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;
  void fast() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  vector<string> vec_splitter(string s) {
    s += ',';
    vector<string> res;
    while (!s.empty()) {
      res.push_back(s.substr(0, s.find(',')));
      s = s.substr(s.find(',') + 1);
    }
    return res;
  }
  void debug_out(vector<string> __attribute__((unused)) args,
                 __attribute__((unused)) int idx,
                 __attribute__((unused)) int LINE_NUM) {
    cerr << endl;
  }
  template <typename Head, typename... Tail>
  void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
    if (idx > 0)
      cerr << ", ";
    else
      cerr << "Line(" << LINE_NUM << ") ";
    stringstream ss;
    ss << H;
    cerr << args[idx] << " = " << ss.str();
    debug_out(args, idx + 1, LINE_NUM, T...);
  }
  const int INF = 1E9 + 5;
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> p(n), pos(n);
    for (int i = 0; i < n; i++) {
      cin >> p[i];
      --p[i];
      pos[p[i]] = i;
    }
    vector<int> cnt(n);
    for (int i = 0; i < n; i++) {
      int k = (pos[i] - i + n) % n;
      cnt[k]++;
    }
    auto check = [&](int shift) {
      vector<int> q;
      for (int i = n - shift; i < n; i++) q.push_back(i);
      for (int i = 0; i < n - shift; i++) q.push_back(i);
      int ret = n;
      vector<int> vis(n);
      for (int i = 0; i < n; i++) {
        if (!vis[i]) {
          ret--;
          int j = i;
          while (!vis[j]) {
            vis[j] = true;
            j = pos[q[j]];
          }
        }
      }
      if (ret <= m) return true;
      return false;
    };
    vector<int> res;
    for (int i = 0; i < n; i++) {
      if (cnt[i] >= n / 3)
        if (check(i)) res.push_back(i);
    }
    cout << (int)res.size() << ' ';
    for (int x : res) cout << x << ' ';
    cout << '\n';
  }
  int main() {
    fast();
    int t;
    cin >> t;
    while (t--) solve();
  }
  