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;
  vector<int> vec;
  int p[300005], to[300005], cnt[300005];
  int read() {
    int A;
    bool K;
    char C;
    C = A = K = 0;
    while (C < '0' || C > '9') K |= C == '-', C = getchar();
    while (C > '/' && C < ':') A = (A << 3) + (A << 1) + (C ^ 48), C = getchar();
    return (K ? -A : A);
  }
  void write(int X) {
    if (X < 0) putchar('-'), X = -X;
    if (X > 9) write(X / 10);
    putchar(X % 10 | 48);
  }
  int main() {
    int t, n, m, s, j, i, h, k;
    t = read();
    while (t--) {
      n = read(), m = read();
      for (i = 1; i <= (n); i++) cnt[i] = 0;
      for (i = 1; i <= (n); i++) p[i] = read(), cnt[(p[i] - i + n) % n]++;
      vec.clear();
      for (k = 0; k <= (n - 1); k++)
        if (cnt[k] >= n - (m << 1)) {
          for (i = 1; i <= (n); i++) to[((i + k) % n ? (i + k) % n : n)] = p[i];
          s = n;
          for (i = 1; i <= (n); i++)
            if (to[i]) {
              j = i;
              while (to[j]) {
                h = to[j];
                to[j] = 0;
                j = h;
              }
              s--;
            }
          if (s <= m) vec.push_back(k);
        }
      write(vec.size());
      if (!vec.empty() && !vec[0]) vec.erase(vec.begin()), vec.push_back(0);
      while (!vec.empty()) {
        putchar(' ');
        write((n - vec.back()) % n);
        vec.pop_back();
      }
      putchar('\n');
    }
    return 0;
  }
  