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;
  const long long mod = 1e9 + 7;
  template <typename T, typename U>
  ostream &operator<<(ostream &os, pair<T, U> p) {
    os << "(" << p.first << "," << p.second << ")";
    return os;
  }
  template <typename T, typename U>
  istream &operator>>(istream &is, pair<T, U> &p) {
    is >> p.first >> p.second;
    return is;
  }
  template <typename T>
  istream &operator>>(istream &is, vector<T> &v) {
    for (int i = 0; i < int(v.size()); ++i) is >> v[i];
    return is;
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<T> v) {
    for (T x : v) os << x << " ";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, set<T> s) {
    for (T x : s) os << x << " ";
    return os;
  }
  template <typename T>
  ostream &operator<<(ostream &os, multiset<T> s) {
    for (T x : s) os << x << " ";
    return os;
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, map<T, U> m) {
    for (auto x : m) os << x << " ";
    return os;
  }
  template <typename T>
  T sum(vector<T> v) {
    T summ = 0;
    for (int i = 0; i < int(int(v.size())); ++i) summ += v[i];
    return summ;
  }
  long long mod_pow(long long a, long long b) {
    long long sol = 1;
    while (b) {
      if (b & 1) {
        sol = (sol * a) % mod;
        b--;
      } else {
        a = (a * a) % mod;
        b /= 2;
      }
    }
    return sol;
  }
  long long rem(long long a, long long b) {
    long long res = a % b;
    return res < 0 ? res + b : res;
  }
  void test_case() {
    int n, m;
    cin >> n >> m;
    vector<int> v(n);
    for (int i = 0; i < int(n); ++i) {
      cin >> v[i];
      v[i]--;
    }
    vector<pair<int, int> > ks;
    for (int i = 0; i < int(n); ++i) {
      ks.push_back({0, i});
    }
    for (int i = 0; i < int(n); ++i) {
      ks[rem(i - v[i], n)].first++;
    }
    sort(ks.rbegin(), ks.rend());
    set<int> sol;
    for (int i = 0; i < int(min(3, n)); ++i) {
      vector<int> perm;
      for (int j = (n - ks[i].second) % n; j < int((n - ks[i].second) % n + n);
           ++j) {
        perm.push_back(j % n);
      }
      map<int, int> g;
      for (int j = 0; j < int(n); ++j) {
        g[v[j]] = perm[j];
      }
      int k = 0;
      set<int> vis;
      for (int j = 0; j < int(n); ++j) {
        if (vis.count(j) > 0) continue;
        int from = j;
        while (!vis.count(from)) {
          vis.insert(from);
          from = g[from];
        }
        k++;
      }
      if (n - k <= m) {
        sol.insert(ks[i].second);
      }
    }
    cout << int(sol.size()) << " " << sol << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    for (int i = 0; i < int(t); ++i) test_case();
    return 0;
  }
  