#include <bits/stdc++.h>
using namespace std;
const long long int MOD = (long long int)1e9 + 7;
long long int powmod(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b = b >> 1;
  }
  return res % MOD;
}
long long int num_swaps(vector<long long int> &arr, long long int k) {
  long long int n = arr.size();
  vector<long long int> pos(n);
  for (long long int i = 0; i < n; i++) {
    long long int val = (n + arr[i] + k) % n;
    pos[val] = i;
  }
  vector<long long int> vis(n, 0);
  long long int cnt_cycles = 0;
  for (long long int i = 0; i < n; i++) {
    if (vis[i] == 0) {
      cnt_cycles++;
      long long int cur_node = i;
      while (1) {
        cur_node = pos[cur_node];
        if (vis[cur_node] == 1) {
          break;
        }
        vis[cur_node] = 1;
      }
    }
  }
  return (n - cnt_cycles);
}
void solve(long long int testnum) {
  long long int n, m;
  cin >> n >> m;
  vector<long long int> a(n), b(n, 0), ans;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (long long int i = 0; i < n; i++) {
    long long int tmp = (n + i - a[i]) % n;
    b[tmp]++;
  }
  long long int cnt = 0;
  for (long long int i = 0; i < n; i++) {
    if (b[i] >= n - 2 * m && num_swaps(a, i) <= m) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << " ";
  for (long long int val : ans) {
    cout << val << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve(i);
  }
}

### Prompt

Generate 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;
const long long int MOD = (long long int)1e9 + 7;
long long int powmod(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b = b >> 1;
  }
  return res % MOD;
}
long long int num_swaps(vector<long long int> &arr, long long int k) {
  long long int n = arr.size();
  vector<long long int> pos(n);
  for (long long int i = 0; i < n; i++) {
    long long int val = (n + arr[i] + k) % n;
    pos[val] = i;
  }
  vector<long long int> vis(n, 0);
  long long int cnt_cycles = 0;
  for (long long int i = 0; i < n; i++) {
    if (vis[i] == 0) {
      cnt_cycles++;
      long long int cur_node = i;
      while (1) {
        cur_node = pos[cur_node];
        if (vis[cur_node] == 1) {
          break;
        }
        vis[cur_node] = 1;
      }
    }
  }
  return (n - cnt_cycles);
}
void solve(long long int testnum) {
  long long int n, m;
  cin >> n >> m;
  vector<long long int> a(n), b(n, 0), ans;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (long long int i = 0; i < n; i++) {
    long long int tmp = (n + i - a[i]) % n;
    b[tmp]++;
  }
  long long int cnt = 0;
  for (long long int i = 0; i < n; i++) {
    if (b[i] >= n - 2 * m && num_swaps(a, i) <= m) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << " ";
  for (long long int val : ans) {
    cout << val << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve(i);
  }
}

```