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 E {
  
  	public static void main(String[] args) throws NumberFormatException, IOException {
  		// TODO Auto-generated method stub
  
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  		
  		int t = Integer.parseInt(br.readLine());
  		
  		while(t --> 0) {
  			
  			StringTokenizer vals = new StringTokenizer(br.readLine());
  			int n = Integer.parseInt(vals.nextToken());
  			int m = Integer.parseInt(vals.nextToken());
  			
  			int[] pos = new int[n+1];
  			int[] cnts = new int[n];
  			int[] act = new int[n];
  			
  			vals = new StringTokenizer(br.readLine());
  			
  			for(int i = 0; i < n; i++) {
  				act[i] = Integer.parseInt(vals.nextToken());
  				pos[act[i]] = i;
  			}
  			
  			for(int i = 0; i < n; i++) {
  				
  				int loc = pos[i+1];
  				
  				if(loc == i)
  					cnts[0]++;
  				else if(loc > i)
  					cnts[loc-i]++;
  				else 
  					cnts[(loc+n)-i]++;
  				
  			}
  			
  			int gc = 0;
  			ArrayList<Integer> o = new ArrayList<Integer>();
  			
  			for(int i = 0; i < n; i++) {
  				
  				if(cnts[i] >= n-2*m) {
  					
  					//System.out.println("POT: " + i);
  					
  					int[] na = new int[n];
  					PriorityQueue<Integer> remaining = new PriorityQueue<Integer>();
  					
  					for(int j = 0; j < n; j++) {
  						
  						na[(j+i)%n] = j+1;
  						if(na[(j+i)%n] != act[(j+i)%n])
  							remaining.add((j+i)%n);
  						
  					}
  					
  					//System.out.println(remaining);
  					
  					int sc = 0;
  					
  					while(remaining.size() > 0) {
  						
  						int ind = remaining.remove();
  						int nind = pos[na[ind]];
  						int cnt = 1;
  						
  						while(nind != ind) {
  							
  							cnt++;
  							remaining.remove(new Integer(nind));
  							nind = pos[na[nind]];
  							
  						}
  						
  						sc += cnt-1;	
  						
  						//System.out.println(sc);
  						//System.out.println(remaining);
  						
  					}
  										
  					if(sc <= m) {
  						gc++;
  						o.add(i);
  					}
  					
  				}
  				
  			}
  			
  			pw.print(gc);
  			for(int i = 0; i < o.size(); i++)
  				pw.print(" " + o.get(i));
  			pw.println();
  			pw.flush();
  			
  		}
  		
  		pw.close();
  		
  	}
  
  }
  