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 _Harbour {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int tt = sc.nextInt();
        while (tt-- > 0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int [] p = new int[n];
            int [] freq = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = sc.nextInt() - 1 - i;
                if (p[i] < 0) p[i] += n;
                freq[p[i]]++;
            }
            ArrayList<Integer> a = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                if (freq[i] + 2 * m >= n) {
                    a.add(i);
                }
            }
            ArrayList<Integer> ans = new ArrayList<>();
            for (Integer f: a) {
                UF dsu = new UF(n);
                for (int i = 0; i < n; i++) {
                    if (p[i] != f) {
                        int val = (p[i] + i) % n;
                        if (val < 0) val += n;
                        int want = (val - f) % n;
                        if (want < 0) want += n;
                        dsu.union(want, i);
                    }
                }
                long minSwaps = 0;
                Set<Integer> par = new HashSet<>();
                for (int i = 0; i < n; i++) {
                    par.add(dsu.find(i));
                }
                for (Integer i: par) {
                    minSwaps += dsu.size[i] - 1;
                }
                if (minSwaps <= m) {
                    int here = (n - f) % n;
                    if (here < 0) here += n;
                    ans.add(here);
                }
            }
            out.print(ans.size() + " ");
            Collections.sort(ans);
            for (int i: ans) out.print(i + " ");
            out.println();
        }
        out.close();
    }

    static class UF {
        private int[] parent;  // parent[i] = parent of i
        private byte[] rank;   // rank[i] = rank of subtree rooted at i (never more than 31)
        private int count;     // number of components
        private int[] size;
        public UF(int n) {
            if (n < 0) throw new IllegalArgumentException();
            count = n;
            parent = new int[n];
            rank = new byte[n];
            size = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                rank[i] = 0;
                size[i] = 1;
            }
        }

        public int find(int p) {
            while (p != parent[p]) {
                parent[p] = parent[parent[p]];    // path compression by halving
                p = parent[p];
            }
            return p;
        }

        public int count() {
            return count;
        }

        public boolean connected(int p, int q) {
            return find(p) == find(q);
        }

        public void union(int p, int q) {
            int rootP = find(p);
            int rootQ = find(q);
            if (rootP == rootQ) return;

            // make root of smaller rank point to root of larger rank
            if (rank[rootP] < rank[rootQ]) {
                parent[rootP] = rootQ;
                size[rootQ] = size[rootQ] + size[rootP];
            }
            else if (rank[rootP] > rank[rootQ]) {
                parent[rootQ] = rootP;
                size[rootP] = size[rootP] + size[rootQ];
            }
            else {
                parent[rootQ] = rootP;
                size[rootP] = size[rootP] + size[rootQ];
                rank[rootP]++;
            }
            count--;
        }
    }


    static void sort(int[] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i : a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }

    static void sort(long[] a) {
        ArrayList<Long> q = new ArrayList<>();
        for (long i : a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


    }

}