequent test cases.

Let's consider the following aspects to create a challenging test input:
1. **Multiple Test Cases**: Ensure that the test input contains multiple test cases to highlight the issue with the global `count` variable.
2. **Permutations with Specific Properties**: Create permutations that have specific properties, such as elements that are more likely to satisfy the divisibility condition, to increase the number of valid pairs.
3. **Edge Cases**: Include edge cases like permutations of length 1, and permutations that are the reverse of the natural order.

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

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

### Explanation:
1. **First Test Case (n=1)**:
   - Permutation: [1]
   - Expected Output: 0 (No pairs)

2. **Second Test Case (n=2)**:
   - Permutation: [1, 2]
   - Expected Output: 1 (Pair (1, 2))

3. **Third Test Case (n=3)**:
   - Permutation: [3, 1, 2]
   - Expected Output: 1 (Pair (2, 3))

### Why This Input is Challenging:
- **First Test Case**: This is a simple edge case that ensures the code handles permutations of length 1 correctly.
- **Second Test Case**: This is a straightforward case with one valid pair, ensuring the code can correctly identify and count pairs.
- **Third Test Case**: This permutation has a valid pair (2, 3) but is not in the natural order, which can help expose issues in the logic for identifying valid pairs.

### Expected Output:
```plaintext
0
1
1
```

This input should expose the bug in the code where the `count` variable is not reset for each test case, leading to incorrect results for the subsequent test cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Problem (Hard Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is the hard version of the problem. The only difference is that in this version $n \leq 5 \cdot 10^5$ and the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

You are given a permutation $p$ of length $n$. Calculate the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

A permutation is a sequence of $n$ integers, in which each integer from $1$ to $n$ occurs exactly once. For example, $[1]$, $[3,5,2,1,4]$, $[1,3,2]$ are permutations, while $[2,3,2]$, $[4,3,1]$, $[0]$ are not.

Input: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains a single integer $n$ ($1 \leq n \leq 5 \cdot 10^5$) — the length of the permutation $p$.

The second line of each set of input data contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \leq p_i \leq n$) — the permutation $p$.

It is guaranteed that the sum of $n$ for all sets of input data does not exceed $5 \cdot 10^5$.

Output: For each set of input data, output the number of index pairs $1 \leq i < j \leq n$ such that $p_i \cdot p_j$ is divisible by $i \cdot j$ without remainder.

Examples:
input:
6
1
1
2
1 2
3
2 3 1
5
2 4 1 3 5
12
8 9 7 12 1 10 6 3 2 4 11 5
15
1 2 4 6 8 10 12 14 3 9 15 5 7 11 13
output:
0
1
1
3
9
3

Note: In the first set of input data, there are no index pairs, as the size of the permutation is $1$.

In the second set of input data, there is one index pair $(1, 2)$ and it is valid.

In the third set of input data, the index pair $(1, 2)$ is valid.

In the fourth set of input data, the index pairs $(1, 2)$, $(1, 5)$, and $(2, 5)$ are valid.

Buggy Code:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <array>
#include <numeric>
#include <map>
#include <unordered_map>
using namespace std;

const int N = 5e5 + 5;

int n, t;
array<int, N> a, b;
array<vector<int>, N> ve;
unordered_map<int, int> f[N];

int main()
{
	ios::sync_with_stdio(0), cin.tie(0);
	for (int i = 1; i < N; i++)
	{
		for (int j = i; j < N; j += i) ve[j].emplace_back(i);
	}
	cin >> t;
	while (t--)
	{
		cin >> n;
		long long ans = 0ll;
		for (int i = 1; i <= n; i++) f[i].clear();
		for (int i = 1; i <= n; i++)
		{
			cin >> a[i];
			b[i] = i;
			int x = gcd(a[i], b[i]);
			ans = ans - ((1ll * a[i] * a[i] % (1ll * i * i)) == 0);
			a[i] /= x, b[i] /= x;
			for (auto& j : ve[a[i]])
			{
				f[j][b[i]]++;
			}
		}
		for (int i = 1; i <= n; i++)
		{
			unordered_map<int, int> g;
			for (auto& k : f[i])
			{
				for (int x = k.first; x <= n; x += k.first) g[x] += k.second;
			}
			f[i] = g;
		}
		for (int i = 1; i <= n; i++)
		{
			ans = ans + f[b[i]][a[i]];
		}
		cout << (ans / 2) << "\n";
	}
	return 0;
}

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

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

Think step by step.

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