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 Solution
  {
      static class FastReader 
      { 
          BufferedReader br; 
          StringTokenizer st; 
    
          public FastReader() 
          { 
              br = new BufferedReader(new
                       InputStreamReader(System.in)); 
          } 
    
          String next() 
          { 
              while (st == null || !st.hasMoreElements()) 
              { 
                  try
                  { 
                      st = new StringTokenizer(br.readLine()); 
                  } 
                  catch (IOException  e) 
                  { 
                      e.printStackTrace(); 
                  } 
              } 
              return st.nextToken(); 
          } 
    
          int nextInt() 
          { 
              return Integer.parseInt(next()); 
          } 
    
          long nextLong() 
          { 
              return Long.parseLong(next()); 
          } 
    
          double nextDouble() 
          { 
              return Double.parseDouble(next()); 
          } 
    
          String nextLine() 
          { 
              String str = ""; 
              try
              { 
                  str = br.readLine(); 
              } 
              catch (IOException e) 
              { 
                  e.printStackTrace(); 
              } 
              return str; 
          } 
      }
      //static final long mod=(long)1e9+7;
      static final long mod=998244353L;
      public static long pow(long a,long p)
      {
          long res=1;
          while(p>0)
          {
              if(p%2==1)
              {
                  p--;
                  res*=a;
                  res%=mod;
              }
              else
              {
                  a*=a;
                  a%=mod;
                  p/=2;
              }
          }
          return res;
      }
      static class Pair
      {
          int u,v;
          Pair(int u,int v)
          {
              this.u=u;
              this.v=v;
          }
      }
      /*static class Pair implements Comparable<Pair>
      {
          int v,l;
          Pair(int v,int l)
          {
              this.v=v;
              this.l=l;
          }
          public int compareTo(Pair p)
          {
              return l-p.l; 
          }
      }*/
      static long gcd(long a,long b)
      {
          if(b%a==0)
          return a;
          return gcd(b%a,a);
      }
      public static class comp implements Comparator<Pair>
      {
          public int compare(Pair p1,Pair p2)
          {
              if(p1.v>p2.v)
              return 1;
              return -1;
          }
      }
      public static void upd1(int a[],int i)
      {
          for(;i<a.length;i+=(i&(-i)))
          a[i]+=1;
      }
      public static void upd2(long a[],int up)
      {
          for(int i=up;i<a.length;i+=(i&(-i)))
          a[i]=a[i]+1L*up;
      }
      public static int q1(int a[],int i)
      {
          int s=0;
          for(;i>0;i-=(i&(-i)))
          s+=a[i];
          return s;
      }
      public static long q2(long a[],int i)
      {
          long s=0;
          for(;i>0;i-=(i&(-i)))
          s+=a[i];
          return s;
      }
      public static void main(String args[])throws Exception
      {
          FastReader fs=new FastReader();
          PrintWriter pw=new PrintWriter(System.out);
          int tc=fs.nextInt();
          while(tc-->0)
          {
              int n=fs.nextInt();
              int m=fs.nextInt();
              int a[]=new int[n];
              ArrayList<Integer> ans=new ArrayList<>();
              for(int i=0;i<n;i++)
              a[i]=fs.nextInt()-1;
              int cnt[]=new int[n];
              for(int i=0;i<n;i++)
              cnt[(i-a[i]+n)%n]++;
              for(int i=0;i<n;i++)
              {
                  if(cnt[i]+2*m>=n)
                  {
                      int ta[]=new int[n];
                      int in=-1;
                      for(int j=n-i;j<n;j++)
                      ta[++in]=j;
                      for(int j=0;j<n-i;j++)
                      ta[++in]=j;
                      boolean vis[]=new boolean[n];
                      int cc=0;
                      int edge[]=new int[n+1];
                      for(int j=0;j<n;j++)
                      edge[a[j]]=ta[j];
                      for(int j=0;j<n;j++)
                      {
                          if(vis[j])continue;
                          cc++;
                          int k=j;
                          while(!vis[k])
                          {
                              vis[k]=true;
                              k=edge[k];
                          }
                      }
                      if(n-cc<=m)
                      ans.add(i);
                  }
              }
              pw.print(ans.size()+" ");
              for(int i:ans)
              pw.print(i+" ");
              pw.println();
          }
          pw.flush();
          pw.close();
      }
  }