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 int maxn = 3e5 + 2;
  const int maxm = 1e5 + 5;
  const int inf = 0x3f3f3f3f;
  int read() {
    int s = 0, w = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) {
      if (c == '-') w = -1;
    }
    for (; isdigit(c); c = getchar()) {
      s = (s << 3) + (s << 1) + (c ^ 48);
    }
    return (w == 1 ? s : -s);
  }
  int n, m, cir;
  int a[maxn], cnt[maxn], p[maxn];
  bool vis[maxn];
  vector<int> ans;
  int fa[maxn];
  int getfa(int x) {
    if (fa[x] != x) fa[x] = getfa(fa[x]);
    return fa[x];
  }
  int main() {
    int t = read();
    while (t--) {
      ans.clear();
      n = read(), m = read();
      for (int i = 1; i <= n; i++) {
        a[i] = read();
        cnt[(i - a[i] + n) % n]++;
      }
      for (int k = 0; k < n; k++) {
        if (n - cnt[k] <= 2 * m) {
          int sum = 0;
          for (int i = 1; i <= n; i++) fa[i] = i;
          for (int i = 1; i <= n; i++) {
            int u = i, v = a[(i - 1 + k + n) % n + 1];
            int x = getfa(u), y = getfa(v);
            if (x != y) {
              sum++;
              fa[x] = y;
            }
          }
          if (sum <= m) ans.push_back(k);
        }
      }
      printf("%d", ans.size());
      for (auto k : ans) {
        printf(" %d", k);
      }
      puts("");
    }
    return 0;
  }
  