#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc;
  cin >> tc;
  while (tc--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> p(n);
    for (long long z = 0; (1 > 0 and z < n) or (1 < 0 and z > n); z += 1)
      cin >> p[z];
    ;
    function<long long(long long)> num_swaps = [&](long long k) {
      vector<long long> b(n);
      for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
        b[i] = (i - k + n) % n + 1;
      vector<vector<long long> > g(n, vector<long long>());
      ;
      for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1) {
        g[p[i] - 1].push_back(b[i] - 1);
        g[b[i] - 1].push_back(p[i] - 1);
      }
      vector<bool> v(n, false);
      function<void(long long)> dfs = [&](long long rt) {
        v[rt] = true;
        for (auto ch : g[rt])
          if (!v[ch]) dfs(ch);
      };
      long long c = 0;
      for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
        if (!v[i]) {
          dfs(i);
          ++c;
        }
      return n - c;
    };
    vector<long long> cnt(n, 0);
    for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
      ++cnt[(i + 1 - p[i] + n) % n];
    vector<long long> ans;
    for (long long k = 0; (1 > 0 and k < n) or (1 < 0 and k > n); k += 1)
      if (cnt[k] >= n - m * 2 and num_swaps(k) <= m) ans.push_back(k);
    cout << ans.size();
    for (auto x : ans) cout << " " << x;
    cout << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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);
  cout.tie(NULL);
  long long tc;
  cin >> tc;
  while (tc--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> p(n);
    for (long long z = 0; (1 > 0 and z < n) or (1 < 0 and z > n); z += 1)
      cin >> p[z];
    ;
    function<long long(long long)> num_swaps = [&](long long k) {
      vector<long long> b(n);
      for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
        b[i] = (i - k + n) % n + 1;
      vector<vector<long long> > g(n, vector<long long>());
      ;
      for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1) {
        g[p[i] - 1].push_back(b[i] - 1);
        g[b[i] - 1].push_back(p[i] - 1);
      }
      vector<bool> v(n, false);
      function<void(long long)> dfs = [&](long long rt) {
        v[rt] = true;
        for (auto ch : g[rt])
          if (!v[ch]) dfs(ch);
      };
      long long c = 0;
      for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
        if (!v[i]) {
          dfs(i);
          ++c;
        }
      return n - c;
    };
    vector<long long> cnt(n, 0);
    for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
      ++cnt[(i + 1 - p[i] + n) % n];
    vector<long long> ans;
    for (long long k = 0; (1 > 0 and k < n) or (1 < 0 and k > n); k += 1)
      if (cnt[k] >= n - m * 2 and num_swaps(k) <= m) ans.push_back(k);
    cout << ans.size();
    for (auto x : ans) cout << " " << x;
    cout << endl;
  }
  return 0;
}

```