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>
  const bool debugging = 1;
  const long long mod = 1e9 + 7;
  using namespace std;
  template <class T>
  ostream &operator<<(ostream &os, vector<T> V) {
    for (auto v : V) {
      os << v << " ";
    }
    return os << "";
  }
  template <class T>
  ostream &operator<<(ostream &os, vector<vector<T>> V) {
    for (long long i = 0; i < V.size(); i++) {
      for (long long j = 0; j < V[i].size(); j++) {
        os << V[i][j] << " ";
      }
      os << "\n";
    }
    return os << "";
  }
  template <class T>
  ostream &operator<<(ostream &os, set<T> S) {
    for (auto s : S) {
      os << s << " ";
    }
    return os << "";
  }
  template <class L, class R>
  ostream &operator<<(ostream &os, pair<L, R> P) {
    return os << P.first << " " << P.second;
  }
  template <class L, class R>
  ostream &operator<<(ostream &os, map<L, R> M) {
    os << "\n";
    for (auto m : M) {
      os << m << "\n";
    }
    return os << "";
  }
  class disjoint_sets {
   private:
    long long N;
    vector<long long> ancestor;
    vector<long long> children;
  
   public:
    disjoint_sets(long long size) {
      N = size;
      ancestor.resize(N + 1);
      children.resize(N + 1, 1);
      for (long long i = 1; i <= N; i++) {
        ancestor[i] = i;
      }
    }
    long long find_root(long long X) {
      if (ancestor[X] == X) {
        return X;
      }
      ancestor[X] = find_root(ancestor[X]);
      return ancestor[X];
    }
    void merge_sets(long long X, long long Y) {
      long long root_X = find_root(X);
      long long root_Y = find_root(Y);
      if (root_X == root_Y) {
        return;
      }
      if (children[root_X] > children[root_Y]) {
        swap(root_X, root_Y);
      }
      ancestor[root_X] = root_Y;
      children[root_Y] += children[root_X];
      children[root_X] = 0;
    }
  };
  void solve() {
    long long N, M;
    cin >> N >> M;
    map<long long, long long> pos;
    vector<long long> arr(N);
    for (long long i = 0; i < N; i++) {
      cin >> arr[i];
      pos[arr[i]] = i;
    }
    vector<long long> shift(N);
    for (long long i = 0; i < N; i++) {
      long long index = ((i + 1) - arr[i] + N) % N;
      shift[index]++;
    }
    auto func = [&](long long shift_len) -> long long {
      disjoint_sets ds(N + 1);
      for (long long i = 0; i < N; i++) {
        long long current = (i - shift_len + N) % N + 1;
        if (current != arr[i]) {
          ds.merge_sets(pos[current], i);
        }
      }
      set<long long> unique;
      for (long long i = 0; i < N; i++) {
        unique.insert(ds.find_root(i));
      }
      if (N - (long long)unique.size() <= M) {
        return 1;
      }
      return 0;
    };
    vector<long long> vals;
    long long ans = 0;
    for (long long i = 0; i < N; i++) {
      if (N - 2 * M <= shift[i]) {
        if (func(i)) {
          ans++;
          vals.push_back(i);
        }
      }
    }
    cout << ans << " ";
    cout << vals << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long no_of_testcases;
    cin >> no_of_testcases;
    while (no_of_testcases-- > 0) {
      solve();
    }
    return 0;
  }
  