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: from collections import Counter, defaultdict, deque
  import bisect
  from sys import stdin, stdout
  from itertools import repeat, permutations
  import math
  import heapq
   
  MOD = 998244353
  def inp(force_list=False):
      re = map(int, raw_input().split())
      if len(re) == 1 and not force_list:
          return re[0]
      return re
   
  def inpf(force_list=False):
      re = map(float, raw_input().split())
      if len(re) == 1 and not force_list:
          return re[0]
      return re
   
  def inst():
      return raw_input().strip()
   
   
  def gcd(x, y):
     while(y):
         x, y = y, x % y
     return x
   
  def lcm(x, y):
      return x*y/(gcd(x,y))
   
  def qmod(a, b, mod=MOD):
      res = 1
      while b:
          if b&1:
              res = (res*a)%mod
          b >>= 1
          a = (a*a)%mod
          # print b
      return res
   
  def inv(a):
      return qmod(a, MOD-2)
   
  def query(num, k):
      print tmk(num, k)
      stdout.flush()
      return inp()
   
  def my_main():
      kase = inp()
      pans = []
      for _ in range(kase):
          n, m = inp()
          da = inp(True)
          ans = []
          diff = [i-da[i]+1 for i in range(n)]
          ct = sorted(Counter(diff).items(), key=lambda v:-v[-1])
          for k, v in ct[:6]:
              ta = [-1] * n
              if k < 0:
                  pos = 0
                  ta[pos] = -k
              else:
                  pos = n-1
                  ta[pos] = n-1-k
              last = ta[pos]
              pos = (pos+1)%n
              while ta[pos] == -1:
                  last = (last+1)%n
                  ta[pos] = last
                  pos = (pos+1)%n
              pos = [0] * n
              for i in range(n):
                  pos[ta[i]] = i
              flag = [0]*n
              tm = 0
              for i in range(n):
                  if flag[i]:
                      continue
                  now = pos[da[i]-1]
                  flag[i] = 1
                  while flag[now]==0:
                      tm += 1
                      flag[now] = 1
                      now = pos[da[now]-1]
              # print pos, ta, da, tm
              if tm <= m:
                  ans.append((n-ta[0])%n)
                  # print ta, k, tm
          ans=list(set(ans))
          ans.sort()
          ans.insert(0, len(ans))
          pans.append(' '.join(map(str, ans)))
          # print '\n'.join(map(str, ans))
  
  
   
      print '\n'.join(pans)
  
  my_main()