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: ###pyrival template for fast IO
  import os
  import sys
  from io import BytesIO, IOBase
  # 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)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  q=int(input())
  from collections import deque
  def bfs(graph,currnode,visited):
      stack=deque()
      stack.append(currnode)
      while stack:
          currnode=stack.popleft()
          if visited[currnode]==False:
              visited[currnode]=True
  
          for neigbour in graph[currnode]:
              if visited[neigbour]==False:
                  visited[neigbour]=True
                  stack.append(neigbour)
      return ans
  
  
  def find_minswaps(a,b):
      graph={}
      for i in range(len(a)):
          if a[i]==b[i]:
              graph[a[i]]=[]
          if a[i] not in graph:
              graph[a[i]]=[b[i]]
          else:
              graph[a[i]].append(b[i])
          if b[i] not in graph:
              graph[b[i]]=[a[i]]
          else:
              graph[b[i]].append(a[i])
      visited=[False for x in range(n)]
      count=0
      for i in range(n):
          if visited[i]==False:
              count+=1
              bfs(graph,i,visited)
      return count
  while q:
      q-=1
      n,m=[int(x) for x in input().split()]
      arr=[int(x)-1 for x in input().split()]
      ##count number of intergers for each cyclic shift k
      k=[0 for x in range(n)] ##cyclic shift by i
      ##for an i in arr there is only one k such tha arr[i]=(i-k)mod n
      for i in range(n):
          x=arr[i];y=arr[i]-n
          if i-x>=0:
              k[i-x]+=1
          else:
              k[i-y]+=1
      #get the valid k"s using unswaped indices as we have no idea about swaped ones
      tocheck=[]
      ###changed<=2*m; -chnaged>=-2*m
      ##n-changed>=n-2*m
      for i in range(n):
          if k[i]>=n-2*m:
              tocheck.append(i)
      ans=[]
      ##for each k shift permutation find min swaps required to get arr(final arr )
      while tocheck:
          k=tocheck.pop()
          a=[(i-k)%n for i in range(n)]
          ##make the new kshofted arr
          s=n-find_minswaps(a,arr)
          if s<=m:
              ans.append(k)
      ans.sort()
      sys.stdout.write(str(len(ans))+" ")
      for i in range(len(ans)-1):
          sys.stdout.write(str(ans[i])+" ")
      if ans:
          sys.stdout.write(str(ans[-1])+"\n")
      else:
          sys.stdout.write("\n")
  
  
  
  