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: //make sure to make new file!
  import java.io.*;
  import java.util.*;
  
  public class EHS{
     
     public static void main(String[] args)throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        
        int t = Integer.parseInt(f.readLine());
        
        for(int q = 1; q <= t; q++){
        
           StringTokenizer st = new StringTokenizer(f.readLine());
        
           int n = Integer.parseInt(st.nextToken());
           int m = Integer.parseInt(st.nextToken());
           
           st = new StringTokenizer(f.readLine());
           
           int[] array = new int[n+1];
           for(int k = 1; k <= n; k++){
              array[k] = Integer.parseInt(st.nextToken());
           }
           
           int[] rotsolves = new int[n];
           
           for(int k = 1; k <= n; k++){
              int i = k-array[k];
              if(i < 0) i += n;
              rotsolves[i]++;
           }
           
           ArrayList<Integer> answer = new ArrayList<Integer>();
           
           int thresh = n-m*2;
           for(int k = 0; k < n; k++){
              if(rotsolves[k] >= thresh){
                 //see if it works in O(N), will be done ~3 times
                 
                 //rotate array by k and count the number of swaps
                 int[] rot = new int[n+1];
                 for(int j = 1; j <= n; j++){
                    int i = j+k;
                    if(i > n) i-=n;
                    rot[j] = array[i];
                 }
                 
                 HashSet<Integer> seen = new HashSet<Integer>();
                 
                 int swaps = 0;
                 for(int j = 1; j <= n; j++){
                    if(seen.contains(j)) continue;
                    seen.add(j);
                    
                    int i = j;
                    while(!seen.contains(rot[i])){
                       swaps++;
                       seen.add(rot[i]);
                       i = rot[i];
                    }
                 }
                 
                 if(swaps <= m){
                    answer.add(k);
                 }
                    
              }
           }
           
           StringJoiner sj = new StringJoiner(" ");
           sj.add("" + answer.size());
           for(int i : answer){
              sj.add("" + i);
           }
           out.println(sj.toString());
        
  
        }
        
        
        
        
        out.close();
     }
     
        
  }