#include <bits/stdc++.h>
#pragma GCC optimize "trapv"
using namespace std;
const int INF = 1e9;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
void dfs(int i, vector<vector<int> > &adj, vector<int> &vis) {
  vis[i] = 1;
  for (auto &x : adj[i]) {
    if (!vis[x]) {
      dfs(x, adj, vis);
    }
  }
}
bool check(vector<int> &x, int m, vector<int> &a) {
  int n = x.size();
  vector<vector<int> > adj(n);
  map<int, int> make_pair;
  for (int i = 0; i < n; i++) {
    make_pair[a[i]] = i;
  };
  for (int i = 0; i < n; i++) {
    if (a[i] == x[i]) continue;
    int ori = make_pair[x[i]];
    adj[x[i] - 1].push_back(x[ori] - 1);
    adj[x[ori] - 1].push_back(x[i] - 1);
  };
  vector<int> vis(n, 0);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (vis[i] == 0) {
      ans++;
      dfs(i, adj, vis);
    }
  }
  return n - ans <= m;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (auto &x : a) cin >> x;
  vector<int> shift(n, 0);
  map<int, int> pos;
  for (int i = 0; i < n; i++) {
    int ori = a[i] - 1;
    shift[(i - ori + n) % n]++;
    pos[a[i]] = i;
  };
  vector<int> ans;
  vector<int> x(n);
  for (int i = 0; i < n; i++) {
    if (shift[i] < n / 3) continue;
    for (int j = 1; j <= n; j++) {
      x[(j - 1 + i) % n] = j;
    };
    ;
    if (check(x, m, a)) ans.push_back(i);
  }
  cout << ans.size() << " ";
  for (auto &x : ans) cout << x << " ";
  cout << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}

### 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>
#pragma GCC optimize "trapv"
using namespace std;
const int INF = 1e9;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
void dfs(int i, vector<vector<int> > &adj, vector<int> &vis) {
  vis[i] = 1;
  for (auto &x : adj[i]) {
    if (!vis[x]) {
      dfs(x, adj, vis);
    }
  }
}
bool check(vector<int> &x, int m, vector<int> &a) {
  int n = x.size();
  vector<vector<int> > adj(n);
  map<int, int> make_pair;
  for (int i = 0; i < n; i++) {
    make_pair[a[i]] = i;
  };
  for (int i = 0; i < n; i++) {
    if (a[i] == x[i]) continue;
    int ori = make_pair[x[i]];
    adj[x[i] - 1].push_back(x[ori] - 1);
    adj[x[ori] - 1].push_back(x[i] - 1);
  };
  vector<int> vis(n, 0);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (vis[i] == 0) {
      ans++;
      dfs(i, adj, vis);
    }
  }
  return n - ans <= m;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (auto &x : a) cin >> x;
  vector<int> shift(n, 0);
  map<int, int> pos;
  for (int i = 0; i < n; i++) {
    int ori = a[i] - 1;
    shift[(i - ori + n) % n]++;
    pos[a[i]] = i;
  };
  vector<int> ans;
  vector<int> x(n);
  for (int i = 0; i < n; i++) {
    if (shift[i] < n / 3) continue;
    for (int j = 1; j <= n; j++) {
      x[(j - 1 + i) % n] = j;
    };
    ;
    if (check(x, m, a)) ans.push_back(i);
  }
  cout << ans.size() << " ";
  for (auto &x : ans) cout << x << " ";
  cout << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}

```