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]. 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Arrays;

public class E_PermShift {
	public static void main(String[] args) throws NumberFormatException, IOException
	{
		BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);

		String[] in;
		int tt = Integer.parseInt(scan.readLine());
		
		for(int t = 0; t < tt; t++)
		{
			in = scan.readLine().split(" ");
			int n = parse(in[0]), m = parse(in[1]);
			
			in = scan.readLine().split(" ");
			int[] ray = new int[n];
			for(int i = 0; i < n; i++)
				ray[i] = parse(in[i]);
			
			int[] delta = new int[n];
			Arrays.fill(delta, n);
			
			for(int i = 0; i < n; i++)
				delta[shifter(i, ray[i], n)] --;
			
			//manual check btwn m and 2*m
			ArrayDeque<Integer> ans = new ArrayDeque<Integer>();
			count(delta, m, n, ray, ans);
			out.print(ans.size());
			
			for(int a : ans)
			{
				out.print(" " + a);
			}
			
			out.println();
		}

		out.flush();
	}

	private static void count(int[] delta, int m, int n, int[] ray, ArrayDeque<Integer> ans) 
	{
		for(int i = 0; i < n; i++)
		{
			if(delta[i] + 1 <= m)
				ans.add(i);
			else if(delta[i] <= m * 2)
			{
				if(valid(i, delta, m, n, ray))
					ans.add(i);
			}
		}
	}

	private static boolean valid(int shift, int[] delta, int m, int n, int[] ray) 
	{
		boolean[] visited = new boolean[n];
		int count = 0;
		
		for(int i = 0; i < n; i++)
		{
			if(visited[i]) continue;
			visited[i] = true;
			
			int group = 1;
			for(int j = swap(ray[i], shift, n); j != i; j = swap(ray[j], shift, n), group++)
				visited[j] = true;;
			
			count += group - 1;
		}
		
		return count <= m;
	}


	private static int swap(int val, int shift, int n) 
	{
		return ((val - 1) + shift) % n;
	}

	private static int shifter(int pos, int val, int n) 
	{
		// where is one supposed to be based on this
		return ((pos + n) - (val - 1)) % n;
	}

	public static int parse(String num)
	{
		return Integer.parseInt(num);
	}
}
