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<long long> vis(3e5 + 10, 0);
  vector<long long> temp;
  class graph {
   private:
    long long v;
    list<long long>* adj;
  
   public:
    graph(long long x) {
      v = x + 1;
      adj = new list<long long>[v];
    }
    void adedge(long long a, long long b) {
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    void dfs(long long s);
    void bfs(long long s);
  };
  void graph::dfs(long long s) {
    vis[s] = 1;
    temp.push_back(s);
    list<long long>::iterator it;
    for (it = adj[s].begin(); it != adj[s].end(); it++) {
      if (!vis[*it]) {
        dfs(*it);
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      long long i, j, k, l, m, n;
      cin >> n >> m;
      long long arr[n + 1];
      long long cnt[n];
      memset(cnt, 0, sizeof cnt);
      for (i = 1; i <= n; i++) {
        cin >> arr[i];
        long long pos = i, num = arr[i];
        if (pos >= num)
          cnt[pos - num] += 1;
        else
          cnt[n - num + pos] += 1;
      }
      vector<long long> ans;
      for (i = 0; i < n; i++) {
        long long a1 = cnt[i];
        long long rem = n - cnt[i];
        if (rem - 1 <= m)
          ans.push_back(i);
        else {
          if (cnt[i] >= n - 2 * m) {
            vector<long long> v2(n + 1, 0);
            vector<long long> v1(n + 1, 0);
            graph g(n);
            for (j = 0; j <= n; j++) vis[j] = 0;
            for (j = 1; j <= n; j++) {
              long long x = n - i + 1;
              if (j <= i)
                v1[j] = n - i + j;
              else
                v1[j] = j - i;
              if (arr[j] > n - i)
                v2[j] = i - (n - arr[j]);
              else
                v2[j] = arr[j] + i;
              g.adedge(j, v2[j]);
            }
            long long tot = 0;
            for (j = 1; j <= n; j++) {
              if (!vis[j]) {
                temp.clear();
                g.dfs(j);
                tot += (long long)temp.size() - 1;
              }
            }
            if (tot <= m) ans.push_back(i);
          }
        }
      }
      cout << (long long)ans.size() << " ";
      for (auto it : ans) cout << it << " ";
      cout << "\n";
    }
    return 0;
  }
  