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;
  template <typename T>
  ostream& operator<<(ostream& os, vector<T>& v) {
    if (v.size() == 0) {
      os << "empty vector\n";
      return os;
    }
    for (auto element : v) os << element << " ";
    return os;
  }
  template <typename T, typename second>
  ostream& operator<<(ostream& os, pair<T, second>& p) {
    os << "(" << p.first << ", " << p.second << ")";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, set<T>& v) {
    if (v.size() == 0) {
      os << "empty set\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "[";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << *it;
      if (it != endit) os << ", ";
    }
    os << "]";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, multiset<T>& v) {
    if (v.size() == 0) {
      os << "empty multiset\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "[";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << *it;
      if (it != endit) os << ", ";
    }
    os << "]";
    return os;
  }
  template <typename T, typename second>
  ostream& operator<<(ostream& os, map<T, second>& v) {
    if (v.size() == 0) {
      os << "empty map\n";
      return os;
    }
    auto endit = v.end();
    endit--;
    os << "{";
    for (auto it = v.begin(); it != v.end(); it++) {
      os << "(" << (*it).first << " : " << (*it).second << ")";
      if (it != endit) os << ", ";
    }
    os << "}";
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, vector<vector<T>>& v) {
    for (auto& subv : v) {
      for (auto& e : subv) os << e << " ";
      os << "\n";
    }
    return os;
  }
  bool do_debug = false;
  vector<long long> par(300100), siz(300100);
  long long find_root(long long node) {
    if (par[node] == node)
      return node;
    else
      return par[node] = find_root(par[node]);
  }
  void merge_sets(long long a, long long b) {
    long long x = find_root(a);
    long long y = find_root(b);
    if (x == y) return;
    if (siz[x] < siz[y]) swap(x, y);
    par[y] = x;
    siz[x] += siz[y];
  }
  long long calc(vector<long long>& before, vector<long long>& after) {
    long long n = before.size();
    for (long long i = 0; i < n + 10; i++) {
      par[i] = i;
      siz[i] = 1;
    }
    for (long long i = 0; i < before.size(); i++) {
      merge_sets(before[i], after[i]);
    }
    set<long long> s;
    for (long long i = 1; i < n + 1; i++) s.insert(find_root(i));
    ;
    return (n - s.size());
  }
  void Runtime_Terror() {
    long long n, m;
    cin >> n >> m;
    vector<long long> v(n + 1);
    for (long long i = 1; i < n + 1; i++) {
      cin >> v[i];
    };
    vector<long long> cnt(n + 1);
    for (long long i = 1; i < n + 1; i++) {
      if (i >= v[i]) {
        cnt[i - v[i] + 1]++;
      } else {
        long long tem = v[i] - i;
        cnt[n - tem + 1]++;
      }
    };
    long long mini = n - 2 * m;
    vector<long long> ans;
    for (long long i = 1; i < n + 1; i++) {
      if (cnt[i] >= mini) {
        vector<long long> before(n), after(n);
        for (long long j = 1; j < n + 1; j++) before[j - 1] = v[j];
        long long start = i;
        start--;
        long long cur = 1;
        for (long long j = 0; j < n; j++) {
          after[start] = cur;
          start++;
          cur++;
          if (start == n) start = 0;
        };
        ;
        ;
        ;
        if (calc(before, after) <= m) ans.push_back(i - 1);
      }
    }
    if (ans.size()) {
      cout << ans.size() << " ";
      cout << ans << endl;
    } else {
      cout << 0 << endl;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    for (long long i = 0; i < t; i++) Runtime_Terror();
    return 0;
  }
  