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 javax.xml.transform.Templates;

import java.io.*;

public class Main {
	public static int solve(int[]a) {
		boolean []vis=new boolean[a.length];
		int ans=0;
		for(int i=0;i<a.length;i++) {
			if(vis[i])continue;
			int b=a[i];
			vis[i]=true;
			int temp=0;
			while(b!=i+1) {
				vis[b-1]=true;
				b=a[b-1];
				temp++;
			}
			ans+=temp;
		}
		return ans;
	}
	public static void main(String[] args) throws Exception {

		
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			int k=sc.nextInt();
			int[]a=sc.nextIntArray(n);
			if(n<=20) {
				ArrayList<Integer>ans=new ArrayList<Integer>();
				for(int i=0;i<n;i++) {
					int[]c=new int[n];
					for(int j=0;j<n;j++) {
						c[j]=a[(j+i)%n];
					}
					int x=solve(c);
					if(x<=k) {
						
						ans.add(i);
					}
				}
				pw.print(ans.size()+" ");
				for(int i:ans)pw.print(i+" ");
				pw.println();
			}else {
				int[][]b=new int[n][2];
				for(int i=0;i<n;i++) {
					b[(i-a[i]+n)%n][0]++;
					b[i][1]=i;
				}
				Arrays.sort(b,(x,y)->y[0]-x[0]);
//				pw.println(Arrays.deepToString(b));
				ArrayList<Integer>ans=new ArrayList<Integer>();
				for(int ii=0;ii<3;ii++) {
				int[]c=new int[n];
				for(int i=0;i<n;i++) {
					c[i]=a[(i+b[ii][1])%n];
				}
				int x=solve(c);
				if(x<=k) {
					ans.add(b[ii][1]);
				}
				}
				Collections.sort(ans);
				pw.print(ans.size()+" ");
				for(int i:ans)pw.print(i+" ");
				pw.println();
			}
		}
		
		
		
		
		
		pw.close();
	}
	

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader r) {
			br = new BufferedReader(r);
		}

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

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public long[] nextlongArray(int n) throws IOException {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public Long[] nextLongArray(int n) throws IOException {
			Long[] a = new Long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public Integer[] nextIntegerArray(int n) throws IOException {
			Integer[] a = new Integer[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}

	static class pair implements Comparable<pair> {
		long x;
		long y;

		public pair(long x, long y) {
			this.x = x;
			this.y = y;
		}

		public String toString() {
			return x + " " + y;
		}

		public boolean equals(Object o) {
			if (o instanceof pair) {
				pair p = (pair) o;
				return p.x == x && p.y == y;
			}
			return false;
		}

		public int hashCode() {
			return new Long(x).hashCode() * 31 + new Long(y).hashCode();
		}

		public int compareTo(pair other) {
			if (this.x == other.x) {
				return Long.compare(this.y, other.y);
			}
			return Long.compare(this.x, other.x);
		}
	}

	static class tuble implements Comparable<tuble> {
		int x;
		int y;
		int z;

		public tuble(int x, int y, int z) {
			this.x = x;
			this.y = y;
			this.z = z;
		}

		public String toString() {
			return x + " " + y + " " + z;
		}

		public int compareTo(tuble other) {
			if (this.x == other.x) {
				if (this.y == other.y) {
					return this.z - other.z;
				}
				return this.y - other.y;
			} else {
				return this.x - other.x;
			}
		}
	}

	static long mod = 1000000007;
	static Random rn = new Random();
	static Scanner sc = new Scanner(System.in);
	static PrintWriter pw = new PrintWriter(System.out);
}