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.util.*;
  
  import static java.util.Arrays.fill;
  
  public class Current {
      FastScanner in;
      PrintWriter out;
  
      private void solve() throws IOException {
          solveE();
      }
  
      private void solveE() throws IOException {
          int n = in.nextInt(), m = in.nextInt();
          int[] p = new int[n];
          for (int i = 0; i < n; i++)
              p[i] = in.nextInt();
  
          int[] b = new int[n * 2];
          System.arraycopy(p, 0, b, 0, n);
          System.arraycopy(p, 0, b, n, n);
          for (int i = 0; i < b.length; i++)
              b[i] -= i + 1;
  
          // out.println(Arrays.toString(b));
  
          boolean[] used = new boolean[n * 2];
          TreeMap<Integer, Integer> map = new TreeMap<>();
          ArrayList<Integer> ans = new ArrayList<>();
  
          for (int k = 0; k < n; k++) {
              if (k == 0) {
                  for (int i = 0; i < n; i++)
                      map.put(b[i], map.getOrDefault(b[i], 0) + 1);
              } else {
                  if (map.get(b[k - 1]) == 1)
                      map.remove(b[k - 1]);
                  else
                      map.put(b[k - 1], map.get(b[k - 1]) - 1);
                  map.put(b[k + n - 1], map.getOrDefault(b[k + n - 1], 0) + 1);
              }
  
              if (map.getOrDefault(-k, 0) >= n - m * 2) { // no more than 3 times
                  /*out.print(-k + (k == 0 ? " " : "") + " :");
                  for (int i = k; i < k + n; i++) {
                      out.print("\t" + (b[i] + k));
                  }
                  out.println();*/
  
                  fill(used, k, k + n, false);
                  int cnt = 0;
                  for (int i = k; i < k + n; i++) {
                      if (!used[i]) {
                          int cur = i;
                          while (!used[cur]) {
                              used[cur] = true;
                              cnt++;
                              cur += b[cur] + k;
                          }
                          cnt--;
                      }
                  }
  
                  if (cnt <= m)
                      ans.add(k);
              }
          }
  
          out.print(ans.size());
          for (int k : ans)
              out.print(" " + k);
          out.println();
      }
  
      private void swap(int[] a, int i, int j) {
          int swap = a[--i];
          a[i] = a[--j];
          a[j] = swap;
      }
  
      class FastScanner {
          StringTokenizer st;
          BufferedReader br;
  
          FastScanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          String next() throws IOException {
              while (st == null || !st.hasMoreTokens())
                  st = new StringTokenizer(br.readLine());
              return st.nextToken();
          }
  
          int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
  
          long nextLong() throws IOException {
              return Long.parseLong(next());
          }
      }
  
      private void run() throws IOException {
          in = new FastScanner(System.in);
          out = new PrintWriter(System.out);
  
          for (int t = in.nextInt(); t-- > 0; )
              solve();
  
          out.flush();
          out.close();
      }
  
      public static void main(String[] args) throws IOException {
          new Current().run();
      }
  }
  