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.util.*;
import java.io.*;

public class Harbour2022 {
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////////   HHHH        HHHH  EEEEEEEEEEEEE   MMMM          MMMM         OOOOOO             SSSSSSS      EEEEEEEEEEEEE  /////////
////////   HHHH        HHHH  EEEEEEEEEEEEE   MMMMMM      MMMMMM      OOO      OOO        SSSS   SSS     EEEEEEEEEEEEE  /////////
////////   HHHH        HHHH  EEEEE           MMMM MMM  MMM MMMM    OOO          OOO    SSSS       SSS   EEEEE          /////////
////////   HHHH        HHHH  EEEEE           MMMM  MMMMMM  MMMM   OOO            OOO   SSSS             EEEEE          /////////
////////   HHHH        HHHH  EEEEE           MMMM          MMMM  OOO              OOO   SSSSSSS         EEEEE          /////////
////////   HHHHHHHHHHHHHHHH  EEEEEEEEEEE     MMMM          MMMM  OOO              OOO      SSSSSS       EEEEEEEEEEE    /////////
////////   HHHHHHHHHHHHHHHH  EEEEEEEEEEE     MMMM          MMMM  OOO              OOO         SSSSSSS   EEEEEEEEEEE    /////////
////////   HHHH        HHHH  EEEEE           MMMM          MMMM   OOO            OOO              SSSS  EEEEE          /////////
////////   HHHH        HHHH  EEEEE           MMMM          MMMM    OOO          OOO     SSS       SSSS  EEEEE          /////////
////////   HHHH        HHHH  EEEEEEEEEEEEE   MMMM          MMMM      OOO      OOO        SSS    SSSS    EEEEEEEEEEEEE  /////////
////////   HHHH        HHHH  EEEEEEEEEEEEE   MMMM          MMMM         OOOOOO             SSSSSSS      EEEEEEEEEEEEE  /////////
////////                                                                                                               /////////
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    static boolean[] vis;

    static int dfs(int idx, int[] edgs) {
        vis[idx] = true;

        int ans = 1;
        if (!vis[edgs[idx]])
            ans += dfs(edgs[idx], edgs);

        return ans;
    }

    static int solve(int[] a) {
        int n = a.length;
        vis = new boolean[n];
        int ans = 0;
        for (int i = 0; i < n; i++) {
            if (!vis[i])
                ans += dfs(i, a) - 1;
        }

        return ans;
    }

    static int[] shift(int[] a, int v) {
        int[] res = new int[a.length];

        for (int i = 0; i < a.length; i++) {
            res[i] = a[(i + v) % a.length];
        }
        return res;
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);


        int tc = sc.nextInt();
        while (tc-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int[] a = sc.nextIntArr(n);
            for (int i = 0; i < n; i++)
                a[i]--;
            int[] count = new int[n + 1];
            for (int i = 0; i < n; i++) {
                if (a[i] <= i) {
                    count[i - a[i]]++;
                } else {
                    count[n - a[i] + i]++;
                }
            }
            ArrayList<Integer> ans = new ArrayList<>();
//            System.err.println(Arrays.toString(count));
            for (int i = 0; i < n; i++) {
                if (count[i] * 3 >= n) {
                    if (solve(shift(a, i)) <= m)
                        ans.add(i);
                }
            }
            pw.print(ans.size());
            for (int x : ans)
                pw.print(" " + x);
            pw.println();
        }
        pw.flush();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

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

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

        public int[] nextIntArr(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }

}