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 _Harbour {
      public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
          int tt = sc.nextInt();
          while (tt-- > 0) {
              int n = sc.nextInt();
              int m = sc.nextInt();
              int [] p = new int[n];
              int [] freq = new int[n];
              for (int i = 0; i < n; i++) {
                  p[i] = sc.nextInt() - 1 - i;
                  if (p[i] < 0) p[i] += n;
                  freq[p[i]]++;
              }
              ArrayList<Integer> a = new ArrayList<>();
              for (int i = 0; i < n; i++) {
                  if (freq[i] + 2 * m >= n) {
                      a.add(i);
                  }
              }
              ArrayList<Integer> ans = new ArrayList<>();
              for (Integer f: a) {
                  UF dsu = new UF(n);
                  for (int i = 0; i < n; i++) {
                      if (p[i] != f) {
                          int val = (p[i] + i) % n;
                          if (val < 0) val += n;
                          int want = (val - f) % n;
                          if (want < 0) want += n;
                          dsu.union(want, i);
                      }
                  }
                  long minSwaps = 0;
                  Set<Integer> par = new HashSet<>();
                  for (int i = 0; i < n; i++) {
                      par.add(dsu.find(i));
                  }
                  for (Integer i: par) {
                      minSwaps += dsu.size[i] - 1;
                  }
                  if (minSwaps <= m) {
                      int here = (n - f) % n;
                      if (here < 0) here += n;
                      ans.add(here);
                  }
              }
              out.print(ans.size() + " ");
              Collections.sort(ans);
              for (int i: ans) out.print(i + " ");
              out.println();
          }
          out.close();
      }
  
      static class UF {
          private int[] parent;  // parent[i] = parent of i
          private byte[] rank;   // rank[i] = rank of subtree rooted at i (never more than 31)
          private int count;     // number of components
          private int[] size;
          public UF(int n) {
              if (n < 0) throw new IllegalArgumentException();
              count = n;
              parent = new int[n];
              rank = new byte[n];
              size = new int[n];
              for (int i = 0; i < n; i++) {
                  parent[i] = i;
                  rank[i] = 0;
                  size[i] = 1;
              }
          }
  
          public int find(int p) {
              while (p != parent[p]) {
                  parent[p] = parent[parent[p]];    // path compression by halving
                  p = parent[p];
              }
              return p;
          }
  
          public int count() {
              return count;
          }
  
          public boolean connected(int p, int q) {
              return find(p) == find(q);
          }
  
          public void union(int p, int q) {
              int rootP = find(p);
              int rootQ = find(q);
              if (rootP == rootQ) return;
  
              // make root of smaller rank point to root of larger rank
              if (rank[rootP] < rank[rootQ]) {
                  parent[rootP] = rootQ;
                  size[rootQ] = size[rootQ] + size[rootP];
              }
              else if (rank[rootP] > rank[rootQ]) {
                  parent[rootQ] = rootP;
                  size[rootP] = size[rootP] + size[rootQ];
              }
              else {
                  parent[rootQ] = rootP;
                  size[rootP] = size[rootP] + size[rootQ];
                  rank[rootP]++;
              }
              count--;
          }
      }
  
  
      static void sort(int[] a) {
          ArrayList<Integer> q = new ArrayList<>();
          for (int i : a) q.add(i);
          Collections.sort(q);
          for (int i = 0; i < a.length; i++) a[i] = q.get(i);
      }
  
      static void sort(long[] a) {
          ArrayList<Long> q = new ArrayList<>();
          for (long i : a) q.add(i);
          Collections.sort(q);
          for (int i = 0; i < a.length; i++) a[i] = q.get(i);
      }
  
  
      //-----------MyScanner class for faster input----------
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public MyScanner() {
              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;
          }
  
  
      }
  
  }