#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
int n, m;
int arr[300005];
int pos[300005];
int distVals[300005];
int graph[300005];
int resPerm[300005];
bool vis[300005];
int dfs(int x) {
  vis[x] = true;
  if (vis[graph[x]]) return 1;
  return 1 + dfs(graph[x]);
}
bool works(int k) {
  for (int i = n - k; i < n; ++i) resPerm[i - (n - k)] = i;
  for (int i = 0; i < n - k; ++i) resPerm[i + k] = i;
  for (int i = 0; i < n; ++i) {
    graph[i] = (pos[resPerm[i]]);
    vis[i] = false;
  }
  vector<int> sizes;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) sizes.push_back(dfs(i));
  }
  int ans = 0;
  for (int x : sizes) ans += (x - 1);
  if (ans <= m) return true;
  return false;
}
void doIt() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    distVals[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    --arr[i];
    pos[arr[i]] = i;
  }
  for (int i = 0; i < n; ++i) {
    int d = (pos[i] - i);
    d = (d + n) % n;
    distVals[d]++;
  }
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    if (distVals[i] < n - 2 * m) continue;
    if (works(i)) ans.push_back(i);
  }
  cout << ans.size() << " ";
  for (int x : ans) cout << x << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) doIt();
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int n, m;
int arr[300005];
int pos[300005];
int distVals[300005];
int graph[300005];
int resPerm[300005];
bool vis[300005];
int dfs(int x) {
  vis[x] = true;
  if (vis[graph[x]]) return 1;
  return 1 + dfs(graph[x]);
}
bool works(int k) {
  for (int i = n - k; i < n; ++i) resPerm[i - (n - k)] = i;
  for (int i = 0; i < n - k; ++i) resPerm[i + k] = i;
  for (int i = 0; i < n; ++i) {
    graph[i] = (pos[resPerm[i]]);
    vis[i] = false;
  }
  vector<int> sizes;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) sizes.push_back(dfs(i));
  }
  int ans = 0;
  for (int x : sizes) ans += (x - 1);
  if (ans <= m) return true;
  return false;
}
void doIt() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    distVals[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    --arr[i];
    pos[arr[i]] = i;
  }
  for (int i = 0; i < n; ++i) {
    int d = (pos[i] - i);
    d = (d + n) % n;
    distVals[d]++;
  }
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    if (distVals[i] < n - 2 * m) continue;
    if (works(i)) ans.push_back(i);
  }
  cout << ans.size() << " ";
  for (int x : ans) cout << x << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) doIt();
  return 0;
}

```