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 solve() {
    long long n, m;
    cin >> n >> m;
    vector<long long> a(n + 1);
    for (long long i = (1); i <= (n); i++) cin >> a[i];
    map<long long, long long> f, f2;
    for (long long i = (1); i <= (n); i++) {
      long long dt;
      if (a[i] == i)
        dt = 0;
      else if (a[i] > i)
        dt = n - a[i] + i;
      else
        dt = i - a[i];
      f[dt]++;
      f2[a[i] - i]++;
    }
    long long biswap = 0;
    for (auto [u, v] : f2) biswap += min(f2[-u], v);
    biswap /= 2;
    vector<long long> ans;
    for (long long k = (0); k <= (n - 1); k++) {
      long long wrong = n - f[k];
      if (wrong <= m + 1)
        ans.emplace_back(k);
      else if (wrong > 2 * m)
        continue;
      else {
        vector<pair<long long, long long>> v;
        for (long long i = (1); i <= (n); i++) {
          long long rsp = (i - k + n - 1) % n + 1;
          v.emplace_back(rsp, a[i]);
        }
        sort(begin(v), end(v));
        vector<long long> vg(n + 1);
        map<long long, long long> pos;
        for (auto [u, v2] : v) pos[v2] = u, vg[u] = v2;
        long long swaps = 0;
        for (long long i = (1); i <= (n); i++)
          if (pos[i] != i) {
            swaps++;
            swap(vg[pos[i]], vg[i]);
            swap(pos[vg[pos[i]]], pos[i]);
          }
        if (swaps <= m) ans.emplace_back(k);
      }
    }
    cout << ans.size() << " ";
    for (auto e : ans) cout << e << " ";
    cout << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  