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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.BitSet;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author sarthakmanna
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        EPermutationShift solver = new EPermutationShift();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class EPermutationShift {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            int i, j, k;

            int N = in.nextInt(), M = in.nextInt();
            int[] P = in.getIntArray(N);

            ArrayList<Integer>[] correctIndexes = new ArrayList[N];
            for (i = 0; i < N; ++i) correctIndexes[i] = new ArrayList<>();

            for (i = 0; i < N; ++i) {
                int diff = i + 1 - P[i];
                if (diff < 0) diff += N;
                correctIndexes[diff].add(i);
            }

            ArrayList<Integer> ans = new ArrayList<>();
            for (i = 0; i < N; ++i) {
                if (M >= N - correctIndexes[i].size() - 1) {
                    ans.add(i);
                } else if (check(i, P, M, N - correctIndexes[i].size())) {
                    ans.add(i);
                }
            }

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

        boolean check(int shift, int[] ar, int swaps, int displaced) {
            int idx = 0;
            BitSet fixed = new BitSet();
            for (; idx < ar.length && swaps >= (displaced + 1 >> 1) && !fixed.get(idx); ++idx) {
                if (ar[idx] - 1 != (ar.length - shift) % ar.length) {
                    int p = idx;
                    while (!fixed.get(idx)) {
                        p = (ar[p] - 1 + shift) % ar.length;
                        fixed.set(p);
                        --displaced;
                        --swaps;
                    }
                    ++swaps;
                }
            }
            return swaps >= (displaced + 1 >> 1);
        }

    }

    static class FastReader {
        static final int BUFSIZE = 1 << 20;
        static byte[] buf;
        static int index;
        static int total;
        static InputStream in;

        public FastReader(InputStream is) {
            try {
                in = is;
                buf = new byte[BUFSIZE];
            } catch (Exception e) {
            }
        }

        private int scan() {
            try {
                if (index >= total) {
                    index = 0;
                    total = in.read(buf);
                    if (total <= 0)
                        return -1;
                }
                return buf[index++];
            } catch (Exception | Error e) {
                System.err.println(e.getMessage());
                return 13 / 0;
            }
        }

        public String next() {
            int c;
            for (c = scan(); c <= 32; c = scan()) ;
            StringBuilder sb = new StringBuilder();
            for (; c > 32; c = scan())
                sb.append((char) c);
            return sb.toString();
        }

        public int nextInt() {
            int c, val = 0;
            for (c = scan(); c <= 32; c = scan()) ;
            boolean neg = c == '-';
            if (c == '-' || c == '+')
                c = scan();
            for (; c >= '0' && c <= '9'; c = scan())
                val = (val << 3) + (val << 1) + (c & 15);
            return neg ? -val : val;
        }

        public int[] getIntArray(int size) {
            int[] ar = new int[size];
            for (int i = 0; i < size; ++i) ar[i] = nextInt();
            return ar;
        }

    }
}

