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.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.lang.reflect.Array;
  import java.util.ArrayDeque;
  import java.util.Arrays;
  
  public class ProblemE {
  
      public BufferedReader scanner = new BufferedReader(new InputStreamReader(System.in));
      public PrintWriter out = new PrintWriter(System.out);
      public int n, m;
      public int[] p;
      public String[] in;
  
      void solve() {
          int[] cnt = new int[n];
          Arrays.fill(cnt, n);
          for (int i = 0; i < n; i++) {
              int d = ((i + n) - (p[i] - 1)) % n;
              // pos is shift to true postion
              cnt[d] --;
          }
          ArrayDeque<Integer> ans = new ArrayDeque<Integer>();
          for (int i = 0; i < n; i++) {
              if (cnt[i] + 1 <= m) {
                  ans.push(i);
              } else {
                   if (cnt[i] <= 2*m) {
                       if (valid(i, p)) {
                           ans.add(i);
                       }
                   }
              }
          }
          out.print(ans.size());
          for (int a: ans) {
               out.print(" " + a);
          }
          out.println();
      }
  
      public boolean valid(int k, int[] a) {
          boolean[] visited = new boolean[n];
          int count = 0;
          for (int i = 0; i < n; i++) {
               if (visited[i]) {
                   continue;
               }
               visited[i] = true;
               int num = 1;
               for (int j = swap(a[i], k); j != i; j = swap(a[j], k), num ++) {
                   visited[j] = true;
               }
               count += num -1;
          }
          return  count <= m;
      }
  
      public int  swap(int val, int k) {
           return ((val - 1) + k) % n;
      }
  
      public void process() throws Exception {
          int t;
          t = Integer.parseInt(scanner.readLine());
          while (t-- != 0) {
              in = scanner.readLine().split(" ");
              n = Integer.parseInt(in[0]);
              m = Integer.parseInt(in[1]);
              in = scanner.readLine().split(" ");
              p = new int[n];
              for (int i = 0; i < n; i++) {
                  p[i] = Integer.parseInt(in[i]);
              }
              solve();
          }
          out.flush();
      }
  
      public static void main(String[] args) throws Exception {
          ProblemE app = new ProblemE();
          app.process();
      }
  }
  