#include <bits/stdc++.h>
using namespace std;
template <class T>
void read(T &x) {
  cin >> x;
}
template <class H, class T>
void read(pair<H, T> &p) {
  cin >> p.first >> p.second;
}
template <class T, size_t S>
void read(array<T, S> &a) {
  for (T &i : a) read(i);
}
template <class T>
void read(vector<T> &v) {
  for (T &i : v) read(i);
}
template <class H, class... T>
void read(H &h, T &...t) {
  read(h);
  read(t...);
}
template <class H, class T>
ostream &operator<<(ostream &o, pair<H, T> &p) {
  o << p.first << " " << p.second;
  return o;
}
template <class T, size_t S>
ostream &operator<<(ostream &o, array<T, S> &a) {
  string s;
  for (T i : a) o << s << i, s = " ";
  return o;
}
template <class T>
ostream &operator<<(ostream &o, vector<T> &v) {
  string s;
  for (T i : v) o << s << i, s = " ";
  return o;
}
template <class T>
void write(T x) {
  cout << x;
}
template <class H, class... T>
void write(const H &h, const T &...t) {
  write(h);
  write(t...);
}
void print() { write('\n'); }
template <class H, class... T>
void print(const H &h, const T &...t) {
  write(h);
  if (sizeof...(t)) write(' ');
  print(t...);
}
template <class T>
bool ckmin(T &a, const T b) {
  return (b < a) ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T &a, const T b) {
  return (a < b) ? a = b, 1 : 0;
}
using ll = long long;
void solve() {
  int n, m;
  read(n, m);
  vector<int> a(n), cnt(n);
  read(a);
  for (int &i : a) --i;
  for (int i = 0; i < n; ++i) {
    int here;
    if (a[i] > i)
      here = n + i - a[i];
    else
      here = i - a[i];
    ++cnt[here];
  }
  vector<int> cand, ans;
  for (int i = 0; i < n; ++i)
    if (cnt[i] + m * 2 >= n) cand.push_back(i);
  for (int x : cand) {
    vector<int> b(n), pos(n);
    iota((b).begin(), (b).end(), 0);
    rotate(b.begin(), b.begin() + (n - x), b.end());
    for (int i = 0; i < n; ++i) pos[b[i]] = i;
    int c = 0;
    for (int i = 0; i < n; ++i) {
      if (a[i] != b[i]) {
        int need = pos[a[i]];
        swap(b[i], b[need]);
        pos[b[i]] = i, pos[b[need]] = need;
        ++c;
      }
    }
    if (c <= m) ans.push_back(x);
  }
  write(ans.size());
  for (int i : ans) write(' ', i);
  print();
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  read(t);
  while (t--) solve();
}

### Prompt

Please formulate a CPP solution 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;
template <class T>
void read(T &x) {
  cin >> x;
}
template <class H, class T>
void read(pair<H, T> &p) {
  cin >> p.first >> p.second;
}
template <class T, size_t S>
void read(array<T, S> &a) {
  for (T &i : a) read(i);
}
template <class T>
void read(vector<T> &v) {
  for (T &i : v) read(i);
}
template <class H, class... T>
void read(H &h, T &...t) {
  read(h);
  read(t...);
}
template <class H, class T>
ostream &operator<<(ostream &o, pair<H, T> &p) {
  o << p.first << " " << p.second;
  return o;
}
template <class T, size_t S>
ostream &operator<<(ostream &o, array<T, S> &a) {
  string s;
  for (T i : a) o << s << i, s = " ";
  return o;
}
template <class T>
ostream &operator<<(ostream &o, vector<T> &v) {
  string s;
  for (T i : v) o << s << i, s = " ";
  return o;
}
template <class T>
void write(T x) {
  cout << x;
}
template <class H, class... T>
void write(const H &h, const T &...t) {
  write(h);
  write(t...);
}
void print() { write('\n'); }
template <class H, class... T>
void print(const H &h, const T &...t) {
  write(h);
  if (sizeof...(t)) write(' ');
  print(t...);
}
template <class T>
bool ckmin(T &a, const T b) {
  return (b < a) ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T &a, const T b) {
  return (a < b) ? a = b, 1 : 0;
}
using ll = long long;
void solve() {
  int n, m;
  read(n, m);
  vector<int> a(n), cnt(n);
  read(a);
  for (int &i : a) --i;
  for (int i = 0; i < n; ++i) {
    int here;
    if (a[i] > i)
      here = n + i - a[i];
    else
      here = i - a[i];
    ++cnt[here];
  }
  vector<int> cand, ans;
  for (int i = 0; i < n; ++i)
    if (cnt[i] + m * 2 >= n) cand.push_back(i);
  for (int x : cand) {
    vector<int> b(n), pos(n);
    iota((b).begin(), (b).end(), 0);
    rotate(b.begin(), b.begin() + (n - x), b.end());
    for (int i = 0; i < n; ++i) pos[b[i]] = i;
    int c = 0;
    for (int i = 0; i < n; ++i) {
      if (a[i] != b[i]) {
        int need = pos[a[i]];
        swap(b[i], b[need]);
        pos[b[i]] = i, pos[b[need]] = need;
        ++c;
      }
    }
    if (c <= m) ans.push_back(x);
  }
  write(ans.size());
  for (int i : ans) write(' ', i);
  print();
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  read(t);
  while (t--) solve();
}

```