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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
  using namespace std;
  using ll = long long;
  using ull = unsigned long long;
  using db = double;
  using ld = long double;
  template <typename T>
  using V = vector<T>;
  template <typename T>
  using VV = vector<vector<T>>;
  template <typename T>
  using PQ = priority_queue<T>;
  template <typename T>
  using PQR = priority_queue<T, vector<T>, greater<T>>;
  constexpr ll mod = 1000000007;
  constexpr ll INF = 1000000099;
  constexpr ll LINF = (ll)(1e18 + 99);
  const ld PI = acos((ld)-1);
  constexpr ll dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
  template <typename T, typename U>
  inline bool chmin(T& t, const U& u) {
    if (t > u) {
      t = u;
      return 1;
    }
    return 0;
  }
  template <typename T, typename U>
  inline bool chmax(T& t, const U& u) {
    if (t < u) {
      t = u;
      return 1;
    }
    return 0;
  }
  template <typename T>
  inline T gcd(T a, T b) {
    return b ? gcd(b, a % b) : a;
  }
  inline void Yes() { cout << "Yes" << '\n'; }
  inline void No() { cout << "No" << '\n'; }
  inline void YES() { cout << "YES" << '\n'; }
  inline void NO() { cout << "NO" << '\n'; }
  template <typename T, typename Y>
  inline T mpow(T a, Y n) {
    T res = 1;
    for (; n; n >>= 1) {
      if (n & 1) res = res * a;
      a = a * a;
    }
    return res;
  }
  template <typename T>
  vector<T> finddivisor(T x) {
    vector<T> divisor;
    for (T i = 1; (i * i) <= x; i++) {
      if (x % i == 0) {
        divisor.push_back(i);
        if (i * i != x) {
          divisor.push_back(x / i);
        }
      }
    }
    sort(divisor.begin(), divisor.end());
    return divisor;
  }
  template <typename T>
  V<T> prefix_sum(const V<T>& v) {
    int n = v.size();
    V<T> ret(n + 1);
    for (ll i = 0; i < (ll)(n); ++i) ret[i + 1] = ret[i] + v[i];
    return ret;
  }
  template <typename T>
  T rand(T l, T r) {
    static random_device rd;
    static mt19937 g(rd());
    return uniform_int_distribution<T>(l, r)(g);
  }
  template <typename T>
  istream& operator>>(istream& is, vector<T>& vec) {
    for (auto&& x : vec) is >> x;
    return is;
  }
  template <typename T, typename Y>
  ostream& operator<<(ostream& os, const pair<T, Y>& p) {
    return os << "{" << p.first << "," << p.second << "}";
  }
  template <typename T>
  ostream& operator<<(ostream& os, const V<T>& v) {
    os << "{";
    for (auto e : v) os << e << ",";
    return os << "}";
  }
  template <typename... Args>
  void debug(Args&... args) {
    for (auto const& x : {args...}) {
      cerr << x << ' ';
    }
    cerr << '\n';
  }
  template <typename T>
  struct UnionFind {
    vector<T> Parent;
    UnionFind(T N) { Parent = vector<T>(N, -1); }
    T root(T A) {
      if (Parent[A] < 0) return A;
      return Parent[A] = root(Parent[A]);
    }
    T size(T A) { return -Parent[root(A)]; }
    bool connect(T A, T B) {
      A = root(A);
      B = root(B);
      if (A == B) return false;
      if (size(A) < size(B)) swap(A, B);
      Parent[A] += Parent[B];
      Parent[B] = A;
      return true;
    }
    bool same(T A, T B) { return root(A) == root(B); }
  };
  ll f(int rot, V<ll> p) {
    rotate(p.begin(), p.begin() + rot, p.end());
    ll n = (ll)(p).size();
    UnionFind<int> u(n);
    ll ans = 0;
    for (ll i = 0; i < (ll)(n); ++i)
      if (u.connect(i, p[i])) ans++;
    return ans;
  }
  signed main() {
    cin.tie(0);
    cerr.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    int qq;
    cin >> qq;
    while (qq--) {
      ll n, m;
      cin >> n >> m;
      V<ll> p(n);
      cin >> p;
      for (ll i = 0; i < (ll)(n); ++i) p[i]--;
      V<ll> cnt(n, 0);
      for (ll i = 0; i < (ll)(n); ++i) {
        cnt[(i - p[i] + n) % n]++;
      }
      V<ll> ans(0);
      for (ll i = 0; i < (ll)(n); ++i) {
        if (cnt[i] == n)
          ans.push_back(i);
        else if ((n - cnt[i] + 1) / 2 <= m) {
          if (f(i, p) <= m) ans.push_back(i);
        }
      }
      cout << (ll)(ans).size() << " ";
      for (auto&& i : ans) cout << i << " ";
      cout << '\n';
    }
  }
  