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]. 
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.StringTokenizer;

/**
 *
 * @author is2ac
 */
public class D_CF {

    public static void main(String[] args) throws IOException {
        FastScanner58 fs = new FastScanner58();
        //Reader fs = new Reader();
        PrintWriter pw = new PrintWriter(System.out);
        int t = fs.ni();
        //int t = 1;
        for (int tc = 0; tc < t; tc++) {
            int n = fs.ni();
            int m = fs.ni();
            int[] a = fs.intArray(n);
            for (int i = 0; i < n; i++) {
                a[i]--;
            }
            HashMap<Integer, Integer> map = new HashMap();
            for (int i = 0; i < n; i++) {
                int d = -1;
                if (i >= a[i]) {
                    d = i - a[i];
                } else {
                    d = i + n - a[i];
                }
                map.put(d, map.getOrDefault(d, 0) + 1);
            }
            int ct = 0;
            List<Integer> res = new ArrayList();
            for (int i = 0; i < n; i++) {
                int good = map.getOrDefault(i, 0);
                int bad = n - good;
                if (bad > m * 2) {
                    continue;
                }
                int c = 0;
                Set<Integer> set = new HashSet();
                for (int j = 0; j < n; j++) {
                    if ((a[j] + i) % n != j) {
                        if (set.contains(j)) {
                            continue;
                        }
                        c++;
                        int p = j;
                        while (!set.contains(p)) {
                            set.add(p);
                            p = (a[p] + i) % n;
                        }
                    }
                }
                c = bad - c;
                if (c <= m) {
                    ct++;
                    res.add(i);
                }
            }
            pw.print(ct + " ");
            for (int i = 0; i < res.size(); i++) {
                pw.print(res.get(i) + " ");
            }
            pw.println();
        }
        pw.close();
    }

    public static boolean isPrime(int n) {
        if (n == 1) {
            return false;
        }
        for (int i = 2; i <= (int) Math.sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static int primes(int a) {
        int b = a;
        int ct = 0;
        for (int i = 2; i <= (int) Math.sqrt(b); i++) {
            if (a % i != 0) {
                continue;
            }
            if (!isPrime(i)) {
                continue;
            }
            int j = a / i;
            while (a % i == 0) {
                a /= i;
                ct++;
            }
            if (!isPrime(j)) {
                continue;
            }
            if (j == 1) {
                continue;
            }
            while (a % j == 0) {
                a /= j;
                ct++;
            }
        }
        return ct;
    }

    public static int dis(int a, int b, int n) {
        //for a to get to b
        if (a < b) {
            return b - a;
        }
        b += n;
        return b - a;
    }

    public static long gcd(long n1, long n2) {
        if (n2 == 0) {
            return n1;
        }
        return gcd(n2, n1 % n2);
    }
}

class GFG {

    // A utility function to get the
    // middle index of given range.
    static int getMid(int s, int e) {
        return s + (e - s) / 2;
    }

    /*
    * A recursive function to get the sum
    of values in given range of the array.
    * The following are parameters
      for this function.
    *
    * st -> Pointer to segment tree
    * node -> Index of current node in
    *         the segment tree.
    * ss & se -> Starting and ending indexes
    *         of the segment represented
    *         by current node, i.e., st[node]
    * l & r -> Starting and ending indexes
    *         of range query
     */
    static long MinUtil(long[] st, int ss,
            int se, int l,
            int r, int node) {

        // If segment of this node is completely
        // part of given range, then return
        // the max of segment
        if (l <= ss && r >= se) {
            return st[node];
        }

        // If segment of this node does not
        // belong to given range
        if (se < l || ss > r) {
            return (long) (1e11);
        }

        // If segment of this node is partially
        // the part of given range
        int mid = getMid(ss, se);

        return Math.min(
                MinUtil(st, ss, mid, l, r,
                        2 * node + 1),
                MinUtil(st, mid + 1, se, l, r,
                        2 * node + 2));
    }

    /*
    * A recursive function to update the
    nodes which have the given index in their
    * range. The following are parameters
    st, ss and se are same as defined above
    * index -> index of the element to be updated.
     */
    static void updateValue(long arr[], long[] st, int ss,
            int se, int index,
            long value,
            int node) {
        if (index < ss || index > se) {
            //System.out.println("Invalid Input");
            return;
        }

        if (ss == se) {

            // update value in array and in
            // segment tree
            arr[index] = value;
            st[node] = value;
        } else {
            int mid = getMid(ss, se);

            if (index >= ss && index <= mid) {
                updateValue(arr, st, ss, mid,
                        index, value,
                        2 * node + 1);
            } else {
                updateValue(arr, st, mid + 1, se, index,
                        value, 2 * node + 2);
            }

            st[node] = Math.min(st[2 * node + 1],
                    st[2 * node + 2]);
        }
        return;
    }

    // Return max of elements in range from
    // index l (query start) to r (query end).
    static long getMin(long[] st, int n, int l, int r) {

        // Check for erroneous input values
        if (l < 0 || r > n - 1 || l > r) {
            //System.out.printf("Invalid Input\n");
            return (long) (1e11);
        }

        return MinUtil(st, 0, n - 1, l, r, 0);
    }

    // A recursive function that constructs Segment
    // Tree for array[ss..se]. si is index of
    // current node in segment tree st
    static long constructSTUtil(long arr[],
            int ss, int se,
            long[] st, int si) {

        // If there is one element in array, store
        // it in current node of segment tree and return
        if (ss == se) {
            st[si] = arr[ss];
            return arr[ss];
        }

        // If there are more than one elements, then
        // recur for left and right subtrees and
        // store the max of values in this node
        int mid = getMid(ss, se);

        st[si] = Math.min(
                constructSTUtil(arr, ss, mid,
                        st, si * 2 + 1),
                constructSTUtil(arr, mid + 1,
                        se, st,
                        si * 2 + 2));

        return st[si];
    }

    /*
    * Function to construct segment tree from
    given array. This function allocates
    * memory for segment tree.
     */
    static long[] constructST(long arr[], int n) {

        // Height of segment tree
        int x = (int) Math.ceil(Math.log(n) / Math.log(2));

        // Maximum size of segment tree
        int max_size = 2 * (int) Math.pow(2, x) - 1;

        // Allocate memory
        long[] st = new long[max_size];

        // Fill the allocated memory st
        constructSTUtil(arr, 0, n - 1, st, 0);

        // Return the constructed segment tree
        return st;
    }
}

class BIT16 {

    long[] bit;

    public BIT16(int size) {
        bit = new long[size];
    }

    public void update(int ind, long delta) {
        while (ind < bit.length) {
            bit[ind] += delta;
            ind = ind + (ind & (-1 * ind));
        }
    }

    public long sum(int ind) {
        long s = 0;
        while (ind > 0) {
            s += bit[ind];
            ind = ind - (ind & (-1 * ind));
        }
        return s;
    }

    public long query(int l, int r) {
        return sum(r) - sum(l);
    }
}

class UnionFind18 {

    int[] id;

    public UnionFind18(int size) {
        id = new int[size];
        for (int i = 0; i < size; i++) {
            id[i] = i;
        }
    }

    public int find(int p) {
        int root = p;
        while (root != id[root]) {
            root = id[root];
        }
        while (p != root) {
            int next = id[p];
            id[p] = root;
            p = next;
        }
        return root;
    }

    public void union(int p, int q) {
        int a = find(p), b = find(q);
        if (a == b) {
            return;
        }
        id[b] = a;
    }
}

class Reader {

    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader() {
        din = new DataInputStream(System.in);
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException {
        din = new DataInputStream(new FileInputStream(file_name));
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public String readLine() throws IOException {
        byte[] buf = new byte[64];
        int cnt = 0, c;
        while ((c = read()) != -1) {
            if (c == '\n') {
                break;
            }
            buf[cnt++] = (byte) c;
        }
        return new String(buf, 0, cnt);
    }

    public int ni() throws IOException {
        int ret = 0;
        byte c = read();
        while (c <= ' ') {
            c = read();
        }
        boolean neg = (c == '-');
        if (neg) {
            c = read();
        }
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0' && c <= '9');
        if (neg) {
            return -ret;
        }
        return ret;
    }

    public long nextLong() throws IOException {
        long ret = 0;
        byte c = read();
        while (c <= ' ') {
            c = read();
        }
        boolean neg = (c == '-');
        if (neg) {
            c = read();
        }
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0' && c <= '9');
        if (neg) {
            return -ret;
        }
        return ret;
    }

    public double nextDouble() throws IOException {
        double ret = 0, div = 1;
        byte c = read();
        while (c <= ' ') {
            c = read();
        }
        boolean neg = (c == '-');
        if (neg) {
            c = read();
        }
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0' && c <= '9');
        if (c == '.') {
            while ((c = read()) >= '0' && c <= '9') {
                ret += (c - '0') / (div *= 10);
            }
        }
        if (neg) {
            return -ret;
        }
        return ret;
    }

    private void fillBuffer() throws IOException {
        bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
        if (bytesRead == -1) {
            buffer[0] = -1;
        }
    }

    private byte read() throws IOException {
        if (bufferPointer == bytesRead) {
            fillBuffer();
        }
        return buffer[bufferPointer++];
    }

    public void close() throws IOException {
        if (din == null) {
            return;
        }
        din.close();
    }
}

class FastScanner58 {

    BufferedReader br;
    StringTokenizer st;

    public FastScanner58() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

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

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

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++) {
            ret[i] = ni();
        }
        return ret;
    }

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

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++) {
            ret[i] = nl();
        }
        return ret;
    }

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

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