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]. 
// Don't place your source in a package


import javax.swing.*;
import java.lang.reflect.Array;
import java.text.DecimalFormat;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
import java.util.stream.Stream;



// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

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

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        //reading /writing file
        //Scanner sc=new Scanner(new File("input.txt"));
        //PrintWriter pr=new PrintWriter("output.txt");



        int T=Int();
        for(int t=0;t<T;t++){
            //int n=Int();
            //int k=Int();
            int n=Int();int m=Int();
            int A[]=Arr(n);

            Solution sol1=new Solution(out);
            sol1.solution(A,m);
        }
        out.close();

    }


    public static int[] Arr(int n){
        int A[]=new int[n];
        for(int i=0;i<n;i++)A[i]=Int();
        return A;
    }
    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}








class Solution {
    PrintWriter out;
    int INF = Integer.MAX_VALUE;
    int MOD = 1000000007;
    long mod = 1000000007;

    public Solution(PrintWriter out) {
        this.out = out;
    }


    public void solution(int A[],int m) {

        for(int i=0;i<A.length;i++){
            A[i]--;
        }

        int change[]=new int[A.length];
        for(int i=0;i<A.length;i++){
            int cur = A[i];
            int k = -1;
            if(cur>=i){
                k = A.length-(cur - i);
            }
            else{
                k = i -cur;
            }

            change[i]=k;
        }



        //System.out.println(Arrays.toString(change));


        Map<Integer,Integer>f=new HashMap<>();
        for(int k:change){
            if(!f.containsKey(k))f.put(k,0);
            f.put(k,f.get(k)+1);
        }

        List<Integer>res=new ArrayList<>();
        for(Integer k:f.keySet()){
            if(m*2+f.get(k)<A.length)continue;
            if(check(A,k,m)){
                res.add(k%A.length);
            }
        }
        Collections.sort(res);


        out.print(res.size()+" ");
        for(int i:res)out.print((i % A.length)+" ");
        out.println();
    }


    public boolean check(int A[],int k,int m){
        int B[]=new int[A.length];
        int n = A.length;
        for(int i=0;i<B.length;i++){
            B[i]=A[i];
        }

        int swap = 0;
        for(int i=0;i<B.length;i++){
            while(true){
                int cur =B[i];
                int pos = (cur + k) % n;
                if(B[pos]==cur)break;
                int t=B[pos];
                B[pos]=cur;
                B[i]=t;
                swap++;
            }
        }
        return swap<=m;
    }

}


//0 1 2 3
//3 0 1 2
//2 3 0 1
//1 2 3 0