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;
  inline int read() {
    int f = 1, ans = 0;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      ans = ans * 10 + c - '0';
      c = getchar();
    }
    return f * ans;
  }
  const int MAXN = 3e5 + 11;
  int cas, N, M, p[MAXN], S[MAXN], tmp[MAXN], vis[MAXN], Ans[MAXN], A[MAXN];
  int main() {
    cas = read();
    while (cas--) {
      N = read(), M = read();
      for (int i = 1; i <= N; i++) A[read()] = i, S[i] = 0;
      for (int i = 1; i <= N; i++) {
        int x = i - A[i] + 1;
        if (x >= 1)
          S[x]++;
        else
          S[x + N]++;
      }
      bool ff = 0;
      Ans[0] = 0;
      for (int i = 1; i <= N; i++)
        if (S[i] >= N - 2 * M) {
          tmp[0] = 0;
          for (int j = i; j <= N; j++) tmp[++tmp[0]] = A[j];
          for (int j = 1; j < i; j++) tmp[++tmp[0]] = A[j];
          int cc = 0;
          for (int j = 1; j <= N; j++) vis[j] = 0;
          for (int j = 1; j <= N; j++)
            if (!vis[j]) {
              ++cc;
              int cur = j;
              while (!vis[cur]) vis[cur] = 1, cur = tmp[cur];
            }
          if (cc >= N - M) {
            Ans[++Ans[0]] = ((i != 1) ? N - i + 1 : 0);
          }
        }
      printf("%d ", Ans[0]);
      sort(Ans + 1, Ans + Ans[0] + 1);
      for (int i = 1; i <= Ans[0]; i++) printf("%d ", Ans[i]);
      printf("\n");
    }
    return 0;
  }
  