**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    public static List<List<Integer>> facs() {
        int n = 1_000_001;
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) ans.add(new ArrayList<>());
        boolean[] visited = new boolean[n];
        List<Integer> list = new ArrayList<>();
        for (int i = 2; i < n; ++i) {
            if (!visited[i]) {
                for (int j = i; j < n; j += i) ans.get(j).add(i);
                list.add(i);
            }

            for (int v : list) {
                if (v * i >= n) break;
                visited[v * i] = true;
                if (i % v == 0) break;
            }
        }
        return ans;
    }

    public static int find(int[] parent, int p) {
        if (parent[p] != p) parent[p] = find(parent, parent[p]);
        return parent[p];
    }
    public static void union(int[] parent, int p, int q) {
        parent[find(parent, p)] = find(parent, q);
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        List<List<Integer>> allFacs = facs();
        StringBuilder sb = new StringBuilder();
        for (int ig = 0; ig < tt; ++ig) {
            int n = nextInt(), k = nextInt();
            int[] as = new int[n];
            for (int i = 0; i < n; ++i) as[i] = nextInt();
            int m = 2 * n - 1;
            int[] bs = new int[m];
            for (int i = 0; i < n - 1; ++i) bs[i] = as[i + 1];
            for (int i = 0; i < n; ++i) bs[i + n - 1] = as[i];
            int[] parent = new int[m];
            for (int i = 0; i < m; ++i) parent[i] = i;
            Map<Integer, List<Integer>> map = new HashMap<>();
            for (int i = 0; i < m; ++i) {
                for (int v : allFacs.get(bs[i])) {
                    List<Integer> tmpList = map.getOrDefault(v, new ArrayList<>());
                    tmpList.add(i);
                    map.put(v, tmpList);
                }
            }
            for (List<Integer> list : map.values()) {
                for (int i = 0; i < list.size() - 1; ++i) {
                    if (list.get(i + 1) - list.get(i) <= k) union(parent, list.get(i), list.get(i + 1));
                }
            }
            Set<Integer> groups = new HashSet<>();
            for (int i = 0; i < m; ++i) groups.add(find(parent, i));
            sb.append(groups.size()).append("\n");

        }
        System.out.println(sb);
    }
}

**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 scenarios that are likely to cause issues. The code constructs a graph based on the greatest common divisor (GCD) of elements in a cyclically shifted matrix. The main issues to consider are:

1. **Edge Cases for `k`**: The value of `k` determines the range of edges that can be drawn. Setting `k` to a value close to `n` can make the graph very dense, which might expose bugs in the union-find logic.

2. **Repeating Elements**: If the array `a` contains many repeated elements, it can create a lot of connections, which might lead to issues in the union-find algorithm.

3. **Large Prime Numbers**: Using large prime numbers in the array `a` can create fewer connections, making it easier to isolate connected components, which might expose bugs in the GCD calculation.

4. **Small Arrays**: Small arrays can help in verifying the correctness of the algorithm more easily.

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

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

### Explanation:
1. **Test Case 1:**
   - `n = 3`, `k = 3`
   - Array: `[2, 4, 6]`
   - Matrix:
     ```
     2 4 6
     6 2 4
     4 6 2
     ```
   - This test case has a large `k`, which means many edges can be drawn. The array contains elements with common factors, which should create multiple connected components.

2. **Test Case 2:**
   - `n = 3`, `k = 2`
   - Array: `[2, 4, 6]`
   - Matrix:
     ```
     2 4 6
     6 2 4
     4 6 2
     ```
   - This test case has a smaller `k`, which should create fewer connections. The array still contains elements with common factors.

3. **Test Case 3:**
   - `n = 5`, `k = 3`
   - Array: `[2, 4, 6, 8, 10]`
   - Matrix:
     ```
     2 4 6 8 10
     10 2 4 6 8
     8 10 2 4 6
     6 8 10 2 4
     4 6 8 10 2
     ```
   - This test case has a larger `n` and a moderate `k`. The array contains elements with common factors, which should create multiple connected components.

4. **Test Case 4:**
   - `n = 5`, `k = 1`
   - Array: `[2, 4, 6, 8, 10]`
   - Matrix:
     ```
     2 4 6 8 10
     10 2 4 6 8
     8 10 2 4 6
     6 8 10 2 4
     4 6 8 10 2
     ```
   - This test case has a small `k`, which should create very few connections. The array still contains elements with common factors.

These test cases should help expose any bugs in the provided code by challenging the union-find logic and the GCD calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

const int N = 2123456;

int prime[N], psum, minp[N];
bool isprime[N];
int p[N], r[N];
int n, k;
int a[N];
bool st[N];

std::vector<int> vi[N];

int find(int u) {
    if (u - p[u]) p[u] = find(p[u]);
    return p[u];
}

void merge(int u, int v) {
    int x = find(u), y = find(v);
    if (x - y) {
        p[x] = y;
    }
}

void cal_prime(int n) {
    for (int i = 2; i <= n; i ++) {
        if (isprime[i] == false) prime[psum ++] = i, minp[i] = i;
        for (int j = 0; j < psum && i * prime[j] <= n; j ++) {
            isprime[i * prime[j]] = true;
            minp[i * prime[j]] = prime[j];
            if (i % prime[j] == 0) break;
        }
    }
}

int main() {
    cal_prime(1001000);
    int t;
    std::cin >> t;
    while (t --) {
        int ans = 0;
        std::cin >> n >> k;
        for (int i = 0; i <= 2 * n; i ++) p[i] = i, st[i] = false;
        for (int i = 1; i <= n; i ++) {
            std::cin >> a[i];
            if (a[i] == 1) {
                ans += n - i + 1;
                if (i - 2 >= 0) ans += i - 1;
            }
            while (a[i] != 1) {
                int now = minp[a[i]];
                while (a[i] % now == 0) a[i] /= now;
                // std::cout << now << ' ';
                // std::cout << i + n - 2 << std::endl;
                // std::cout << i - 2 << std::endl;
                vi[now].push_back(i + n - 2);
                if (i - 2 >= 0) vi[now].push_back(i - 2);
            }
            // std::cout << std::endl;
        }

        for (int i = 0; i < psum; i ++) {
            int now = prime[i];
            // std::cout << now << std::endl;
            // std::cout << "visize = " << vi[now].size() << std::endl;
            if (vi[now].size()) {
                sort(vi[now].begin(), vi[now].end());
                // for (int i = 0; i < vi[2].size(); i ++) std::cout << vi[2][i] << " ";
                for (int j = 0; j < vi[now].size() - 1; j ++) {
                    if (vi[now][j + 1] - vi[now][j] <= k) {
                        merge(vi[now][j], vi[now][j + 1]);
                    }
                }
                vi[now].clear();
            }
        }
        for (int i = 0; i < 2 * n - 1; i ++) {
            int now = find(i);
            if (st[now] == false) {
                st[now] = true;
                ans ++;
            }
        }
        std::cout << ans << std::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 provided buggy code, we need to consider edge cases and scenarios that might cause the code to fail. Here are some steps to think through:

1. **Edge Cases for `n` and `k`**:
   - `n` and `k` are both at their minimum values (2).
   - `n` and `k` are both at their maximum values (10^6 and 2 * 10^6, respectively).
   - `n` is at its maximum value, and `k` is small (e.g., 2).

2. **Special Values for `a`**:
   - All elements of `a` are the same (e.g., all 2s).
   - All elements of `a` are 1.
   - Elements of `a` are prime numbers.
   - Elements of `a` are composite numbers with common factors.

3. **Large Number of Test Cases**:
   - A large number of test cases (e.g., 10^5) to test the time limit.

4. **Special Graph Structures**:
   - A graph where all vertices are connected.
   - A graph where no vertices are connected.
   - A graph with a specific number of connected components.

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

```plaintext
3
2 2
2 8
3 3
3 4 9
5 3
8 27 5 4 3
```

This test input includes:
1. A test case with `n = 2` and `k = 2`, where all elements of `a` are 2.
2. A test case with `n = 3` and `k = 3`, where `a` contains a mix of 3, 4, and 9.
3. A test case with `n = 5` and `k = 3`, where `a` contains a mix of 8, 27, 5, 4, and 3.

These test cases cover a range of scenarios and should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
typedef pair<int, int> pii;
#define endl "\n"
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f // 注意，memset赋值负无穷是0xC0,得出的值是-0x3f-1
#define eps 1e-8
// double 型memset最大127，最小128,int memset赋值-1为255
//---------------------------------------------------------------------------------------------------------------------//
//---------------------------------------------------------------------------------------------------------------------//
const int mod = 998244353; //(ll)1e18,double浮点数只有15位，1e15后面+的小数会被省略
const int N = 2e6 + 10;

int n, k;
int a[N], fa[N], last[N];
vector<int> b[N >> 1];
int find(int x)
{
  if (fa[x] != x) fa[x] = find(fa[x]);
  return fa[x];
}

void mysolve()
{
  cin >> n >> k;
  for (int i = n; i < (n << 1); ++i)
    {
      cin >> a[i];
      a[i - n] = a[i];
      for (auto v : b[a[i]]) last[v] = 0;
    }

  for (int i = 1; i < (n << 1); ++i) fa[i] = i;
  for (int i = 1; i < (n << 1); ++i)
    {
      for (auto v : b[a[i]])
        {
          if (last[v] && i - last[v] <= k) fa[find(last[v])] = find(i);
          last[v] = i;
        }
    }
  ll ans = 0;
  for (int i = 1; i < (n << 1); ++i)
    if (find(i) == i)
      {
        if (a[i] > 1)
          ans++;
        else if (i < n)
          ans += i - 1;
        else
          ans += 2 * n - i;
      }
  cout << ans << endl;
}

int32_t main()
{
#ifdef DEBUG
  freopen("in.in", "r", stdin);
  freopen("out.out", "w", stdout);
#endif
  std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); // 使用read请把解绑注释了
  int t = 1;
  cin >> t;
  const int lim = 1e6;
  int cnt = 0;
  for (int i = 2; i < lim; ++i)
    if (b[i].empty())
      {
        for (int j = i; j <= lim; j += i)
          b[j].push_back(i), cnt++;
      }
  while (t--)
    {
      mysolve();
    }
  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
4 2
2 4 6 8
```

### Explanation:
1. **Array and Matrix Construction**:
   - Given `a = [2, 4, 6, 8]`, the matrix `b` will be:
     ```
     2 4 6 8
     8 2 4 6
     6 8 2 4
     4 6 8 2
     ```

2. **Graph Construction**:
   - The graph will have vertices corresponding to each element in the matrix.
   - Edges will be drawn between vertices if the Manhattan distance between them is at most `2` and the GCD of the corresponding matrix elements is greater than `1`.

3. **GCD Conditions**:
   - All elements in the array `a` are even, so the GCD condition `gcd(b[i1, j1], b[i2, j2]) > 1` will always be true for any pair of vertices.
   - The Manhattan distance condition `|i1 - i2| + |j1 - j2| <= 2` will also be satisfied for many pairs of vertices.

4. **Connected Components**:
   - Since all elements are even and the GCD condition is always true, all vertices will be connected to each other within the constraints of the Manhattan distance.
   - Therefore, the graph will have only one connected component.

### Expected Output:
- The expected output for this test case is `1`, as there is only one connected component in the graph.

This test input is designed to expose the bug in the buggy code by ensuring that the GCD condition is always true and the Manhattan distance condition is also easily satisfied, leading to a single connected component.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Yes')
def df():
    print('No')
G = 10 ** 6 + 3
p = [[] for i in range(G)]
for i in range(2, G):
    if len(p[i]) == 0:
        for j in range(i, G, i):
            p[j].append(i)
for ___ in range(fg()):
    n, k = fgh()
    a = fgh()
    ans = 0
    d = {}
    for i in range(n):
        if a[i] == 1:
            ans += n
    ks = 0
    frs = []
    lst = []
    e = {}
    for i in range(n):
        if i > k:
            for j in p[a[i - k - 1]]:
                d[j] -= 1
        fl = 1
        for kl in p[a[i]]:
            if kl not in d:
                d[kl] = 1
            else:
                if d[kl]:
                    lst[e[kl]] = i
                    fl = 0
                d[kl] += 1
        if fl:
            for kl in p[a[i]]:
                e[kl] = ks
            if a[i] - 1:
                frs.append(i)
                lst.append(i)
                ks += 1
    ans += ks
    for i in range(ks):
        if (frs[i] or frs[i] - lst[i]) and frs[i] + n - lst[i] > k:
            ans += 1
    print(ans)

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

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

Think step by step.

Response:
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 can cause the code to fail. The key points to consider are:

1. **Edge Cases for \( k \)**: Since \( k \) affects the connectivity of the graph, we should test with \( k \) values that are close to the limits (i.e., \( k = 2 \) and \( k = 2 \cdot 10^6 \)).
2. **Special Values in Array \( a \)**: Test with arrays that contain values that have many common divisors, as this can affect the number of connected components.
3. **Large \( n \)**: Test with large values of \( n \) to ensure the code handles large inputs correctly.
4. **Special Patterns in Array \( a \)**: Test with arrays that have specific patterns, such as all elements being the same or having a specific cyclic pattern.

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

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

### Explanation:

1. **Test Case 1: \( n = 3 \), \( k = 2 \), \( a = [1, 2, 3] \)**
   - This test case includes a small \( n \) and \( k \) with distinct values in \( a \). The matrix \( b \) will be:
     \[
     b = \begin{bmatrix} 1 & 2 & 3 \\\ 3 & 1 & 2 \\\ 2 & 3 & 1 \end{bmatrix}
     \]
   - The connected components are:
     - \((1, 1)\), \((2, 2)\), \((3, 3)\) (all 1s)
     - \((1, 2)\), \((2, 3)\), \((3, 1)\) (all 2s)
     - \((1, 3)\), \((2, 1)\), \((3, 2)\) (all 3s)
   - Expected output: 3

2. **Test Case 2: \( n = 4 \), \( k = 3 \), \( a = [2, 4, 6, 8] \)**
   - This test case includes a larger \( n \) with values that have common divisors. The matrix \( b \) will be:
     \[
     b = \begin{bmatrix} 2 & 4 & 6 & 8 \\\ 8 & 2 & 4 & 6 \\\ 6 & 8 & 2 & 4 \\\ 4 & 6 & 8 & 2 \end{bmatrix}
     \]
   - The connected components are:
     - All vertices are connected because all values are even and have a common divisor of 2.
   - Expected output: 1

3. **Test Case 3: \( n = 5 \), \( k = 1 \), \( a = [1, 1, 1, 1, 1] \)**
   - This test case includes a large \( n \) with all elements being 1. The matrix \( b \) will be:
     \[
     b = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\\ 1 & 1 & 1 & 1 & 1 \\\ 1 & 1 & 1 & 1 & 1 \\\ 1 & 1 & 1 & 1 & 1 \\\ 1 & 1 & 1 & 1 & 1 \end{bmatrix}
     \]
   - The connected components are:
     - All vertices are connected because all values are 1.
   - Expected output: 1

4. **Test Case 4: \( n = 3 \), \( k = 3 \), \( a = [2, 4, 6] \)**
   - This test case includes a small \( n \) with values that have common divisors. The matrix \( b \) will be:
     \[
     b = \begin{bmatrix} 2 & 4 & 6 \\\ 6 & 2 & 4 \\\ 4 & 6 & 2 \end{bmatrix}
     \]
   - The connected components are:
     - All vertices are connected because all values are even and have a common divisor of 2.
   - Expected output: 1

This test input covers a variety of scenarios and should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
import math
from collections import defaultdict

def gcd(x, y):
    while y:
        x, y = y, x % y
    return x

def build_matrix(n, a):
    b = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(n):
            b[i][j] = a[(j + i) % n]
    return b

def add_edge(graph, i1, j1, i2, j2, k, b):
    if abs(i1 - i2) + abs(j1 - j2) <= k and gcd(b[i1][j1], b[i2][j2]) > 1:
        graph[(i1, j1)].add((i2, j2))
        graph[(i2, j2)].add((i1, j1))

def dfs(node, visited, graph):
    stack = [node]
    while stack:
        u = stack.pop()
        if u not in visited:
            visited.add(u)
            for v in graph[u]:
                if v not in visited:
                    stack.append(v)

def count_connected_components(n, k, a):
    b = build_matrix(n, a)
    graph = defaultdict(set)

    for i in range(n):
        for j in range(n):
            for di in range(-k+1, k):
                for dj in range(-k+1, k):
                    ni, nj = i + di, j + dj
                    if 0 <= ni < n and 0 <= nj < n:
                        add_edge(graph, i, j, ni, nj, k, b)

    visited = set()
    num_components = 0

    for i in range(n):
        for j in range(n):
            if (i, j) not in visited:
                dfs((i, j), visited, graph)
                num_components += 1

    return num_components

# Read input
input_data = """6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2"""

lines = input_data.split('\n')
t = int(lines[0].strip())
index = 1
results = []

for _ in range(t):
    n, k = map(int, lines[index].strip().split())
    a = list(map(int, lines[index + 1].strip().split()))
    index += 2
    results.append(count_connected_components(n, k, a))

# Print results
for result in results:
    print(result)

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

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

Think step by step.

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

1. **Edge Case with Small `n` and `k`:** Test with the minimum values of `n` and `k` to see if the code handles small graphs correctly.
2. **Edge Case with Large `k`:** Test with `k` values that are close to or equal to `n` to ensure the code can handle large connectivity ranges.
3. **Edge Case with Repeated Elements:** Test with arrays that have repeated elements to check if the graph construction and connectivity are handled correctly.
4. **Edge Case with Prime Numbers:** Test with arrays that have prime numbers to ensure the GCD condition is correctly applied.
5. **Edge Case with Mixed Values:** Test with arrays that have a mix of prime and composite numbers to ensure the graph construction and connectivity are handled correctly.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a mix of these scenarios:

```plaintext
5
2 2
2 3
3 3
2 2 2
4 4
3 6 9 12
5 5
2 3 5 7 11
6 6
2 4 6 8 10 12
```

### Explanation:
1. **Test Case 1:**
   - `n = 2`, `k = 2`
   - Array: `[2, 3]`
   - This test case checks the smallest possible values of `n` and `k` with a mix of a prime and a composite number.

2. **Test Case 2:**
   - `n = 3`, `k = 3`
   - Array: `[2, 2, 2]`
   - This test case checks a small graph with repeated elements to ensure the connectivity is handled correctly.

3. **Test Case 3:**
   - `n = 4`, `k = 4`
   - Array: `[3, 6, 9, 12]`
   - This test case checks a larger graph with multiples of 3 to ensure the GCD condition is correctly applied.

4. **Test Case 4:**
   - `n = 5`, `k = 5`
   - Array: `[2, 3, 5, 7, 11]`
   - This test case checks a larger graph with prime numbers to ensure the GCD condition is correctly applied.

5. **Test Case 5:**
   - `n = 6`, `k = 6`
   - Array: `[2, 4, 6, 8, 10, 12]`
   - This test case checks a larger graph with multiples of 2 to ensure the GCD condition is correctly applied.

This test input should help identify any issues in the buggy code, especially in handling edge cases and ensuring correct graph construction and connectivity.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

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

public class Main {

    static List<Integer> primes;
    static boolean[] isPrime;

    static {
        int maxn = (int) (1e6 + 5);
        primes = new ArrayList<>();
        isPrime = new boolean[maxn];
        Arrays.fill(isPrime, true);
        for (int i = 2; i < maxn; i++) {
            if (!isPrime[i]) continue;
            primes.add(i);
            for (int j = i + i; j < maxn; j += i) isPrime[j] = false;
        }
    }

    public static int[] parent;

    public static int getRoot(int root) {
        return root == parent[root] ? root : (parent[root] = getRoot(parent[root]));
    }

    public static void solve() throws IOException {
        int n = in.nextInt();
        int k = in.nextInt();
        int[] a = new int[2*n];
        long ans = 0;
        for (int i = 1; i <= n; i++) {
            a[i+n-1] = in.nextInt();
            if (a[i] == 1) ans += n;
        }
        for (int i = 1; i < n; i++) a[i] = a[i+n];
        List<Integer>[] groups = new ArrayList[primes.size()];
        Arrays.setAll(groups, e -> new ArrayList<>());
        parent = new int[2*n];
        Arrays.setAll(parent, i -> i);
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 1; i < 2 * n; i++) {
            int x = a[i];
            for (int j = 0; j < primes.size() && primes.get(j) <= x; j++) {
                if (x % primes.get(j) != 0) continue;
                int pre = map.getOrDefault(j, Integer.MIN_VALUE / 10);
                if (i - pre <= k) parent[getRoot(i)] = getRoot(pre);
                map.put(j, i);
                while (x % primes.get(j) == 0) x /= primes.get(j);
            }
        }
        HashSet<Integer> set = new HashSet<>();
        for (int i = 1; i < 2 * n; i++) {
            if (a[i] == 1) continue;
            set.add(getRoot(i));
        }
        out.println(ans + set.size());
    }

    static boolean MULTI_CASE = true;

    public static void main(String[] args) throws IOException {
        if (MULTI_CASE) {
            int T = in.nextInt();
            for (int i = 0; i < T; ++i) {
                solve();
            }
        } else {
            solve();
        }
        out.close();
    }

    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;

        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

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

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

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

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

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

/*
 */

**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
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

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

public class Main {

    static int[] primes;
    static boolean[] isPrime;
    static int N;

    static {
        int maxn = (int) (1e6 + 5);
        primes = new int[maxn];
        isPrime = new boolean[maxn];
        for (int i = 2; i < maxn; i++) {
            if (isPrime[i]) continue;
            primes[N++] = i;
            for (int j = i + i; j < maxn; j += i) isPrime[j] = true;
        }
    }

    public static int[] parent;

    public static int getRoot(int root) {
        return root == parent[root] ? root : (parent[root] = getRoot(parent[root]));
    }

    public static void solve() throws IOException {
        int n = in.nextInt();
        int k = in.nextInt();
        int[] a = new int[2*n];
        long ans = 0;
        for (int i = 1; i <= n; i++) {
            a[i+n-1] = in.nextInt();
            if (a[i+n-1] == 1) ans += n;
        }
        for (int i = 1; i < n; i++) a[i] = a[i+n];
        parent = new int[2*n];
        Arrays.setAll(parent, i -> i);
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 1; i < 2 * n; i++) {
            int x = a[i];
            for (int j = 0; j < N && primes[j] <= x; j++) {
                if (x % primes[j] != 0) continue;
                int pre = map.getOrDefault(j, Integer.MIN_VALUE / 10);
                if (i - pre <= k) parent[getRoot(i)] = getRoot(pre);
                map.put(j, i);
                while (x % primes[j] == 0) x /= primes[j];
            }
        }
        for (int i = 1; i < 2 * n; i++) {
            if (a[i] == 1) continue;
            if (getRoot(i) == i) ans++;
        }
        out.println(ans);
    }

    static boolean MULTI_CASE = true;

    public static void main(String[] args) throws IOException {
        if (MULTI_CASE) {
            int T = in.nextInt();
            for (int i = 0; i < T; ++i) {
                solve();
            }
        } else {
            solve();
        }
        out.close();
    }

    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;

        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

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

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

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

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

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

/*
 */

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

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

Think step by step.

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

1. **Edge Cases for `n` and `k`**: The values of `n` and `k` can be at their limits (e.g., `n = 10^6` and `k = 2 * 10^6`).
2. **Special Values in Array `a`**: The array `a` can contain values that are either 1 or have specific properties (e.g., all elements are the same, all elements are prime, or all elements are composite with specific prime factors).
3. **Cyclic Shifts and GCD Conditions**: The cyclic shifts in the matrix `b` and the GCD conditions can lead to complex connected components, especially when `k` is large.

Let's construct a test case that maximizes the probability of exposing the bug:

- **Test Case 1**: All elements in `a` are the same and greater than 1.
- **Test Case 2**: All elements in `a` are 1.
- **Test Case 3**: Elements in `a` are prime numbers.
- **Test Case 4**: Elements in `a` are composite numbers with specific prime factors.

Here is a small-scale test input that includes these edge cases:

```plaintext
4
3 3
2 2 2
3 3
1 1 1
3 3
2 3 5
3 3
4 6 9
```

Explanation of the test cases:
1. **Test Case 1**: All elements are 2. This should result in one connected component.
2. **Test Case 2**: All elements are 1. This should result in 9 connected components (since 1 does not connect to any other vertex).
3. **Test Case 3**: All elements are prime and different. This should result in 3 connected components (each prime number forms its own component).
4. **Test Case 4**: All elements are composite with specific prime factors. This should result in 2 connected components (4 and 6 form one, and 9 forms another).

This test input is designed to cover a variety of scenarios that can expose potential 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: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
// https://codeforces.com/contest/1978/problem/F
#include <bits/stdc++.h>
#define st first
#define nd second
#define pb push_back
#define all(x) x.begin(), x.end()
#define rep(i, a) for (int i = 0; i < int(a); i++)
#define rep1(i, a) for (int i = 1; i <= int(a); i++)
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
typedef long long ll;
typedef pair<int, int> pi;
const int MAXN = 2e5 + 5;
int fac[1'000'005];
int re[2 * MAXN];
int res[2 * MAXN];
int cas = 1;
void sieve() {
    for (int i = 2; i <= 1e6; i++) {
        if (fac[i])
            continue;
        fac[i] = i;
        if ((ll)i * i > 1e6)
            continue;
        int j = i * i;
        while (j <= 1e6) {
            fac[j] = i;
            j += i;
        }
    }
    // rep(i, 25) cerr << fac[i] << ' ';
    // cerr << "\n";
}
int fint(int a) {
    if (re[a] == a)
        return a;
    else
        re[a] = fint(re[a]);
    return re[a];
}
void onion(int a, int b) {
    if (fint(a) != fint(b))
        re[fint(a)] = fint(b);
}
void sol() {
    cas++;
    int n, k;
    cin >> n >> k;
    iota(re, re + 2 * n, 0);
    vector<int> v(n - 1);
    int x1;
    cin >> x1;
    rep(i, n - 1) cin >> v[i];
    v.pb(x1);
    v.insert(v.end(), all(v));
    v.pop_back();
    // for (auto i : v)
    //     cerr << i << ' ';
    // cerr << '\n';
    map<int, basic_string<int>> m;
    ll rPlus = 0;
    rep(j, v.size()) {
        if (v[j] == 1) {
            rPlus += min(j + 1, (int)size(v) - j) - 1;
            continue;
        }
        int x = v[j];
        while (x > 1) {
            m[fac[x]].pb(j);
            while (fac[x] == fac[x / fac[x]])
                x /= fac[x];
            x /= fac[x];
        }
    }
    // cerr << "c\n";
    for (auto [x, y] : m) {
        // cerr << x << ": ";
        rep(i, y.size() - 1) if (y[i + 1] - y[i] <= k) onion(y[i + 1], y[i]);
        // cerr << '\n';
    }
    int re = 0;
    rep(i, v.size()) {
        if (res[fint(i)] != cas)
            re++;
        res[fint(i)] = cas;
    }
    cout << re + rPlus << '\n';
    // cerr << "r: ";
    // for (auto i : res)
    //     cerr << i << ' ';
    // cerr << '\n';
}
int main() {
    cin.tie(0)->sync_with_stdio(0);
    sieve();
    int t;
    cin >> t;
    cerr << fac[t] << '\n';
    rep(i, t) sol();
}
/*
1
3 3
3 4 9

*/

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause issues in the algorithm. Here are the steps to consider:

1. **Small-scale Input**: Ensure the input is small enough to be valid but large enough to reveal potential bugs.
2. **Edge Cases**: Consider cases where the array `a` contains elements that have common divisors, and the matrix `b` has elements that are cyclically shifted.
3. **Special Values**: Use values that are prime, composite, and have specific properties (e.g., multiples of small primes).

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

- **Array `a`**: Use a small array with elements that have common divisors.
- **Parameter `k`**: Set `k` to a value that allows for multiple connections but is not too large.

Here is a small-scale test input that can expose potential bugs:

```plaintext
3
4 3
2 4 6 8
3 2
2 3 6
5 4
10 15 20 25 30
```

### Explanation:
1. **Test Case 1**:
   - **Array `a`**: `[2, 4, 6, 8]`
   - **Parameter `k`**: `3`
   - **Matrix `b`**:
     ```
     2 4 6 8
     8 2 4 6
     6 8 2 4
     4 6 8 2
     ```
   - **Expected Behavior**: This test case should expose issues with handling multiple common divisors and cyclic shifts.

2. **Test Case 2**:
   - **Array `a`**: `[2, 3, 6]`
   - **Parameter `k`**: `2`
   - **Matrix `b`**:
     ```
     2 3 6
     6 2 3
     3 6 2
     ```
   - **Expected Behavior**: This test case should test the handling of mixed prime and composite numbers and their cyclic shifts.

3. **Test Case 3**:
   - **Array `a`**: `[10, 15, 20, 25, 30]`
   - **Parameter `k`**: `4`
   - **Matrix `b`**:
     ```
     10 15 20 25 30
     30 10 15 20 25
     25 30 10 15 20
     20 25 30 10 15
     15 20 25 30 10
     ```
   - **Expected Behavior**: This test case should test the handling of larger numbers with multiple common divisors and their cyclic shifts.

This input should help in identifying any issues in the buggy code by covering a range of scenarios with different properties of the array elements and the parameter `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
#include <bits/stdc++.h>
    #include <ext/pb_ds/assoc_container.hpp>
    using namespace std;
    using namespace __gnu_pbds;
    #define F first
    #define S second
    #define FOR(i,a,b) for (auto i = (a); i <= (b); ++i)
    #define NFOR(i,a,b) for(auto i = (a); i >= (b); --i)
    #define all(x) (x).begin(), (x).end()
    #define sz(x) lli(x.size())
    #define mp(i,a) make_pair(i,a)
    #define pb(a) push_back(a)
    #define bit(x,b) (x&(1LL<<b))
    typedef long long int lli;
    typedef pair <lli,lli> ii;
    typedef pair <ii,lli> iii;
    typedef pair <ii,ii> iiii;
    typedef vector <lli> vi;
    typedef tree<ii,null_type,less<ii>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
    vi al[2000001],p[1000001];
    int vis[1000001];
    void dfs(int node)
    {
      vis[node]=1;
      for(int i=0;i<al[node].size();++i)
      {
        if(!vis[al[node][i]])
        {
          dfs(al[node][i]);
        }
      }
    }
    int main()
    {
      ios_base::sync_with_stdio(false);
      cin.tie(NULL);
      int tests=1;
      cin>>tests;
      for(int i=2;i<=1000000;++i)
      {
        if(!vis[i])
        {
          for(int j=i;j<=1000000;j+=i)
          {
            vis[j]=1;
            p[j].pb(i);
          }
        }
      }
      for(;tests;--tests)
      {
        int n,kt;
        cin>>n>>kt;
        int a[n];
        for(int i=0;i<2*n-1;++i)
        {
          al[i].clear();
          vis[i]=0;
        }
        unordered_map<int,int> m;
        int co=0;
        for(int i=0;i<n;++i)
        {
          cin>>a[i];
          if(a[i]==1)
          {
            int d=n-2;
            if(i!=0)
              co+=d;
            else
              co+=n-1;
          }
        }
        for(int i=0,j;i<2*n-1;++i)
        {
          if(i<=n-2)
            j=a[i+1];
          else
            j=a[i-(n-1)];
          for(int k=0;k<p[j].size();++k)
          {
            if(m[p[j][k]]!=0)
            {
              if(i-(m[p[j][k]]-1)<=kt)
              {
                al[i].pb(m[p[j][k]]-1);
                al[m[p[j][k]]-1].pb(i);
              }
            }
            m[p[j][k]]=i+1;
          }
        }
        for(int i=0;i<2*n-1;++i)
        {
          if(!vis[i])
          {
            dfs(i);
            co++;
          }
        }
        cout<<co<<endl;
      }
      return(0);
    }

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

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

Think step by step.

Response:
```plaintext
1
4 2
2 4 6 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
#vis false after recursion
#prefix sum
#comp bit wise bmq
#print instead of insertion
#powers to log
#reverse dp
#2^ , 10^ bitm
#list prod- use MOD

from collections import *
from math import *
from heapq import *
from functools import lru_cache
from bisect import *
import sys
from bisect import *

def bins(array, x, l, h):
    while l <= h:
        m = l + (h - l)//2
        if array[m] == x:
            return m
        elif array[m] < x:
            l = m + 1
        else:
            h = m - 1
    return l

def quicksort(array):
    def sort(arr, l, r):
        if l < r:
            p = part(arr, l, r)
            sort(arr, l, p-1)
            sort(arr, p+1, r)

    def part(arr, l, r):
        pivot = arr[r]
        a = l
        for i in range(l,r):
            if arr[i] < pivot:
                arr[i], arr[a] = arr[a], arr[i]
                a += 1
        arr[r], arr[a] = arr[a], arr[r]
        return a

    sort(array, 0, len(array)-1)
    return array

def countXYZ(a, l,m,r):
    ans = 0
    x = m+1
    for i in range(l,m+1):
        ans += x-(m+1)
        for j in range(x, r):
            if a[i] <= a[j]:
                ans += 1
                x += 1
            else:
                break
    return ans

def partZY(arr, l , m , r):
    ans = countXYZ(arr, l,m,r)
    res = []
    x,y = l ,m+1
    while x <= m and y <= r:
        if arr[x] < arr[y]:
            res += [arr[x]]
            x +=1
        else:
            res += [arr[y]]
            y +=1
    while x <=m:
        res += [arr[x]]
        x +=1
    while y <=r:
        res += [arr[y]]
        y +=1

    j = 0
    for i in range(l,r+1):
        arr[i] = res[j]
        j+=1
    return ans

def msort(arr,l,r):
    if l>= r:
        return 0
    m = l + (r-l)//2
    x = msort(arr,l,m)
    y = msort(arr, m+1, r)
    return x+y + partZY(arr, l, m, r)

inf = float('inf')
mod = 10**9 + 7
input = lambda: sys.stdin.readline().rstrip("\r\n")
inps = lambda: map(int,input().split())
inpl = lambda: list(map(int, input().split()))
inpt = lambda: int(input())
yes = lambda: print("YES")
no = lambda: print("NO")
draw = lambda: print("DRAW")
win = lambda:print("WIN")
lose = lambda:print("LOSE")
imps = lambda:print("IMPOSSIBLE")
infi = lambda:print("INFINITY")
#vis = defaultdict(lambda:False)
#grph = defaultdict(list)
#dict = defaultdict(lambda:-1)

def solve():
    def gccd(a,b):
        a,b = min(a,b), max(a,b)
        if dp[(a,b)] == 0:
            dp[(a,b)] = gcd(a,b)
        return dp[(a,b)]

    def nocg(graph):
        def bfs(node):
            ans = 0
            for adj in graph[node]:
                if not vis[adj]:
                    vis[adj] = True
                    ans += 1+bfs(adj)
            return ans

        res = 0
        for node in graph.keys():
            if not vis[node]:
                vis[node] = True
                res += bfs(node)
        return res

    def dfs(i,k,num):
        for n in range(i+1, min(i+k+1, len(grph))):
            if vis[n] ==0 and gccd(grph[n],grph[i]) >1:
                vis[n] = num
                dfs(n, k, num)
                continue
            if vis[n] == num:
                continue
            if vis[n] != 0 and gccd(grph[n], grph[i])>1:
                edges[num] += [vis[n]]
                edges[vis[n]] += [edges[num]]

    n,k = inps()
    a = inpl()
    grph = []
    ones = 0
    tones = 0
    for i in range(1, n):
        grph += [a[i]]
        if a[i] ==1:
            ones +=1
    for i in range(n):
        grph += [a[i]]
        if a[i] ==1:
            ones +=1
            tones += 1

    dp = defaultdict(lambda:0)
    vis = [0]* (len(grph))
    edges = defaultdict(list)
    num = 1

    # 2n
    for i in range(len(grph)):
        if vis[i] == 0:
            vis[i] = num
            dfs(i,k, num)
            num +=1
    num -=1
    vis = defaultdict(lambda:False)
    print(num + nocg(edges) - ones + tones*n)
n =1
n = inpt()
for _ in range(n):
    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:
```plaintext
1
3 2
2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
class UnionFind:
    def __init__(self,n):
        self.root=[i for i in range(n)]
    def find(self,x):
        if x==self.root[x]:
            return x
        self.root[x]=self.find(self.root[x])
        return self.root[x]
    def union(self,x,y):
        rx,ry=self.find(x),self.find(y)
        self.root[rx]=ry

class PrimeFactor:
    def __init__(self, n):
        self.n = n
        # calculate the minimum prime factor for all numbers from 1 to self.n
        self.min_prime = [0] * (self.n + 1)
        # determine whether all numbers from 1 to self.n are prime numbers
        self.is_prime = [0] * (self.n + 1)
        # calculate the prime factorization of all numbers from 1 to self.n
        self.prime_factor = [[] for _ in range(self.n + 1)]
        # calculate all factors of all numbers from 1 to self.n, including 1 and the number itself
        self.build()
        return

    def build(self):
        # complexity is O(nlogn)
        for i in range(2, self.n + 1):
            if not self.min_prime[i]:
                self.is_prime[i] = 1
                self.min_prime[i] = i
                for j in range(i * i, self.n + 1, i):
                    if not self.min_prime[j]:
                        self.min_prime[j] = i

        # complexity is O(nlogn)
        for num in range(2, self.n + 1):
            i = num
            while num > 1:
                p = self.min_prime[num]
                cnt = 0
                while num % p == 0:
                    num //= p
                    cnt += 1
                self.prime_factor[i].append((p, cnt))
        return
def solve():
    n, k = map(int,input().split())
    nums = list(map(int,input().split()))
    dct = dict()
    uf = UnionFind(2 * n -1)
    for i in range(1, 2 * n):
        num = nums[i % n]
        for p, _ in pf.prime_factor[num]:
            if p in dct and i - dct[p] <= k and p > 1:
                uf.union(dct[p] - 1, i - 1)
            dct[p] = i
    ans = len(set([uf.find(i) for i in range(2*n-1)]))
    for i in range(n):
        if nums[i] == 1:
            ans += n - 2 if i else n - 1
    print(ans)

pf = PrimeFactor(10 ** 6)
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 test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The key aspects to consider are:

1. **Small-scale test input**: The input should be small enough to be manageable and fit within the output length limit.
2. **Edge cases**: Consider cases where the array `a` contains values that have specific properties, such as multiple prime factors, repeated values, or values that are all 1.
3. **Parameter `k`**: Choose a value for `k` that can create or break connections in the graph based on the gcd condition.

### Step-by-Step Analysis

1. **Array `a` with Repeated Values**:
   - If the array `a` contains repeated values, the union-find operations might not work correctly if the indices are not handled properly.

2. **Array `a` with Prime Numbers**:
   - If the array `a` contains prime numbers, the gcd condition will be less likely to create connections, which can help in testing the union-find operations.

3. **Array `a` with Values of 1**:
   - If the array `a` contains values of 1, the gcd condition will always fail, and the number of connected components should be `n^2`.

4. **Parameter `k`**:
   - Choose a value for `k` that is large enough to potentially create connections but small enough to avoid creating too many connections, which can help in testing the union-find operations.

### Test Input

Let's create a small-scale test input with the following properties:
- The array `a` contains repeated values and some prime numbers.
- The parameter `k` is set to a value that can create some connections but not too many.

```plaintext
3
5 3
2 3 5 7 11
5 2
2 2 3 3 5
5 3
1 1 1 1 1
```

### Explanation

1. **Test Case 1**:
   - `a = [2, 3, 5, 7, 11]`
   - `k = 3`
   - This test case contains distinct prime numbers. The gcd condition will create some connections, but not too many.

2. **Test Case 2**:
   - `a = [2, 2, 3, 3, 5]`
   - `k = 2`
   - This test case contains repeated values. The gcd condition will create more connections due to the repeated values.

3. **Test Case 3**:
   - `a = [1, 1, 1, 1, 1]`
   - `k = 3`
   - This test case contains only values of 1. The gcd condition will always fail, and the number of connected components should be `25` (since there are 5 rows and 5 columns).

This test input should help in exposing any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Large Graph
time_limit_ms: 4000
memory_limit_mb: 512
Description: Given an array $a$ of length $n$. Let's construct a square matrix $b$ of size $n \times n$, in which the $i$-th row contains the array $a$ cyclically shifted to the right by $(i - 1)$. For example, for the array $a = [3, 4, 5]$, the obtained matrix is

$$b = \begin{bmatrix} 3 & 4 & 5 \\\ 5 & 3 & 4 \\\ 4 & 5 & 3 \end{bmatrix}$$

Let's construct the following graph:

  * The graph contains $n^2$ vertices, each of which corresponds to one of the elements of the matrix. Let's denote the vertex corresponding to the element $b_{i, j}$ as $(i, j)$.   * We will draw an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ if $|i_1 - i_2| + |j_1 - j_2| \le k$ and $\gcd(b_{i_1, j_1}, b_{i_2, j_2}) > 1$, where $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers $x$ and $y$.

Your task is to calculate the number of connected components$^{\dagger}$ in the obtained graph.

$^{\dagger}$A connected component of a graph is a set of vertices in which any vertex is reachable from any other via edges, and adding any other vertex to the set violates this rule.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $n$ and $k$ ($2 \le n \le 10^6$, $2 \le k \le 2 \cdot 10^6$) — the length of the array and the parameter $k$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array $a$.

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

Output: For each test case, output a single integer — the number of connected components in the obtained graph.

Examples:
input:
6
3 3
3 4 5
3 3
3 4 9
3 2
3 4 9
2 2
2 8
5 3
8 27 5 4 3
4 10
2 2 2 2
output:
3
2
3
1
4
1

Note: In the first test case, the matrix $b$ is given in the statement. The first connected component contains the vertices $(1, 1)$, $(2, 2)$, and $(3, 3)$. The second connected component contains the vertices $(1, 2)$, $(2, 3)$, and $(3, 1)$. The third connected component contains the vertices $(1, 3)$, $(2, 1)$, and $(3, 2)$. Thus, the graph has $3$ connected components.

In the second test case, the following matrix is obtained:

$$b = \begin{bmatrix} 3 & 4 & 9 \\\ 9 & 3 & 4 \\\ 4 & 9 & 3 \end{bmatrix}$$

The first connected component contains all vertices corresponding to elements with values $3$ and $9$. The second connected component contains all vertices corresponding to elements with the value $4$.

In the fourth test case, all vertices are in one connected component.

Buggy Code:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.Set;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.TreeMap;
import java.io.Writer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Code generated by CHelperX
 *
 * @author Nipuna Samarasekara
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        FastPrinter out = new FastPrinter(outputStream);
        FLargeGraph solver = new FLargeGraph();
        solver.solve(1, in, out);
        out.close();
    }

    static class FLargeGraph {
        public void solve(int testNumber, FastScanner in, FastPrinter out) {
            int t = in.nextInt();
            int nlim = 1000009;
            int[] div = IntegerUtils.generateDivisorTable(nlim);

            while (t-- > 0) {
                int n = in.nextInt();
                int k = in.nextInt();
                int[] a = in.readIntArray(n);
                int nnn = 2 * n - 1;
                int[] aa = new int[nnn];

                int ans = 0;

                for (int i = 0; i < n - 1; i++) {
                    aa[i] = a[i + 1];

                    if (a[i + 1] == 1) {
                        ans += i;
                    }
                }

                for (int i = 0; i < n; i++) {
                    aa[n - 1 + i] = a[i];

                    if (a[i] == 1) {
                        int cc = n - 1 - i;
                        ans += cc;
                    }
                }

                int no = n;
                n = nnn;
                a = aa;

                TreeMap<Integer, TreeSet<Integer>> tm = new TreeMap<>();
                for (int i = 0; i < no; i++) {
                    int nn = a[i];
                    while (nn > 1) {
                        int dd = div[nn];
                        while (nn % dd == 0)
                            nn /= dd;
                        if (!tm.containsKey(dd)) {
                            tm.put(dd, new TreeSet<>());
                        }
                        tm.get(dd).add(i);
                        if (i + no < n)
                            tm.get(dd).add(i + no);

                    }
                }

                DisjointSetUnion dsu = new DisjointSetUnion(n + 10);
                for (TreeSet<Integer> pos : tm.values()) {
                    int ff = pos.first();
                    int ll = pos.last();
//                if ( n + ff - ll<= k)
//                    dsu.union(ff, ll);
                    int prev = ff;
                    for (Integer po : pos) {
                        if (po - prev <= k) {
                            dsu.union(po, prev);
                        }
                        prev = po;
                    }
                }

                tm = new TreeMap<>();
                for (int i = 0; i < n; i++) {
                    int par = dsu.get(i);
                    if (!tm.containsKey(par)) {
                        tm.put(par, new TreeSet<>());
                    }
                    tm.get(par).add(i);
                }

                ans += tm.keySet().size();

//            long ans = 0;
//            for (TreeSet<Integer> pos : tm.values()) {
//                if (pos.size() == 1){
//                    ans += n;
//                    continue;
//                }
//                int ff = pos.first();
//                int ll = pos.last();
//                int gg = n+ff-ll;
//                int prev = ff;
//                for (Integer po : pos) {
//                    gg = MathUtils.gcd(gg, po - prev);
//                    prev = po;
//                }
//
//                ans += gg;
//            }
//
                out.println(ans);

            }
        }

    }

    static class DisjointSetUnion {
        int[] p;
        int[] count;

        public DisjointSetUnion(int n) {
            p = new int[n];
            count = new int[n];
            for (int i = 0; i < n; i++) {
                count[i] = 1;
            }
            clear();
        }

        public void clear() {
            for (int i = 0; i < p.length; i++) {
                p[i] = i;
            }
        }

        public int get(int x) {
            return x != p[x] ? p[x] = get(p[x]) : x;
        }

        public boolean union(int a, int b) {
            a = get(a);
            b = get(b);
            p[a] = b;
            if (a != b) {
                count[b] += count[a];
                count[a] = 0;
            }
            return a != b;
        }

    }

    static class IntegerUtils {
        public static int[] generateDivisorTable(int upTo) {
            int[] divisor = new int[upTo];
            for (int i = 1; i < upTo; i++)
                divisor[i] = i;
            for (int i = 2; i * i < upTo; i++) {
                if (divisor[i] == i) {
                    for (int j = i * i; j < upTo; j += i)
                        divisor[j] = i;
                }
            }
            return divisor;
        }

    }

    static class FastScanner extends BufferedReader {
        public FastScanner(InputStream is) {
            super(new InputStreamReader(is));
        }

        public int read() {
            try {
                int ret = super.read();
//            if (isEOF && ret < 0) {
//                throw new InputMismatchException();
//            }
//            isEOF = ret == -1;
                return ret;
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }

        static boolean isWhiteSpace(int c) {
            return c >= 0 && c <= 32;
        }

        public int nextInt() {
            int c = read();
            while (isWhiteSpace(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int ret = 0;
            while (c >= 0 && !isWhiteSpace(c)) {
                if (c < '0' || c > '9') {
                    throw new NumberFormatException("digit expected " + (char) c
                            + " found");
                }
                ret = ret * 10 + c - '0';
                c = read();
            }
            return ret * sgn;
        }

        public String readLine() {
            try {
                return super.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        public int[] readIntArray(int n) {
            int[] ret = new int[n];
            for (int i = 0; i < n; i++) {
                ret[i] = nextInt();
            }
            return ret;
        }

    }

    static class FastPrinter extends PrintWriter {
        public FastPrinter(OutputStream out) {
            super(out);
        }

        public FastPrinter(Writer out) {
            super(out);
        }

        public void close() {
            super.close();
        }

    }
}

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

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

Think step by step.

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

1. **Edge Cases for `k` and `n`:**
   - Test with `k` and `n` values that are at the boundary of the allowed range.
   - Test with `k` values that are just above the typical small values to ensure the cyclic shifting and edge connections are handled correctly.

2. **Special Values in Array `a`:**
   - Include values that have common divisors, especially small primes, to test the GCD condition.
   - Include values that are 1, as these should not form any connections.

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

4. **Small-Scale Test Input:**
   - Keep the test input small to ensure it fits within the output length limit.

Here is a small-scale test input that meets these criteria:

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

### Explanation:
1. **Test Case 1:**
   - `n = 3`, `k = 2`
   - Array: `2, 3, 4`
   - This case tests the cyclic shifting and the GCD condition with small values.

2. **Test Case 2:**
   - `n = 3`, `k = 3`
   - Array: `1, 1, 1`
   - This case tests the scenario where all elements are 1, which should not form any connections.

3. **Test Case 3:**
   - `n = 4`, `k = 2`
   - Array: `2, 2, 3, 3`
   - This case tests the scenario with repeated values and different prime numbers.

4. **Test Case 4:**
   - `n = 5`, `k = 3`
   - Array: `2, 3, 4, 5, 6`
   - This case tests a larger array with a mix of values and a moderate `k` value.

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