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.util.StringTokenizer;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          FastScanner in = new FastScanner(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          TaskE solver = new TaskE();
          solver.solve(1, in, out);
          out.close();
      }
  
      static class TaskE {
          public void solve(int testNumber, FastScanner in, PrintWriter out) {
              int numTests = in.nextInt();
              for (int test = 0; test < numTests; test++) {
                  int n = in.nextInt();
                  int m = in.nextInt();
                  int[] a = new int[n];
                  int[] inPlace = new int[n];
                  for (int i = 0; i < n; i++) {
                      a[i] = in.nextInt() - 1;
                      ++inPlace[(i - a[i] + n) % n];
                  }
                  boolean[] good = new boolean[n];
                  int[] p = new int[n];
                  for (int k = 0; k < n; k++) {
                      if (inPlace[k] * 3 < n) {
                          continue;
                      }
                      for (int i = 0; i < n; i++) {
                          p[a[i]] = (i - k + n) % n;
                      }
                      good[k] = n - countCycles(p) <= m;
                  }
  
                  int numGood = 0;
                  for (int i = 0; i < n; i++) {
                      if (good[i]) {
                          ++numGood;
                      }
                  }
                  out.print(numGood);
                  for (int i = 0; i < n; i++) {
                      if (good[i]) {
                          out.print(" " + i);
                      }
                  }
                  out.println();
              }
          }
  
          private int countCycles(int[] a) {
              int n = a.length;
              boolean[] was = new boolean[n];
              int numCycles = 0;
              for (int i = 0; i < n; i++) {
                  if (was[i]) {
                      continue;
                  }
                  int len = 0;
                  for (int j = i; !was[j]; j = a[j]) {
                      was[j] = true;
                      ++len;
                  }
                  ++numCycles;
              }
              return numCycles;
          }
  
      }
  
      static class FastScanner {
          private BufferedReader in;
          private StringTokenizer st;
  
          public FastScanner(InputStream stream) {
              in = new BufferedReader(new InputStreamReader(stream));
          }
  
          public String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(in.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return st.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
      }
  }
  
  