1553_E. Permutation Shift_101
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{
    public static void main(String[] args) {
        FastReader s = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        Main main = new Main();
        int t = s.nextInt();
        while(t > 0) {
            int n = s.nextInt();
            int m = s.nextInt();
            int[] arr = new int[n];
            for(int i=0; i<n; i++) {
                arr[i] = s.nextInt();
            }
            int[] res = main.check(arr, m);
            StringBuilder sb = new StringBuilder();
            sb.append(res.length);

            for(int i=0; i<res.length; i++) {
                sb.append(" ");
                sb.append(res[i]);
            }

            out.println(sb.toString());

            t--;
        }
        out.flush();
    }

    private int[] check(int[] check, int m) {
        int n = check.length;

        List<Integer> ans = new ArrayList<>();

        int[] cnt = new int[n];
        for(int i=0; i<n; i++) {
            int offset = i+1-check[i];
            if(offset < 0) offset += n;
            cnt[offset]++;
        }

        for(int i=0; i<n; i++) {
            if(cnt[i] + 2*m >= n) {
                if(n - getCycle(check, n-i) <= m) ans.add(i);
            }
        }

        int[] res = new int[ans.size()];
        for(int i=0; i<res.length; i++) {
            res[i] = ans.get(i);
        }
        return res;
    }

    private int getCycle(int[] check, int s)
    {
        int n = check.length;
        boolean[] used = new boolean[n];
        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            if(used[i]) continue;
            int j = i;
            while(!used[j])
            {
                used[j] = true;
                int sj = j-s;
                if(sj < 0) sj+=n;
                j = check[sj]-1;
            }
            ans++;
        }
        return ans;
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;
    InputStreamReader input;//no buffer
    public FastReader()
    {
        br=new BufferedReader(new InputStreamReader(System.in));
    }

    public FastReader(boolean isBuffer)
    {
        if(!isBuffer){
            input=new InputStreamReader(System.in);
        }else{
            br=new BufferedReader(new InputStreamReader(System.in));
        }
    }

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

    public String next()
    {
        if(input!=null){
            try {
                StringBuilder sb=new StringBuilder();
                int ch=input.read();
                while(ch=='\n'||ch=='\r'||ch==32){
                    ch=input.read();
                }
                while(ch!='\n'&&ch!='\r'&&ch!=32){
                    sb.append((char)ch);
                    ch=input.read();
                }
                return sb.toString();
            }catch (Exception e){
                e.printStackTrace();
            }
        }
        while(st==null || !st.hasMoreElements())//回车，空行情况
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    public int nextInt()
    {
        return (int)nextLong();
    }

    public long nextLong() {
        try {
            if(input!=null){
                long ret=0;
                int b=input.read();
                while(b<'0'||b>'9'){
                    b=input.read();
                }
                while(b>='0'&&b<='9'){
                    ret=ret*10+(b-'0');
                    b=input.read();
                }
                return ret;
            }
        }catch (Exception e){
            e.printStackTrace();
        }
        return Long.parseLong(next());
    }

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