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 java.io.*;
  import java.util.*;
  public class Main {
  	static int func(int[] temp,int k){
  //		System.out.println(k);
  		int[] l=temp.clone();
  		int n=l.length;
  		int[] to=new int[n];
  		int curr=1;
  		for(int i=k;i<n;i++) {
  			to[i]=curr++;
  		}
  		for(int i=0;i<k;i++) {
  			to[i]=curr++;
  		}
  //		System.out.println(Arrays.toString(to));
  		int[] change=new int[n];
  		for(int i=0;i<n;i++) {
  			change[to[i]-1]=i+1;
  		}
  		for(int i=0;i<n;i++) {
  			l[i]=change[l[i]-1];
  		}
  //		System.out.println(Arrays.toString(change));
  //		System.out.println(Arrays.toString(l));
  		int c=0;
  		boolean[] v=new boolean[n];
  		for(int i=0;i<n;i++) {
  			if(!v[i]) {
  //				System.out.println("i "+i);
  				curr=i;
  				while(!v[curr]) {
  //					System.out.println("curr "+curr);
  					v[curr]=true;
  					curr=l[curr]-1;
  				}
  				c++;
  			}
  		}
  		return n-c;
  	}
  	public static void main(String[] args) throws IOException 
  	{ 
  		FastScanner f = new FastScanner(); 
  		int ttt=1;
  		ttt=f.nextInt();
  		PrintWriter out=new PrintWriter(System.out);
  		outer:for(int tt=0;tt<ttt;tt++) {
  			int n=f.nextInt(),m=f.nextInt();
  			int[] l=f.readArray(n);
  			int[] right=new int[n];
  			for(int i=0;i<n;i++) {
  				int curr=i;
  				int shouldbe=l[i]-1;
  				right[(curr-shouldbe+n)%n]++;
  //				System.out.println(Arrays.toString(right));
  			}
  			ArrayList<Integer> ans=new ArrayList<>();
  			for(int i=0;i<n;i++) {
  				if(right[i]>=n/3) {
  					int num=func(l,i);
  //					System.out.println(i+" num "+num);
  					if(num<=m) {
  						ans.add(i);
  					}
  				}
  			}
  			System.out.print(ans.size()+" ");
  			for(int i:ans) System.out.print(i+" ");
  			System.out.println();
  			
  		}
  		out.close();
  	} 
  	static void sort(int[] p) {
          ArrayList<Integer> q = new ArrayList<>();
          for (int i: p) q.add( i);
          Collections.sort(q);
          for (int i = 0; i < p.length; i++) p[i] = q.get(i);
      }
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		long[] readLongArray(int n) {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  	}
  }