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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.io.PrintStream;
  import java.util.BitSet;
  import java.util.ArrayList;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   *
   * @author sarthakmanna
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          FastReader in = new FastReader(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          EPermutationShift solver = new EPermutationShift();
          int testCount = Integer.parseInt(in.next());
          for (int i = 1; i <= testCount; i++)
              solver.solve(i, in, out);
          out.close();
      }
  
      static class EPermutationShift {
          public void solve(int testNumber, FastReader in, PrintWriter out) {
              int i, j, k;
  
              int N = in.nextInt(), M = in.nextInt();
              int[] P = in.getIntArray(N);
  
              ArrayList<Integer>[] correctIndexes = new ArrayList[N];
              for (i = 0; i < N; ++i) correctIndexes[i] = new ArrayList<>();
  
              for (i = 0; i < N; ++i) {
                  int diff = i + 1 - P[i];
                  if (diff < 0) diff += N;
                  correctIndexes[diff].add(i);
              }
  
              ArrayList<Integer> ans = new ArrayList<>();
              for (i = 0; i < N; ++i) {
                  if (M >= N - correctIndexes[i].size() - 1) {
                      ans.add(i);
                  } else if (check(i, P, M, N - correctIndexes[i].size())) {
                      ans.add(i);
                  }
              }
  
              out.print(ans.size() + " ");
              for (int itr : ans) out.print(itr + " ");
              out.println();
          }
  
          boolean check(int shift, int[] ar, int swaps, int displaced) {
              BitSet fixed = new BitSet();
              for (int idx = 0; idx < ar.length && swaps >= (displaced + 1 >> 1); ++idx) {
                  if (!fixed.get(idx) && ar[idx] - 1 != (ar.length - shift + idx) % ar.length) {
                      int p = idx;
                      while (!fixed.get(idx)) {
                          p = (ar[p] - 1 + shift) % ar.length;
                          fixed.set(p);
                          --displaced;
                          --swaps;
                      }
                      ++swaps;
                  }
              }
              return swaps >= (displaced + 1 >> 1);
          }
  
      }
  
      static class FastReader {
          static final int BUFSIZE = 1 << 20;
          static byte[] buf;
          static int index;
          static int total;
          static InputStream in;
  
          public FastReader(InputStream is) {
              try {
                  in = is;
                  buf = new byte[BUFSIZE];
              } catch (Exception e) {
              }
          }
  
          private int scan() {
              try {
                  if (index >= total) {
                      index = 0;
                      total = in.read(buf);
                      if (total <= 0)
                          return -1;
                  }
                  return buf[index++];
              } catch (Exception | Error e) {
                  System.err.println(e.getMessage());
                  return 13 / 0;
              }
          }
  
          public String next() {
              int c;
              for (c = scan(); c <= 32; c = scan()) ;
              StringBuilder sb = new StringBuilder();
              for (; c > 32; c = scan())
                  sb.append((char) c);
              return sb.toString();
          }
  
          public int nextInt() {
              int c, val = 0;
              for (c = scan(); c <= 32; c = scan()) ;
              boolean neg = c == '-';
              if (c == '-' || c == '+')
                  c = scan();
              for (; c >= '0' && c <= '9'; c = scan())
                  val = (val << 3) + (val << 1) + (c & 15);
              return neg ? -val : val;
          }
  
          public int[] getIntArray(int size) {
              int[] ar = new int[size];
              for (int i = 0; i < size; ++i) ar[i] = nextInt();
              return ar;
          }
  
      }
  }
  
  