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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.ArrayDeque;
  import java.util.Arrays;
  
  public class E_PermShift {
  	public static void main(String[] args) throws NumberFormatException, IOException
  	{
  		BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
  		PrintWriter out = new PrintWriter(System.out);
  
  		String[] in;
  		int tt = Integer.parseInt(scan.readLine());
  		
  		for(int t = 0; t < tt; t++)
  		{
  			in = scan.readLine().split(" ");
  			int n = parse(in[0]), m = parse(in[1]);
  			
  			in = scan.readLine().split(" ");
  			int[] ray = new int[n];
  			for(int i = 0; i < n; i++)
  				ray[i] = parse(in[i]);
  			
  			int[] delta = new int[n];
  			Arrays.fill(delta, n);
  			
  			for(int i = 0; i < n; i++)
  				delta[shifter(i, ray[i], n)] --;
  			
  			//manual check btwn m and 2*m
  			ArrayDeque<Integer> ans = new ArrayDeque<Integer>();
  			count(delta, m, n, ray, ans);
  			out.print(ans.size());
  			
  			for(int a : ans)
  			{
  				out.print(" " + a);
  			}
  			
  			out.println();
  		}
  
  		out.flush();
  	}
  
  	private static void count(int[] delta, int m, int n, int[] ray, ArrayDeque<Integer> ans) 
  	{
  		for(int i = 0; i < n; i++)
  		{
  			if(delta[i] + 1 <= m)
  				ans.add(i);
  			else if(delta[i] <= m * 2)
  			{
  				if(valid(i, delta, m, n, ray))
  					ans.add(i);
  			}
  		}
  	}
  
  	private static boolean valid(int shift, int[] delta, int m, int n, int[] ray) 
  	{
  		boolean[] visited = new boolean[n];
  		int count = 0;
  		
  		for(int i = 0; i < n; i++)
  		{
  			if(visited[i]) continue;
  			visited[i] = true;
  			
  			int group = 1;
  			for(int j = swap(ray[i], shift, n); j != i; j = swap(ray[j], shift, n), group++)
  				visited[j] = true;;
  			
  			count += group - 1;
  		}
  		
  		return count <= m;
  	}
  
  
  	private static int swap(int val, int shift, int n) 
  	{
  		return ((val - 1) + shift) % n;
  	}
  
  	private static int shifter(int pos, int val, int n) 
  	{
  		// where is one supposed to be based on this
  		return ((pos + n) - (val - 1)) % n;
  	}
  
  	public static int parse(String num)
  	{
  		return Integer.parseInt(num);
  	}
  }
  