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: 
  def naiveSolve():
      
      
      
      return
  
  class UnionFind:
      def __init__(self, n):
          self.parent = list(range(n))
      def find(self, a): #return parent of a. a and b are in same set if they have same parent
          acopy = a
          while a != self.parent[a]:
              a = self.parent[a]
          while acopy != a: #path compression
              self.parent[acopy], acopy = a, self.parent[acopy]
          return a
      def union(self, a, b): #union a and b
          self.parent[self.find(b)] = self.find(a)
  
  def main():
      
      t=int(input())
      allans=[]
      for _ in range(t):
          n,m=readIntArr()
          p=readIntArr()
          
          for i in range(n):
              p[i]-=1 # 0-index
          
          # m<=n//3
          # Best-case, if m==n//3, we can fix at most 2n//3 elements.
          # So, at least n//3 elements must be correct.
          # Need to consider at most 3 permutations.
          
          kMatches=[0]*n
          for i in range(n):
              kMatches[(i-p[i]+n)%n]+=1
          
          matches=[]
          minMatchCnt=n-2*m
          origPerm=list(range(n))
          for k in range(n):
              matchCnts=kMatches[k]
              if matchCnts>=minMatchCnt: # check
                  arr=origPerm[n-k:]+origPerm[:n-k]
                  
                  targetMap=dict() # {value:index}
                  for i in range(n):
                      targetMap[arr[i]]=i
                  adj=[-1]*n # adj[i] is i's next destination
                  for i in range(n):
                      adj[i]=targetMap[p[i]]
                  
                  # count number of cycles which is number of groups
                  uf=UnionFind(n)
                  visited=[False]*n
                  groups=0
                  for i in range(n):
                      if i==adj[i]: continue # already matching
                      if visited[i]: continue
                      groups+=1
                      visited[i]=True
                      node=adj[i]
                      while uf.find(i)!=uf.find(node):
                          visited[node]=True
                          uf.union(i,node)
                          node=adj[node]
                  misMatchCnts=n-matchCnts
                  # cycle+moves=mismatch counts
                  minMoves=misMatchCnts-groups
                  if minMoves<=m:
                      matches.append(k)
                  # print('p:{}'.format(p))
                  # print('permutation:{}'.format(arr))
                  # print('groups:{} matchCnts:{} minMoves:{}'.format(groups,matchCnts,minMoves))
          ans=[len(matches)]
          ans.extend(matches)
          allans.append(ans)
              
      multiLineArrayOfArraysPrint(allans)
      
      return
  
  
  import sys
  # input=sys.stdin.buffer.readlin,e #FOR READING PURE INTEGER INPUTS (space separation ok)
  input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
  
  def oneLineArrayPrint(arr):
      print(' '.join([str(x) for x in arr]))
  def multiLineArrayPrint(arr):
      print('\n'.join([str(x) for x in arr]))
  def multiLineArrayOfArraysPrint(arr):
      print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
   
  def readIntArr():
      return [int(x) for x in input().split()]
  # def readFloatArr():
  #     return [float(x) for x in input().split()]
   
  def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
      dv=defaultValFactory;da=dimensionArr
      if len(da)==1:return [dv() for _ in range(da[0])]
      else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
   
  def queryInteractive(l,r):
      print('? {} {}'.format(l,r))
      sys.stdout.flush()
      return int(input())
   
  def answerInteractive(x):
      print('! {}'.format(x))
      sys.stdout.flush()
   
  inf=float('inf')
  # MOD=10**9+7
  MOD=998244353
  
  from math import gcd,floor,ceil
  # from math import floor,ceil # for Python2
   
  for _abc in range(1):
      main()