#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<int> p(n);
    vector<int> f(n);
    for (int i = 0; i < n; i++) {
      cin >> p[i];
      p[i]--;
      f[(i - p[i] + n) % n]++;
    }
    vector<int> v;
    for (int k = 0; k < n; k++) {
      if (f[k] < n / 3) continue;
      vector<int> q(n);
      for (int j = 0; j < n; j++) q[j] = p[(j + k) % n];
      vector<int> vis(n);
      int c = 0;
      for (int i = 0; i < n; i++)
        if (!vis[i]) {
          while (!vis[i]) {
            vis[i] = true;
            i = q[i];
          }
          c++;
        }
      if (c >= n - m) v.push_back(k);
    }
    cout << v.size() << '\n';
    for (auto k : v) cout << k << " ";
    cout << '\n';
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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]. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<int> p(n);
    vector<int> f(n);
    for (int i = 0; i < n; i++) {
      cin >> p[i];
      p[i]--;
      f[(i - p[i] + n) % n]++;
    }
    vector<int> v;
    for (int k = 0; k < n; k++) {
      if (f[k] < n / 3) continue;
      vector<int> q(n);
      for (int j = 0; j < n; j++) q[j] = p[(j + k) % n];
      vector<int> vis(n);
      int c = 0;
      for (int i = 0; i < n; i++)
        if (!vis[i]) {
          while (!vis[i]) {
            vis[i] = true;
            i = q[i];
          }
          c++;
        }
      if (c >= n - m) v.push_back(k);
    }
    cout << v.size() << '\n';
    for (auto k : v) cout << k << " ";
    cout << '\n';
  }
}

```