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;
  long long int power(long long int x, unsigned long long int y) {
    long long int res = 1;
    while (y > 0) {
      if (y & 1) {
        res = res * x;
      }
      y = y >> 1;
      x = x * x;
    }
    return res;
  }
  long long int powermod(long long int x, unsigned long long int y,
                         long long int p) {
    long long int res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) {
        res = (res * x) % p;
      }
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long int dfs(vector<long long int> &p, vector<long long int> &v) {
    long long int n = v.size() - 1;
    vector<long long int> vis(n + 1), idx(n + 1);
    for (long long int i = (long long int)(1); i < (long long int)(n + 1); i++) {
      idx[p[i]] = i;
    }
    long long int ans = 0;
    for (long long int i = (long long int)(1); i < (long long int)(n + 1); i++) {
      if (vis[i]) continue;
      long long int c = 0;
      long long int node = i;
      while (!vis[node]) {
        vis[node] = 1;
        long long int go_to = v[node];
        node = idx[go_to];
        c++;
      }
      ans += (c - 1);
    }
    return ans;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int T;
    cin >> T;
    while (T--) {
      long long int n, m;
      cin >> n >> m;
      vector<long long int> v, f(n + 1), diff[n + 1];
      v.push_back(0);
      for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
        long long int t;
        cin >> t;
        v.push_back(t);
        f[t] = i + 1;
      }
      for (long long int i = (long long int)(1); i < (long long int)(n + 1);
           i++) {
        long long int val = v[i];
        if (i >= v[i]) {
          diff[i - v[i]].push_back(v[i]);
        } else {
          long long int d = (n - v[i]);
          d += i;
          diff[d].push_back(v[i]);
        }
      }
      vector<long long int> ans;
      for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
        long long int val = diff[i].size();
        if (val < n / 3) continue;
        vector<long long int> p(n + 1);
        for (long long int j = (long long int)(0); j < (long long int)(n); j++)
          p[(j + i) % n + 1] = j + 1;
        long long int rem = (n - val);
        if (dfs(p, v) <= m) ans.push_back(i);
      }
      cout << ans.size() << " ";
      for (const auto itr : ans) {
        cout << itr << ' ';
      }
      cout << "\n";
      ;
    }
    return 0;
  }
  