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.util.*;
import java.util.Map.Entry;

import java.math.*;
import java.io.*;

public class Main {
	public static void main(String[] args) throws FileNotFoundException {
		InputReader in = new InputReader(System.in);
		// Scanner in = new Scanner(System.in);
		// Scanner in = new Scanner(new BufferedReader(new
		// InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(System.out);
		// InputReader in = new InputReader(new
		// File("ethan_traverses_a_tree.txt"));
		// PrintWriter out = new PrintWriter(new
		// File("ethan_traverses_a_tree-output.txt"));

		int pi = in.nextInt();
		for (int qi = 0; qi < pi; qi++) {
			int n = in.nextInt();
			int m = in.nextInt();

			int[] a = new int[n];
			int[] countbase = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = in.nextInt() - 1;

				int nowbase = (i - a[i] + n) % n;
				countbase[nowbase]++;
			}

			List<Integer> ans = new ArrayList<Integer>();
			for (int i = 0; i < n; i++) {
				if (countbase[i] >= n / 3) {
					if (check(a, n, i, m) == true) {
						ans.add(i);
					}
				}
			}

			out.printf("%d", ans.size());
			for (int i = 0; i < ans.size(); i++) {
				out.printf(" %d", ans.get(i));
			}
			out.printf("\n");
		}

		out.close();
	}

	static public boolean check(int[] ori, int n, int shift, int m) {
		int[] a = new int[n];
		int[] mapa = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = ori[i];
			mapa[ori[i]] = i;
		}

		int[] base = new int[n];
		for (int i = 0; i < n; i++) {
			base[(i + shift) % n] = i;
		}

		int count = 0;
		for (int i = 0; i < n; i++) {
			if (base[i] != a[i]) {
				count++;

				int i1 = i;
				int v1 = a[i1];
				int i2 = mapa[base[i]];
				int v2 = a[i2];

				a[i1] = v2;
				a[i2] = v1;
				mapa[v1] = i2;
				mapa[v2] = i1;
			}
		}

		return count <= m;
	}

	static class InputReader {
		BufferedReader br;
		StringTokenizer st;

		public InputReader(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public InputReader(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		public boolean hasNext() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return false;
				st = new StringTokenizer(s);
			}
			return true;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}