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 modd = (1000LL * 1000LL * 1000LL + 7LL);
  bool is_possible(vector<int> a, int shift, int m) {
    for (auto& x : a) {
      x -= shift;
      if (x < 0) {
        x += a.size();
      }
    }
    set<int> done;
    int a2 = 0;
    for (int i = (0); i < (0) + (a.size()); ++i) {
      if ((done.find((i)) != done.end())) {
        continue;
      }
      done.insert(i);
      int curr = a[i], ct = 0;
      while (curr != i) {
        done.insert(curr);
        curr = a[curr];
        ++ct;
      }
      a2 += ct;
    }
    return (a2 <= m);
  }
  int main(int argc, char* argv[]) {
    mt19937 rang(chrono::steady_clock::now().time_since_epoch().count());
    uniform_int_distribution<int> rand_gen(0, modd);
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.precision(12);
    int test_cases;
    cin >> test_cases;
    for (int t = (1); t < (1) + (test_cases); ++t) {
      int n;
      cin >> n;
      int m;
      cin >> m;
      vector<int> p;
      for (int i = (0); i < (0) + (n); ++i) {
        int aa;
        cin >> aa;
        p.push_back(aa - 1);
      }
      vector<int> ans;
      if (n < 30) {
        for (int shift = (0); shift < (0) + (n); ++shift) {
          if (is_possible(p, shift, m)) {
            ans.push_back(shift);
          }
        }
      } else {
        map<int, int> freq;
        for (int i = (0); i < (0) + (p.size()); ++i) {
          int shift = p[i] - i;
          if (shift < 0) {
            shift += n;
          }
          shift %= n;
          ++freq[shift];
        }
        map<int, vector<int>> freq_invert;
        for (auto x : freq) {
          freq_invert[x.second].push_back(x.first);
        }
        set<int> candidates;
        auto it = freq_invert.end();
        int cc = 0;
        while ((it != freq_invert.begin()) && (cc < 4)) {
          --it;
          for (auto sh : it->second) {
            ++cc;
            candidates.insert(sh);
            candidates.insert((n - sh) % n);
            if (cc == 4) {
              break;
            }
          }
        }
        for (auto x : candidates) {
          if (is_possible(p, x, m)) {
            ans.push_back(x);
          }
        }
      }
      for (auto& x : ans) {
        x = (n - x) % n;
      }
      sort(ans.begin(), ans.end());
      cout << ans.size() << " ";
      for (auto x : ans) {
        cout << x << " ";
      }
      cout << endl;
    }
    return 0;
  }
  