#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  if (v.size() == 0) {
    os << "empty vector\n";
    return os;
  }
  for (auto element : v) os << element << " ";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, pair<T, second>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& v) {
  if (v.size() == 0) {
    os << "empty set\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, multiset<T>& v) {
  if (v.size() == 0) {
    os << "empty multiset\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, map<T, second>& v) {
  if (v.size() == 0) {
    os << "empty map\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "{";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << "(" << (*it).first << " : " << (*it).second << ")";
    if (it != endit) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<vector<T>>& v) {
  for (auto& subv : v) {
    for (auto& e : subv) os << e << " ";
    os << "\n";
  }
  return os;
}
bool do_debug = false;
vector<long long> par(300100), siz(300100);
long long find_root(long long node) {
  if (par[node] == node)
    return node;
  else
    return par[node] = find_root(par[node]);
}
void merge_sets(long long a, long long b) {
  long long x = find_root(a);
  long long y = find_root(b);
  if (x == y) return;
  if (siz[x] < siz[y]) swap(x, y);
  par[y] = x;
  siz[x] += siz[y];
}
long long calc(vector<long long>& before, vector<long long>& after) {
  long long n = before.size();
  for (long long i = 0; i < n + 10; i++) {
    par[i] = i;
    siz[i] = 1;
  }
  for (long long i = 0; i < before.size(); i++) {
    merge_sets(before[i], after[i]);
  }
  set<long long> s;
  for (long long i = 1; i < n + 1; i++) s.insert(find_root(i));
  ;
  return (n - s.size());
}
void Runtime_Terror() {
  long long n, m;
  cin >> n >> m;
  vector<long long> v(n + 1);
  for (long long i = 1; i < n + 1; i++) {
    cin >> v[i];
  };
  vector<long long> cnt(n + 1);
  for (long long i = 1; i < n + 1; i++) {
    if (i >= v[i]) {
      cnt[i - v[i] + 1]++;
    } else {
      long long tem = v[i] - i;
      cnt[n - tem + 1]++;
    }
  };
  long long mini = n - 2 * m;
  vector<long long> ans;
  for (long long i = 1; i < n + 1; i++) {
    if (cnt[i] >= mini) {
      vector<long long> before(n), after(n);
      for (long long j = 1; j < n + 1; j++) before[j - 1] = v[j];
      long long start = i;
      start--;
      long long cur = 1;
      for (long long j = 0; j < n; j++) {
        after[start] = cur;
        start++;
        cur++;
        if (start == n) start = 0;
      };
      ;
      ;
      ;
      if (calc(before, after) <= m) ans.push_back(i - 1);
    }
  }
  if (ans.size()) {
    cout << ans.size() << " ";
    cout << ans << endl;
  } else {
    cout << 0 << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) Runtime_Terror();
  return 0;
}

### Prompt

Create a solution in CPP for 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;
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  if (v.size() == 0) {
    os << "empty vector\n";
    return os;
  }
  for (auto element : v) os << element << " ";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, pair<T, second>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& v) {
  if (v.size() == 0) {
    os << "empty set\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, multiset<T>& v) {
  if (v.size() == 0) {
    os << "empty multiset\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "[";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << *it;
    if (it != endit) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T, typename second>
ostream& operator<<(ostream& os, map<T, second>& v) {
  if (v.size() == 0) {
    os << "empty map\n";
    return os;
  }
  auto endit = v.end();
  endit--;
  os << "{";
  for (auto it = v.begin(); it != v.end(); it++) {
    os << "(" << (*it).first << " : " << (*it).second << ")";
    if (it != endit) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<vector<T>>& v) {
  for (auto& subv : v) {
    for (auto& e : subv) os << e << " ";
    os << "\n";
  }
  return os;
}
bool do_debug = false;
vector<long long> par(300100), siz(300100);
long long find_root(long long node) {
  if (par[node] == node)
    return node;
  else
    return par[node] = find_root(par[node]);
}
void merge_sets(long long a, long long b) {
  long long x = find_root(a);
  long long y = find_root(b);
  if (x == y) return;
  if (siz[x] < siz[y]) swap(x, y);
  par[y] = x;
  siz[x] += siz[y];
}
long long calc(vector<long long>& before, vector<long long>& after) {
  long long n = before.size();
  for (long long i = 0; i < n + 10; i++) {
    par[i] = i;
    siz[i] = 1;
  }
  for (long long i = 0; i < before.size(); i++) {
    merge_sets(before[i], after[i]);
  }
  set<long long> s;
  for (long long i = 1; i < n + 1; i++) s.insert(find_root(i));
  ;
  return (n - s.size());
}
void Runtime_Terror() {
  long long n, m;
  cin >> n >> m;
  vector<long long> v(n + 1);
  for (long long i = 1; i < n + 1; i++) {
    cin >> v[i];
  };
  vector<long long> cnt(n + 1);
  for (long long i = 1; i < n + 1; i++) {
    if (i >= v[i]) {
      cnt[i - v[i] + 1]++;
    } else {
      long long tem = v[i] - i;
      cnt[n - tem + 1]++;
    }
  };
  long long mini = n - 2 * m;
  vector<long long> ans;
  for (long long i = 1; i < n + 1; i++) {
    if (cnt[i] >= mini) {
      vector<long long> before(n), after(n);
      for (long long j = 1; j < n + 1; j++) before[j - 1] = v[j];
      long long start = i;
      start--;
      long long cur = 1;
      for (long long j = 0; j < n; j++) {
        after[start] = cur;
        start++;
        cur++;
        if (start == n) start = 0;
      };
      ;
      ;
      ;
      if (calc(before, after) <= m) ans.push_back(i - 1);
    }
  }
  if (ans.size()) {
    cout << ans.size() << " ";
    cout << ans << endl;
  } else {
    cout << 0 << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) Runtime_Terror();
  return 0;
}

```