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.lang.*;
  import java.io.*;
  
  public class Main
  {
  	PrintWriter out = new PrintWriter(System.out);
  	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer tok = new StringTokenizer("");
      String next() throws IOException {
          if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
          return tok.nextToken();
      }
      int ni() throws IOException { return Integer.parseInt(next()); }
      long nl() throws IOException { return Long.parseLong(next()); }
      
      long mod=1000000007;
      int[]A,C,P;
      int n,m;
      
      void solve() throws IOException {
          for (int tc=ni();tc>0;tc--) {
              n=ni();
              m=ni();
              C=new int[n];
              P=new int[n+1];
              A=new int[n+1];
              for (int i=1;i<=n;i++) {
                  int u=ni();
                  A[i]=u;
                  P[u]=i;
                  int p=(i-u+n)%n;
                  C[p]++;
              }
              
              ArrayList<Integer>Z=new ArrayList();
              for (int i=0;i<n;i++) {
                  if (n-C[i]>m*2) continue;
                  if (test(i)<=m) Z.add(i);
              }
              
              out.print(Z.size()+" ");
              for (int i:Z) out.print(i+" ");
              out.println();
          }
          out.flush();
      }
      
      int test(int p) {
          int[]B=new int[n+1];
          int ct=0;
          for (int i=1;i<=n;i++) {
              B[i]=i-p;
              if (B[i]<=0) B[i]+=n;
          }
          
          for (int i=1;i<=n;i++) {
              if (B[i]==A[i]) continue;
              int u=P[B[i]];
              int d=B[i];
              while (u!=i) {
                  int v=u;
                  int e=d;
                  u=P[B[v]];
                  d=B[v];
                  B[v]=e;
                  ct++;
              }
          }
          
          return ct;
      }
      
      int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); }
      long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); }
      long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; }
      
      public static void main(String[] args) throws IOException {
          new Main().solve();
      }
  }