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.*;
import java.util.*;
public class Main {
	static int func(int[] temp,int k){
//		System.out.println(k);
		int[] l=temp.clone();
		int n=l.length;
		int[] to=new int[n];
		int curr=1;
		for(int i=k;i<n;i++) {
			to[i]=curr++;
		}
		for(int i=0;i<k;i++) {
			to[i]=curr++;
		}
//		System.out.println(Arrays.toString(to));
		int[] change=new int[n];
		for(int i=0;i<n;i++) {
			change[to[i]-1]=i+1;
		}
		for(int i=0;i<n;i++) {
			l[i]=change[l[i]-1];
		}
//		System.out.println(Arrays.toString(change));
//		System.out.println(Arrays.toString(l));
		int c=0;
		boolean[] v=new boolean[n];
		for(int i=0;i<n;i++) {
			if(!v[i]) {
//				System.out.println("i "+i);
				curr=i;
				while(!v[curr]) {
//					System.out.println("curr "+curr);
					v[curr]=true;
					curr=l[curr]-1;
				}
				c++;
			}
		}
		return n-c;
	}
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int ttt=1;
		ttt=f.nextInt();
		PrintWriter out=new PrintWriter(System.out);
		outer:for(int tt=0;tt<ttt;tt++) {
			int n=f.nextInt(),m=f.nextInt();
			int[] l=f.readArray(n);
			int[] right=new int[n];
			for(int i=0;i<n;i++) {
				int curr=i;
				int shouldbe=l[i]-1;
				right[(curr-shouldbe+n)%n]++;
//				System.out.println(Arrays.toString(right));
			}
			ArrayList<Integer> ans=new ArrayList<>();
			for(int i=0;i<n;i++) {
				if(right[i]>=n/3) {
					int num=func(l,i);
//					System.out.println(i+" num "+num);
					if(num<=m) {
						ans.add(i);
					}
				}
			}
			System.out.print(ans.size()+" ");
			for(int i:ans) System.out.print(i+" ");
			System.out.println();
			
		}
		out.close();
	} 
	static void sort(int[] p) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: p) q.add( i);
        Collections.sort(q);
        for (int i = 0; i < p.length; i++) p[i] = q.get(i);
    }
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
}