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: /*
  stream Butter!
  eggyHide eggyVengeance
  I need U
  xiao rerun when
   */
  import static java.lang.Math.*;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  
  public class x1553E2
  {
      public static void main(String hi[]) throws Exception
      {
          BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer(infile.readLine());
          int T = Integer.parseInt(st.nextToken());
          StringBuilder sb = new StringBuilder();
          while(T-->0)
          {
              st = new StringTokenizer(infile.readLine());
              N = Integer.parseInt(st.nextToken());
              int M = Integer.parseInt(st.nextToken());
              int[] arr = readArr(N, infile, st);
              for(int i=0; i < N; i++)
                  arr[i]--;
              //small N
              if(N <= 6)
              {
                  ArrayList<Integer> ls = new ArrayList<Integer>();
                  for(int k=0; k < N; k++)
                  {
                      int[] temp = new int[N];
                      for(int i=0; i < N; i++)
                          temp[i] = arr[(i+k)%N];
                      int cnt = N;
                      boolean[] seen = new boolean[N];
                      for(int v=0; v < N; v++)
                          if(!seen[v])
                          {
                              int curr = v;
                              while(!seen[curr])
                              {
                                  seen[curr] = true;
                                  curr = temp[curr];
                              }
                              cnt--;
                          }
                      if(cnt <= M)
                          ls.add(k);
                  }
                  sb.append(ls.size()+" ");
                  for(int x: ls)
                      sb.append(x+" ");
              }
              else
              {
                  //bruh
                  int[] cnt = new int[N];
                  for(int i=0; i < N; i++)
                  {
                      int k = arr[i]-i;
                      if(k < 0)
                          k += N;
                      k = (N-k)%N;
                      cnt[k]++;
                  }
                  ArrayList<Integer> candidates = new ArrayList<Integer>();
                  for(int k=0; k < N; k++)
                      if(cnt[k] >= N-2*M)
                          candidates.add(k);
                  ArrayList<Integer> ls = new ArrayList<Integer>();
                  for(int k: candidates)
                  {
                      int[] temp = new int[N];
                      for(int i=0; i < N; i++)
                          temp[i] = arr[(i+k)%N];
                      int cntS = N;
                      boolean[] seen = new boolean[N];
                      for(int v=0; v < N; v++)
                          if(!seen[v])
                          {
                              int curr = v;
                              while(!seen[curr])
                              {
                                  seen[curr] = true;
                                  curr = temp[curr];
                              }
                              cntS--;
                          }
                      if(cntS <= M)
                          ls.add(k);
                  }
                  sb.append(ls.size()+" ");
                  for(int x: ls)
                      sb.append(x+" ");
              }
              sb.append("\n");
          }
          System.out.print(sb);
      }
      static int N;
      public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
      {
          int[] arr = new int[N];
          st = new StringTokenizer(infile.readLine());
          for(int i=0; i < N; i++)
              arr[i] = Integer.parseInt(st.nextToken());
          return arr;
      }
  }
  /*
  1 0 4 3 5 2
  5 0 1 2 3 4
  
  1 0 4 3 5 2
  4 5 0 1 2 3
  
  something something cycles?
  BRUH
   */