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 E {
      static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
      static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
      private static int nextInt() {
          try {
              in.nextToken();
          } catch (IOException e) {
              e.printStackTrace();
          }
          return (int) in.nval;
      }
      private static long nextLong() {
          try {
              in.nextToken();
          } catch (IOException e) {
              e.printStackTrace();
          }
          return (long) in.nval;
      }
      private static String next(){
          try {
              in.nextToken();
          } catch (IOException e) {
              e.printStackTrace();
          }
          return in.sval;
      }
  
      public static void main(String[] args) {
          int t = nextInt();
          for (int c = 0; c < t; c++) {
              int n = nextInt();
              int m = nextInt();
              int[] nums = new int[n+1];
              for (int i = 1; i <= n; i++) {
                  nums[i] = nextInt();
              }
              int[] state = new int[n];
              for (int i = 1; i <= n; i++) {
                  state[(n-nums[i]+i)%n]++;
              }
              arr = new int[n+1];
              //System.out.println(Arrays.toString(state));
              List<Integer> candidate = new ArrayList<>();
              for (int i = 0; i < n; i++) {
                  int dif = n - state[i];
                  if(((dif+1)/2)<=m)candidate.add(i);
              }
              List<Integer> ans = new ArrayList<>();
              for (Integer an : candidate) {
                  if(check(an,nums,m,n))ans.add(an);
              }
              System.out.print(ans.size());
              for (Integer an : ans) {
                  System.out.print(" "+an);
              }
              System.out.println();
          }
      }
      static int[] arr;
      public static boolean check(int bias,int[] target,int m,int n){
          for (int i = 1; i <= n; i++) {
              int t = i + bias;
              if(t>n)t-=n;
              arr[t] = i;
          }
          return can(m,arr,target,n);
      }
      public static boolean can(int m,int[] arr,int[] target,int n){
          Map<Integer,Integer> map = new HashMap<>();
          for (int i = 1; i <= n; i++) {
              map.put(arr[i],i);
          }
          for (int i = 1; i <= n; i++) {
              if(arr[i]!=target[i]){
                  m--;
                  int y = map.get(target[i]);
                  swap(arr,i,y);
                  map.put(arr[y],y);
              }
          }
          return m>=0;
      }
  
      public static void swap(int[] a,int x,int y){
          int t = a[x];
          a[x] = a[y];
          a[y] = t;
      }
  }
  