#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <typename T, typename U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (int i = 0; i < int(v.size()); ++i) is >> v[i];
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> v) {
  for (T x : v) os << x << " ";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, set<T> s) {
  for (T x : s) os << x << " ";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, multiset<T> s) {
  for (T x : s) os << x << " ";
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, map<T, U> m) {
  for (auto x : m) os << x << " ";
  return os;
}
template <typename T>
T sum(vector<T> v) {
  T summ = 0;
  for (int i = 0; i < int(int(v.size())); ++i) summ += v[i];
  return summ;
}
long long mod_pow(long long a, long long b) {
  long long sol = 1;
  while (b) {
    if (b & 1) {
      sol = (sol * a) % mod;
      b--;
    } else {
      a = (a * a) % mod;
      b /= 2;
    }
  }
  return sol;
}
long long rem(long long a, long long b) {
  long long res = a % b;
  return res < 0 ? res + b : res;
}
void test_case() {
  int n, m;
  cin >> n >> m;
  vector<int> v(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> v[i];
    v[i]--;
  }
  vector<pair<int, int> > ks;
  for (int i = 0; i < int(n); ++i) {
    ks.push_back({0, i});
  }
  for (int i = 0; i < int(n); ++i) {
    ks[rem(i - v[i], n)].first++;
  }
  set<int> sol;
  for (int i = 0; i < int(n); ++i) {
    if (ks[i].first < n / 3) continue;
    vector<int> perm;
    for (int j = (n - ks[i].second) % n; j < int((n - ks[i].second) % n + n);
         ++j) {
      perm.push_back(j % n);
    }
    map<int, int> g;
    for (int j = 0; j < int(n); ++j) {
      g[v[j]] = perm[j];
    }
    int k = 0;
    set<int> vis;
    for (int j = 0; j < int(n); ++j) {
      if (vis.count(j) > 0) continue;
      int from = j;
      while (!vis.count(from)) {
        vis.insert(from);
        from = g[from];
      }
      k++;
    }
    if (n - k <= m) {
      sol.insert(ks[i].second);
    }
  }
  cout << int(sol.size()) << " " << sol << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int i = 0; i < int(t); ++i) test_case();
  return 0;
}

### 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;
const long long mod = 1e9 + 7;
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <typename T, typename U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (int i = 0; i < int(v.size()); ++i) is >> v[i];
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> v) {
  for (T x : v) os << x << " ";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, set<T> s) {
  for (T x : s) os << x << " ";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, multiset<T> s) {
  for (T x : s) os << x << " ";
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, map<T, U> m) {
  for (auto x : m) os << x << " ";
  return os;
}
template <typename T>
T sum(vector<T> v) {
  T summ = 0;
  for (int i = 0; i < int(int(v.size())); ++i) summ += v[i];
  return summ;
}
long long mod_pow(long long a, long long b) {
  long long sol = 1;
  while (b) {
    if (b & 1) {
      sol = (sol * a) % mod;
      b--;
    } else {
      a = (a * a) % mod;
      b /= 2;
    }
  }
  return sol;
}
long long rem(long long a, long long b) {
  long long res = a % b;
  return res < 0 ? res + b : res;
}
void test_case() {
  int n, m;
  cin >> n >> m;
  vector<int> v(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> v[i];
    v[i]--;
  }
  vector<pair<int, int> > ks;
  for (int i = 0; i < int(n); ++i) {
    ks.push_back({0, i});
  }
  for (int i = 0; i < int(n); ++i) {
    ks[rem(i - v[i], n)].first++;
  }
  set<int> sol;
  for (int i = 0; i < int(n); ++i) {
    if (ks[i].first < n / 3) continue;
    vector<int> perm;
    for (int j = (n - ks[i].second) % n; j < int((n - ks[i].second) % n + n);
         ++j) {
      perm.push_back(j % n);
    }
    map<int, int> g;
    for (int j = 0; j < int(n); ++j) {
      g[v[j]] = perm[j];
    }
    int k = 0;
    set<int> vis;
    for (int j = 0; j < int(n); ++j) {
      if (vis.count(j) > 0) continue;
      int from = j;
      while (!vis.count(from)) {
        vis.insert(from);
        from = g[from];
      }
      k++;
    }
    if (n - k <= m) {
      sol.insert(ks[i].second);
    }
  }
  cout << int(sol.size()) << " " << sol << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int i = 0; i < int(t); ++i) test_case();
  return 0;
}

```