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 E {
    static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    private static int nextInt() {
        try {
            in.nextToken();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return (int) in.nval;
    }
    private static long nextLong() {
        try {
            in.nextToken();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return (long) in.nval;
    }
    private static String next(){
        try {
            in.nextToken();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return in.sval;
    }

    public static void main(String[] args) {
        int t = nextInt();
        for (int c = 0; c < t; c++) {
            int n = nextInt();
            int m = nextInt();
            int[] nums = new int[n+1];
            for (int i = 1; i <= n; i++) {
                nums[i] = nextInt();
            }
            int[] state = new int[n];
            for (int i = 1; i <= n; i++) {
                state[(n-nums[i]+i)%n]++;
            }
            //System.out.println(Arrays.toString(state));
            List<Integer> ans = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                int dif = n - state[i];
                if(((dif+1)/2)<=m)ans.add(i);
            }
            System.out.print(ans.size());
            for (Integer an : ans) {
                System.out.print(" "+an);
            }
            System.out.println();
        }
    }
    public static boolean can(int m,List<Integer> arr,int[] target){
        Map<Integer,Integer> map = new HashMap<>();
        int n = target.length;
        for (int i = 0; i < n; i++) {
            map.put(arr.get(i),i);
        }
        for (int i = 0; i < n; i++) {
            if(arr.get(i)!=target[i]){
                swap(arr,i,map.get(target[i]));
            }
        }
        return m>=0;
    }

    public static void swap(List<Integer> a,int x,int y){
        int t = a.get(x);
        a.set(x,y);
        a.set(y,t);
    }
}
