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 max_n = 300005;
  inline int read() {
    int x = 0;
    bool w = 0;
    char c = getchar();
    while (c < '0' || c > '9') w |= c == '-', c = getchar();
    while (c >= '0' && c <= '9')
      x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return w ? -x : x;
  }
  inline void write(int x) {
    if (x < 0) x = -x, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 ^ 48);
  }
  struct graph {
    int ct, hd[max_n], nx[max_n * 2], to[max_n * 2];
    graph() { ct = 1; }
    inline void clear(int n) {
      ct = 1;
      for (register int i = 1; i <= n; ++i) hd[i] = 0;
    }
    inline void add(int u, int v) { nx[++ct] = hd[u], hd[u] = ct, to[ct] = v; }
  } e;
  int n, m, a[max_n], t[max_n];
  bool vis[max_n];
  inline void dfs(int x) {
    vis[x] = 1;
    for (register int i = e.hd[x]; i; i = e.nx[i]) {
      int v = e.to[i];
      if (vis[v]) continue;
      dfs(v);
    }
  }
  signed main() {
    int T = read();
    while (T--) {
      n = read(), m = read();
      for (register int i = 1; i <= n; ++i) {
        int x = read();
        a[x] = i;
        ++t[(i - x + n) % n];
      }
      int s = 0, ans[5];
      for (register int i = 0; i < n; ++i) {
        if (t[i] < n - 2 * m) continue;
        for (register int j = 1; j <= n; ++j) vis[j] = 0;
        e.clear(n);
        for (register int j = 1; j <= n; ++j) {
          int p = (i + j - 1) % n + 1;
          if (a[j] != p) e.add(p, a[j]);
        }
        int cnt = 0;
        for (register int j = 1; j <= n; ++j) {
          int p = (i + j - 1) % n + 1;
          if (a[j] != p && !vis[p]) {
            ++cnt;
            dfs(p);
          }
        }
        if (n - t[i] - cnt <= m) ans[++s] = i;
      }
      write(s), putchar(' ');
      for (register int i = 1; i <= s; ++i) write(ans[i]), putchar(' ');
      putchar('\n');
      for (register int i = 0; i <= n; ++i) t[i] = 0;
    }
    return 0;
  }
  