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.util.*;
  import java.io.*;
  
  public class Harbour2022 {
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  ////////                                                                                                               /////////
  ////////                                                                                                               /////////
  ////////   HHHH        HHHH  EEEEEEEEEEEEE   MMMM          MMMM         OOOOOO             SSSSSSS      EEEEEEEEEEEEE  /////////
  ////////   HHHH        HHHH  EEEEEEEEEEEEE   MMMMMM      MMMMMM      OOO      OOO        SSSS   SSS     EEEEEEEEEEEEE  /////////
  ////////   HHHH        HHHH  EEEEE           MMMM MMM  MMM MMMM    OOO          OOO    SSSS       SSS   EEEEE          /////////
  ////////   HHHH        HHHH  EEEEE           MMMM  MMMMMM  MMMM   OOO            OOO   SSSS             EEEEE          /////////
  ////////   HHHH        HHHH  EEEEE           MMMM          MMMM  OOO              OOO   SSSSSSS         EEEEE          /////////
  ////////   HHHHHHHHHHHHHHHH  EEEEEEEEEEE     MMMM          MMMM  OOO              OOO      SSSSSS       EEEEEEEEEEE    /////////
  ////////   HHHHHHHHHHHHHHHH  EEEEEEEEEEE     MMMM          MMMM  OOO              OOO         SSSSSSS   EEEEEEEEEEE    /////////
  ////////   HHHH        HHHH  EEEEE           MMMM          MMMM   OOO            OOO              SSSS  EEEEE          /////////
  ////////   HHHH        HHHH  EEEEE           MMMM          MMMM    OOO          OOO     SSS       SSSS  EEEEE          /////////
  ////////   HHHH        HHHH  EEEEEEEEEEEEE   MMMM          MMMM      OOO      OOO        SSS    SSSS    EEEEEEEEEEEEE  /////////
  ////////   HHHH        HHHH  EEEEEEEEEEEEE   MMMM          MMMM         OOOOOO             SSSSSSS      EEEEEEEEEEEEE  /////////
  ////////                                                                                                               /////////
  ////////                                                                                                               /////////
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  ////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
      static boolean[] vis;
  
      static int dfs(int idx, int[] edgs) {
          vis[idx] = true;
  
          int ans = 1;
          if (!vis[edgs[idx]])
              ans += dfs(edgs[idx], edgs);
  
          return ans;
      }
  
      static int solve(int[] a) {
          int n = a.length;
          vis = new boolean[n];
          int ans = 0;
          for (int i = 0; i < n; i++) {
              if (!vis[i])
                  ans += dfs(i, a) - 1;
          }
  
          return ans;
      }
  
      static int[] shift(int[] a, int v) {
          int[] res = new int[a.length];
  
          for (int i = 0; i < a.length; i++) {
              res[i] = a[(i + v) % a.length];
          }
          return res;
      }
  
      public static void main(String[] args) throws IOException {
          Scanner sc = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
  
  
          int tc = sc.nextInt();
          while (tc-- > 0) {
              int n = sc.nextInt();
              int m = sc.nextInt();
              int[] a = sc.nextIntArr(n);
              for (int i = 0; i < n; i++)
                  a[i]--;
              int[] count = new int[n + 1];
              for (int i = 0; i < n; i++) {
                  if (a[i] <= i) {
                      count[i - a[i]]++;
                  } else {
                      count[n - a[i] + i]++;
                  }
              }
              ArrayList<Integer> ans = new ArrayList<>();
  //            System.err.println(Arrays.toString(count));
              for (int i = 0; i < n; i++) {
                  if (count[i] * 3 >= n) {
                      if (solve(shift(a, i)) <= m)
                          ans.add(i);
                  }
              }
              pw.print(ans.size());
              for (int x : ans)
                  pw.print(" " + x);
              pw.println();
          }
          pw.flush();
      }
  
      static class Scanner {
          StringTokenizer st;
          BufferedReader br;
  
          public Scanner(FileReader r) {
              br = new BufferedReader(r);
          }
  
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          public String next() throws IOException {
              while (st == null || !st.hasMoreTokens())
                  st = new StringTokenizer(br.readLine());
              return st.nextToken();
          }
  
          public int[] nextIntArr(int n) throws IOException {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) {
                  a[i] = nextInt();
              }
              return a;
          }
  
          public int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
  
          public long nextLong() throws IOException {
              return Long.parseLong(next());
          }
  
          public String nextLine() throws IOException {
              return br.readLine();
          }
  
          public double nextDouble() throws IOException {
              String x = next();
              StringBuilder sb = new StringBuilder("0");
              double res = 0, f = 1;
              boolean dec = false, neg = false;
              int start = 0;
              if (x.charAt(0) == '-') {
                  neg = true;
                  start++;
              }
              for (int i = start; i < x.length(); i++)
                  if (x.charAt(i) == '.') {
                      res = Long.parseLong(sb.toString());
                      sb = new StringBuilder("0");
                      dec = true;
                  } else {
                      sb.append(x.charAt(i));
                      if (dec)
                          f *= 10;
                  }
              res += Long.parseLong(sb.toString()) / f;
              return res * (neg ? -1 : 1);
          }
  
          public boolean ready() throws IOException {
              return br.ready();
          }
      }
  
  }