#include <bits/stdc++.h>
using namespace std;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const int INF = 1E9 + 5;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> p(n), pos(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    --p[i];
    pos[p[i]] = i;
  }
  vector<int> cnt(n);
  for (int i = 0; i < n; i++) {
    int k = (pos[i] - i + n) % n;
    cnt[k]++;
  }
  auto check = [&](int shift) {
    vector<int> q;
    for (int i = n - shift; i < n; i++) q.push_back(i);
    for (int i = 0; i < n - shift; i++) q.push_back(i);
    int ret = n;
    vector<int> vis(n);
    for (int i = 0; i < n; i++) {
      if (!vis[i]) {
        ret--;
        int j = i;
        while (!vis[j]) {
          vis[j] = true;
          j = pos[q[j]];
        }
      }
    }
    if (ret <= m) return true;
    return false;
  };
  vector<int> res;
  for (int i = 0; i < n; i++) {
    if (cnt[i] >= n / 3)
      if (check(i)) res.push_back(i);
  }
  cout << (int)res.size() << ' ';
  for (int x : res) cout << x << ' ';
  cout << '\n';
}
int main() {
  fast();
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please create a solution in cpp to the following problem:

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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const int INF = 1E9 + 5;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> p(n), pos(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    --p[i];
    pos[p[i]] = i;
  }
  vector<int> cnt(n);
  for (int i = 0; i < n; i++) {
    int k = (pos[i] - i + n) % n;
    cnt[k]++;
  }
  auto check = [&](int shift) {
    vector<int> q;
    for (int i = n - shift; i < n; i++) q.push_back(i);
    for (int i = 0; i < n - shift; i++) q.push_back(i);
    int ret = n;
    vector<int> vis(n);
    for (int i = 0; i < n; i++) {
      if (!vis[i]) {
        ret--;
        int j = i;
        while (!vis[j]) {
          vis[j] = true;
          j = pos[q[j]];
        }
      }
    }
    if (ret <= m) return true;
    return false;
  };
  vector<int> res;
  for (int i = 0; i < n; i++) {
    if (cnt[i] >= n / 3)
      if (check(i)) res.push_back(i);
  }
  cout << (int)res.size() << ' ';
  for (int x : res) cout << x << ' ';
  cout << '\n';
}
int main() {
  fast();
  int t;
  cin >> t;
  while (t--) solve();
}

```