#include <bits/stdc++.h>
using namespace std;
const int N = 200 + 5;
class Node {
 public:
  int rnk, parent;
  Node() {
    rnk = -1;
    parent = -1;
  }
};
class Disjoint_Set {
 public:
  vector<Node> node;
  int n;
  int comps;
  Disjoint_Set(int nn) {
    n = nn;
    node.resize(n);
    for (int i = 0; i < n; i++) {
      make_set(i);
    }
    comps = n;
  }
  void make_set(int x) {
    node[x].rnk = 0;
    node[x].parent = x;
  }
  int find_set(int x) {
    if (node[x].parent != x) {
      node[x].parent = find_set(node[x].parent);
    }
    return node[x].parent;
  }
  void Union(int u, int v) {
    int p = find_set(u), q = find_set(v);
    if (p == q) return;
    comps--;
    if (node[p].rnk > node[q].rnk)
      node[q].parent = p;
    else {
      node[p].parent = q;
      if (node[p].rnk == node[q].rnk) {
        node[q].rnk += 1;
      }
    }
  }
};
void TEST_CASES() {
  int n, m;
  cin >> n >> m;
  vector<int> b(n);
  vector<pair<int, int> > cnt;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
    int x = ((i - b[i]) % n + n) % n;
    mp[x]++;
    ;
  }
  for (auto it : mp) {
    cnt.push_back({it.second, it.first});
  }
  sort(cnt.rbegin(), cnt.rend());
  ;
  vector<int> ans;
  int x = 0;
  for (int i = 0; i < cnt.size(); i++) {
    int k = cnt[i].second;
    Disjoint_Set d(n);
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      a[i] = ((i - k) % n + n) % n;
    }
    for (int i = 0; i < n; i++) {
      d.Union(b[i], a[i]);
    }
    if (n - d.comps <= m) {
      ans.push_back(k);
      x = 0;
    } else {
      x++;
    }
    if (x >= 10) break;
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << " ";
  for (int i : ans) {
    cout << i << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) {
    TEST_CASES();
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const int N = 200 + 5;
class Node {
 public:
  int rnk, parent;
  Node() {
    rnk = -1;
    parent = -1;
  }
};
class Disjoint_Set {
 public:
  vector<Node> node;
  int n;
  int comps;
  Disjoint_Set(int nn) {
    n = nn;
    node.resize(n);
    for (int i = 0; i < n; i++) {
      make_set(i);
    }
    comps = n;
  }
  void make_set(int x) {
    node[x].rnk = 0;
    node[x].parent = x;
  }
  int find_set(int x) {
    if (node[x].parent != x) {
      node[x].parent = find_set(node[x].parent);
    }
    return node[x].parent;
  }
  void Union(int u, int v) {
    int p = find_set(u), q = find_set(v);
    if (p == q) return;
    comps--;
    if (node[p].rnk > node[q].rnk)
      node[q].parent = p;
    else {
      node[p].parent = q;
      if (node[p].rnk == node[q].rnk) {
        node[q].rnk += 1;
      }
    }
  }
};
void TEST_CASES() {
  int n, m;
  cin >> n >> m;
  vector<int> b(n);
  vector<pair<int, int> > cnt;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
    int x = ((i - b[i]) % n + n) % n;
    mp[x]++;
    ;
  }
  for (auto it : mp) {
    cnt.push_back({it.second, it.first});
  }
  sort(cnt.rbegin(), cnt.rend());
  ;
  vector<int> ans;
  int x = 0;
  for (int i = 0; i < cnt.size(); i++) {
    int k = cnt[i].second;
    Disjoint_Set d(n);
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      a[i] = ((i - k) % n + n) % n;
    }
    for (int i = 0; i < n; i++) {
      d.Union(b[i], a[i]);
    }
    if (n - d.comps <= m) {
      ans.push_back(k);
      x = 0;
    } else {
      x++;
    }
    if (x >= 10) break;
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << " ";
  for (int i : ans) {
    cout << i << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) {
    TEST_CASES();
  }
  return 0;
}

```