#include <bits/stdc++.h>
using namespace std;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
long long GCD(long long a, long long b) { return (a) ? GCD(b % a, a) : b; }
long long LCM(long long a, long long b) { return a * b / GCD(a, b); }
long long fastpow(long long b, long long p) {
  if (!p) return 1;
  long long ret = fastpow(b, p >> 1);
  ret *= ret;
  if (p & 1) ret *= b;
  return ret;
}
int n, m;
int a[300005];
void solve() {
  cin >> n >> m;
  unordered_map<int, int> mp;
  int x;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mp[(i - a[i] + n) % n]++;
  }
  vector<int> ans;
  for (int k = 0; k < n; k++) {
    if (n - mp[k] > m * 2) continue;
    vector<int> vec(n + 1, 0);
    int s, cur, mm = 0, temp;
    for (int i = 1; i <= n; i++) {
      if (a[(i + k - 1) % n + 1] == i) vec[i] = 1;
      if (vec[i]) continue;
      s = i;
      cur = s;
      temp = (cur + k - 1) % n + 1;
      while (a[temp] != s)
        vec[cur] = 1, cur = a[temp], temp = (cur + k - 1) % n + 1;
      mm++;
      vec[cur] = 1;
    }
    if (n - mp[k] - mm <= m) ans.push_back(k);
  }
  cout << ans.size() << ' ';
  for (auto xx : ans) cout << xx << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int TT = 1;
  cin >> TT;
  for (int tc = 1; tc <= TT; tc++) {
    solve();
  }
}

### Prompt

In Cpp, your task is to solve 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;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
long long GCD(long long a, long long b) { return (a) ? GCD(b % a, a) : b; }
long long LCM(long long a, long long b) { return a * b / GCD(a, b); }
long long fastpow(long long b, long long p) {
  if (!p) return 1;
  long long ret = fastpow(b, p >> 1);
  ret *= ret;
  if (p & 1) ret *= b;
  return ret;
}
int n, m;
int a[300005];
void solve() {
  cin >> n >> m;
  unordered_map<int, int> mp;
  int x;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mp[(i - a[i] + n) % n]++;
  }
  vector<int> ans;
  for (int k = 0; k < n; k++) {
    if (n - mp[k] > m * 2) continue;
    vector<int> vec(n + 1, 0);
    int s, cur, mm = 0, temp;
    for (int i = 1; i <= n; i++) {
      if (a[(i + k - 1) % n + 1] == i) vec[i] = 1;
      if (vec[i]) continue;
      s = i;
      cur = s;
      temp = (cur + k - 1) % n + 1;
      while (a[temp] != s)
        vec[cur] = 1, cur = a[temp], temp = (cur + k - 1) % n + 1;
      mm++;
      vec[cur] = 1;
    }
    if (n - mp[k] - mm <= m) ans.push_back(k);
  }
  cout << ans.size() << ' ';
  for (auto xx : ans) cout << xx << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int TT = 1;
  cin >> TT;
  for (int tc = 1; tc <= TT; tc++) {
    solve();
  }
}

```