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>
  void chkmax(T &first, T second) {
    first < second ? first = second : T();
  }
  template <typename T>
  void chkmin(T &first, T second) {
    second < first ? first = second : T();
  }
  template <typename T>
  void readint(T &first) {
    first = 0;
    int f = 1;
    char c;
    for (c = getchar(); !isdigit(c); c = getchar())
      if (c == '-') f = -1;
    for (; isdigit(c); c = getchar()) first = first * 10 + (c - '0');
    first *= f;
  }
  const int MAXN = 300005;
  int n, m, p[MAXN], q[MAXN], cnt[MAXN], fa[MAXN];
  vector<int> res;
  int getfa(int u) { return u == fa[u] ? u : fa[u] = getfa(fa[u]); }
  void check(int k) {
    for (int i = 1; i <= k; ++i) q[i] = n - k + i;
    for (int i = k + 1; i <= n; ++i) q[i] = i - k;
    for (int i = 1; i <= n; ++i) fa[i] = i;
    for (int i = 1; i <= n; ++i) fa[getfa(p[i])] = getfa(q[i]);
    int cnt = n;
    for (int i = 1; i <= n; ++i)
      if (i == getfa(i)) --cnt;
    if (cnt <= m) res.push_back(k);
  }
  void solve() {
    readint(n), readint(m);
    memset(cnt, 0, sizeof(int) * n);
    for (int i = 1; i <= n; ++i) readint(p[i]), ++cnt[(i - p[i] + n) % n];
    res.clear();
    for (int i = 0; i < n; ++i)
      if (cnt[i] >= n - 2 * m) check(i);
    printf("%d", res.size());
    for (auto i : res) printf(" %d", i);
    putchar('\n');
  }
  int main() {
    int T;
    readint(T);
    while (T--) solve();
    return 0;
  }
  