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;
  namespace DBH {
  const int DIR4[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
  const int DIR8[8][2] = {{1, 0},   {-1, 0}, {0, 1},  {0, -1},
                          {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
  const int MOD197 = 1e9 + 7;
  const int MOD19 = 1e9;
  const int MOD353 = 998244353;
  const int INF = 0x3f3f3f3f;
  const int NINF = 0xc0c0c0c0;
  inline int LT(int x) { return x << 1; }
  inline int RT(int x) { return (x << 1) | 1; }
  inline int LB(int x) { return x & -x; }
  inline int MSK(int x) { return 1 << x; }
  inline int HF(int x) { return x >> 1; }
  inline int DB(int x) { return x << 1; }
  inline int GT(int x, int i) { return (x >> i) & 1; }
  inline int ODD(int x) { return x & 1; }
  template <typename T>
  inline void qread(T &X) {
    X = 0;
    bool flag = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') flag = 0;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      X = (X << 1) + (X << 3) + ch - '0';
      ch = getchar();
    }
    if (flag) return X;
    return ~(X - 1);
  }
  }  // namespace DBH
  using namespace DBH;
  int n, m;
  int arr[300005];
  int cnt[300005];
  int pre[300005];
  int conn;
  int find(int x) { return x == pre[x] ? x : pre[x] = find(pre[x]); }
  void unite(int u, int v) {
    int ur = find(u);
    int vr = find(v);
    if (ur != vr) {
      conn--;
      pre[ur] = vr;
    }
  }
  int subsolve(int k) {
    conn = n;
    for (int i = 0; i < n; i++) {
      pre[i] = i;
    }
    for (int i = 0; i < n; i++) {
      unite(i, (arr[i] - 1 + k) % n);
    }
    return n - conn;
  }
  void solve() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      cnt[(i - arr[i] + 1 + n) % n]++;
    }
    vector<int> ans;
    for (int i = 0; i <= n - 1; i++) {
      if (cnt[i] < n - 2 * m) continue;
      int k = subsolve(i);
      if (k <= m) ans.push_back(i);
    }
    cout << ans.size() << " ";
    for (int i : ans) cout << i << " ";
    cout << endl;
  }
  int main() {
    ;
    int T;
    cin >> T;
    while (T--) {
      solve();
    }
  }
  