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 + 10;
  long long T, n, m, p[MAXN], f[MAXN], g[MAXN];
  map<long long, long long> cnt;
  vector<long long> ret;
  long long fa[MAXN], vis[MAXN];
  int Find(int x) {
    if (fa[x] == x) return x;
    return fa[x] = Find(fa[x]);
  }
  int main() {
    cin >> T;
    while (T--) {
      ret.clear();
      cin >> n >> m;
      long long val = 0;
      for (long long i = (1); i <= (n); i++) {
        long long tmp;
        cin >> tmp;
        p[tmp] = i;
      }
      for (long long i = (1); i <= (n); i++) {
        if (i != p[i]) val++;
      }
      f[0] = g[0] = f[n + 1] = g[n + 1] = 0;
      cnt.clear();
      for (long long i = (1); i <= (n); i++) {
        cnt[p[i] - i]++;
        f[i] = i - cnt[n - i];
      }
      cnt.clear();
      for (long long i = (n); i >= (1); i--) {
        cnt[i - p[i]]++;
        g[i] = (n - i + 1) - cnt[i - 1];
      }
      for (long long i = (1); i <= (n); i++) {
        if ((n - f[i] - g[i + 1]) < (n / 3)) continue;
        for (long long j = (1); j <= (n); j++) fa[j] = j, vis[j] = 0;
        for (long long j = (1); j <= (n); j++) {
          if (j <= i) {
            fa[Find(j + n - i)] = Find(p[j]);
          } else {
            fa[Find(j - i)] = Find(p[j]);
          }
        }
        long long num = 0;
        for (long long j = (1); j <= (n); j++) {
          if (vis[Find(j)]) continue;
          vis[Find(j)] = 1;
          num++;
        }
        if (n - num <= m) ret.push_back(n - i);
      }
      sort(ret.begin(), ret.end());
      long long sz = ret.size();
      printf("%lld ", sz);
      for (long long j = (0); j <= (sz - 1); j++) printf("%lld ", ret[j]);
      printf("\n");
    }
    return 0;
  }
  