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: import sys, math, os
  from math import factorial, inf, gcd, pi, sqrt, ceil
  from heapq import *
  from functools import *
  from itertools import *
  from collections import *
  sys.setrecursionlimit(10**5)
  
  if os.environ.get('HOME') == '/Users/clarity':
      os.chdir(os.path.dirname(os.path.abspath(__file__)))
      inp = open('E_test.txt', 'r').readline
  
      def debug(*args):
          print(*args, file=sys.stderr)
  else:
      inp = sys.stdin.readline
  
      def debug(*args):
          pass
  
  
  def line(): return list(map(int, inp().rstrip().split()))
  def fline(): return list(map(float, inp().rstrip().split()))
  def number(): return int(inp())
  def print_array(l): print(' '.join(list(map(str, l))))
  
  def testcase(t):
      for p in range(t):
          solve()
  
  def check_if_shift_is_achievable_in_m_pers(s, shift, m):
      m_left = m
      s = [(x-shift-1)%len(s) for x in s]
      debug(s)
      # check if we can arrange s to match position in m moves.
      to_update = {}
      for i in range(len(s)):
          while s[i] in to_update: 
              s[i] = to_update[s[i]]
          if s[i] != i:
              #put i in s[i] place. But where is i? 
              # remember to update it later
              to_update[i] = s[i]
              s[i] = i
              m_left -=1 
              if m_left < 0:
                  return 0
      return 1
  
              
              
  
  def solve():
      n, m = line()
      s = line()
  
      dist = [ (s[i] -i -1 + n)%n for i in range(n)]
      c = Counter(dist)
      potential_shifts = [x[0] for x in c.most_common(3)]
      debug(s, dist,  potential_shifts)
      ans = []
      for shift in potential_shifts:
          if check_if_shift_is_achievable_in_m_pers(s, shift, m):
              ans.append((n - shift ) %n )
  
      print(len(ans),' '.join(list(map(str, sorted(ans)))))
  
  def main():
      testcase(number())
  
  if __name__ == "__main__":
      main()
  