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;
  using ll = long long;
  ll MODULO_007 = 1000000007;
  ll MODULO_009 = 1000000009;
  ll MOD = MODULO_007;
  int main() {
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    for (int t = 0; t < T; t++) {
      int N;
      cin >> N;
      int posOp;
      cin >> posOp;
      vector<int> arr(N);
      vector<int> pos(N);
      for (int i = 0; i < N; i++) {
        cin >> arr[i];
        pos[arr[i] - 1] = i;
      }
      vector<int> shiftNeeded(N);
      for (int i = 0; i < N; i++) {
        shiftNeeded[i] = pos[i] - i;
        shiftNeeded[i] = (shiftNeeded[i] + N) % N;
      }
      vector<int> votesForShifts(N, 0);
      {
        for (int e : shiftNeeded) {
          votesForShifts[e]++;
        }
      }
      vector<int> resultingKs;
      for (int i = 0; i < N; i++) {
        if (votesForShifts[i] + 2 * posOp >= N) {
          vector<int> aftershifted(N);
          vector<int> reference(N);
          for (int j = 0; j < N; j++) {
            aftershifted[j] = ((j + N - i) % N);
            reference[j] = arr[j] - 1;
          }
          vector<pair<int, int>> vec(N, pair<int, int>(-1, -1));
          for (int j = 0; j < N; j++) {
            vec[aftershifted[j]].first = j;
            vec[reference[j]].second = j;
          }
          int needOps = 0;
          vector<bool> visited(N + 5, false);
          vector<vector<int>> neis(N + 5);
          for (int j = 0; j < N; j++) {
            if (vec[j].first == vec[j].second) continue;
            neis[vec[j].first].push_back(vec[j].second);
            neis[vec[j].second].push_back(vec[j].first);
          }
          for (int j = 0; j < N; j++) {
            if (visited[j]) {
              continue;
            }
            int compSize = 0;
            queue<int> q;
            q.push(j);
            while (!q.empty()) {
              int cur = q.front();
              q.pop();
              if (visited[cur]) {
                continue;
              }
              visited[cur] = true;
              compSize += 1;
              for (int a : neis[cur]) {
                if (visited[a]) {
                  continue;
                }
                q.push(a);
              }
            }
            if (compSize > 1) {
              needOps += compSize - 1;
            }
          }
          if (needOps <= posOp) {
            resultingKs.push_back(i);
          }
        }
      }
      cout << resultingKs.size() << " ";
      sort(resultingKs.begin(), resultingKs.end());
      for (int e : resultingKs) {
        cout << e << " ";
      }
      cout << endl;
    }
    return 0;
  }
  