**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
import java.util.Scanner;

public class cf_1982F {

    public static int[] res(int[] a, int ind, int rep) {
        a[ind] = rep;
        return findUnsortedSubarray(a, ind);
    }

    public static void main(String[] args) {
        // long start = System.currentTimeMillis();
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int j = 0; j < n; j++) {
                a[j] = sc.nextInt();
            }
            int q = sc.nextInt();
            int[][] b = new int[q][2];
            int[] prev = findUnsortedSubarray(a, -1);
            System.out.println(prev[0] + " " + prev[1]);
            for (int j = 0; j < q; j++) {
                b[j][0] = sc.nextInt();
                b[j][1] = sc.nextInt();
                int[] ress = res(a, b[j][0] - 1, b[j][1]);
                System.out.println(ress[0] + " " + ress[1]);
            }
        }
        // System.out.println("total time: "+(int)(System.currentTimeMillis() - start));
        sc.close();
    }

    public static int[] findUnsortedSubarray(int[] nums, int modifiedIndex) {
        if (nums == null || nums.length == 0)
            return new int[] { -1, -1 };

        int n = nums.length;
        int left = 0, right = n - 1;

        // Find the first out of order element from the left
        while (left < n - 1 && nums[left] <= nums[left + 1]) {
            left++;
        }

        // If the array is already sorted
        if (left == n - 1)
            return new int[] { -1, -1 };

        // Find the first out of order element from the right
        while (right > 0 && nums[right] >= nums[right - 1]) {
            right--;
        }

        // Find the minimum and maximum values in the subarray nums[left..right]
        int subarrayMin = Integer.MAX_VALUE, subarrayMax = Integer.MIN_VALUE;
        for (int i = left; i <= right; i++) {
            subarrayMin = Math.min(subarrayMin, nums[i]);
            subarrayMax = Math.max(subarrayMax, nums[i]);
        }

        // Extend the left boundary to include any elements greater than subarrayMin
        while (left > 0 && nums[left - 1] > subarrayMin) {
            left--;
        }

        // Extend the right boundary to include any elements less than subarrayMax
        while (right < n - 1 && nums[right + 1] < subarrayMax) {
            right++;
        }

        return new int[] { left + 1, right + 1 };
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
import java.util.Scanner;

public class cf_1982F {

    public static int[] res(int a[],int ind, int rep){
        a[ind] = rep;
        return findUnsortedSubarray(a);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t= sc.nextInt();
        for(int i=0;i<t;i++){
            // System.out.println("n: ");
            int n = sc.nextInt();
            // System.out.println("enter n elements of a: ");
            int a[] = new int[n];
            for(int j=0;j<n;j++){
                a[j] = sc.nextInt();
            }
            // System.out.println("q: ");
            int q = sc.nextInt();
            // System.out.println("enter q elements of b[][]: ");
            int b[][] = new int[q][2];
            // System.out.println("call prev and: ");
            int prev[] = findUnsortedSubarray(a);
            System.out.println(prev[0]+" "+prev[1]);
            for(int j=0;j<q;j++){
                // System.out.println("enter b[0], b[1]: ");
                b[j][0] = sc.nextInt();
                b[j][1] = sc.nextInt();
                int[] ress = res(a, b[j][0]-1, b[j][1]);
                System.out.println(ress[0]+" "+ress[1]);
            }

        }
    }

    public static int[] findUnsortedSubarray(int[] nums) {
        if (nums == null || nums.length == 0) return new int[]{-1, -1};

        int n = nums.length;
        int left = 0, right = n - 1;

        // Find the first out of order element from the left
        while (left < n - 1 && nums[left] <= nums[left + 1]) {
            left++;
        }

        // If the array is already sorted
        if (left == n - 1) return new int[]{-1, -1};

        // Find the first out of order element from the right
        while (right > 0 && nums[right] >= nums[right - 1]) {
            right--;
        }

        // Find the minimum and maximum values in the subarray nums[left..right]
        int subarrayMin = Integer.MAX_VALUE, subarrayMax = Integer.MIN_VALUE;
        for (int i = left; i <= right; i++) {
            subarrayMin = Math.min(subarrayMin, nums[i]);
            subarrayMax = Math.max(subarrayMax, nums[i]);
        }

        // Extend the left boundary to include any elements greater than subarrayMin
        while (left > 0 && nums[left - 1] > subarrayMin) {
            left--;
        }

        // Extend the right boundary to include any elements less than subarrayMax
        while (right < n - 1 && nums[right + 1] < subarrayMax) {
            right++;
        }

        return new int[]{left+1, right+1};
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's focus on edge cases and scenarios that can maximize the failure probability. The code's `findUnsortedSubarray` function has a potential issue in how it handles the boundaries of the subarray. Specifically, the function might not correctly handle cases where the array is already sorted or where the subarray boundaries need to be extended.

### Test Case 1: Already Sorted Array
- **Description:** The array is already sorted. After modifications, the array should still be considered sorted or the subarray to sort should be minimized.
- **Input:**
  ```plaintext
  1
  5
  1 2 3 4 5
  3
  2 2
  3 3
  4 4
  ```

### Test Case 2: Single Element Array
- **Description:** The array has only one element. Any modification should not change the subarray boundaries.
- **Input:**
  ```plaintext
  1
  1
  1
  3
  1 2
  1 3
  1 4
  ```

### Test Case 3: Array with All Elements Equal
- **Description:** The array has all elements equal. Any modification should not change the subarray boundaries.
- **Input:**
  ```plaintext
  1
  5
  2 2 2 2 2
  3
  3 1
  4 3
  5 2
  ```

### Test Case 4: Subarray with Repeated Elements
- **Description:** The array has repeated elements in a specific pattern. Modifications should correctly identify the subarray to sort.
- **Input:**
  ```plaintext
  1
  5
  1 2 2 3 4
  3
  2 3
  3 1
  4 2
  ```

### Test Case 5: Subarray with Outliers
- **Description:** The array has outliers that can affect the subarray boundaries. Modifications should correctly extend the boundaries if necessary.
- **Input:**
  ```plaintext
  1
  5
  1 3 2 4 5
  3
  2 1
  3 5
  4 2
  ```

### Test Case 6: Subarray with Negative Elements
- **Description:** The array has negative elements. Modifications should correctly handle negative values.
- **Input:**
  ```plaintext
  1
  5
  -3 -2 -1 0 1
  3
  2 0
  3 -1
  4 -2
  ```

### Combined Test Input
```plaintext
6
5
1 2 3 4 5
3
2 2
3 3
4 4
1
1
3
1 2
1 3
1 4
5
2 2 2 2 2
3
3 1
4 3
5 2
5
1 2 2 3 4
3
2 3
3 1
4 2
5
1 3 2 4 5
3
2 1
3 5
4 2
5
-3 -2 -1 0 1
3
2 0
3 -1
4 -2
```

This combined test input covers a variety of edge cases and should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

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

pair<int, int> findSubarrayToSort(const vector<int>& a) {
    int n = a.size();
    int l = 0, r = n - 1;

    while (l < r && a[l] <= a[l + 1]) l++;
    if (l == r) return {-1, -1};

    while (r > l && a[r] >= a[r - 1]) r--;

    int min_val = *min_element(a.begin() + l, a.begin() + r + 1);
    int max_val = *max_element(a.begin() + l, a.begin() + r + 1);

    while (l > 0 && a[l - 1] > min_val) l--;

    while (r < n - 1 && a[r + 1] < max_val) r++;

    return {l + 1, r + 1};
}

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

    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n);

        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }

        int q;
        cin >> q;
        vector<pair<int, int>> modifications(q);

        for (int i = 0; i < q; ++i) {
            int pos, val;
            cin >> pos >> val;
            modifications[i] = {pos - 1, val};
        }

        auto result = findSubarrayToSort(a);
        cout << result.first << " " << result.second << "\n";

        for (auto& mod : modifications) {
            a[mod.first] = mod.second;
            result = findSubarrayToSort(a);
            cout << result.first << " " << result.second << "\n";
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
2
5
1 2 3 4 5
5
1 5
2 4
3 3
4 2
5 1
5
1 2 3 4 5
1
3 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <limits.h>
using namespace std;

#define ll long long

void solve() {
    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        vector<ll> a(n + 2);
        for(int i = 1; i <= n; ++i) {
            cin >> a[i];
        }
        a[0] = LLONG_MIN;
        a[n+1] = LLONG_MAX;
        int q;
        cin >> q;
        set<int> inversions;
        for(int i = 1; i < n; ++i) {
            if(a[i] > a[i+1]) inversions.insert(i);
        }
        int s = -1, e = -1;
        if(!inversions.empty()) {
            s = *inversions.begin();
            e = *inversions.rbegin() +1;
            // Find min and max in a[s..e]
            ll min_val = a[s], max_val = a[s];
            for(int i = s; i <= e; ++i) {
                min_val = min(min_val, a[i]);
                max_val = max(max_val, a[i]);
            }
            // Expand s and e
            while(s > 1 && a[s-1] > min_val) {
                s--;
                min_val = min(min_val, a[s]);
            }
            while(e < n && a[e+1] < max_val) {
                e++;
                max_val = max(max_val, a[e]);
            }
        }
        if(s == -1) cout << "-1 -1\n";
        else cout << s << " " << e << "\n";
        while(q--) {
            int pos; ll val;
            cin >> pos >> val;
            a[pos] = val;
            // Update inversions at pos-1 and pos
            vector<int> positions;
            if(pos -1 >= 1) positions.push_back(pos -1);
            if(pos >=1 && pos < n) positions.push_back(pos);
            for(int i : positions) {
                if(a[i] > a[i+1]) inversions.insert(i);
                else inversions.erase(i);
            }
            // Update s and e
            int new_s = -1, new_e = -1;
            if(!inversions.empty()) {
                new_s = *inversions.begin();
                new_e = *inversions.rbegin() +1;
                // Find min and max in a[new_s..new_e]
                ll min_val = a[new_s], max_val = a[new_s];
                for(int i = new_s; i <= new_e; ++i) {
                    min_val = min(min_val, a[i]);
                    max_val = max(max_val, a[i]);
                }
                // Expand s and e
                int temp_s = new_s;
                while(temp_s > 1 && a[temp_s-1] > min_val) {
                    temp_s--;
                    min_val = min(min_val, a[temp_s]);
                }
                int temp_e = new_e;
                while(temp_e < n && a[temp_e+1] < max_val) {
                    temp_e++;
                    max_val = max(max_val, a[temp_e]);
                }
                new_s = temp_s;
                new_e = temp_e;
            }
            s = new_s;
            e = new_e;
            if(s == -1) cout << "-1 -1\n";
            else cout << s << " " << e << "\n";
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    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
2 2 3 4 5
3
2 1
4 1
1 1
```


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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

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

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

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

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

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

Buggy Code:
#include<bits/stdc++.h>
#define N 200005
#define int long long
using namespace std;

int T,n,m;
int k;
int a[N];
int b[N];
int use[N];

int max(int x,int y){
	return x > y ? x : y;
}

int min(int x,int y){
	return x < y ? x : y;
}

int Gcd(int x,int y){
	return y==0 ? x :Gcd(y,x%y);
}

void Solve(){
	cin>>n;
	int g,ans=0;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		b[i]=i;
		g=Gcd(a[i],b[i]);
		a[i]/=g,b[i]/=g;
		use[i]=0;
	}
	unordered_map<int,int[100005]>M;
	for(int i=1;i<=n;i++){
		for(int j=b[i];j<=n;j+=b[i]){
			if(use[j]){
				for(int k=1;k<=sqrt(a[i]);k++){
					if(a[i]%k==0){
						ans+=M[j][k];
						if(a[i]!=k*k){
							ans+=M[j][a[i]/k];
						}
					}
				}
			}
		}
		use[a[i]]=1;
		M[a[i]][b[i]]++;
	}
	cout<<ans<<'\n';
	return;
}

signed main()
{
	cin.tie(0), cout.tie(0), ios::sync_with_stdio(0);
	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
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 (Simple Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is a simple version of the problem. The only difference is that in this version $n \leq 10^5$ and the sum of $n$ for all sets of input data does not exceed $10^5$.

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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 <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <vector>
#include <fstream>
using namespace std;
#define ll long long
#define endl "\n"
#define inf (1LL<<61)
//#define inf (1<<30)
#define all(x) x.begin(), x.end()
#define intPow(n, p) (ll)(pow(n, p) + 0.5)
#define InTheNameofAllah ios_base::sync_with_stdio(0); cin.tie(NULL);

template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }

void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#ifdef NEAL_DEBUG
#define dbg(...) cerr << '[' << __FILE__ << ':' << __LINE__ << "] (" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif

const int N = 1e5+3;

vector<int> ase[N];

void solve() {
    int n;
    cin>> n;
    ll ans = 0;
    vector<int> alld;
    for(int i=1; i<=n; i++) {
        int p;
        cin>> p;
        int j = i;
        int g = __gcd(p, i);
        p /= g;
        j /= g;
        vector<int> divs;
        for(int d=1; d*d <= p; d++) {
            if(p % d) continue;
            divs.push_back(d);
            ans += upper_bound(all(ase[j]), d) - lower_bound(all(ase[j]), d);
            if(d*d != p) {
                divs.push_back(p/d);
                ans += upper_bound(all(ase[j]), p/d) - lower_bound(all(ase[j]), p/d);
            }
        }
        for(int d: divs) {
            ase[d].push_back(j);
            alld.push_back(d);
        }
    }
    cout<< ans<< endl;
    for(int d: alld) {
        ase[d].clear();
    }
}

signed main()
{
    InTheNameofAllah
    int t = 1;
    cin>> t;
    for(int i=1; i<=t; i++) {
        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 expose the bug in the provided code, let's consider the following edge cases and scenarios:

1. **Small Permutations with Specific Patterns:**
   - Permutations where the elements are such that the condition \( p_i \cdot p_j \) is divisible by \( i \cdot j \) is rarely or never true.
   - Permutations where the elements are such that the condition is always true.

2. **Permutations with Large Numbers:**
   - Permutations where the elements are large and the gcd calculation might be more complex.

3. **Permutations with Repeated Patterns:**
   - Permutations that repeat certain patterns to test the accumulation of results.

Given these considerations, let's create a small-scale test input that can help expose the bug:

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

Explanation:
- **First Test Case (n=3):** The permutation is `[1, 3, 2]`. The only valid pair is `(1, 2)` because \( p_1 \cdot p_2 = 1 \cdot 3 = 3 \) is divisible by \( 1 \cdot 2 = 2 \). This should output `1`.
- **Second Test Case (n=4):** The permutation is `[4, 3, 2, 1]`. The valid pairs are `(1, 4)`, `(2, 4)`, and `(3, 4)`. This should output `3`.
- **Third Test Case (n=5):** The permutation is `[5, 4, 3, 2, 1]`. The valid pairs are `(1, 5)`, `(2, 5)`, `(3, 5)`, `(4, 5)`, and `(2, 4)`. This should output `5`.

This test input is designed to cover different patterns and edge cases to help identify any issues in the buggy code.


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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

int main() {
    ll t;
    cin >> t;
    vector<int> ans;
    while (t--) {
        ll n;
        cin >> n;
        vector<ll> p(n);
        for (int i = 0; i < n; i++) {
            cin >> p[i];
        }

        ll count = 0;
        unordered_set<ll> seenProducts;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                ll product = p[i] * p[j];
                ll div = (i + 1) * (j + 1);
                if (product % div == 0 && seenProducts.find(product) == seenProducts.end()) {
                    seenProducts.insert(product);
                    count++;
                }
            }
        }

        ans.push_back(count);
    }

    for (auto &a : ans) {
        cout << a << 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 buggy code, let's analyze the code and identify potential issues:

### Issues in the Buggy Code:
1. **Use of `unordered_set` for `seenProducts`:**
   - The code uses an `unordered_set` to track seen products. However, the problem is that the product `p[i] * p[j]` should be checked for each pair `(i, j)` independently, and using a set to avoid counting duplicates is not necessary and can lead to incorrect results.
   - For example, if `p[i] * p[j]` and `p[k] * p[l]` are the same for different pairs `(i, j)` and `(k, l)`, the code will only count one of them, which is incorrect.

2. **Incorrect use of indices:**
   - The code uses `i + 1` and `j + 1` for the divisor, which is correct because indices in the problem are 1-based. However, the use of `i` and `j` directly in the loop is fine as long as the product is calculated correctly.

### Challenging Test Input:
To expose the bug, we need a test case where the same product `p[i] * p[j]` appears multiple times but should be counted separately. A good example is a permutation where the product `p[i] * p[j]` is the same for multiple pairs.

### Example Test Input:
Let's create a permutation where the product `p[i] * p[j]` is the same for multiple pairs:

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

### Explanation:
- For the permutation `[2, 4, 1, 3, 5]`, the valid pairs are:
  - `(1, 2)`: `p[1] * p[2] = 2 * 4 = 8` and `1 * 2 = 2` (8 % 2 == 0)
  - `(1, 5)`: `p[1] * p[5] = 2 * 5 = 10` and `1 * 5 = 5` (10 % 5 == 0)
  - `(2, 5)`: `p[2] * p[5] = 4 * 5 = 20` and `2 * 5 = 10` (20 % 10 == 0)

However, the product `8` and `10` are different, so they should be counted separately. The correct answer should be `3`, but the buggy code might only count one of the pairs with the same product, leading to an incorrect result.

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


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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:
from math import lcm
for _ in range(int(input())):
    n = int(input())
    p = [int(_) for _ in input().split(' ')]
    counter = 0
    for i in range(n):
        lcm_i = lcm(p[i], (i+1))
        k = lcm_i //p[i]
        if k == 0 : k = 1
        for j in range(i+1, n, k):
            if p[i]*p[j] % ((i+1)*(j+1)) == 0:
                counter += 1
    print(counter)

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

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

Think step by step.

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


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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

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

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

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

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

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

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

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

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

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

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

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

def simple_solve():
  return []

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

#0:提出用、1:与えられたテスト用、2:ストレステスト用
main(0)

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can highlight potential issues. Here are some considerations:

1. **Edge Cases:**
   - Permutations of length 1.
   - Permutations with small lengths (2, 3, etc.).
   - Permutations with large lengths (close to the upper limit of 100,000).

2. **Specific Scenarios:**
   - Permutations where the condition \( p_i \cdot p_j \) is divisible by \( i \cdot j \) is easily satisfied.
   - Permutations where the condition is never satisfied.
   - Permutations with repeated patterns or specific structures.

3. **Special Values:**
   - Permutations where \( p_i = i \) for all \( i \).
   - Permutations where \( p_i \) and \( i \) have specific relationships, such as \( p_i = 2i \) or \( p_i = i + 1 \).

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

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

### Explanation:

1. **Test Case 1:**
   - Length: 1
   - Permutation: [1]
   - No pairs, so the expected output is 0.

2. **Test Case 2:**
   - Length: 2
   - Permutation: [2, 1]
   - One pair (1, 2), and \( p_1 \cdot p_2 = 2 \cdot 1 = 2 \) is divisible by \( 1 \cdot 2 = 2 \).
   - Expected output: 1.

3. **Test Case 3:**
   - Length: 3
   - Permutation: [3, 1, 2]
   - One pair (1, 2), and \( p_1 \cdot p_2 = 3 \cdot 1 = 3 \) is divisible by \( 1 \cdot 2 = 2 \).
   - Expected output: 1.

4. **Test Case 4:**
   - Length: 10
   - Permutation: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   - Multiple pairs where \( p_i \cdot p_j \) is divisible by \( i \cdot j \).
   - Expected output: 9 (all pairs (i, j) where \( i < j \)).

This test input covers a range of permutation lengths and structures, which should help in identifying any issues in the buggy code.


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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:
N = 10000
divisors = [[] for i in range(N+1)]
multiples = [[] for i in range(N+1)]
for i in range(1, N+1):
    for j in range(i, N+1, i):
        divisors[j].append(i)
        multiples[i].append(j)

def solve() -> None:
    n = int(input())
    v = [int(x) for x in input().split()]
    pairs = sum([1 if (v[i] % (i+1) == 0) else 0 for i in range(n)])
    ans = (pairs * (pairs-1)) // 2
    cnt = [0 for i in range(n+1)]
    for i in range(0, n):
        cnt[v[i]]+=1
    for i in range(0, n):
        idx = (i+1)
        if (v[i] % idx == 0):
            continue
        divs = divisors[v[i]]
        for j in divs:
            if v[j-1] % idx == 0:
                ans += 1

    print(ans)

    #fix i. I want to find j st (p[i] * p[j]) % (i * j) = 0.

t = int(input())

for _ in range(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 generate a challenging test input that exposes the bug in the provided code, we need to consider the logic and potential flaws in the algorithm. The code is intended to count pairs of indices \((i, j)\) such that \(p_i \cdot p_j\) is divisible by \(i \cdot j\). The buggy code has a few issues:

1. It incorrectly calculates the number of valid pairs.
2. It does not handle the case where \(v[i] % (i+1) != 0\) correctly.
3. It does not correctly count pairs for all possible valid \(j\) values.

Given these issues, a good test case would be one where the permutation has many elements that do not satisfy \(v[i] % (i+1) == 0\), and where the correct pairs are not easily identified by the given logic.

Let's create a test case with a permutation of length 5, which is small enough to be manageable but large enough to show the issues:

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

### Explanation:
- **Input:**
  - \(t = 1\) (one test case)
  - \(n = 5\) (permutation length)
  - \(p = [2, 4, 1, 3, 5]\) (permutation)

- **Expected Output:**
  - The valid pairs are:
    - \((1, 2)\): \(p_1 = 2\), \(p_2 = 4\), \(2 \cdot 4 = 8\) is divisible by \(1 \cdot 2 = 2\)
    - \((1, 5)\): \(p_1 = 2\), \(p_5 = 5\), \(2 \cdot 5 = 10\) is divisible by \(1 \cdot 5 = 5\)
    - \((2, 5)\): \(p_2 = 4\), \(p_5 = 5\), \(4 \cdot 5 = 20\) is divisible by \(2 \cdot 5 = 10\)
  - Total valid pairs: 3

This test case should expose the bug in the code because the current logic does not correctly count all valid pairs. The code might miss some pairs or incorrectly count others.


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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

        int t = sc.nextInt();
        sc.nextLine();

        while (t-- > 0) {
            int n = sc.nextInt();
            int [] arr=new int[n];
            for(int i=0;i<n;i++){
                arr[i]=sc.nextInt();
            }
            int count=0;
            if(n==1){
                System.out.println(count);
                continue;
            }
            for(int i=0;i<n/2;i++){
                for(int j=i+1;j<n;j++){
                    int divident = arr[i]*arr[j];
                    int divisor=(i+1)*(j+1);
                    if(divident%divisor==0){
                        count++;
                    }
                }
            }
            System.out.println(count);
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has a nested loop that iterates up to `n/2` for the outer loop, which might not cover all possible pairs. We need to create a test case that includes such scenarios where the pairs `(i, j)` might be missed.

Here's a step-by-step approach to generate the test input:

1. **Small Permutation with Edge Cases**:
   - A permutation of length 2 or 3 can be simple but effective.
   - For length 2, the only possible pair is (1, 2).
   - For length 3, the pairs are (1, 2), (1, 3), and (2, 3).

2. **Larger Permutation**:
   - A permutation of length 5 or 6 can test more pairs.
   - Ensure the permutation has pairs that should be counted.

3. **Edge Case with All Pairs**:
   - A permutation where all pairs are valid can expose the bug if the code misses any pairs.

4. **Edge Case with No Valid Pairs**:
   - A permutation where no pairs are valid can also be useful to ensure the code handles such cases correctly.

Based on these considerations, here is a small-scale test input that should expose the bug:

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

Explanation:
- The first set of input data has a permutation of length 2, with the pair (1, 2) being valid.
- The second set of input data has a permutation of length 3, with the pair (1, 2) being valid.
- The third set of input data has a permutation of length 5, with pairs (1, 2), (1, 5), and (2, 5) being valid.
- The fourth set of input data has a permutation of length 6, where all pairs are valid.

This test input should help expose any issues in the code's logic, particularly the nested loop that might not cover all possible pairs.


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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

public class G {
    public static void main(String[] args) throws IOException {
        FastReader s = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int tt = s.nextInt();
        outer: while(tt-->0){
            int n = s.nextInt();
            int[] arr = s.nextArrayInt(n);
            int[] a = new int[n];
            for(int i = 0; i < n; i++){
                a[arr[i]-1] = i + 1;
            }
            long ans = 0;
            long count = 0;
            for(int i = 1; i <= n; i++){
                if(arr[i-1] % i == 0){
                    count++;
                }
                for(int j = i; j <= n; j += i){
                    if(a[j-1] == i){
                        continue;
                    }
                    if((((long)arr[i-1] * j) % ((long)i * a[j-1]) == 0)){
                        if(!((a[j-1] > i && arr[i-1] % a[j-1] == 0))){
                            if(!(arr[i-1] % i == 0 && (j % a[j-1]) == 0)) {
                                ans++;
                            }
                        }
                    }
                }
            }
            out.println(ans + count * (count -1) / 2);
        }
        out.flush();
    }

    public static class Pair<T, E> {
        T p1;
        E p2;
        Pair(T a, E b) {
            this.p1 = a;
            this.p2 = b;
        }
        void setValue(T a, E b) {
            this.p1 = a;
            this.p2 = b;
        }
        void setA(T a){
            this.p1 = a;
        }
        void setB(E b){
            this.p2 = b;
        }
        Pair getValue() {
            return this;
        }
        @Override
        public int hashCode() {
            return Objects.hash(p1, p2);
        }
        @Override
        public boolean equals(Object startObject) {
            if(this == startObject){
                return true;
            }
            if(startObject == null || getClass() != startObject.getClass()){
                return false;
            }
            Pair<T, E> pair = (Pair<T, E>) startObject;
            return p1.equals(pair.p1) && p2.equals(pair.p2);
        }
    }
    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public String next() throws IOException {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        long nextLong() throws IOException {
            return Long.parseLong(next());
        }
        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
        String nextLine() throws IOException {
            String str = "";
            if (st != null && st.hasMoreTokens()) {
                str = st.nextToken("\n");
            } else {
                str = br.readLine();
            }
            return str;
        }
        int[] nextArrayInt(int n) throws IOException {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }
        long[] nextArrayLong(int n) throws IOException {
            long[] arr = new long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }
        double[] nextArrayDouble(int n) throws IOException {
            double[] arr = new double[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextDouble();
            }
            return arr;
        }
        String[] nextArrayString(int n) throws IOException {
            String[] arr = new String[n];
            for (int i = 0; i < n; i++) {
                arr[i] = next();
            }
            return arr;
        }
        char[] nextArrayChar() throws IOException {
            return next().toCharArray();
        }
        static void shuffleArray(int[] ar){
            Random rnd = new Random();
            for (int i = ar.length - 1; i > 0; i--)
            {
                int index = rnd.nextInt(i + 1);
                int a = ar[index];
                ar[index] = ar[i];
                ar[i] = a;
            }
        }
        static void shuffleArray(long[] ar){
            Random rnd = new Random();
            for (int i = ar.length - 1; i > 0; i--)
            {
                int index = rnd.nextInt(i + 1);
                long a = ar[index];
                ar[index] = ar[i];
                ar[i] = a;
            }
        }
        public int[] nextSortedArrayInt(int n) throws IOException {
            int[] arr = nextArrayInt(n);
            shuffleArray(arr);
            Arrays.sort(arr);
            return arr;
        }
        public long[] nextSortedArrayLong(int n) throws IOException {
            long[] arr = nextArrayLong(n);
            shuffleArray(arr);
            Arrays.sort(arr);
            return arr;
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are the steps to think through this:

1. **Understand the Bug**: The provided code seems to have issues with the logic for checking the condition `((long)arr[i-1] * j) % ((long)i * a[j-1]) == 0`. It also has a complex condition that might not be necessary or correctly implemented.

2. **Identify Edge Cases**:
   - **Single Element Permutations**: Permutations of length 1 should return 0.
   - **Permutations with Specific Patterns**: Permutations where the product of elements and their indices have specific divisibility properties.
   - **Large Permutations**: Large permutations that might cause performance issues or incorrect logic to fail.

3. **Construct a Test Case**:
   - A permutation that has a specific pattern where the product of elements and their indices is divisible by the product of the indices.
   - A permutation that is a bit more complex to ensure the logic handles it correctly.

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

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

**Explanation**:
1. **Test Case 1**: A permutation of length 1. Expected output: 0.
2. **Test Case 2**: A permutation of length 2. Expected output: 1.
3. **Test Case 3**: A permutation of length 3. Expected output: 1.
4. **Test Case 4**: A permutation of length 4. Expected output: 2.

This test input is designed to cover a range of scenarios, including the simplest case (a single element), a small permutation with a valid pair, and a slightly more complex permutation to ensure the logic is correctly implemented.


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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:
t=int(input())
for c in range(t):
    som=0
    n=int(input())
    ch=input()
    permut=[int(x) for x in ch.split()]
    if n==1:
        print(0)
    else:
        i=0
        while i<=n-2:
            j=i+1
            while j<=n-1:
                x=permut[i]*permut[j]
                if (i+1)*(j+1)<=x:
                    if x%((i+1)*(j+1))==0:
                        som=som+1
                j=j+1
            i=i+1
        print(som)

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


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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:
def valid_permutation(n,seq):
    count = 0
    for i in range(n):
        for j in range(i+1,n):
            res = (seq[i]*seq[j]) % ((i+1)*(j+1))
            if res == 0:
                count += 1
    return count

t = int(input())
res = []
for _ in range(t):
    seq = int(input())
    array = input()
    arr = list(map(int, array.split()))
    res.append(valid_permutation(seq,arr))

for r in res:
    print(r)

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

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

Think step by step.

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

1. **Edge Cases**: Permutations of small size, especially those with only one element, or those that have specific patterns.
2. **Large Input**: A permutation that is large enough to potentially expose performance issues or logic errors.
3. **Specific Patterns**: Permutations that have specific patterns in their values and indices that might cause the condition `(seq[i]*seq[j]) % ((i+1)*(j+1)) == 0` to fail.

Given the constraints, a small-scale test input can be constructed as follows:

1. **Single Element Permutation**: This should return 0 because there are no pairs.
2. **Permutation with a Single Valid Pair**: This should return 1.
3. **Permutation with Multiple Valid Pairs**: This should return a higher count.
4. **Permutation with No Valid Pairs**: This should return 0.
5. **Permutation with a Mixed Pattern**: This should return a moderate count.

Here is a small-scale test input that covers these scenarios:

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

Explanation:
- **Test Case 1**: A permutation of length 1. Expected output: 0.
- **Test Case 2**: A permutation of length 2 with one valid pair (1, 2). Expected output: 1.
- **Test Case 3**: A permutation of length 3 with one valid pair (2, 3). Expected output: 1.
- **Test Case 4**: A permutation of length 5 with three valid pairs (2, 4), (2, 5), (4, 5). Expected output: 3.
- **Test Case 5**: A permutation of length 6 with multiple valid pairs. Expected output: 2 (pairs (3, 6) and (4, 6)).

This test input should help expose any bugs in the provided code, especially in handling edge cases and larger permutations.


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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 Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            int[] a=new int[n];
            for(int i=0;i<n;i++){
                a[i]=sc.nextInt();
            }
            int[] c=new int[n+1];
            for(int num:a){
                c[num]++;
            }
            long count=0;
            for(int i=1;i<=n;i++){
                for(int j=i+1;j<=n;j++){
                    if(a[i-1]*(long)a[j-1] % (i*j)==0){
                        count++;
                    }
                }
            }
            System.out.println(count);
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The bug in the code is related to the way it checks the condition for the pairs. Specifically, the code incorrectly uses the indices of the array `a` without considering the actual permutation values.

### Step-by-Step Analysis:
1. **Edge Case: Small Permutations**:
   - Test with the smallest possible permutations to ensure basic functionality.
   - Example: `1\n1\n2\n1 2`

2. **Edge Case: Permutations with Repeated Values**:
   - Although the problem guarantees that the input is a valid permutation, we can still test with permutations that might cause unexpected behavior.
   - Example: `1\n3\n1 2 3`

3. **Edge Case: Large Permutations**:
   - Test with a large permutation to ensure the algorithm can handle large inputs efficiently.
   - Example: `1\n100000\n1 2 3 ... 100000`

4. **Edge Case: Permutations with Specific Patterns**:
   - Test with permutations that have specific patterns to ensure the condition `p_i * p_j % (i * j) == 0` is checked correctly.
   - Example: `1\n5\n2 4 1 3 5`

5. **Edge Case: Permutations with High Frequency of Valid Pairs**:
   - Test with permutations that have a high frequency of valid pairs to ensure the counting mechanism is accurate.
   - Example: `1\n10\n2 4 6 8 10 1 3 5 7 9`

### Chosen Test Input:
To maximize the failure probability, we will use a permutation that has a high frequency of valid pairs and is large enough to cause issues if the code is not optimized correctly.

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

This test input:
- Is a valid permutation of length 10.
- Has a high frequency of valid pairs, which should expose any issues in the counting mechanism.
- Is small enough to be easily verifiable.

### Expected Output:
The expected output for this test input is:
```
10
```

This is because there are 10 valid pairs in this permutation that satisfy the condition `p_i * p_j % (i * j) == 0`.

### Test Input:
```plaintext
1
10
2 4 6 8 10 1 3 5 7 9
```


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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 Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t =sc.nextInt();
        for(int k=0;k<t;k++){
            long cnt=0;
            int s = sc.nextInt();
            long a[]=new long[s];
            for(int i=0;i<s;i++)
                a[i]=sc.nextLong();
            for(int i=0;i<s;i++){
                for(int j=i+1;j<s;j++){
                    if((a[i]*a[j])%((i+1)*(j+1))==0)
                        cnt+=1;
                }
            }
            System.out.println(cnt);
        }
}
    }

**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
1 2
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 (Simple Version)
time_limit_ms: 3000
memory_limit_mb: 128
Description: This is a simple version of the problem. The only difference is that in this version $n \leq 10^5$ and the sum of $n$ for all sets of input data does not exceed $10^5$.

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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;

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t;cin >> t;
    while(t--){
        int n;cin >>n;
        vector<int>p(n);
        for (int i=0;i<n;++i)cin >> p[i];
        long long ans=0;
        for (int i=0;i<n;++i){
            for (int j=i+1;j<n;++j){
                long long pi_pj=(long long)p[i]*p[j],ij=(long long)(i+1)*(j+1);
                if (pi_pj % ij == 0) {
                    ans++;}}}
        cout << ans << "\n";
    }
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - **Single Element Permutation:** A permutation with only one element.
   - **Two Element Permutation:** A permutation with exactly two elements.
   - **Permutations with Repeated Patterns:** Permutations that have specific patterns or repetitions that might cause issues in the loop.

2. **Maximizing Failure Probability:**
   - **Large Permutations:** Large permutations can help identify issues with the time complexity of the code.
   - **Special Permutations:** Permutations where the condition \( p_i \cdot p_j \) is divisible by \( i \cdot j \) is more likely to be true for many pairs.

3. **Combining Edge Cases:**
   - **Mixed Test Cases:** Include a mix of small and large permutations to ensure the code handles both efficiently.

Given these considerations, here is a small-scale test input that includes a mix of edge cases and special permutations:

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

This test input includes:
- A single element permutation.
- A two-element permutation.
- A three-element permutation.
- A four-element permutation in reverse order.
- A five-element permutation.
- A six-element permutation in reverse order.

These cases should help expose any issues in the code, especially in the nested loop and the condition checking.


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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 <unordered_map>

using namespace std;

int main()
{
    int t;
    cin >> t;

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

        vector<int> p(n);
        for (int i = 0; i < n; ++i)
        {
            cin >> p[i];
        }
        unordered_map<int, int> countMap;

        long long validPairs = 0;
        for (int i = 0; i < n; ++i)
        {
            int pi = p[i];
            for (int j = i + 1; j < n; ++j)
            {
                int pj = p[j];
                int idx = i + 1;
                long long product1 = 1LL * pi * pj;
                long long product2 = 1LL * idx * (j + 1);
                if (product1 % product2 == 0)
                {
                    validPairs++;
                    countMap[product2]++;
                }
            }
        }

        cout << validPairs << "\n";
    }

    return 0;
}

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

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

Think step by step.

Response:
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 maximize the failure probability. The key issue in the code is that it does not correctly handle the condition for the divisibility of \( p_i \cdot p_j \) by \( i \cdot j \). Specifically, the code should check if \( p_i \cdot p_j \) is divisible by \( (i + 1) \cdot (j + 1) \), not \( i \cdot (j + 1) \).

Here's a step-by-step approach to generate a small-scale test input:

1. **Edge Case with Small Permutation:**
   - Use a small permutation to ensure the code's logic is tested thoroughly.
   - A permutation of length 3 or 4 can be sufficient to test basic conditions.

2. **Include a Valid Pair:**
   - Ensure there is at least one valid pair to check if the code correctly counts it.

3. **Include an Invalid Pair:**
   - Ensure there is at least one invalid pair to check if the code correctly ignores it.

4. **Multiple Test Cases:**
   - Use multiple test cases to ensure the code handles different permutations correctly.

Based on these considerations, here is a small-scale test input:

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

This input includes:
1. A permutation of length 3 with one valid pair.
2. A permutation of length 4 with one valid pair.
3. A permutation of length 5 with multiple valid pairs.

This should help expose any issues in the buggy code, particularly the incorrect calculation of the divisibility condition.


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

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, where 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 multiple 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 $n$ ($1 \leq n \leq 10^5$) — 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 $10^5$.

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>
using namespace std;

void solved() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> p(n + 1);  // p is 1-indexed

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

        int count = 0;
        // Check pairs (i, j) where 1 <= i < j <= n
        for (int i = 1; i < n; ++i) {
            for (int j = i + 1; j <= n; ++j) {
                if ((p[i] * p[j]) % (i * j) == 0) {
                    count++;
                }
            }
        }
        cout << count << endl;
    }
}

int main() {
    solved();
    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. **Case 1 (n = 1):**
   - This is a trivial case with no pairs, so it should output `0`.

2. **Case 2 (n = 2):**
   - The permutation is `[2, 1]`.
   - The only pair is `(1, 2)`, and `p[1] * p[2] = 2 * 1 = 2` is not divisible by `1 * 2 = 2` without remainder.
   - Therefore, the output should be `0`.

3. **Case 3 (n = 3):**
   - The permutation is `[3, 1, 2]`.
   - The pairs are:
     - `(1, 2)`: `p[1] * p[2] = 3 * 1 = 3` is not divisible by `1 * 2 = 2`.
     - `(1, 3)`: `p[1] * p[3] = 3 * 2 = 6` is divisible by `1 * 3 = 3`.
     - `(2, 3)`: `p[2] * p[3] = 1 * 2 = 2` is not divisible by `2 * 3 = 6`.
   - Therefore, the output should be `1`.

This test input is designed to cover a trivial case, a small permutation with no valid pairs, and a small permutation with one valid pair, which should help in exposing any potential bugs in the code.