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;
  vector<int> edge[300005];
  bool visit[300005];
  int dfs(int n) {
    int ck = 1;
    visit[n] = 1;
    for (auto it : edge[n]) {
      if (visit[it] == 0) ck += dfs(it);
    }
    return ck;
  }
  void solve() {
    int n;
    int m;
    scanf("%d%d", &n, &m);
    int a[n + 2], b[n + 1];
    memset(a, 0, sizeof(a));
    for (int i = 1; i <= n; i++) {
      int a1;
      scanf("%d", &a1);
      b[i] = a1;
      int x = n - (a1 - i);
      if (x >= n) x -= n;
      a[x]++;
    }
    vector<int> v;
    for (int i = 0; i < n; i++) {
      int x = n - a[i];
      if (x <= 2 * m) {
        v.push_back(i);
      }
    }
    vector<int> ans;
    for (auto it : v) {
      int k = it;
      int ok = 0;
      int per[n + 2];
      for (int i = 0; i < n; i++) {
        int x = (i - k + n) % n + 1;
        per[x] = i + 1;
      }
      for (int i = 1; i <= n; i++) {
        int x = b[i];
        edge[x].push_back(b[per[x]]);
      }
      for (int i = 1; i <= n; i++) {
        ok += dfs(i) - 1;
      }
      for (int i = 1; i <= n; i++) {
        edge[i].clear();
        visit[i] = 0;
      }
      if (ok <= m) ans.push_back(k);
    }
    cout << ans.size();
    for (auto it : ans) {
      cout << " " << it;
    }
    cout << "\n";
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      solve();
    }
  }
  