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: /**
   * author: derrick20
   * created: 7/23/21 11:29 PM
   */
  import java.io.*;
  import java.util.*;
  import static java.lang.Math.*;
  
  public class E_PermutationShift {
      static FastScanner sc = new FastScanner();
      static PrintWriter out = new PrintWriter(System.out);
      
      public static void main(String[] args) {
          int T = sc.nextInt();
          while (T-->0) {
              int N = sc.nextInt();
              int M = sc.nextInt();
              int[] perm = new int[N];
              Arrays.setAll(perm, i -> sc.nextInt() - 1);
              int[] free = new int[N];
              for (int i = 0; i < N; i++) {
                  int shift = (i - perm[i] + N) % N; // goal - curr
                  free[shift]++;
              }
              int possShifts = 0;
              StringBuilder ans = new StringBuilder();
              for (int shift = 0; shift < N; shift++) {
                  int toFix = N - free[shift];
                  if (2 * M < toFix) continue;
                  id = new int[N];
                  sz = new int[N];
                  Arrays.setAll(id, i -> i);
                  Arrays.fill(sz, 1);
                  for (int i = 0; i < N; i++) {
                      int shiftedElem = (i - shift + N) % N;
                      unite(shiftedElem, perm[i]);
                  }
                  int tangles = 0;
                  for (int i = 0; i < N; i++) {
                      int root = find(i);
                      if (i == root) {
                          tangles += sz[root] - 1;
                      }
                  }
                  if (tangles <= M) {
                      possShifts++;
                      ans.append(shift).append(" ");
                  }
              }
              out.println(possShifts + " " + ans);
          }
          out.close();
      }
  
      static int[] id, sz;
  
      static int find(int u) {
          return id[u] == u ? u : (id[u] = find(id[u]));
      }
  
      static void unite(int u, int v) {
          if ((u = find(u)) != (v = find(v))) {
              if (sz[u] < sz[v]) {
                  int temp = u;
                  u = v;
                  v = temp;
              }
              sz[u] += sz[v];
              id[v] = u;
          }
      }
  
      static int gcd(int a, int b) {
          return b == 0 ? a : gcd(b, a % b);
      }
  
      static class FastScanner {
          private int BS = 1 << 16;
          private char NC = (char) 0;
          private byte[] buf = new byte[BS];
          private int bId = 0, size = 0;
          private char c = NC;
          private double cnt = 1;
          private BufferedInputStream in;
      
          public FastScanner() {
              in = new BufferedInputStream(System.in, BS);
          }
      
          public FastScanner(String s) {
              try {
                  in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
              } catch (Exception e) {
                  in = new BufferedInputStream(System.in, BS);
              }
          }
      
          char getChar() {
              while (bId == size) {
                  try {
                      size = in.read(buf);
                  } catch (Exception e) {
                      return NC;
                  }
                  if (size == -1) return NC;
                  bId = 0;
              }
              return (char) buf[bId++];
          }
      
          int nextInt() {
              return (int) nextLong();
          }
      
          long nextLong() {
              cnt = 1;
              boolean neg = false;
              if (c == NC) c = getChar();
              for (; (c < '0' || c > '9'); c = getChar()) {
                  if (c == '-') neg = true;
              }
              long res = 0;
              for (; c >= '0' && c <= '9'; c = getChar()) {
                  res = (res << 3) + (res << 1) + c - '0';
                  cnt *= 10;
              }
              return neg ? -res : res;
          }
      
          double nextDouble() {
              boolean neg = false;
              if (c == NC) c = getChar();
              for (; (c < '0' || c > '9'); c = getChar()) {
                  if (c == '-') neg = true;
              }
              double cur = nextLong();
              if (c != '.') {
                  return neg ? -cur : cur;
              } else {
                  double frac = nextLong() / cnt;
                  return neg ? -cur - frac : cur + frac;
              }
          }
      
          String next() {
              StringBuilder res = new StringBuilder();
              while (c <= 32) c = getChar();
              while (c > 32) {
                  res.append(c);
                  c = getChar();
              }
              return res.toString();
          }
      
          String nextLine() {
              StringBuilder res = new StringBuilder();
              while (c <= 32) c = getChar();
              while (c != '\n') {
                  res.append(c);
                  c = getChar();
              }
              return res.toString();
          }
      
          boolean hasNext() {
              if (c > 32) return true;
              while (true) {
                  c = getChar();
                  if (c == NC) return false;
                  else if (c > 32) return true;
              }
          }
      }
      
      static void ASSERT(boolean assertion, String message) {
          if (!assertion) throw new AssertionError(message);
      }
      
      static void ASSERT(boolean assertion) {
          if (!assertion) throw new AssertionError();
      }
  }