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.*;
  
  public class PermutationShift {
  
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		Scanner sc = new Scanner(System.in);
  		int t = sc.nextInt();
  		int i;
  		for(i=0; i<t; i++) {
  			int n = sc.nextInt();
  			int m = sc.nextInt();
  			int[] arr = new int[n];
  			int j;
  			for(j=0; j<n; j++) {
  				arr[j] = sc.nextInt()-1;
  			}
  			solve(arr, m);
  		}
  		sc.close();
  	}
  	static void solve(int[] arr, int m) {
  		int n = arr.length;
  		int[] kValue = new int[n];
  		int[] copyArr = new int[n];
  		int i;
  		for(i=0; i<n; i++) {
  			if(arr[i] <= i) {
  				kValue[i] = i - arr[i];
  			}
  			else {
  				kValue[i] = i + n - arr[i];
  			}
  		}
  		ArrayList<Integer> list = new ArrayList<>();
  		int[] freq = new int[n];
  		for(i=0; i<n; i++) {
  			freq[kValue[i]]++;
  		}
  		for(i=0; i<n; i++) {
  			if(freq[i] >= n/3) {
  				copyArray(arr, copyArr);
  				if(checkForK(copyArr, m, i)) {
  					list.add(i);
  				}
  			}
  		}
  		System.out.print(list.size() + " ");
  		for(i=0; i<list.size(); i++) {
  			System.out.print(list.get(i) + " ");
  		}
  		System.out.println();
  	}
  	static void copyArray(int[] arr, int[] toCopy) {
  		int n = arr.length;
  		int i;
  		for(i=0; i<n; i++) {
  			toCopy[i] = arr[i];
  		}
  	}
  	static boolean checkForK(int[] arr, int m, int k) {
  		int n = arr.length;
  		int i=0;
  		int temp;
  		while(i<n) {
  			if((arr[i]+k)%n != i) {
  				temp = arr[i];
  				arr[i] = arr[(temp+k)%n];
  				arr[(temp+k)%n] = temp;
  				m--;
  			}
  			else {
  
  				i++;
  			}
  		}
  		if(m>=0) {
  			return true;
  		}
  		else return false;
  	}
  }
  