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.*;

import static java.util.Arrays.fill;

public class Current {
    FastScanner in;
    PrintWriter out;

    private void solve() throws IOException {
        solveE();
    }

    private void solveE() throws IOException {
        int n = in.nextInt(), m = in.nextInt();
        int[] p = new int[n];
        for (int i = 0; i < n; i++)
            p[i] = in.nextInt();

        int[] b = new int[n * 2];
        System.arraycopy(p, 0, b, 0, n);
        System.arraycopy(p, 0, b, n, n);
        for (int i = 0; i < b.length; i++)
            b[i] -= i + 1;

        // out.println(Arrays.toString(b));

        boolean[] used = new boolean[n * 2];
        TreeMap<Integer, Integer> map = new TreeMap<>();
        ArrayList<Integer> ans = new ArrayList<>();

        for (int k = 0; k < n; k++) {
            if (k == 0) {
                for (int i = 0; i < n; i++)
                    map.put(b[i], map.getOrDefault(b[i], 0) + 1);
            } else {
                if (map.get(b[k - 1]) == 1)
                    map.remove(b[k - 1]);
                else
                    map.put(b[k - 1], map.get(b[k - 1]) - 1);
                map.put(b[k + n - 1], map.getOrDefault(b[k + n - 1], 0) + 1);
            }

            if (map.getOrDefault(-k, 0) >= n - m * 2) { // no more than 3 times
                /*out.print(-k + (k == 0 ? " " : "") + " :");
                for (int i = k; i < k + n; i++) {
                    out.print("\t" + (b[i] + k));
                }
                out.println();*/

                fill(used, k, k + n, false);
                int cnt = 0;
                for (int i = k; i < k + n; i++) {
                    if (!used[i]) {
                        int cur = i;
                        while (!used[cur]) {
                            used[cur] = true;
                            cnt++;
                            cur += b[cur] + k;
                        }
                        cnt--;
                    }
                }

                if (cnt <= m)
                    ans.add(k);
            }
        }

        out.print(ans.size());
        for (int k : ans)
            out.print(" " + k);
        out.println();
    }

    private void swap(int[] a, int i, int j) {
        int swap = a[--i];
        a[i] = a[--j];
        a[j] = swap;
    }

    class FastScanner {
        StringTokenizer st;
        BufferedReader br;

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

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

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

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

    private void run() throws IOException {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);

        for (int t = in.nextInt(); t-- > 0; )
            solve();

        out.flush();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Current().run();
    }
}
