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.util.*;
  import java.io.*;
  
  public class A1553 {
  	public static int[] rotateLeft(int[] arr, int k) {
  		int n = arr.length;
  		int[] res = new int[n];
  		for (int i = 0; i < n; i++) {
  			res[i] = arr[(i + k) % n];
  		}
  		return res;
  	}
  
  	public static int solve(int[] arr) {
  		int comp = 0;
  		boolean[] vis = new boolean[arr.length];
  		for (int i = 0; i < arr.length; i++) {
  			if (!vis[i]) {
  				comp++;
  				int cur = i;
  				while (!vis[cur]) {
  					vis[cur] = true;
  					cur = arr[cur];
  				}
  			}
  		}
  //		System.out.println(Arrays.toString(arr) + " " + (arr.length - comp));
  
  		return arr.length - comp;
  	}
  
  	public static void main(String[] args) throws IOException {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int m = sc.nextInt();
  			int[] arr = sc.nextIntArr(n);
  			for (int i = 0; i < arr.length; i++) {
  				arr[i]--;
  			}
  			int[] cnt = new int[n];
  			for (int i = 0; i < arr.length; i++) {
  				cnt[(i - arr[i] + n) % n]++;
  			}
  			int still = n - 2 * (n / 3);
  			ArrayList<Integer> ans = new ArrayList<Integer>();
  			for (int i = 0; i < n; i++) {
  				if (cnt[i] >= still) {
  					if (solve(rotateLeft(arr, i)) <= m) {
  						ans.add(i);
  					}
  				}
  			}
  			pw.print(ans.size());
  			for (int x : ans) {
  				pw.print(" " + x);
  			}
  			pw.println();
  		}
  		pw.close();
  	}
  
  	static class Scanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public Scanner(InputStream s) {
  			br = new BufferedReader(new InputStreamReader(s));
  		}
  
  		public Scanner(FileReader f) {
  			br = new BufferedReader(f);
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public int[] nextIntArr(int n) throws IOException {
  			int[] arr = new int[n];
  			for (int i = 0; i < n; i++) {
  				arr[i] = Integer.parseInt(next());
  			}
  			return arr;
  		}
  
  	}
  
  }
  