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]. 
/*
stream Butter!
eggyHide eggyVengeance
I need U
xiao rerun when
 */
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1553E2
{
    public static void main(String hi[]) throws Exception
    {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int T = Integer.parseInt(st.nextToken());
        StringBuilder sb = new StringBuilder();
        while(T-->0)
        {
            st = new StringTokenizer(infile.readLine());
            N = Integer.parseInt(st.nextToken());
            int M = Integer.parseInt(st.nextToken());
            int[] arr = readArr(N, infile, st);
            for(int i=0; i < N; i++)
                arr[i]--;
            //small N
            if(N <= 6)
            {
                ArrayList<Integer> ls = new ArrayList<Integer>();
                for(int k=0; k < N; k++)
                {
                    int[] temp = new int[N];
                    for(int i=0; i < N; i++)
                        temp[i] = arr[(i+k)%N];
                    int cnt = N;
                    boolean[] seen = new boolean[N];
                    for(int v=0; v < N; v++)
                        if(!seen[v])
                        {
                            int curr = v;
                            while(!seen[curr])
                            {
                                seen[curr] = true;
                                curr = temp[curr];
                            }
                            cnt--;
                        }
                    if(cnt <= M)
                        ls.add(k);
                }
                sb.append(ls.size()+" ");
                for(int x: ls)
                    sb.append(x+" ");
            }
            else
            {
                //bruh
                int[] cnt = new int[N];
                for(int i=0; i < N; i++)
                {
                    int k = arr[i]-i;
                    if(k < 0)
                        k += N;
                    k = (N-k)%N;
                    cnt[k]++;
                }
                ArrayList<Integer> candidates = new ArrayList<Integer>();
                for(int k=0; k < N; k++)
                    if(cnt[k] >= N-2*M)
                        candidates.add(k);
                ArrayList<Integer> ls = new ArrayList<Integer>();
                for(int k: candidates)
                {
                    int[] temp = new int[N];
                    for(int i=0; i < N; i++)
                        temp[i] = arr[(i+k)%N];
                    int cntS = N;
                    boolean[] seen = new boolean[N];
                    for(int v=0; v < N; v++)
                        if(!seen[v])
                        {
                            int curr = v;
                            while(!seen[curr])
                            {
                                seen[curr] = true;
                                curr = temp[curr];
                            }
                            cntS--;
                        }
                    if(cntS <= M)
                        ls.add(k);
                }
                sb.append(ls.size()+" ");
                for(int x: ls)
                    sb.append(x+" ");
            }
            sb.append("\n");
        }
        System.out.print(sb);
    }
    static int N;
    public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
    {
        int[] arr = new int[N];
        st = new StringTokenizer(infile.readLine());
        for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
        return arr;
    }
}
/*
1 0 4 3 5 2
5 0 1 2 3 4

1 0 4 3 5 2
4 5 0 1 2 3

something something cycles?
BRUH
 */