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 maxn = 3e5 + 10;
  const long long mod = 1e9 + 7;
  const long long m2 = 1e9 + 7;
  const long long INF64 = long long(1e18);
  const long long max2 = 1e3 + 10;
  const long long N = 1000001;
  const long long MAXN = 2e5 + 10;
  const long long ALPHABET_SIZE = 2;
  int cnt = 0;
  int vis[maxn];
  vector<int> adj[maxn];
  vector<int> cyclic_shift(vector<int> v, int k) {
    int ind = (k);
    int n = v.size();
    vector<int> v2(n);
    int i1 = k;
    int i;
    for (i = 0; i < n; ++i, ++i1) {
      v2[i1 % n] = v[i];
    }
    return (v2);
  }
  void dfs(int p) {
    vis[p] = 1;
    ++cnt;
    for (auto x : adj[p]) {
      if (vis[x] == 0) {
        dfs(x);
      }
    }
  }
  int find(vector<int> v, vector<int> v2) {
    int i;
    int n = v.size();
    for (i = 0; i < n; ++i) {
      adj[i].clear();
      vis[i] = 0;
    }
    for (i = 0; i < n; ++i) {
      adj[v[i]].push_back(v2[i]);
      adj[v2[i]].push_back(v[i]);
    }
    int ans = 0;
    for (i = 0; i < n; ++i) {
      cnt = 0;
      if (vis[i] == 0) {
        dfs(i);
        ans += (cnt - 1);
      }
    }
    return (ans);
  }
  vector<vector<int>> vf2;
  void printArr(vector<int>& a, int n) {
    vector<int> v;
    for (int i = 0; i < n; i++) v.push_back(a[i]);
    vf2.push_back(v);
  }
  void heapPermutation(vector<int>& a, int size, int n) {
    if (size == 1) {
      printArr(a, n);
      return;
    }
    for (int i = 0; i < size; i++) {
      heapPermutation(a, size - 1, n);
      if (size % 2 == 1)
        swap(a[0], a[size - 1]);
      else
        swap(a[i], a[size - 1]);
    }
  }
  int main() {
    ios_base::sync_with_stdio(NULL);
    cin.tie(0);
    cout.tie(0);
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      vector<int> v(n);
      int i;
      int ind[n];
      for (i = 0; i < n; ++i) {
        cin >> v[i];
        --v[i];
        ind[v[i]] = i;
      }
      vector<int> vf(n);
      for (i = 0; i < n; ++i) {
        vf[i] = i;
      }
      int ans[n];
      memset(ans, 0, sizeof(ans));
      for (i = 0; i < n; ++i) {
        int x = ind[i] - i;
        if (x < 0) {
          x += n;
        }
        ans[x]++;
      }
      vector<int> ans2;
      for (i = 0; i < n; ++i) {
        if (n - ans[i] <= 2 * k) {
          vector<int> v2 = cyclic_shift(vf, i);
          int x = find(v, v2);
          if (x <= k) {
            ans2.push_back(i);
          }
        }
      }
      cout << ans2.size() << " ";
      for (auto x : ans2) cout << x << " ";
      cout << "\n";
    }
    return (0);
  }
  