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 MAXN = 3e5 + 5;
  struct UN {
    int fa[MAXN], siz, n;
    void init(int _n) {
      siz = _n;
      n = _n;
      for (int i = 1; i <= n; ++i) fa[i] = i;
    }
    int find(int x) { return fa[x] == x ? fa[x] : fa[x] = find(fa[x]); }
    void merge(int x, int y) {
      x = find(x);
      y = find(y);
      if (x == y) return;
      fa[x] = y;
      siz--;
    }
    int count() { return n - siz; }
  } un;
  int n, m;
  int b[MAXN], cnt[MAXN];
  int a[MAXN];
  bool check(int n, int m, int k) {
    for (int i = 1; i <= n; ++i) {
      a[i] = (((i - 1) - k) % n + n) % n;
      a[i]++;
    }
    un.init(n);
    for (int i = 1; i <= n; ++i) un.merge(a[i], b[i]);
    return un.count() <= m;
  }
  void solve() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
      cin >> b[i];
    }
    for (int i = 0; i <= n; ++i) cnt[i] = 0;
    for (int i = 1; i <= n; ++i) {
      int k = i - b[i];
      if (k < 0) k += n;
      cnt[k]++;
    }
    vector<int> ans;
    for (int k = 0; k < n; ++k) {
      if (cnt[k] >= n - 2 * m && check(n, m, k)) ans.push_back(k);
    }
    cout << ans.size() << " ";
    for (auto c : ans) cout << c << " ";
    cout << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  