**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>
using namespace std;
const int N = 5e5 + 10;
int n, p[N];
vector<pair<int, int> > vec;
vector<int> mul[N];
vector<int> pri[N];
map<pair<int, int>, int> ans;
void solve(){
    ans.clear();
    vec.clear();
    cin >> n;
    for(int i = 1 ; i <= n ; i ++)
        mul[i].clear(), pri[i].clear();
    for (int i = 1; i <= n; i ++) cin >> p[i];
    for (int i = 1; i <= n; i ++){
        int g = __gcd(i, p[i]);
        vec.push_back({p[i] / g, i / g});
        pri[vec[i - 1].first].push_back(i - 1);
    }
    for (int d = 1; d <= n; d ++)
        for (int num = d ; num <= n; num += d)
            for (int x : pri[num])
                mul[d].push_back(x);
    int final = 0;
    for (int i = 0; i < n; i ++){
        int num = vec[i].first;
        int d = vec[i].second;
        if (ans.find({num, d}) == ans.end()){
            for (int j : mul[d]){
                if (i != j and num % vec[j].second == 0)
                    ans[{num, d}]++;
            }
        }
        final += ans[{num, d}];
    }
    cout << final / 2 << endl;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

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


**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>
using namespace std;
const int N = 5e5 + 5;
int a[N],b[N],c[N],sp[105][105];
int n, m,k,ans;
vector<int>op1[N],op2[N],yin[N];  //a【i】，b【i】
void solve(){
	scanf("%d", &n);
	ans=0;
	for(int i=1;i<=n;i++) {
		op1[i].clear(); op2[i].clear(); yin[i].clear();
	}
	for(int i=1;i<=n;i++) {
		scanf("%d",&a[i]);
		b[i]=i;
		int op=__gcd(a[i],b[i]);
		a[i]/=op; b[i]/=op;
		op1[a[i]].push_back(b[i]);
		op2[b[i]].push_back(a[i]);
		if(a[i]%b[i]==0) ans--;//消除影响
	}//预处理
	for(int i=1;i<=n;i++) for(int j=i;j<=n;j+=i) yin[j].push_back(i);//摄取因数
	for(int i=1;i<=n;i++){//枚举每个b【i】
		for(int j=i;j<=n;j+=i){//对于每一个可以把i当因子的数
			for(auto k:op1[j])//以j为a的数，取这里面的b
				c[k]++;//获得至关因子，bj
			for(auto k:op2[i]){//以i为b的所有a
				for(auto s:yin[k]) ans+=c[s];
			}
			for(auto k:op1[j])//以j为a的数，取这里面的b
				c[k]--;//获得至关因子，bj
		}
	}
	cout<<ans/2<<endl;
	return;
}
signed main()
{
    int T = 1;
    cin >> T;
    for (int i = 1; i <= T; i++)
    {
        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
5
1 2 3 4 5
```

**Explanation:**
This test case is designed to challenge the algorithm by providing a permutation that is simply the sequence of numbers from 1 to 5. This input should be straightforward and easy to handle, but it can still expose issues in the implementation, especially if the code does not correctly handle the basic properties of permutations and the divisibility condition.

For this input, the expected output is `4`, as the valid pairs are:
- (1, 2) where \(1 \cdot 2\) is divisible by \(1 \cdot 2\)
- (1, 3) where \(1 \cdot 3\) is divisible by \(1 \cdot 3\)
- (1, 4) where \(1 \cdot 4\) is divisible by \(1 \cdot 4\)
- (1, 5) where \(1 \cdot 5\) is divisible by \(1 \cdot 5\)

This test case is small and manageable, making it easy to verify the correctness of the output. If the code fails to produce the correct result, it may indicate issues with how the algorithm handles the basic divisibility conditions or the permutation properties.


**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:
# Here I can't use a list of dictionaries to store the results
# num1//den1 = (den2//num1) * k
# since num1,den1 and num2,den2 are co-prime pairs => num1%den2==0 and num2%den1==0
# If I iterate over den1==> num2 is multiple of den1, I only need to check if num1%den2==0
# For each den1s, there will be multiple num1s associated so I need to mark their factors
# For each multiple of den1 (basically num2): we check if den2 is marked:-
# (it should be marked if it divides num1 since if den2 divides num1, it should be equal to a divisor of num1)
import math
from collections import defaultdict
import sys
from functools import lru_cache

sys.setrecursionlimit(31)

import sys, os, io
#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
input = sys.stdin.readline
print = sys.stdout.write

# O(N)
N = 5*10**5 + 5
spf = [0] * N
primes = []

def build():
    global spf, primes
    for i in range(2, N):
        if spf[i] == 0:
            spf[i] = i
            primes.append(i)
        for j in range(len(primes)):
            if i * primes[j] >= N:
                break
            spf[i * primes[j]] = primes[j]
            if primes[j] == spf[i]:
                break

# O(d(n^p))
def gen_divisors(pfs):
    divs = [1]

    def f(x, i):
        if i >= len(pfs):
            return
        f(x, i + 1)
        for j in range(pfs[i][1]):
            x *= pfs[i][0]
            divs.append(x)
            f(x, i + 1)

    f(1, 0)
    return divs

def memodict(f):
    """memoization decorator for a function taking a single argument"""
    class memodict(dict):
        def __missing__(self, key):
            ret = self[key] = f(key)
            return ret

    return memodict().__getitem__

# O(log(n) + d(n^p))
@memodict
def divisors(n, p=1):
    if n == 1:
        return [1]
    if not primes:
        build()

    pfs = [[spf[n], p]]
    n //= spf[n]
    while n != 1:
        if pfs[-1][0] == spf[n]:
            pfs[-1][1] += p
        else:
            pfs.append([spf[n], p])
        n //= spf[n]

    return gen_divisors(pfs)

# Example usage
build()

def get(s):
    n = len(s)
    i = 0
    arr = []
    while i<n:
        j = i+1
        while j<n and s[j]!=s[i]:
            j+=1
        arr.append(int(s[i:j]))
        i = j
    return arr

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

    n = int(input())
    p = list(map(int,input().split()))

    a = [""]*(n+1)
    b = [""]*(n+1)
    ans,c = 0,0
    btimes = [0]*(n+1)
    for i in range(n):
        g = math.gcd(i+1,p[i])
        num = p[i]//g
        den = (i+1)//g
        sd,sn = "+","+"
        sd+=str(den)
        sn+=str(num)
        td = b[den]
        btimes[den]+=1
        td+=sn
        b[den] = td
        tn = a[num]
        tn+=sd
        a[num] = tn

    #print(a)
    #print(b)

    vis = [0]*(n+1)
    ans = 0
    #print(a)
    for i in range(1,n+1): # Iterating over den1
        temp = []
        for j in get(b[i]): # Marking all the divisors of num1
            for k in divisors(j):
                vis[k]+=1
                temp.append(k)
                if i==1: ans+=btimes[k]-int(k==1)

        for j in range(i,n+1,i): # Iterate over all multiples of den2 (basically num1) and check if any den2 is marked
            for k in get(a[j]):
                if i>1: ans+=vis[k]

        for k in temp: vis[k]-=1

    print(str(ans//2) + "\n")

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

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

Think step by step.

Response:
```plaintext
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
```


**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 <vector>
#include <array>
#include <algorithm>
#include <queue>
#include <numeric>
#include <functional>
#include <string>
#include <map>
#include <cmath>
#include <set>
#include <bitset>
#include <climits>
#include <cstring>
#include <cinttypes>
#include <cstdio>
#include <bit>

using namespace std;

//#define int long long
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

template<typename T>
istream &operator>>(istream &is, vector<T> &v) {
    for (auto &i: v) is >> i;
    return is;
}

template<typename T, typename Q>
istream &operator>>(istream &is, vector<pair<T, Q>> &v) {
    for (auto &[x, y]: v) is >> x >> y;
    return is;
}

template<typename T, typename Q>
ostream &operator<<(ostream &os, vector<pair<T, Q>> &v) {
    for (auto &[x, y] : v) os << x << " " << y << '\n';
    return os;
}

template<typename T>
ostream &operator<<(ostream &os, vector<T> v) {
    for (auto &i: v) os << i << ' ';
    return os;
}

int gcd(int a, int b) {
    return (b == 0 ? a : gcd(b, a % b));
}

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

    vector<int> p(n);
    cin >> p;

    vector<int> a(n + 1), b(n + 1);
    vector pa(n + 1, vector<int>()), pb(n + 1, vector<int>());

    for(int i = 0; i < n; i++) {
        int z = gcd(p[i], i + 1);

        a[i] = p[i] / z;
        b[i] = (i + 1) / z;

        pa[a[i]].push_back(i);
        pb[b[i]].push_back(i);
    }

    vector<int> cnt(n + 1);
    int ans = 0;
    for(int need_i = 1; need_i <= n; need_i++) {
        for(int i : pb[need_i])
            cnt[a[i]]++;

        for(int x = need_i; x <= n; x += need_i) {
            for(int i : pa[x]) {
                int need_j = b[i];

                for(int y = need_j; y <= n; y += need_j) {
                    ans += cnt[y];
                }
            }
        }

        for(int i : pb[need_i])
            cnt[a[i]]--;
    }

    for(int i = 0; i < n; i++) {
        if(b[i] == 1) ans--;
    }

    cout << ans / 2 << '\n';
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int t = 1;
    cin >> t;
    while(t--)
        solve();

}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that are likely to cause issues. Here are the steps to think through this problem:

1. **Understanding the Bug**:
   - The code uses a combination of GCD calculations and counting to determine valid pairs.
   - The main issue often arises when the GCD calculations or the counting logic is incorrect, especially for large or specific values.

2. **Edge Cases**:
   - **Small Permutations**: Test with small permutations to ensure the basic logic is correct.
   - **Large Permutations**: Test with large permutations to check for performance and correctness.
   - **Special Values**: Test with permutations that have specific values that might cause issues, such as all even numbers, all odd numbers, or specific patterns.

3. **Test Input**:
   - We will create a test input that includes a permutation of a specific size that might cause issues in the GCD calculations or the counting logic.

Given the constraints and the need for a small-scale test input, let's create a test input with a permutation of length 10, which includes a mix of values that might cause issues in the GCD calculations.

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

This test input includes a permutation of length 10, which is a small-scale input but includes all values from 1 to 10. This will help in verifying if the GCD calculations and the counting logic are correctly implemented in the buggy code.

Let's output the test input:

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


**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:
ans = []
def gcd(x,y):
	if y < x: return gcd(y,x)
	if x == 0: return y
	return gcd(y%x, x)
def divisors(n):
	divs = [1]
	for (p,e) in prime_factorization(n):
		L = len(divs)
		for _ in range(L*e):
			divs.append( divs[-L]*p )
	#divs.sort()
	return divs
primes = [2,3,5,7,11]
def generate_primes(max_n):
	P = primes[-1] + 1
	if max_n > P:
		L = max_n - P + 1
		sieve = [0]*L
		for p in primes:
			b = P%p
			if b: b = p-b
			while b < L:
				sieve[b] = 1
				b += p
		for a in range(L):
			if not sieve[a]:
				p = a+P
				b = a+p
				while b < L:
					sieve[b] = 1
					b += p
				primes.append(p)
def prime_factorization(n): #O(len(primes))
	generate_primes(1+int(n**0.5))
	factors = []
	a = n
	for p in primes:
		if a%p == 0:
			c = 0
			while a%p == 0:
				a //= p
				c += 1
			factors.append( (p,c) )
		if p*p > a:
			break
	if a > 1:
		factors.append( (a,1) )
	return factors

for _ in range(int(input())):
	n = int(input())
	p = [int(x) for x in input().split()]

	S = 0
	#M = [[0]*(n+1) for _ in range(n+1)]
	M = {}

	for i in range(n):
		g = gcd(p[i],i+1)
		a,b = p[i]//g, (i+1)//g
		if b == 1:
			S -= 1

		for d in divisors(a):
			if (d,b) in M:
				M[(d,b)] += 1
			else:
				M[(d,b)] = 1

	for i in range(n):
		g = gcd(p[i],i+1)
		a,b = p[i]//g, (i+1)//g
		for d in divisors(a):
			if (b,d) in M:
				S += M[(b,d)]
	ans.append( S//2 )

for item in ans:
	print(item)

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

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

Think step by step.

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

1. **Edge Cases**: Test with permutations of length 1, 2, and small values of \( n \).
2. **Large Input**: Test with a permutation of a large \( n \) (up to \( 5 \cdot 10^5 \)).
3. **Special Patterns**: Test with permutations that have specific patterns, such as a sequence where \( p_i = i \) or a reverse sequence where \( p_i = n - i + 1 \).
4. **Randomized Input**: Test with a random permutation to ensure the algorithm handles general cases.

Given the constraints and the need to generate a small-scale test input, let's focus on a permutation that is likely to expose the bug. A permutation where \( p_i = i \) is a good candidate because it should have many valid pairs, and it is a simple pattern to check.

Here's a small-scale test input:

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

Explanation:
- The first set of input data has a permutation of length 1, which should result in 0 valid pairs.
- The second set of input data has a permutation of length 2, which should result in 1 valid pair (1, 2).
- The third set of input data has a permutation of length 5, which should result in 10 valid pairs (since every pair \( (i, j) \) where \( 1 \leq i < j \leq 5 \) is valid).

This input should be sufficient to test the algorithm and expose any issues 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 java.util.*;

public class Que4Div4 {

    private static List<Map<Long, Long>> g;

    // Method to clear the global map
    private static void clearMap(int n) {
        for (int i = 0; i <= n + 5; i++) {
            g.get(i).clear();
        }
    }

    // Method to process each test case
    private static void processTestCase(Scanner sc) {
        int n = sc.nextInt();
        clearMap(n);

        long[] v = new long[n];
        for (int i = 0; i < n; i++) {
            v[i] = sc.nextLong();
        }

        long count = 0;
        for (int i = n - 1; i >= 0; i--) {
            long x = gcd(v[i], i + 1);

            long z = v[i] / x;
            long y = (i + 1) / x;

            Set<Long> st = new HashSet<>();
            for (long it = 1; it * it <= z; it++) {
                if (z % it == 0) {
                    st.add(it);
                    st.add(z / it);
                }
            }

            for (long it : st) {
                for (Map.Entry<Long, Long> nu : g.get((int) it).entrySet()) {
                    if (nu.getKey() % y == 0) {
                        count += nu.getValue();
                    }
                }
            }

            g.get((int) y).put(z, g.get((int) y).getOrDefault(z, 0L) + 1);
        }

        System.out.println(count);
    }

    // Method to calculate the gcd of two numbers
    private static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    // Main method to handle input and execute the program
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        // Initialize the list of maps
        g = new ArrayList<>();
        for (int i = 0; i < 200500; i++) {
            g.add(new HashMap<>());
        }

        while (t-- > 0) {
            processTestCase(sc);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Understanding the Bug**: The buggy code initializes a global list of maps (`g`) with a fixed size of 200500. However, the problem constraints allow for `n` to be up to 500,000. This means that if `n` exceeds 200500, the code will access out-of-bounds indices, leading to an `IndexOutOfBoundsException`.

2. **Edge Case**: We need to create a test case where `n` is greater than 200500 to expose this bug.

3. **Valid Test Input**: The test input must be valid according to the problem constraints. It should include a permutation of length `n` where `n` is greater than 200500.

Given these considerations, let's create a test input with `n = 200501` and a valid permutation of that length.

```plaintext
1
200501
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 4


**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>
using namespace std;
#define maxn 500005
int a[maxn];
map<int,int>mp[maxn];

void solve()
{
    int n;cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    long long ans=0;
    for(int i=1;i<=n;i++)
    {
        int k=gcd(i,a[i]);
        int x=a[i]/k,y=i/k;
        for(int j=1;j*j<=x;j++)
        {
            if(x%j!=0)continue;
            mp[j][y]=0;
            if(j*j==x)continue;
            mp[x/j][y]=0;
        }
    }
    for(int i=1;i<=n;i++)
    {
        int k=gcd(i,a[i]);
        int x=a[i]/k,y=i/k;
        for(int j=1;j*j<=x;j++)
        {
            if(x%j!=0)continue;
            if(mp[j].count(y))ans+=mp[j][y];
            if(j*j==x)continue;
            if(mp[x/j].count(y))ans+=mp[x/j][y];
        }
        for(int j=1;j*j<=x;j++)
        {
            if(x%j!=0)continue;
            if(mp[y].count(j))mp[y][j]++;
            if(j*j==x)continue;
            if(mp[y].count(x/j))mp[y][x/j]++;
        }
    }
    for(int i=1;i<=n;i++)mp[i].clear();
    cout<<ans<<'\n';
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    int T;cin>>T;
    while(T--)solve();
    return 0;
}

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

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

Think step by step.

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

**Explanation:**

1. **First Test Case:**
   - **Input:** `1` (n = 1)
   - **Permutation:** `[1]`
   - **Output:** `0`
   - **Reason:** There are no pairs (i, j) such that 1 ≤ i < j ≤ 1.

2. **Second Test Case:**
   - **Input:** `2` (n = 2)
   - **Permutation:** `[2, 1]`
   - **Output:** `1`
   - **Reason:** The pair (1, 2) is valid because \( p_1 \cdot p_2 = 2 \cdot 1 = 2 \) and \( i \cdot j = 1 \cdot 2 = 2 \), and 2 is divisible by 2 without a remainder.

3. **Third Test Case:**
   - **Input:** `3` (n = 3)
   - **Permutation:** `[3, 1, 2]`
   - **Output:** `1`
   - **Reason:** The pair (2, 3) is valid because \( p_2 \cdot p_3 = 1 \cdot 2 = 2 \) and \( i \cdot j = 2 \cdot 3 = 6 \), and 2 is not divisible by 6 without a remainder, but the pair (1, 3) is valid because \( p_1 \cdot p_3 = 3 \cdot 2 = 6 \) and \( i \cdot j = 1 \cdot 3 = 3 \), and 6 is divisible by 3 without a remainder.

This test input is designed to cover edge cases and small permutations to ensure the code handles them correctly. It also includes a permutation that is not in the standard order to test the algorithm's ability to handle non-sequential permutations.


**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>
using namespace std;
#define ll long long
#define fori(i,a,b) for(int i=a;i<b;i++)
#define vi vector<int>
#define vvi vector<vector<int>>
const int N = 1e5+2;

vvi fac(N);

void comp(){

    for(int i=1;i<N;i++){

        for(int j=i;j<N;j+=i){

            fac[j].push_back(i);

        }

    }

}

int main() {

	ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int t = 1;

    cin>>t;

    comp();

    while(t--){

        int n;
        cin>>n;

        ll ans = 0;

        vector<vector<pair<int,int>>> pos(n+1);
        vector<vector<pair<int,int>>> den(n+1);
        fori(i,1,n+1){

            int x;
            cin>>x;

            int use = gcd(i,x);

            pair<int,int> a = {x/use,i/use};
            pos[x/use].push_back(a);
            den[i/use].push_back(a);
            // cout<<a.first<<" "<<a.second<<endl;

            if(x%i==0){
                ans--;
            }

        }

        for(int b = 1;b<=n;b++){

            unordered_map<int,int> cnt;

            for(int j=b;j<=n;j+=b){

                for(auto use:pos[j]){

                    cnt[use.second]++;

                }

            }

            for(auto a:den[b]){

                int curr = a.first;

                for(auto v:fac[curr]){

                    ans += cnt[v];

                }

            }

        }

        cout<<ans/2<<endl;

    }

}

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

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

Think step by step.

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


**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
import sys
from collections import deque, defaultdict

input = lambda: sys.stdin.readline()
in_str = lambda: input().strip()
in_int = lambda: int(input())
in_ints = lambda: list(map(int, input().split()))
in_intt = lambda: tuple(map(int, input().split()))
print_bool = lambda v: print('YES' if v else 'NO')
print_list = lambda v: print(' '.join(map(str, v)))

def solve():
    n = in_int()
    ps = in_ints()

    ab = []
    ba = []
    for i, p in enumerate(ps):
        g = math.gcd(i + 1, p)
        a = p // g
        b = (i + 1) // g
        ab.append((a, b))
        ba.append((b, a))

    ab.sort()
    ba.sort()

    b2i = [-1] * (n + 1)
    for i, (b, a) in enumerate(ba):
        if b2i[b] == -1:
            b2i[b] = i

    a2i = [-1] * (n + 1)
    for i, (a, b) in enumerate(ab):
        if a2i[a] == -1:
            a2i[a] = i

    last = [-1] * (n + 1)
    cnt = [0] * (n + 1)
    res = [0] * (n + 1)

    ans = 0

    for b1 in range(1, n + 1):
        if b2i[b1] == -1:
            continue

        i1 = b2i[b1]
        while i1 < len(ba) and ba[i1][0] == b1:
            a1 = ba[i1][1]
            cnt[a1] += 1
            i1 += 1

        for a2 in range(b1, n + 1, b1):
            i2 = a2i[a2]
            if i2 == -1:
                continue
            while i2 < len(ab) and ab[i2][0] == a2:
                b2 = ab[i2][1]
                if last[b2] != b1:
                    res[b2] = 0
                    for a1 in range(b2, n + 1, b2):
                        res[b2] += cnt[a1]
                    last[b2] = b1
                ans += res[b2]
                i2 += 1

        i1 = b2i[b1]
        while i1 < len(ba) and ba[i1][0] == b1:
            a1 = ba[i1][1]
            cnt[a1] -= 1
            i1 += 1

    for b, _ in ba:
        if b == 1:
            ans -= 1
    ans //= 2
    print(ans)

for _ in range(in_int()):
    solve()

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Edge Cases:**
   - **Single Element Permutation:** This is already covered in the example.
   - **Two Element Permutation:** This is also covered in the example.
   - **Permutations with specific patterns:** For example, permutations where the elements are in a specific order that might cause issues with the GCD calculations.

2. **Large Permutations:**
   - **Permutations with repeated GCD values:** This can help in testing how the code handles repeated values and their indices.
   - **Permutations with large gaps:** This can test how the code handles large differences in indices and values.

3. **Specific Patterns:**
   - **Permutations where \( p_i \cdot p_j \) is divisible by \( i \cdot j \) for many pairs:** This can help in testing the efficiency and correctness of the code.

Given the constraints and the need for a small-scale test input, let's create a permutation that is likely to expose the bug:

- A permutation where the elements are arranged in a way that the GCD calculations might lead to incorrect results or inefficiencies.
- A permutation that has a few elements where \( p_i \cdot p_j \) is divisible by \( i \cdot j \) to ensure there are some valid pairs.

Here is a test input that fits these criteria:

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

Explanation:
- **Permutation:** [1, 3, 2, 4, 6, 5]
- **Valid Pairs:**
  - (1, 3): \( p_1 = 1 \), \( p_3 = 2 \), \( 1 \cdot 2 \) is divisible by \( 1 \cdot 3 \)
  - (1, 6): \( p_1 = 1 \), \( p_6 = 5 \), \( 1 \cdot 5 \) is divisible by \( 1 \cdot 6 \)
  - (3, 6): \( p_3 = 2 \), \( p_6 = 5 \), \( 2 \cdot 5 \) is divisible by \( 3 \cdot 6 \)

This permutation has a few valid pairs and should help in exposing any issues in the GCD calculations or the way the code handles the indices and values.


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

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

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

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

from bisect import bisect_left as lower_bound
from bisect import bisect_right as upper_bound

class FenwickTree:
    def __init__(self, x):
        bit = self.bit = list(x)
        size = self.size = len(bit)
        for i in range(size):
            j = i | (i + 1)
            if j < size:
                bit[j] += bit[i]

    def update(self, idx, x):
        """updates bit[idx] += x"""
        while idx < self.size:
            self.bit[idx] += x
            idx |= idx + 1

    def __call__(self, end):
        """calc sum(bit[:end])"""
        x = 0
        while end:
            x += self.bit[end - 1]
            end &= end - 1
        return x

    def find_kth(self, k):
        """Find largest idx such that sum(bit[:idx]) <= k"""
        idx = -1
        for d in reversed(range(self.size.bit_length())):
            right_idx = idx + (1 << d)
            if right_idx < self.size and self.bit[right_idx] <= k:
                idx = right_idx
                k -= self.bit[idx]
        return idx + 1, k

class SortedList:
    block_size = 700

    def __init__(self, iterable=()):
        self.macro = []
        self.micros = [[]]
        self.micro_size = [0]
        self.fenwick = FenwickTree([0])
        self.size = 0
        for item in iterable:
            self.insert(item)

    def insert(self, x):
        i = lower_bound(self.macro, x)
        j = upper_bound(self.micros[i], x)
        self.micros[i].insert(j, x)
        self.size += 1
        self.micro_size[i] += 1
        self.fenwick.update(i, 1)
        if len(self.micros[i]) >= self.block_size:
            self.micros[i:i + 1] = self.micros[i][:self.block_size >> 1], self.micros[i][self.block_size >> 1:]
            self.micro_size[i:i + 1] = self.block_size >> 1, self.block_size >> 1
            self.fenwick = FenwickTree(self.micro_size)
            self.macro.insert(i, self.micros[i + 1][0])

    def pop(self, k=-1):
        i, j = self._find_kth(k)
        self.size -= 1
        self.micro_size[i] -= 1
        self.fenwick.update(i, -1)
        return self.micros[i].pop(j)

    def __getitem__(self, k):
        i, j = self._find_kth(k)
        return self.micros[i][j]

    def count(self, x):
        return self.upper_bound(x) - self.lower_bound(x)

    def __contains__(self, x):
        return self.count(x) > 0

    def lower_bound(self, x):
        i = lower_bound(self.macro, x)
        return self.fenwick(i) + lower_bound(self.micros[i], x)

    def upper_bound(self, x):
        i = upper_bound(self.macro, x)
        return self.fenwick(i) + upper_bound(self.micros[i], x)

    def _find_kth(self, k):
        return self.fenwick.find_kth(k + self.size if k < 0 else k)

    def __len__(self):
        return self.size

    def __iter__(self):
        return (x for micro in self.micros for x in micro)

    def __repr__(self):
        return str(list(self))

def all_factors(n):
    """returns a sorted list of all distinct factors of n"""
    small, large = [], []
    for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
        if not n % i:
            small.append(i)
            large.append(n // i)
    if small[-1] == large[-1]:
        large.pop()
    large.reverse()
    small.extend(large)
    return small

def is_prime(n):
    """returns True if n is prime else False"""
    if n < 5 or n & 1 == 0 or n % 3 == 0:
        return 2 <= n <= 3
    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            p = (p * p) % n
            if p == n - 1:
                break
        else:
            return False
    return True

# Solver function
def solve():
    # d = mydict()
    # d = defaultdict(lambda: 0)
    r = randint(10 ** 7, 10 ** 8)
    r2 = randint(10 ** 9, 10 ** 10)
    ans = 0
    l = SortedList()
    for i in range(n):
        num = a[i]
        den = i + 1
        g = gcd(num, den)
        num //= g
        den //= g
        # i = 1
        pf = [1, num]
        if(is_prime(num) == False):
            pf = all_factors(num)
        # pf.append(1)
        # pf.append(num)
        for i in pf:
            if(num % i == 0):
                # ans += d[i][den]
                # ans += d[hash(tuple([i, den]))]
                val = i * r + den
                val ^= r2
                # ans += d[val]
                ans += l.count(val)
            #     if(num // i != i):
            #         # ans += d[num // i][den]
            #         # ans += d[hash(tuple([num // i, den]))]
            #         val = (num // i) * r + den
            #         val ^= r2
            #         # ans += d[val]
            #         ans += l.count(val)
            # i += 1
        # i = 1
        for i in pf:
            if(num % i == 0):
                # d[den][i] += 1
                # d[hash(tuple([den, i]))] += 1
                val = den * r + i
                val ^= r2
                # d[val] += 1
                l.insert(val)
            #     if(num // i != i):
            #         # d[den][num // i] += 1
            #         # d[hash(tuple([den, num // i]))] += 1
            #         val = den * r + (num // i)
            #         val ^= r2
            #         # d[val] += 1
            #         l.insert(val)
            # i += 1
    print(str(ans) + "\n")

# Main
for _ in range(int(input())):
    n = int(input())
    a = [int(x) for x in input().split()]
    solve()

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

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

Think step by step.

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

1. **Understanding the Bug**: The buggy code uses a hash function and a `SortedList` to count and store pairs. The key issue is in how the hash is computed and used to check for pairs. The hash function might cause collisions, leading to incorrect counts.

2. **Edge Cases**:
   - **Single Element Permutations**: These should always return 0.
   - **Permutations with Small n**: These should be straightforward to verify.
   - **Permutations with Large n**: These can help identify performance issues and hash collisions.
   - **Permutations with Specific Patterns**: These can help identify logical errors in the code.

3. **Test Input Generation**:
   - **Single Element Permutation**: This should always return 0.
   - **Small Permutations**: These should be easy to verify manually.
   - **Large Permutations**: These should be designed to maximize the probability of hash collisions.
   - **Permutations with Specific Patterns**: These should be designed to test specific logical conditions.

Given the constraints, let's generate a small-scale test input that includes a permutation with a specific pattern to maximize the failure probability:

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

### Explanation:
- **Test Case 1**: A single element permutation. Expected output: 0.
- **Test Case 2**: A permutation of length 2. Expected output: 1.
- **Test Case 3**: A permutation of length 3 with a specific pattern. Expected output: 1.
- **Test Case 4**: A permutation of length 4 with a specific pattern. Expected output: 3.

This test input is small enough to be easily verified and should expose any issues with the hash function or the logic used 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 java.io.*;
import java.math.*;
import java.util.*;

import static java.lang.Math.max;
import static java.lang.Math.min;

public class A_1986 {

  private static final int MAX = 100005;

  public static void main(String[] args) throws IOException {
    final var cin = new Reader();
    final var sb = new StringBuilder();
    var t = cin.nextInt();
    while (t-- > 0) {
      HashMap<Integer, HashMap<Integer, Long>> m = new HashMap<>();
      int n = cin.nextInt();
      long ans = 0;
      for (int i = 1; i <= n; i++) {
        int x = cin.nextInt();
        int index = i;
        int div = gcd(x, index);
        x = x/div;
        index = index/div;
        var divisors = new ArrayList<Integer>();
        for (int j = 1; j*j <= x ; j++) {
          if(x%j == 0) {
            ans += m.getOrDefault(j, new HashMap<>()).getOrDefault(index, 0L);
            divisors.add(j);

            if(j != x/j) {
              ans += m.getOrDefault(x/j, new HashMap<>()).getOrDefault(index, 0L);
              divisors.add(x/j);
            }
          }
        }
        m.putIfAbsent(index, new HashMap<>());
        for (Integer y : divisors) {
          var list = m.get(index);
          list.put(y, list.getOrDefault(y, 0L) + 1L) ;
        }
      }
      sb.append(ans).append("\n");
    }
    System.out.println(sb);
  }

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

  static class Reader {
    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader() {
      din = new DataInputStream(System.in);
      buffer = new byte[BUFFER_SIZE];
      bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException {
      din = new DataInputStream(new FileInputStream(file_name));
      buffer = new byte[BUFFER_SIZE];
      bufferPointer = bytesRead = 0;
    }

    public String readLine(int size) throws IOException {
      byte[] buf = new byte[size + 1]; // line length
      int cnt = 0, c;
      while ((c = read()) != -1) {
        if (c == '\n') {
          if (cnt != 0) {
            break;
          } else {
            continue;
          }
        }
        buf[cnt++] = (byte) c;
      }
      return new String(buf, 0, size);
    }

    public int nextInt() throws IOException {
      int ret = 0;
      byte c = read();
      while (c <= ' ') {
        c = read();
      }
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');

      if (neg)
        return -ret;
      return ret;
    }

    public long nextLong() throws IOException {
      long ret = 0;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');
      if (neg)
        return -ret;
      return ret;
    }

    public double nextDouble() throws IOException {
      double ret = 0, div = 1;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();

      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');

      if (c == '.') {
        while ((c = read()) >= '0' && c <= '9') {
          ret += (c - '0') / (div *= 10);
        }
      }

      if (neg)
        return -ret;
      return ret;
    }

    private void fillBuffer() throws IOException {
      bytesRead = din.read(buffer, bufferPointer = 0,
        BUFFER_SIZE);
      if (bytesRead == -1)
        buffer[0] = -1;
    }

    private byte read() throws IOException {
      if (bufferPointer == bytesRead)
        fillBuffer();
      return buffer[bufferPointer++];
    }

    public void close() throws IOException {
      if (din == null)
        return;
      din.close();
    }
  }

  static final class Pair implements Comparable<Pair> {
    int x;
    int y;

    Pair(int x, int y) {
      this.x = x;
      this.y = y;
    }

    @Override
    public String toString() {
      return "(" + x + ", " + y + ")";
    }

    @Override
    public int hashCode() {
      return x + y + (y << 16);
    }

    @Override
    public boolean equals(Object obj) {
      if (this == obj)
        return true;
      if (obj == null)
        return false;
      if (getClass() != obj.getClass())
        return false;
      Pair other = (Pair) obj;
      return x == other.x && y == other.y;
    }

    @Override
    public int compareTo(Pair o) {
      if (this.x < o.x) return -1;
      if (this.x > o.x) return 1;
      if (this.y < o.y) return -1;
      if (this.y > o.y) return 1;
      return 0;
    }
  }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can highlight potential issues. The bug in the code is related to the way the `gcd` is calculated and how the divisors are handled. Let's break down the steps to create a suitable test case:

1. **Edge Cases**:
   - **Single Element Permutation**: This should return 0.
   - **Permutations with Small Values**: These can help verify basic functionality.
   - **Permutations with Large Values**: These can help verify the handling of larger inputs.

2. **Specific Scenarios**:
   - **Permutations with Repeated GCDs**: This can expose issues in the `gcd` calculation and the handling of divisors.
   - **Permutations with Specific Patterns**: For example, permutations where `p_i` and `i` have specific relationships.

3. **Combining Scenarios**:
   - **Multiple Test Cases**: To ensure the solution handles multiple inputs correctly.

Given these considerations, let's create a small-scale test input that includes a mix of these scenarios:

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

### Explanation:
- **Test Case 1**: A single element permutation. This should return 0.
- **Test Case 2**: A permutation of length 2. This should return 1.
- **Test Case 3**: A permutation of length 3. This should return 1.
- **Test Case 4**: A permutation of length 4. This should return 2.

This test input is designed to cover basic edge cases and simple permutations, which can help identify issues in the `gcd` calculation and the handling of divisors.


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

public class G {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);
	static int [][] factors;

	public static void main(String[] args) throws IOException {

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		private static void genFactors(int n) {
			factors = new int[n + 1][];
			int [] count = new int [n + 1];

			for (int i = 1; i < factors.length; ++i) {
				for (int j = i; j < factors.length; j += i) {
					count[j]++;
				}
			}

			for (int i = 1; i < factors.length; ++i) {
				factors[i] = new int [count[i]];
			}

			Arrays.fill(count, 0);

			for (int i = 1; i < factors.length; ++i) {
				for (int j = i; j < factors.length; j += i) {
					factors[j][count[j]++] = i;
				}
			}
		}

		public void run() {
			// TODO Auto-generated method stub
			genFactors(500_000);
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void advice() {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			// Source of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
		}

		private void solve(int t) {
			int n = sc.nextInt();

			int [] arr = sc.readArray2(n);

			int num, den;
			int val;

			boolean [] isDen = new boolean [n + 1];

			for (int i = 1; i <= n; ++i) {
				val = gcd(arr[i], i);
				num = arr[i] / val;
				den = i / val;

				isDen[den] = true;
			}

			Map<Integer, Integer> [] map = new Map[n + 1];

			long result = 0;

			for (int i = 1; i <= n; ++i) {
				val = gcd(arr[i], i);
				num = arr[i] / val;
				den = i / val;

				if (map[den] != null) {
					for (int fact : factors[num]) {
						result += map[den].getOrDefault(fact, 0);
					}
				}

				for (int fact : factors[num]) {
					if (!isDen[fact])
						continue;
					else if (map[fact] == null)
						map[fact] = new HashMap<>();

					map[fact].put( den, map[fact].getOrDefault(den, 0) + 1);
				}
			}

			out.println(result);
		}

		private int gcd(int a, int b) {
			if (a % b == 0)
				return b;

			return gcd(b , a % b);
		}

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

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

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

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

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

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

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

      String nextLine(){
          String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
      }

   }

}

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

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

Think step by step.

Response:
To create a test input that maximizes the probability of exposing the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Edge Cases:**
   - Single-element permutation: This is a trivial case where no pairs can exist.
   - Small permutations with known valid pairs: This can help verify if the basic logic is correct.
   - Permutations where the product of elements and their indices are always divisible: This can help verify if the algorithm correctly identifies all valid pairs.
   - Permutations where the product of elements and their indices are never divisible: This can help verify if the algorithm correctly identifies no valid pairs.

2. **Scenarios with High Failure Probability:**
   - Permutations with a large number of elements where the product of elements and their indices are divisible by a large number of factors.
   - Permutations with a large number of elements where the product of elements and their indices are divisible by a small number of factors.
   - Permutations with a mix of elements that are both divisible and not divisible by their indices.

3. **Small-Scale Test Input:**
   - We will create a small-scale test input that includes a mix of the above scenarios to ensure that the algorithm is thoroughly tested.

Here is the test input:

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

This test input includes:
- A single-element permutation (trivial case).
- A two-element permutation with one valid pair.
- A three-element permutation with one valid pair.
- A five-element permutation with three valid pairs.

This should be sufficient to test the basic logic and edge cases of the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

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

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#define ll long long
#define ls k << 1
#define rs k << 1 | 1
#define mid ((l + r) >> 1)
using namespace std;
const int z = 2e5 + 7;
int n, m, k, q,
	a[z], col[z], f[z];
ll res[z];
struct Query {
	int id, l, r;
}ask[z];
struct Tree {
	int l, r, ln, mx, mn, lzy;
	ll sm;
}tr[z<<2];
multiset<int> st;
map<int, int> mp;
bool cmp(Query a, Query b) {
	return a.l > b.l;
}
void _push_up(Tree &a, Tree b, Tree c) {
	a.sm = b.sm + c.sm;
	a.mx = max(b.mx, c.mx);
	a.mn = min(b.mn, c.mn);
}
void push_up(int k) {
	_push_up(tr[k], tr[ls], tr[rs]);
}
void add(int k, int v) {
	tr[k].mx = tr[k].mn = tr[k].lzy = v;
	tr[k].sm = tr[k].ln * v;
}
void push_down(int k) {
	if(!tr[k].lzy) return;
	add(ls, tr[k].lzy), add(rs, tr[k].lzy);
	tr[k].lzy = 0;
}
void build(int k, int l, int r) {
	tr[k] = {l, r, r - l + 1};
	if(l == r) return tr[k].sm = tr[k].mx = tr[k].mn = f[l], void();
	build(ls, l, mid), build(rs, mid + 1, r);
	push_up(k);
}
void update(int k, int x, int y, int v) {
	if(y < tr[k].l || tr[k].r < x || v >= tr[k].mx) return;
	if(x <= tr[k].l && tr[k].r <= y && tr[k].mn >= v) return add(k, v);
	if(tr[k].l == tr[k].r) return;
	push_down(k);
	update(ls, x, y, v), update(rs, x, y, v);
	push_up(k);
}
Tree query(int k, int x, int y) {
	if(x <= tr[k].l && tr[k].r <= y) return tr[k];
	push_down(k);
	if(y <= tr[ls].r) return query(ls, x, y);
	if(tr[rs].l <= x) return query(rs, x, y);
	Tree Tl = query(ls, x, y),
		 Tr = query(rs, x, y),
		 Tq = {Tl.l, Tr.r, Tr.r - Tl.l};
	_push_up(Tq, Tl, Tr);
	push_up(k);
	return Tq;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int T;
	cin >> T;
	while(T--) {
		cin >> n >> k >> q;
		m = n - k + 1;
		for(int i = 1; i <= n; ++i) cin >> a[i];
		for(int i = 1; i <= q; ++i)
			cin >> ask[i].l >> ask[i].r, ask[i].id = i;
		//input
		sort(ask + 1, ask + q + 1, cmp);
		st.clear();
		mp.clear();
		for(int i = 1; i <= 2*n; ++i) st.insert(0);
		for(int i = 1; i <= k; ++i) {
			int col = a[i] - i + n;
			st.erase(st.find(mp[col]));
			++mp[col];
			st.insert(mp[col]);
		}
		f[1] = k - *st.rbegin();
		for(int i = k + 1; i <= n; ++i) {
			int col1 = a[i - k] - (i - k) + n,
				col2 = a[i] - i + n;
			st.erase(st.find(mp[col1])), st.erase(st.find(mp[col2]));
			--mp[col1], ++mp[col2];
			st.insert(mp[col1]), st.insert(mp[col2]);
			f[i - k + 1] = k - *st.rbegin();
		}
		build(1, 1, m);
		//reset
		int x = m - 1;
		for(int i = 1; i <= q; ++i) {
			int l = ask[i].l, r = ask[i].r - k + 1;

			while(x >= l)
				update(1, x + 1, m, f[x]), --x;
			res[ask[i].id] = query(1, l, r).sm;
		}
		for(int i = 1; i <= q; ++i)
			cout << res[i] << "\n";
		//solve
	}
	return 0;
}
/*
1
14 13 1
6 4 4 9 8 9 8 7 11 9 3 6 10 5
2 15

11
*/

**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
14 13 1
6 4 4 9 8 9 8 7 11 9 3 6 10 5
2 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

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

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
#include <iostream>
#include <cstdio>
#include <map>
#include <set>
#define pii pair<int,int>
#include <array>
#include <algorithm>
#define int long long
using namespace std;
const int maxn=2e5+5;
typedef long long ll;

int T,n,k,q;
int a[maxn],c[maxn];

struct query{
	int l,r;
	int idx;
	bool operator<(const query a)const{
		return l<a.l;
	}
}qry[maxn];
ll ans[maxn];

void init(){
	map<int,int> mp;
	multiset<int> S;
	for(int i=1;i<k;i++){
		if(mp.find(a[i])!=mp.end())S.erase(S.find(mp[a[i]]));
		mp[a[i]]++;
		S.insert(mp[a[i]]);
	}
	for(int l=1,r=k;r<=n;l++,r++){
		if(r<=n){
			if(mp.find(a[r])!=mp.end())S.erase(S.find(mp[a[r]]));
			mp[a[r]]++;
			S.insert(mp[a[r]]);
		}

		c[l]=k-*S.rbegin();

		S.erase(S.find(mp[a[l]]));
		mp[a[l]]--;
		if(mp[a[l]]==0)mp.erase(a[l]);
		else S.insert(mp[a[l]]);
	}

	for(int i=1;i<=q;i++){
		auto &[l,r,id]=qry[i];
		cin>>l>>r;
		id=i;
	}
	sort(qry+1,qry+q+1);
}

signed main(){
	cin>>T;
	while(T--){
		cin>>n>>k>>q;
		for(int i=1;i<=n;i++)cin>>a[i],a[i]-=i;
		init();
		int last=n-k+1;
//		for(int i=1;i<=last;i++)cout<<c[i]<<" ";
//		cout<<endl;
//		for(int i=1;i<=q;i++){
//			auto [l,r,id]=qry[i];
//			cout<<l<<" "<<r<<" "<<id<<endl;
//		}

		set<array<ll,3> > S;//位置，c值，后缀和
		for(int i=q,head=last+1;i>=1;i--){
			auto [l,r,id]=qry[i];
//			cout<<l<<" "<<r<<" "<<id<<endl;
			while(l<head){
				head--;
				while(!S.empty() && (*S.begin())[1]>=c[head])S.erase(S.begin());
				int tail=S.empty()?last:(*S.begin())[0]-1;
				ll tailsum=S.empty()?0:(*S.begin())[2];
				S.insert({head,c[head],tailsum+1ll*(tail-head+1)*c[head]});
			}
//			cout<<"****"<<endl;
//			for(auto a:S){
//				cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;
//			}
			ll y=r-k+1;
			auto pre=(*S.rbegin())[0]<y?*S.rbegin():*S.lower_bound({y,-1,-1});
			ans[id]=(*S.begin())[2]-pre[2]+1ll*pre[1]*(y-pre[0]+1);
		}
		for(int i=1;i<=q;i++)cout<<ans[i]<<" ";
		cout<<endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that might cause the code to fail. Here are some considerations:

1. **Edge Cases for Array Length and Query Bounds**:
   - Ensure the array length `n` and the subarray length `k` are at their limits.
   - Ensure the query bounds `l` and `r` are valid and push the limits of the algorithm.

2. **Special Values in the Array**:
   - Use values that might cause unexpected behavior, such as repeated values or values that are close to each other.

3. **Queries with Specific Conditions**:
   - Queries where `r - l + 1` is exactly `k`.
   - Queries where `r - l + 1` is much larger than `k`.

4. **Multiple Test Cases**:
   - Include multiple test cases to ensure the code handles different scenarios.

Given these considerations, here is a small-scale test input that should challenge the code:

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

This input includes:
- Multiple test cases with different array lengths and subarray lengths.
- Queries that cover different ranges and push the limits of the algorithm.
- Arrays with values that might cause unexpected behavior.

This should help in identifying 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: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

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

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair <int, int> P;
const int N = 2e5 + 10;
int n, k, Q, a[N];
unordered_map <int, int> cnt;
multiset <int> s;
int stk[N], rt[N], ans[N];
struct qu {
    int l, r, id;
} q[N];
bool cmp(qu a, qu b) {
    return a.l > b.l;
}
ll sum[N], lz[N];
void pushup(int i) {
    sum[i] = sum[i * 2] + sum[i * 2 + 1];
}
void build(int i, int l, int r) {
    sum[i] = lz[i] = 0;
    if(l == r) return;
    int mid = (l + r) / 2;
    build(i * 2, l, mid);
    build(i * 2 + 1, mid + 1, r);
}
void pushdown(int i, int l, int r) {
    if(lz[i]) {
        int mid = (l + r) / 2;
        sum[i * 2] = lz[i] * (mid - l + 1);
        sum[i * 2 + 1] = lz[i] * (r - mid);
        lz[i * 2] = lz[i * 2 + 1] = lz[i];
        lz[i] = 0;
    }
}
void modify(int i, int l, int r, int ql, int qr, ll val) {
    // printf("modify %d %d: i = %d, l = %d, r = %d\n", ql, qr, i, l, r);
    if(ql <= l && qr >= r) {
        sum[i] = (r - l + 1) * val;
        lz[i] = val;
        return;
    }
    pushdown(i, l, r);
    int mid = (l + r) / 2;
    if(ql <= mid) modify(i * 2, l, mid, ql, qr, val);
    if(qr > mid) modify(i * 2 + 1, mid + 1, r, ql, qr, val);
    pushup(i);
}
ll query(int i, int l, int r, int ql, int qr) {
    if(ql <= l && qr >= r) return sum[i];
    pushdown(i, l, r);
    int mid = (l + r) / 2;
    ll ans = 0;
    if(ql <= mid) ans += query(i * 2, l, mid, ql, qr);
    if(qr > mid) ans += query(i * 2 + 1, mid + 1, r, ql, qr);
    return ans;
}
ll Ans[N];
void solve() {
    scanf("%d%d%d", &n, &k, &Q);
    cnt.clear(), s.clear();
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    for(int i = 1; i <= n; i++) {
        int d = a[i] - i;
        if(cnt[d]) s.erase(s.find(cnt[d]));
        cnt[d]++;
        s.insert(cnt[d]);
        // printf("cnt[%d] = %d\n", d, cnt[d]);
        if(i > k) {
            d = a[i - k] - (i - k);
            s.erase(s.find(cnt[d]));
            cnt[d]--;
            if(cnt[d]) s.insert(cnt[d]);
        }
        ans[i] = k - *s.rbegin();
        // printf("ans[%d] = %d\n", i, ans[i]);
    }
    int top = 0;
    for(int i = n; i; i--) {
        while(top && ans[stk[top]] >= ans[i]) top--;
        rt[i] = stk[top];
        if(!rt[i]) rt[i] = n + 1;
        // printf("rt[%d] = %d\n", i, rt[i]);
        stk[++top] = i;
    }
    for(int i = 1; i <= Q; i++) {
        scanf("%d%d", &q[i].l, &q[i].r);
        q[i].id = i;
    }
    sort(q + 1, q + Q + 1, cmp);
    build(1, 1, n);
    for(int i = q[1].l + k - 1; i <= n; i = rt[i]) {
        modify(1, 1, n, i, rt[i] - 1, ans[i]);
    }
    for(int i = 1; i <= Q; i++) {
        int l = q[i].l + k - 1;
        if(i > 1) {
            int t = q[i - 1].l + k - 1;
            for(int j = l; j < t; j++) modify(1, 1, n, j, rt[j] - 1, ans[j]);
        }
        // printf("l = %d, r = %d\n", l, q[i].r);
        Ans[q[i].id] = query(1, 1, n, l, q[i].r);
    }
    for(int i = 1; i <= Q; i++) {
        printf("%lld\n", Ans[i]);
    }

}
int main() {
    int T = 1;
    scanf("%d", &T);
    while(T--) solve();

    system("pause");
    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
10 5 2
1 2 3 4 5 6 7 8 9 10
1 10
5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

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

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

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

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
t = int(input())
for _ in range(t):
    n, k, q = map(int, input().split())
    a = list(map(int, input().split()))
    prefix_sum = [0]*(n+1)
    for _ in range(q):
        l, r = map(int, input().split())
        l -= 1  # Convert to 0-based index
        total_f = 0
        freq = {}
        window = []
        max_freq = 0
        for j in range(l, r):
            c_j = a[j] - (j - l)
            window.append(c_j)
            freq[c_j] = freq.get(c_j, 0) + 1
            if freq[c_j] > max_freq:
                max_freq = freq[c_j]
            # Maintain window size
            if len(window) > k:
                left_c_j = window.pop(0)
                freq[left_c_j] -= 1
                if freq[left_c_j] == 0:
                    del freq[left_c_j]
                # Recalculate max_freq if necessary
                if left_c_j == c_j:
                    max_freq = max(freq.values())
            # Only compute f(b) when window size >= k
            if j - l + 1 >= k:
                total_f += k - max_freq
        print(total_f)

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

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

Think step by step.

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