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]. 
//make sure to make new file!
import java.io.*;
import java.util.*;

public class EHS{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      int t = Integer.parseInt(f.readLine());
      
      for(int q = 1; q <= t; q++){
      
         StringTokenizer st = new StringTokenizer(f.readLine());
      
         int n = Integer.parseInt(st.nextToken());
         int m = Integer.parseInt(st.nextToken());
         
         st = new StringTokenizer(f.readLine());
         
         int[] array = new int[n+1];
         for(int k = 1; k <= n; k++){
            array[k] = Integer.parseInt(st.nextToken());
         }
         
         int[] rotsolves = new int[n];
         
         for(int k = 1; k <= n; k++){
            int i = k-array[k];
            if(i < 0) i += n;
            rotsolves[i]++;
         }
         
         ArrayList<Integer> answer = new ArrayList<Integer>();
         
         int thresh = n-m*2;
         for(int k = 0; k < n; k++){
            if(rotsolves[k] >= thresh){
               //see if it works in O(N), will be done ~3 times
               
               //rotate array by k and count the number of swaps
               int[] rot = new int[n+1];
               for(int j = 1; j <= n; j++){
                  int i = j+k;
                  if(i > n) i-=n;
                  rot[j] = array[i];
               }
               
               HashSet<Integer> seen = new HashSet<Integer>();
               
               int swaps = 0;
               for(int j = 1; j <= n; j++){
                  if(seen.contains(j)) continue;
                  seen.add(j);
                  
                  int i = j;
                  while(!seen.contains(rot[i])){
                     swaps++;
                     seen.add(rot[i]);
                     i = rot[i];
                  }
               }
               
               if(swaps <= m){
                  answer.add(k);
               }
                  
            }
         }
         
         StringJoiner sj = new StringJoiner(" ");
         sj.add("" + answer.size());
         for(int i : answer){
            sj.add("" + i);
         }
         out.println(sj.toString());
      

      }
      
      
      
      
      out.close();
   }
   
      
}