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 long long INF = 2e9 + 9;
  vector<long long> adj[300009];
  long long vis[300009];
  long long n, m;
  void dfs(long long x) {
    vis[x] = 1;
    for (auto e : adj[x]) {
      if (vis[e] == -1) dfs(e);
    }
  }
  long long calculate(vector<long long> need, long long k) {
    long long have[n + 2];
    for (long long i = 0; i < (n); ++i) {
      vis[i] = -1;
      adj[i].clear();
    }
    for (long long i = 0; i < (n); ++i) {
      have[i] = i - k;
      if (have[i] < 0) have[i] += n;
    }
    for (long long i = 0; i < (n); ++i) {
      if (need[i] != have[i]) {
        adj[need[i]].push_back(have[i]);
        adj[have[i]].push_back(need[i]);
      }
    }
    long long comp = 0;
    for (long long i = 0; i < (n); ++i) {
      if (vis[i] == -1) {
        dfs(i);
        comp++;
      }
    }
    return (long long)(need).size() - comp;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      cin >> n >> m;
      vector<long long> vet(n);
      for (long long i = 0; i < (n); ++i) {
        cin >> vet[i];
        vet[i]--;
      }
      long long cont[n + 2];
      vector<long long> ans;
      memset(cont, 0, sizeof cont);
      for (long long i = 0; i < (n); ++i) {
        long long pos = i - vet[i];
        if (pos < 0) pos += n;
        cont[pos]++;
      }
      for (long long i = 0; i < (n); ++i) {
        if (cont[i] >= (n + 2) / 3 && calculate(vet, i) <= m) {
          ans.push_back(i);
        }
      }
      cout << (long long)(ans).size();
      if ((long long)(ans).size() > 0) {
        for (long long i = 0; i < ((long long)(ans).size()); ++i) {
          cout << " " << ans[i];
        }
      }
      cout << "\n";
    }
    return 0;
  }
  