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 int N = 200 + 5;
  class Node {
   public:
    int rnk, parent;
    Node() {
      rnk = -1;
      parent = -1;
    }
  };
  class Disjoint_Set {
   public:
    vector<Node> node;
    int n;
    int comps;
    Disjoint_Set(int nn) {
      n = nn;
      node.resize(n);
      for (int i = 0; i < n; i++) {
        make_set(i);
      }
      comps = n;
    }
    void make_set(int x) {
      node[x].rnk = 0;
      node[x].parent = x;
    }
    int find_set(int x) {
      if (node[x].parent != x) {
        node[x].parent = find_set(node[x].parent);
      }
      return node[x].parent;
    }
    void Union(int u, int v) {
      int p = find_set(u), q = find_set(v);
      if (p == q) return;
      comps--;
      if (node[p].rnk > node[q].rnk)
        node[q].parent = p;
      else {
        node[p].parent = q;
        if (node[p].rnk == node[q].rnk) {
          node[q].rnk += 1;
        }
      }
    }
  };
  void TEST_CASES() {
    int n, m;
    cin >> n >> m;
    vector<int> b(n);
    vector<pair<int, int> > cnt;
    map<int, int> mp;
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      b[i]--;
      int x = ((i - b[i]) % n + n) % n;
      mp[x]++;
      ;
    }
    for (auto it : mp) {
      cnt.push_back({it.second, it.first});
    }
    sort(cnt.rbegin(), cnt.rend());
    ;
    vector<int> ans;
    int x = 0;
    for (int i = 0; i < cnt.size(); i++) {
      int k = cnt[i].second;
      Disjoint_Set d(n);
      vector<int> a(n);
      for (int i = 0; i < n; i++) {
        a[i] = ((i - k) % n + n) % n;
      }
      for (int i = 0; i < n; i++) {
        d.Union(b[i], a[i]);
      }
      if (n - d.comps <= m) {
        ans.push_back(k);
        x = 0;
      } else {
        x++;
      }
      if (x >= 10) break;
    }
    sort(ans.begin(), ans.end());
    cout << ans.size() << " ";
    for (int i : ans) {
      cout << i << " ";
    }
    cout << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    cin >> t;
    while (t--) {
      TEST_CASES();
    }
    return 0;
  }
  