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 Main2{
  	static void main() throws Exception{
  		int n=sc.nextInt(),m=sc.nextInt();
  		int[]in=new int[n];
  		for(int i=0;i<n;i++) {
  			in[sc.nextInt()-1]=i;
  		}
  		int[]difs=new int[n];
  		for(int i=0;i<n;i++) {
  			difs[(in[i]-i+n)%n]++;
  		}
  		LinkedList<Integer>l=new LinkedList<Integer>();
  		int[]cur=new int[n];
  		boolean[]v=new boolean[n];
  		for(int i=0;i<difs.length;i++) {
  			if(difs[i]>=(n/3)) {
  				//at most 3 times
  				int shift=i;
  				for(int j=0;j<n;j++) {
  					v[j]=false;
  					cur[j]=in[(j-shift+n)%n];
  				}
  				int swaps=0;
  				for(int j=0;j<n;j++) {
  					if(v[j] || cur[j]==j)continue;
  					int cnt=0;
  					int x=j;
  					while(!v[x]) {
  						cnt++;
  						v[x]=true;
  						x=cur[x];
  					}
  					swaps+=(cnt-1);
  				}
  				if(swaps<=m) {
  					l.add(i);
  				}
  			}
  		}
  		pw.print(l.size());
  		while(!l.isEmpty())pw.print(" "+l.pollFirst());
  		pw.println();
  	}
  	public static void main(String[] args) throws Exception{
  		sc=new MScanner(System.in);
  		pw = new PrintWriter(System.out);
  		int tc=1;
          tc=sc.nextInt();
  		for(int i=1;i<=tc;i++) {
  //			pw.printf("Case #%d: ", i);
  			main();
  		}
  		
  		pw.flush();
  	}
  	static PrintWriter pw;
  	static MScanner sc;
  	static class MScanner {
  		StringTokenizer st;
  		BufferedReader br;
  		public MScanner(InputStream system) {
  			br = new BufferedReader(new InputStreamReader(system));
  		}
  	 
  		public MScanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  	 
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  		public int[] intArr(int n) throws IOException {
  	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
  	        return in;
  		}
  		public long[] longArr(int n) throws IOException {
  	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
  	        return in;
  		}
  		public int[] intSortedArr(int n) throws IOException {
  	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
  	        shuffle(in);
  	        Arrays.sort(in);
  	        return in;
  		}
  		public long[] longSortedArr(int n) throws IOException {
  	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
  	        shuffle(in);
  	        Arrays.sort(in);
  	        return in;
  		}
  		public Integer[] IntegerArr(int n) throws IOException {
  	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
  	        return in;
  		}
  		public Long[] LongArr(int n) throws IOException {
  	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
  	        return in;
  		}
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  	 
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  	 
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  	 
  		public char nextChar() throws IOException {
  			return next().charAt(0);
  		}
  	 
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  	 
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  	 
  		public void waitForInput() throws InterruptedException {
  			Thread.sleep(3000);
  		}
  		
  	}
  	static void dbg(int[]in) {
  		System.out.println(Arrays.toString(in));
  	}
  	static void dbg(long[]in) {
  		System.out.println(Arrays.toString(in));
  	}
  	static void sort(int[]in) {
  		shuffle(in);
  		Arrays.sort(in);
  	}
  	static void sort(long[]in) {
  		shuffle(in);
  		Arrays.sort(in);
  	}
  	static void shuffle(int[]in) {
  		for(int i=0;i<in.length;i++) {
  			int idx=(int)(Math.random()*in.length);
  			int tmp=in[i];
  			in[i]=in[idx];
  			in[idx]=tmp;
  		}
  	}
  	static void shuffle(long[]in) {
  		for(int i=0;i<in.length;i++) {
  			int idx=(int)(Math.random()*in.length);
  			long tmp=in[i];
  			in[i]=in[idx];
  			in[idx]=tmp;
  		}
  	}
  }