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;
  void err() { cout << endl; }
  template <class T, class... Ts>
  void err(T arg, Ts &...args) {
    cout << arg << ' ';
    err(args...);
  }
  const int N = 3e5 + 10;
  const long long mod = 1e9 + 7;
  mt19937 mt(chrono::steady_clock::now().time_since_epoch().count());
  long long rng(long long l, long long r) {
    uniform_int_distribution<long long> uni(l, r);
    return uni(mt);
  }
  int p[N];
  int cnt[N];
  int f[N];
  int find(int x) { return f[x] < 0 ? x : f[x] = find(f[x]); }
  void merge(int x, int y) {
    int fx = find(x);
    int fy = find(y);
    if (fx == fy) return;
    if (f[fx] > f[fy]) swap(fx, fy);
    f[fx] += f[fy];
    f[fy] = fx;
  }
  void solve(int tCase) {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; ++i) {
      cin >> p[i];
      p[i]--;
      cnt[i] = 0;
    }
    int lim = n - 2 * m;
    set<int> s;
    for (int i = 0; i < n; ++i) {
      cnt[(i - p[i] + n) % n]++;
    }
    for (int i = 0; i < n; ++i) {
      if (cnt[i] >= lim) s.insert(i);
    }
    vector<int> ans;
    for (int k : s) {
      for (int i = 0; i < n; ++i) f[i] = -1;
      for (int i = 0; i < n; ++i) {
        if (i == (k + p[i]) % n)
          continue;
        else
          merge(i, (p[i] + k) % n);
      }
      int sum = 0;
      for (int i = 0; i < n; ++i) {
        if (i == (k + p[i]) % n) continue;
        if (f[i] < 0) sum += (-f[i] - 1);
      }
      if (sum <= m) ans.push_back(k);
    }
    cout << ans.size();
    for (int i : ans) {
      cout << ' ' << i;
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    int T = 1;
    cin >> T;
    for (int t = 1; t <= T; ++t) {
      solve(t);
    }
    return 0;
  }
  