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.*;
  
  public class Main{
      public static void main(String[] args) {
          FastReader s = new FastReader();
          Main main = new Main();
          int t = s.nextInt();
          while(t > 0) {
              int n = s.nextInt();
              int m = s.nextInt();
              int[] arr = new int[n];
              for(int i=0; i<n; i++) {
                  arr[i] = s.nextInt();
              }
              int[] res = main.check(arr, m);
              System.out.print(res.length);
  
              for(int i=0; i<res.length; i++) {
                  System.out.print(" ");
                  System.out.print(res[i]);
              }
  
              System.out.println(" ");
  
              t--;
          }
      }
  
      private int[] check(int[] check, int m) {
          int n = check.length;
  
          List<Integer> ans = new ArrayList<>();
  
          int[] cnt = new int[n];
          for(int i=0; i<n; i++) {
              int offset = i+1-check[i];
              if(offset < 0) offset += n;
              cnt[offset]++;
          }
  
          for(int i=0; i<n; i++) {
              if(cnt[i] + 2*m >= n) {
                  if(n - getCycle(check, n-i) <= m) ans.add(i);
              }
          }
  
          int[] res = new int[ans.size()];
          for(int i=0; i<res.length; i++) {
              res[i] = ans.get(i);
          }
          return res;
      }
  
      private int getCycle(int[] check, int s)
      {
          int n = check.length;
          boolean[] used = new boolean[n];
          int ans = 0;
          for(int i = 0; i < n; i++)
          {
              if(used[i]) continue;
              int j = i;
              while(!used[j])
              {
                  used[j] = true;
                  int sj = j-s;
                  if(sj < 0) sj+=n;
                  j = check[sj]-1;
              }
              ans++;
          }
          return ans;
      }
  }
  
  class FastReader
  {
      BufferedReader br;
      StringTokenizer st;
      InputStreamReader input;//no buffer
      public FastReader()
      {
          br=new BufferedReader(new InputStreamReader(System.in));
      }
  
      public FastReader(boolean isBuffer)
      {
          if(!isBuffer){
              input=new InputStreamReader(System.in);
          }else{
              br=new BufferedReader(new InputStreamReader(System.in));
          }
      }
  
      public boolean hasNext(){
          try{
              String s=br.readLine();
              if(s==null){
                  return  false;
              }
              st=new StringTokenizer(s);
          }catch(IOException e){
              e.printStackTrace();
          }
          return true;
      }
  
      public String next()
      {
          if(input!=null){
              try {
                  StringBuilder sb=new StringBuilder();
                  int ch=input.read();
                  while(ch=='\n'||ch=='\r'||ch==32){
                      ch=input.read();
                  }
                  while(ch!='\n'&&ch!='\r'&&ch!=32){
                      sb.append((char)ch);
                      ch=input.read();
                  }
                  return sb.toString();
              }catch (Exception e){
                  e.printStackTrace();
              }
          }
          while(st==null || !st.hasMoreElements())//回车，空行情况
          {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      public int nextInt()
      {
          return (int)nextLong();
      }
  
      public long nextLong() {
          try {
              if(input!=null){
                  long ret=0;
                  int b=input.read();
                  while(b<'0'||b>'9'){
                      b=input.read();
                  }
                  while(b>='0'&&b<='9'){
                      ret=ret*10+(b-'0');
                      b=input.read();
                  }
                  return ret;
              }
          }catch (Exception e){
              e.printStackTrace();
          }
          return Long.parseLong(next());
      }
  
      public double nextDouble()
      {
          return Double.parseDouble(next());
      }
  }