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.InputStreamReader;
  import java.io.IOException;
  import java.util.*;
  import java.math.BigInteger;
  
  public class Main {
  	static FastScanner fs = new FastScanner();
  	public static void main(String[] args) {
  		int qq = fs.nextInt();
  		while (qq-- > 0) {
  			int n = fs.nextInt();
  			int m = fs.nextInt();
  
  			int [] a = new int[n];
  			int [] p = new int[n];
  			int [] k = new int[n];
  
  			for(int i = 0; i < n; ++i) {
  				a[i] = fs.nextInt();
  				--a[i];
  				p[a[i]] = i;
  			}
  
  			int co = n - 2 * m;
  			for(int i = 0; i < n; ++i) {
  				k[sub(p[i], i, n)]++;
  			}
  
  			ArrayList<Integer> ans = new ArrayList<Integer> ();
  			for(int i = 0; i < n; ++i) {
  				if(k[i] >= co) {
  					int checkRes = check(a, i, n);
  					if(checkRes <= m)
  						ans.add(i);
  				}
  			}
  			System.out.printf("%d", ans.size());
  			for(Integer x : ans) {
  				System.out.printf(" %d", x);
  			}
  			System.out.println();
  		}
  	}
  
  	static class Dsu {
  		int [] par;
  		int [] size;
  		public Dsu (int n) {
  			par = new int[n];
  			size = new int[n];
  			for(int i = 0; i < n; ++i) {
  				par[i] = i;
  				size[i] = 1;
  			}
  		}
  
  		public int get(int a) {
  			if(a == par[a])
  				return a;
  			return par[a] = get(par[a]);
  		}
  
  		public void unite(int a, int b) {
  			a = get(a);
  			b = get(b);
  			if(a != b) {
  				if(size[a] > size[b]) {
  					int t = a;
  					a = b;
  					b = t;
  				}
  				par[b] = a;
  				size[a] += size[b];
  			}
  		}
  
  		public int getNumComp( ) {
  			int count = 0;
  			for(int i = 0; i < size.length; ++i) {
  				if(par[i] == i)
  					++count;
  			}
  			return count;
  		}
  	}
  
  	static int check(int [] a, int k, int n) {
  		Dsu ob = new Dsu(n);
  		for(int i = 0 ; i < n; ++i) {
  			int x = sub(i, k, n);
  			int y = a[i];
  			ob.unite(x, y);
  		}	
  		return n - ob.getNumComp();
  	}
  
  	static int sub (int a, int b, int mod) {
  		return ((a - b) % mod + mod) % mod;
  	}
  
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer token = new StringTokenizer("");
  
  		String next() {
  			while (!token.hasMoreTokens()) {
  				try {
  					token = new StringTokenizer(br.readLine());
  				} catch (IOException ie) {
  					ie.printStackTrace();
  				}
  			}
  			return token.nextToken();
  		}
  
  		Integer nextInt() {
  			return Integer.parseInt(next());
  		}
  		int [] readArray(int n) {
  			int [] arr = new int[n];
  			for (int i = 0; i < n; ++i) {
  				arr[i] = nextInt();
  			}
  			return arr;
  		}
  	}
  }