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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.ArrayDeque;
import java.util.Arrays;

public class ProblemE {

    public BufferedReader scanner = new BufferedReader(new InputStreamReader(System.in));
    public PrintWriter out = new PrintWriter(System.out);
    public int n, m;
    public int[] p;
    public String[] in;

    void solve() {
        int[] cnt = new int[n];
        Arrays.fill(cnt, n);
        for (int i = 0; i < n; i++) {
            int d = ((i + n) - (p[i] - 1)) % n;
            // pos is shift to true postion
            cnt[d] --;
        }
        ArrayDeque<Integer> ans = new ArrayDeque<Integer>();
        for (int i = 0; i < n; i++) {
            if (cnt[i] + 1 <= m) {
                ans.push(i);
            } else {
                 if (cnt[i] <= 2*m) {
                     if (valid(i, p)) {
                         ans.add(i);
                     }
                 }
            }
        }
        out.print(ans.size());
        for (int a: ans) {
             out.print(" " + a);
        }
        out.println();
    }

    public boolean valid(int k, int[] a) {
        boolean[] visited = new boolean[n];
        int count = 0;
        for (int i = 0; i < n; i++) {
             if (visited[i]) {
                 continue;
             }
             visited[i] = true;
             int num = 1;
             for (int j = swap(a[i], k); j != i; j = swap(a[j], k), num ++) {
                 visited[j] = true;
             }
             count += num -1;
        }
        return  count <= m;
    }

    public int  swap(int val, int k) {
         return ((val - 1) + k) % n;
    }

    public void process() throws Exception {
        int t;
        t = Integer.parseInt(scanner.readLine());
        while (t-- != 0) {
            in = scanner.readLine().split(" ");
            n = Integer.parseInt(in[0]);
            m = Integer.parseInt(in[1]);
            in = scanner.readLine().split(" ");
            p = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = Integer.parseInt(in[i]);
            }
            solve();
        }
        out.flush();
    }

    public static void main(String[] args) throws Exception {
        ProblemE app = new ProblemE();
        app.process();
    }
}
