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: # template begins
  #####################################
  from io import BytesIO, IOBase
  import sys
  import math
  import os
  import heapq
  from collections import defaultdict, deque
  from math import ceil
  from bisect import bisect_left, bisect_left
  from time import perf_counter
  
  
  # region fastio
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  def input(): return sys.stdin.readline().rstrip("\r\n")
  def mint(): return map(int, input().split())
  def mfloat(): return map(float, input().split())
  def intin(): return int(input())
  
  
  #####################################
  # template ends
  # Use the recursion snippet if heavy recursion is needed (depth>1000)
  # If constraints are tight, use 1d arrays instead of 2d, like g[i*m+j] instead of g[i][j]
  
  def moves(k, n, a):
      accounted = [False]*n
      ans = 0
      for i in range(n):
          # print(i, accounted[i])
          if accounted[i]:
              continue
          shift = i+1-a[i]
          if shift < 0:
              shift += n
          if shift == k:
              continue
          start = i
          flag = 0
          while not accounted[start]:
              flag = 1
              index = (a[start]-1+k) % n
              accounted[start] = True
              ans += 1
              start = index
          ans -= flag
      return ans
  
  
  def solve():
      n, m = map(int, input().split())
      shifts = [0]*n
      *a, = map(int, input().split())
      for i, x in enumerate(a):
          if x <= i+1:
              shifts[i+1-x] += 1
          else:
              shifts[n-x+i+1] += 1
      # print(shifts)
      # It might not be possible even if there are enough right characters?
      # Is it possible to achieve any permutation of the remaining in m swaps?
      possible = [i for i in range(n) if shifts[i] >= n-2*m]
      # print(shifts)
      # print(len(possible), *possible)
      # These are just potential, check if the remaining can be fit
      # How many will remain?
      # Wait will we have max 3 potentials?????
      # yessss
      # print(moves(0, n, a))
      # for i in possible:
      #     print(i, moves(i, n, a))
      ans = [i for i in possible if moves(i, n, a) <= m]
      print(len(ans), *ans)
  
  
  def main():
      t = int(input())
      # t = 1
      for _ in range(t):
          solve()
  
  
  if __name__ == "__main__":
      start_time = perf_counter()
      main()
      print(perf_counter()-start_time, file=sys.stderr)
  