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: #include <bits/stdc++.h>
  using namespace std;
  class Graph {
   public:
    vector<vector<int> > adj_list;
    Graph(int n) : adj_list(vector<vector<int> >(n)) {}
    void add_edge(int u, int v, bool bidirectional = true) {
      adj_list[u].push_back(v);
      if (bidirectional) {
        adj_list[v].push_back(u);
      }
    }
  };
  void print_err() { cerr << "\n"; }
  template <class T, class... Arg>
  void print_err(T x, Arg &&...args) {
    cerr << x << " ";
    print_err(args...);
  }
  template <class T>
  void print_container(T &cont) {
    for (auto iter : cont) {
      cerr << iter << " ";
    }
    cerr << "\n";
  }
  void in() {}
  template <class T, class... Arg>
  void in(T &x, Arg &&...args) {
    cin >> x;
    in(args...);
  }
  template <class T>
  void in_c(T &container, int start, int end) {
    for (int i = start; i < end; ++i) {
      cin >> container[i];
    }
  }
  inline void kick_temp(int test) { cout << "Case #" << test << ": "; }
  vector<int> d4x{0, 0, 1, -1};
  vector<int> d4y{-1, 1, 0, 0};
  vector<int> d8x{0, 0, 1, -1, 1, 1, -1, -1};
  vector<int> d8y{1, -1, 0, 0, -1, 1, 1, -1};
  int get_swaps(vector<int> P, int k) {
    ;
    ;
    int count = 0;
    int i, j;
    int n = P.size();
    int index[n];
    for (i = 0; i < n; ++i) {
      index[P[i]] = i;
    }
    for (i = 0; i < n; ++i) {
      if (P[i] != (i - k + n) % n) {
        int temp = index[(i - k + n) % n];
        P[temp] = P[i];
        index[P[i]] = temp;
        ++count;
      }
    }
    return count;
  }
  void solve(int test) {
    ;
    int i, j, n, m;
    in(n, m);
    vector<int> A(n, 0);
    vector<int> P;
    for (i = 0; i < n; ++i) {
      cin >> j;
      j--;
      P.push_back(j);
      A[(i - j + n) % n]++;
    }
    int thres = n - 2 * m;
    vector<int> ans;
    ;
    for (i = 0; i < n; ++i) {
      if (A[i] >= thres) {
        if (get_swaps(P, i) <= m) {
          ans.push_back(i);
        }
      }
    }
    cout << ans.size();
    cout << " ";
    for (i = 0; i < ans.size(); ++i) {
      cout << ans[i];
      cout << " ";
    }
    cout << "\n";
  }
  int main() {
    int i = 1;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int test_cases;
    cin >> test_cases;
    for (i = 1; i <= test_cases; ++i) solve(i);
    return 0;
  }
  