#include <bits/stdc++.h>
using namespace std;
const long long modd = (1000LL * 1000LL * 1000LL + 7LL);
bool is_possible(vector<int> a, int shift, int m) {
  for (auto& x : a) {
    x -= shift;
    if (x < 0) {
      x += a.size();
    }
  }
  set<int> done;
  int a2 = 0;
  for (int i = (0); i < (0) + (a.size()); ++i) {
    if ((done.find((i)) != done.end())) {
      continue;
    }
    done.insert(i);
    int curr = a[i], ct = 0;
    while (curr != i) {
      done.insert(curr);
      curr = a[curr];
      ++ct;
    }
    a2 += ct;
  }
  return (a2 <= m);
}
int main(int argc, char* argv[]) {
  mt19937 rang(chrono::steady_clock::now().time_since_epoch().count());
  uniform_int_distribution<int> rand_gen(0, modd);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(12);
  int test_cases;
  cin >> test_cases;
  for (int t = (1); t < (1) + (test_cases); ++t) {
    int n;
    cin >> n;
    int m;
    cin >> m;
    vector<int> p;
    for (int i = (0); i < (0) + (n); ++i) {
      int aa;
      cin >> aa;
      p.push_back(aa - 1);
    }
    vector<int> ans;
    if (n < 30) {
      for (int shift = (0); shift < (0) + (n); ++shift) {
        if (is_possible(p, shift, m)) {
          ans.push_back(shift);
        }
      }
    } else {
      map<int, int> freq;
      for (int i = (0); i < (0) + (p.size()); ++i) {
        int shift = p[i] - i;
        if (shift < 0) {
          shift += n;
        }
        shift %= n;
        ++freq[shift];
      }
      map<int, vector<int>> freq_invert;
      for (auto x : freq) {
        freq_invert[x.second].push_back(x.first);
      }
      set<int> candidates;
      auto it = freq_invert.end();
      int cc = 0;
      while ((it != freq_invert.begin()) && (cc < 4)) {
        --it;
        for (auto sh : it->second) {
          ++cc;
          candidates.insert(sh);
          candidates.insert((n - sh) % n);
          if (cc == 4) {
            break;
          }
        }
      }
      for (auto x : candidates) {
        if (is_possible(p, x, m)) {
          ans.push_back(x);
        }
      }
    }
    for (auto& x : ans) {
      x = (n - x) % n;
    }
    sort(ans.begin(), ans.end());
    cout << ans.size() << " ";
    for (auto x : ans) {
      cout << x << " ";
    }
    cout << endl;
  }
  return 0;
}

### 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>
using namespace std;
const long long modd = (1000LL * 1000LL * 1000LL + 7LL);
bool is_possible(vector<int> a, int shift, int m) {
  for (auto& x : a) {
    x -= shift;
    if (x < 0) {
      x += a.size();
    }
  }
  set<int> done;
  int a2 = 0;
  for (int i = (0); i < (0) + (a.size()); ++i) {
    if ((done.find((i)) != done.end())) {
      continue;
    }
    done.insert(i);
    int curr = a[i], ct = 0;
    while (curr != i) {
      done.insert(curr);
      curr = a[curr];
      ++ct;
    }
    a2 += ct;
  }
  return (a2 <= m);
}
int main(int argc, char* argv[]) {
  mt19937 rang(chrono::steady_clock::now().time_since_epoch().count());
  uniform_int_distribution<int> rand_gen(0, modd);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(12);
  int test_cases;
  cin >> test_cases;
  for (int t = (1); t < (1) + (test_cases); ++t) {
    int n;
    cin >> n;
    int m;
    cin >> m;
    vector<int> p;
    for (int i = (0); i < (0) + (n); ++i) {
      int aa;
      cin >> aa;
      p.push_back(aa - 1);
    }
    vector<int> ans;
    if (n < 30) {
      for (int shift = (0); shift < (0) + (n); ++shift) {
        if (is_possible(p, shift, m)) {
          ans.push_back(shift);
        }
      }
    } else {
      map<int, int> freq;
      for (int i = (0); i < (0) + (p.size()); ++i) {
        int shift = p[i] - i;
        if (shift < 0) {
          shift += n;
        }
        shift %= n;
        ++freq[shift];
      }
      map<int, vector<int>> freq_invert;
      for (auto x : freq) {
        freq_invert[x.second].push_back(x.first);
      }
      set<int> candidates;
      auto it = freq_invert.end();
      int cc = 0;
      while ((it != freq_invert.begin()) && (cc < 4)) {
        --it;
        for (auto sh : it->second) {
          ++cc;
          candidates.insert(sh);
          candidates.insert((n - sh) % n);
          if (cc == 4) {
            break;
          }
        }
      }
      for (auto x : candidates) {
        if (is_possible(p, x, m)) {
          ans.push_back(x);
        }
      }
    }
    for (auto& x : ans) {
      x = (n - x) % n;
    }
    sort(ans.begin(), ans.end());
    cout << ans.size() << " ";
    for (auto x : ans) {
      cout << x << " ";
    }
    cout << endl;
  }
  return 0;
}

```