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: //Utilities
  import java.io.*;
  import java.util.*;
  
  public class A {
  	static int t;
  	static int n, m;
  	static int[] a, tmp;
  	static boolean[] vis;
  	static int[] cntOrdered;
  	static ArrayList<Integer> res;
  	
  	public static void main(String[] args) throws IOException {
  		t = in.iscan();
  		while (t-- > 0) {
  			n = in.iscan(); m = in.iscan(); a = new int[n+1];
  			cntOrdered = new int[n];
  			for (int i = 1; i <= n; i++) {
  				a[i] = in.iscan();
  				if (i >= a[i]) {
  					cntOrdered[i-a[i]]++;
  				}
  				else {
  					cntOrdered[i+n-a[i]]++;
  				}
  			}
  			res = new ArrayList<Integer>();
  			for (int k = 0; k < n; k++) {
  				if (cntOrdered[k] < n-2*m) {
  //					out.println("K: " + k + " " + cntOrdered[k]);
  					continue;
  				}
  				tmp = new int[n+1];
  				for (int i = 1; i <= n; i++) {
  					if (i + k <= n) {
  						tmp[i] = a[i+k];
  					}
  					else {
  						tmp[i] = a[(i+k)%n];
  					}
  				}
  //				out.println("K: " + k);
  //				for (int i = 1; i <= n; i++) {
  //					out.print(tmp[i] + " ");
  //				}
  //				out.println();
  //				out.println("-----");
  				cycleCnt = 0;
  				vis = new boolean[n+1];
  				for (int i = 1; i <= n; i++) {
  					if (!vis[i]) {
  						dfs(i);
  						cycleCnt++;
  					}
  				}
  				int edgesLeft = n-cycleCnt;
  				if (m >= edgesLeft) {
  					res.add(k);
  				}
  			}
  			out.print(res.size() + " ");
  			for (int x : res) {
  				out.print(x + " ");
  			}
  			out.println();
  		}
  		out.close();
  	} 
  	
  	static int cycleCnt;
  	static void dfs(int idx) {
  		vis[idx] = true;
  		if (!vis[tmp[idx]]) {
  			dfs(tmp[idx]);
  		}
  	}
  	
  	static INPUT in = new INPUT(System.in);
  	static PrintWriter out = new PrintWriter(System.out);
  	private static class INPUT {
  
  		private InputStream stream;
  		private byte[] buf = new byte[1024];
  		private int curChar, numChars;
  
  		public INPUT (InputStream stream) {
  			this.stream = stream;
  		}
  
  		public INPUT (String file) throws IOException {
  			this.stream = new FileInputStream (file);
  		}
  
  		public int cscan () throws IOException {
  			if (curChar >= numChars) {
  				curChar = 0;
  				numChars = stream.read (buf);
  			}
  			
  			if (numChars == -1)
  				return numChars;
  
  			return buf[curChar++];
  		}
  
  		public int iscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			int res = 0;
  
  			do {
  				res = (res << 1) + (res << 3);
  				res += c - '0';
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res * sgn;
  		}
  
  		public String sscan () throws IOException {
  			int c = cscan ();
  			
  			while (space (c))
  				c = cscan ();
  
  			StringBuilder res = new StringBuilder ();
  
  			do {
  				res.appendCodePoint (c);
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res.toString ();
  		}
  
  		public double dscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			double res = 0;
  
  			while (!space (c) && c != '.') {
  				if (c == 'e' || c == 'E')
  					return res * UTILITIES.fast_pow (10, iscan ());
  				
  				res *= 10;
  				res += c - '0';
  				c = cscan ();
  			}
  
  			if (c == '.') {
  				c = cscan ();
  				double m = 1;
  
  				while (!space (c)) {
  					if (c == 'e' || c == 'E')
  						return res * UTILITIES.fast_pow (10, iscan ());
  
  					m /= 10;
  					res += (c - '0') * m;
  					c = cscan ();
  				}
  			}
  
  			return res * sgn;
  		}
  
  		public long lscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			long res = 0;
  
  			do {
  				res = (res << 1) + (res << 3);
  				res += c - '0';
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res * sgn;
  		}
  
  		public boolean space (int c) {
  			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
  		}
  	}
  
  	public static class UTILITIES {
  
  		static final double EPS = 10e-6;
  
  		public static int lower_bound (int[] arr, int x) {
  			int low = 0, high = arr.length, mid = -1;
  
  			while (low < high) {
  				mid = (low + high) / 2;
  
  				if (arr[mid] >= x)
  					high = mid;
  				else
  					low = mid + 1;
  			}
  
  			return low;
  		}
  
  		public static int upper_bound (int[] arr, int x) {
  			int low = 0, high = arr.length, mid = -1;
  
  			while (low < high) {
  				mid = (low + high) / 2;
  
  				if (arr[mid] > x)
  					high = mid;
  				else
  					low = mid + 1;
  			}
  
  			return low;
  		}
  
  		public static int gcd (int a, int b) {
  			return b == 0 ? a : gcd (b, a % b);
  		}
  
  		public static int lcm (int a, int b) {
  			return a * b / gcd (a, b);
  		}
  
  		public static long fast_pow_mod (long b, long x, int mod) {
  			if (x == 0) return 1;
  			if (x == 1) return b;
  			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;
  
  			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
  		}
  
  		public static int fast_pow (int b, int x) {
  			if (x == 0) return 1;
  			if (x == 1) return b;
  			if (x % 2 == 0) return fast_pow (b * b, x / 2);
  
  			return b * fast_pow (b * b, x / 2);
  		}
  
  		public static long choose (long n, long k) {
  			k = Math.min (k, n - k);
  			long val = 1;
  
  			for (int i = 0; i < k; ++i)
  				val = val * (n - i) / (i + 1);
  
  			return val;
  		}
  
  		public static long permute (int n, int k) {
  			if (n < k) return 0;
  			long val = 1;
  
  			for (int i = 0; i < k; ++i)
  				val = (val * (n - i));
  
  			return val;
  		}
  		
  		// start of permutation and lower/upper bound template
  		public static void nextPermutation(int[] nums) {
  		    //find first decreasing digit
  		    int mark = -1;
  		    for (int i = nums.length - 1; i > 0; i--) {
  		        if (nums[i] > nums[i - 1]) {
  		            mark = i - 1;
  		            break;
  		        }
  		    }
  		 
  		    if (mark == -1) {
  		        reverse(nums, 0, nums.length - 1);
  		        return;
  		    }
  		 
  		    int idx = nums.length-1;
  		    for (int i = nums.length-1; i >= mark+1; i--) {
  		        if (nums[i] > nums[mark]) {
  		            idx = i;
  		            break;
  		        }
  		    }
  		 
  		    swap(nums, mark, idx);
  		 
  		    reverse(nums, mark + 1, nums.length - 1);
  		}
  		 
  		public static void swap(int[] nums, int i, int j) {
  		    int t = nums[i];
  		    nums[i] = nums[j];
  		    nums[j] = t;
  		}
  		 
  		public static void reverse(int[] nums, int i, int j) {
  		    while (i < j) {
  		        swap(nums, i, j);
  		        i++;
  		        j--;
  		    }
  		}
  		
  		static int lower_bound (int[] arr, int hi, int cmp) {
  			int low = 0, high = hi, mid = -1;
  			while (low < high) {
  				mid = (low + high) / 2;
  				if (arr[mid] >= cmp) high = mid;
  				else low = mid + 1;
  			}
  			return low;
  		}
  	 
  		static int upper_bound (int[] arr, int hi, int cmp) {
  			int low = 0, high = hi, mid = -1;
  			while (low < high) {
  				mid = (low + high) / 2;
  				if (arr[mid] > cmp) high = mid;
  				else low = mid + 1;
  			}
  			return low;
  		}
  		// end of permutation and lower/upper bound template
  	}
  }
  