description: An identity permutation of length n is an array [1, 2, 3, ..., n].
  
  We performed the following operations to an identity permutation of length n:
  
    * firstly, we cyclically shifted it to the right by k positions, where k is unknown to you (the only thing you know is that 0 ≤ k ≤ n - 1). When an array is cyclically shifted to the right by k positions, the resulting array is formed by taking k last elements of the original array (without changing their relative order), and then appending n - k first elements to the right of them (without changing relative order of the first n - k elements as well). For example, if we cyclically shift the identity permutation of length 6 by 2 positions, we get the array [5, 6, 1, 2, 3, 4]; 
    * secondly, we performed the following operation at most m times: pick any two elements of the array and swap them. 
  
  
  
  You are given the values of n and m, and the resulting array. Your task is to find all possible values of k in the cyclic shift operation.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases.
  
  Each test case consists of two lines. The first line contains two integers n and m (3 ≤ n ≤ 3 ⋅ 10^5; 0 ≤ m ≤ n/3).
  
  The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, each integer from 1 to n appears in this sequence exactly once) — the resulting array.
  
  The sum of n over all test cases does not exceed 3 ⋅ 10^5.
  
  Output
  
  For each test case, print the answer in the following way:
  
    * firstly, print one integer r (0 ≤ r ≤ n) — the number of possible values of k for the cyclic shift operation; 
    * secondly, print r integers k_1, k_2, ..., k_r (0 ≤ k_i ≤ n - 1) — all possible values of k in increasing order. 
  
  Example
  
  Input
  
  
  4
  4 1
  2 3 1 4
  3 1
  1 2 3
  3 1
  3 2 1
  6 0
  1 2 3 4 6 5
  
  
  Output
  
  
  1 3
  1 0
  3 0 1 2
  0
  
  Note
  
  Consider the example: 
  
    * in the first test case, the only possible value for the cyclic shift is 3. If we shift [1, 2, 3, 4] by 3 positions, we get [2, 3, 4, 1]. Then we can swap the 3-rd and the 4-th elements to get the array [2, 3, 1, 4]; 
    * in the second test case, the only possible value for the cyclic shift is 0. If we shift [1, 2, 3] by 0 positions, we get [1, 2, 3]. Then we don't change the array at all (we stated that we made at most 1 swap), so the resulting array stays [1, 2, 3]; 
    * in the third test case, all values from 0 to 2 are possible for the cyclic shift: 
      * if we shift [1, 2, 3] by 0 positions, we get [1, 2, 3]. Then we can swap the 1-st and the 3-rd elements to get [3, 2, 1]; 
      * if we shift [1, 2, 3] by 1 position, we get [3, 1, 2]. Then we can swap the 2-nd and the 3-rd elements to get [3, 2, 1]; 
      * if we shift [1, 2, 3] by 2 positions, we get [2, 3, 1]. Then we can swap the 1-st and the 2-nd elements to get [3, 2, 1]; 
    * in the fourth test case, we stated that we didn't do any swaps after the cyclic shift, but no value of cyclic shift could produce the array [1, 2, 3, 4, 6, 5]. 
solution: 
  import java.io.*;
  import java.lang.*;
  import java.util.*;
  
  public final class CodeForces {
      static FastScanner scan = new FastScanner();
      static StringBuffer output = new StringBuffer();
  
      public static void main(String[] args) throws IOException {
          int t = 1;
          t = scan.nextInt();
          while (t-- > 0) {
              solve();
          }
          System.out.println(output);
      }
  
      static void solve() throws IOException {
          int n = scan.nextInt(), m = scan.nextInt();
          int[] groups = new int[n];
          List<Integer> ks = new ArrayList<>();
          int[] arr = new int[n];
          for(int i=0;i<n;i++) {
              arr[i] = scan.nextInt();
              int groupId = (i + 1 - arr[i] + n) % n;
              groups[groupId]++;
          }
          int ans = 0;
          for(int i=0;i<n;i++) {
              if(groups[i] >= n/3) {
                  int minSwaps = getMinSwaps(arr, i);
                  if(minSwaps <= m) {
                      ans++;
                      ks.add(i);
                  }
              }
          }
          output.append(ans).append(" ");
          for(int k: ks) {
              output.append(k).append(" ");
          }
          output.append("\n");
      }
  
      private static int getMinSwaps(int[] arr, int k) {
          int n = arr.length, minSwaps = n;
          boolean[] visited = new boolean[n];
          for(int i=0;i<n;i++) {
              if(!visited[i]) {
                  minSwaps--;
                  dfs(i, arr, k, visited);
              }
          }
          return minSwaps;
      }
  
      private static void dfs(int i, int[] arr, int k, boolean[] visited) {
          visited[i] = true;
          int child = (arr[i] + k - 1) % arr.length;
          if(!visited[child])
              dfs(child, arr, k, visited);
      }
  
      static class FastScanner {
          String line;
          String[] values;
          int position;
          BufferedReader bufferedReader;
  
          FastScanner() {
              bufferedReader = new BufferedReader(new InputStreamReader(System.in));
          }
  
          FastScanner(Boolean test) {
              InputStream inputStream = CodeForces.class.getResourceAsStream("test.txt");
              InputStreamReader inputStreamReader = new InputStreamReader(inputStream);
              bufferedReader = new BufferedReader(inputStreamReader);
          }
  
          void readLine() throws IOException {
              if (values == null || position == values.length) {
                  line = bufferedReader.readLine();
                  values = line.trim().split(" ");
                  position = 0;
              }
          }
  
          String next() throws IOException {
              readLine();
              return values[position++];
          }
  
          Integer nextInt() throws IOException {
              return Integer.parseInt(next());
          }
  
          Long nextLong() throws IOException {
              return Long.parseLong(next());
          }
      }
  }