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;
  template <class T>
  void read(T &x) {
    cin >> x;
  }
  template <class H, class T>
  void read(pair<H, T> &p) {
    cin >> p.first >> p.second;
  }
  template <class T, size_t S>
  void read(array<T, S> &a) {
    for (T &i : a) read(i);
  }
  template <class T>
  void read(vector<T> &v) {
    for (T &i : v) read(i);
  }
  template <class H, class... T>
  void read(H &h, T &...t) {
    read(h);
    read(t...);
  }
  template <class H, class T>
  ostream &operator<<(ostream &o, pair<H, T> &p) {
    o << p.first << " " << p.second;
    return o;
  }
  template <class T, size_t S>
  ostream &operator<<(ostream &o, array<T, S> &a) {
    string s;
    for (T i : a) o << s << i, s = " ";
    return o;
  }
  template <class T>
  ostream &operator<<(ostream &o, vector<T> &v) {
    string s;
    for (T i : v) o << s << i, s = " ";
    return o;
  }
  template <class T>
  void write(T x) {
    cout << x;
  }
  template <class H, class... T>
  void write(const H &h, const T &...t) {
    write(h);
    write(t...);
  }
  void print() { write('\n'); }
  template <class H, class... T>
  void print(const H &h, const T &...t) {
    write(h);
    if (sizeof...(t)) write(' ');
    print(t...);
  }
  template <class T>
  bool ckmin(T &a, const T b) {
    return (b < a) ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T &a, const T b) {
    return (a < b) ? a = b, 1 : 0;
  }
  using ll = long long;
  void solve() {
    int n, m;
    read(n, m);
    vector<int> a(n), cnt(n);
    read(a);
    for (int &i : a) --i;
    for (int i = 0; i < n; ++i) {
      int here;
      if (a[i] > i)
        here = n + i - a[i];
      else
        here = i - a[i];
      ++cnt[here];
    }
    vector<int> cand, ans;
    for (int i = 0; i < n; ++i)
      if (cnt[i] + m * 2 >= n) cand.push_back(i);
    for (int x : cand) {
      vector<int> b(n), pos(n);
      iota((b).begin(), (b).end(), 0);
      rotate(b.begin(), b.begin() + (n - x), b.end());
      for (int i = 0; i < n; ++i) pos[b[i]] = i;
      int c = 0;
      for (int i = 0; i < n; ++i) {
        if (a[i] != b[i]) {
          int need = pos[a[i]];
          swap(b[i], b[need]);
          pos[b[i]] = i, pos[b[need]] = need;
          ++c;
        }
      }
      if (c <= m) ans.push_back(x);
    }
    write(ans.size());
    for (int i : ans) write(' ', i);
    print();
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    int t;
    read(t);
    while (t--) solve();
  }
  